lördag 5 augusti 2023 kl. 02:38:51 UTC+2 skrev Eram semper recta:
> On Friday, 4 August 2023 at 16:52:23 UTC-4, markus...
> > > > > > > On Thursday, 3 August 2023 at 17:41:50 UTC-4, markus.
> > > > > > > > > > > > > > > > https://www.academia.edu/45154026/Teaching_the_fundamental_theorem_of_calculus
> > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > I was the FIRST HUMAN to fully understand the mean value theorem and to prove constructively.
> > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > NONE of the stupid FUCKS who came before me where up to the task:
> > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > https://www.academia.edu/81300370/Mainstream_mathematics_academics_are_arrogant_and_incorrigible_ignoramuses_The_mean_value_theorem_IS_the_fundamental_theorem_of_calculus
> > > > > > > > > > > > > > Your proof is invalid.
> > > > > > > > > > > > > You've said that about a lot of things and every time you have been wrong.
> > > > > > > > > > > > >
> > > > > > > > > > > > > You're wrong here yet again.
> > > > > > > > > > > > > > The real mean value theorem requires real numbers.
> > > > > > > > > > > > > It doesn't have anything to do with an ill-formed object you think of as a "real number".
> > > > > > > > > > > > >
> > > > > > > > > > > > > The mean value theorem is about a level magnitude (what you erroneously call an "arithmetic mean"). It's the reason calculus works at all.
> > > > > > > > > > > > >
> > > > > > > > > > > > > You're right - I don't understand the "limits" of your ignorance and stupidity.
> > > > > > > > > > > > The mean value theorem doesn't work without real numbers. A very elementary counterexample is f(x)=0 if x²>2 and f(x)=1 if x²<2. It is continuous and differentiable everywhere on the rational number line with f'=0. If we consider [a, b]=[0, 2], we have a counterexample to the MVT (mean value theorem). From (b-a)f'(c)=f(b)-f(a) we have
> > > > > > > > > > > >
> > > > > > > > > > > > f'(c)=0-1=-1.
> > > > > > > > > > > >
> > > > > > > > > > > > But there is no such c. If we attempt to explicitly find c, we find that it doesn't exist.
> > > > > > > > > > > >
> > > > > > > > > > > > The mean value theorem requires a real number line.
> > > > > > > > > > > > See a psychiatrist soon.
> > > > > > > > > > >
> > > > > > > > > > > You most definitely do need to see a psychiatrist. You are mentally ill.
> > > > > > > > > > > > I don't need to.
> > > > > > > > > > > > But you can consider f(x)=x³ over [0, 1] and see why the mean value theorem requires real numbers.
> > > > > > > > > > >
> > > > > > > > > > > It doesn't show at all what you claim. In fact, the mvt doesn't give a shit about anything but magnitudes.
> > > > > > > > > > >
> > > > > > > > > > > You obviously don't understand the mvt which is about a level magnitude (aka arithmetic mean in your bullshit mathematics).
> > > > > > > > > > Then try it.
> > > > > > > > > >
> > > > > > > > > > f(x)=x³, a=0 and b=0. Then f'(x)=3x², f(b)=1, f(a)=0.
> > > > > > > > > What?! f(b)=f(0)=0, not 1, you imbecile!
> > > > > > > > >
> > > > > > > > > <scheißen>
> > > > > > > > b=0 was a typo. b=1, so f(b) is indeed one. You will get f'(c)=3c³=1, and this can't be solved with rational numbers only..
> > > > > > > It still is wrong, whether a typo or not. Incoherent gibberish out of your juvenile brain.
> > > > > > Can you solve f'(c)=3c²=1 without real numbers?
> > > > > Yes, idiot! It is the CONSTANT known as the square root of 1/3 and there is NO number that decribes its measure.
> > > > The square root of 1/3 isn't rational.
> > > And so? Stating the obvious incorrectly. There is no number that describes the measure of the constant root(3).
> > > > Thus the mean value theorem fails without real numbers.
> > > It doesn't because there is no such thing as "real number".
> > >
> > > <END OF DISCUSSION>
> > If there is no such c, then the mean value theorem is false. The theorem states there is a such c.
> You'll have to tell me again what is your example. I stop after the first mistake I find and you get ignored.
You have already gotten TWO counterexamples.

f(x)=x³, a=0 and b=1. Then f'(x)=3x², f(b)=1, f(a)=0.

(b-a)f'(c)=f(b)-f(a)=1.

f'(c)=3c²=1.

We end up with something we can't solve only using rational numbers.

On Saturday, 5 August 2023 at 03:35:21 UTC-4, markus...@gmail.com wrote:
> lördag 5 augusti 2023 kl. 02:38:51 UTC+2 skrev Eram semper recta:
> > On Friday, 4 August 2023 at 16:52:23 UTC-4, markus...
> > > > > > > > On Thursday, 3 August 2023 at 17:41:50 UTC-4, markus.
> > > > > > > > > > > > > > > > > https://www.academia.edu/45154026/Teaching_the_fundamental_theorem_of_calculus
> > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > I was the FIRST HUMAN to fully understand the mean value theorem and to prove constructively.
> > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > NONE of the stupid FUCKS who came before me where up to the task:
> > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > https://www.academia.edu/81300370/Mainstream_mathematics_academics_are_arrogant_and_incorrigible_ignoramuses_The_mean_value_theorem_IS_the_fundamental_theorem_of_calculus
> > > > > > > > > > > > > > > Your proof is invalid.
> > > > > > > > > > > > > > You've said that about a lot of things and every time you have been wrong.
> > > > > > > > > > > > > >
> > > > > > > > > > > > > > You're wrong here yet again.
> > > > > > > > > > > > > > > The real mean value theorem requires real numbers.
> > > > > > > > > > > > > > It doesn't have anything to do with an ill-formed object you think of as a "real number".
> > > > > > > > > > > > > >
> > > > > > > > > > > > > > The mean value theorem is about a level magnitude (what you erroneously call an "arithmetic mean"). It's the reason calculus works at all.
> > > > > > > > > > > > > >
> > > > > > > > > > > > > > You're right - I don't understand the "limits" of your ignorance and stupidity.
> > > > > > > > > > > > > The mean value theorem doesn't work without real numbers. A very elementary counterexample is f(x)=0 if x²>2 and f(x)=1 if x²<2. It is continuous and differentiable everywhere on the rational number line with f'=0. If we consider [a, b]=[0, 2], we have a counterexample to the MVT (mean value theorem). From (b-a)f'(c)=f(b)-f(a) we have
> > > > > > > > > > > > >
> > > > > > > > > > > > > f'(c)=0-1=-1.
> > > > > > > > > > > > >
> > > > > > > > > > > > > But there is no such c. If we attempt to explicitly find c, we find that it doesn't exist.
> > > > > > > > > > > > >
> > > > > > > > > > > > > The mean value theorem requires a real number line.
> > > > > > > > > > > > > See a psychiatrist soon.
> > > > > > > > > > > >
> > > > > > > > > > > > You most definitely do need to see a psychiatrist. You are mentally ill.
> > > > > > > > > > > > > I don't need to.
> > > > > > > > > > > > > But you can consider f(x)=x³ over [0, 1] and see why the mean value theorem requires real numbers.
> > > > > > > > > > > >
> > > > > > > > > > > > It doesn't show at all what you claim. In fact, the mvt doesn't give a shit about anything but magnitudes.
> > > > > > > > > > > >
> > > > > > > > > > > > You obviously don't understand the mvt which is about a level magnitude (aka arithmetic mean in your bullshit mathematics).
> > > > > > > > > > > Then try it.
> > > > > > > > > > >
> > > > > > > > > > > f(x)=x³, a=0 and b=0. Then f'(x)=3x², f(b)=1, f(a)=0.
> > > > > > > > > > What?! f(b)=f(0)=0, not 1, you imbecile!
> > > > > > > > > >
> > > > > > > > > > <scheißen>
> > > > > > > > > b=0 was a typo. b=1, so f(b) is indeed one. You will get f'(c)=3c³=1, and this can't be solved with rational numbers only.
> > > > > > > > It still is wrong, whether a typo or not. Incoherent gibberish out of your juvenile brain.
> > > > > > > Can you solve f'(c)=3c²=1 without real numbers?
> > > > > > Yes, idiot! It is the CONSTANT known as the square root of 1/3 and there is NO number that decribes its measure.
> > > > > The square root of 1/3 isn't rational.
> > > > And so? Stating the obvious incorrectly. There is no number that describes the measure of the constant root(3).
> > > > > Thus the mean value theorem fails without real numbers.
> > > > It doesn't because there is no such thing as "real number".
> > > >
> > > > <END OF DISCUSSION>
> > > If there is no such c, then the mean value theorem is false. The theorem states there is a such c.
> > You'll have to tell me again what is your example. I stop after the first mistake I find and you get ignored.
> You have already gotten TWO counterexamples.
>
> f(x)=x³, a=0 and b=1. Then f'(x)=3x², f(b)=1, f(a)=0.
> (b-a)f'(c)=f(b)-f(a)=1.
>
> f'(c)=3c²=1.
> We end up with something we can't solve only using rational numbers.

I have already answered this question. The square root of 1/3 is not described by any number. It is a CONSTANT which is a numeric approximation but not the actual value.

The ratio square root of 1/3 is realised from a right angled triangle with one leg equal to leg (p) of original isosceles right-angled triangle and the other leg (q) to hypotenuse of original isosceles right-angled triangle.

Thus the ratio is p:q and this ratio has no measure. If we attempt to measure it, we end up with the same failed measure that is the constant 0.5773 (which is an approximation or failed measure of p:q).

You fail to understand these things because you don't understand ratio and number.

lördag 5 augusti 2023 kl. 12:25:04 UTC+2 skrev Eram semper recta:
> On Saturday, 5 August 2023 at 03:35:21 UTC-4, markus...@gmail.com wrote:
> > lördag 5 augusti 2023 kl. 02:38:51 UTC+2 skrev Eram semper recta:
> > > On Friday, 4 August 2023 at 16:52:23 UTC-4, markus...
> > > > > > > > > On Thursday, 3 August 2023 at 17:41:50 UTC-4, markus.
> > > > > > > > > > > > > > > > > > https://www.academia.edu/45154026/Teaching_the_fundamental_theorem_of_calculus
> > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > I was the FIRST HUMAN to fully understand the mean value theorem and to prove constructively.
> > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > NONE of the stupid FUCKS who came before me where up to the task:
> > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > https://www.academia.edu/81300370/Mainstream_mathematics_academics_are_arrogant_and_incorrigible_ignoramuses_The_mean_value_theorem_IS_the_fundamental_theorem_of_calculus
> > > > > > > > > > > > > > > > Your proof is invalid.
> > > > > > > > > > > > > > > You've said that about a lot of things and every time you have been wrong.
> > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > You're wrong here yet again.
> > > > > > > > > > > > > > > > The real mean value theorem requires real numbers.
> > > > > > > > > > > > > > > It doesn't have anything to do with an ill-formed object you think of as a "real number".
> > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > The mean value theorem is about a level magnitude (what you erroneously call an "arithmetic mean"). It's the reason calculus works at all.
> > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > You're right - I don't understand the "limits" of your ignorance and stupidity.
> > > > > > > > > > > > > > The mean value theorem doesn't work without real numbers. A very elementary counterexample is f(x)=0 if x²>2 and f(x)=1 if x²<2. It is continuous and differentiable everywhere on the rational number line with f'=0. If we consider [a, b]=[0, 2], we have a counterexample to the MVT (mean value theorem). From (b-a)f'(c)=f(b)-f(a) we have
> > > > > > > > > > > > > >
> > > > > > > > > > > > > > f'(c)=0-1=-1.
> > > > > > > > > > > > > >
> > > > > > > > > > > > > > But there is no such c. If we attempt to explicitly find c, we find that it doesn't exist.
> > > > > > > > > > > > > >
> > > > > > > > > > > > > > The mean value theorem requires a real number line.
> > > > > > > > > > > > > > See a psychiatrist soon.
> > > > > > > > > > > > >
> > > > > > > > > > > > > You most definitely do need to see a psychiatrist.. You are mentally ill.
> > > > > > > > > > > > > > I don't need to.
> > > > > > > > > > > > > > But you can consider f(x)=x³ over [0, 1] and see why the mean value theorem requires real numbers.
> > > > > > > > > > > > >
> > > > > > > > > > > > > It doesn't show at all what you claim. In fact, the mvt doesn't give a shit about anything but magnitudes.
> > > > > > > > > > > > >
> > > > > > > > > > > > > You obviously don't understand the mvt which is about a level magnitude (aka arithmetic mean in your bullshit mathematics).
> > > > > > > > > > > > Then try it.
> > > > > > > > > > > >
> > > > > > > > > > > > f(x)=x³, a=0 and b=0. Then f'(x)=3x², f(b)=1, f(a)=0.
> > > > > > > > > > > What?! f(b)=f(0)=0, not 1, you imbecile!
> > > > > > > > > > >
> > > > > > > > > > > <scheißen>
> > > > > > > > > > b=0 was a typo. b=1, so f(b) is indeed one. You will get f'(c)=3c³=1, and this can't be solved with rational numbers only.
> > > > > > > > > It still is wrong, whether a typo or not. Incoherent gibberish out of your juvenile brain.
> > > > > > > > Can you solve f'(c)=3c²=1 without real numbers?
> > > > > > > Yes, idiot! It is the CONSTANT known as the square root of 1/3 and there is NO number that decribes its measure.
> > > > > > The square root of 1/3 isn't rational.
> > > > > And so? Stating the obvious incorrectly. There is no number that describes the measure of the constant root(3).
> > > > > > Thus the mean value theorem fails without real numbers.
> > > > > It doesn't because there is no such thing as "real number".
> > > > >
> > > > > <END OF DISCUSSION>
> > > > If there is no such c, then the mean value theorem is false. The theorem states there is a such c.
> > > You'll have to tell me again what is your example. I stop after the first mistake I find and you get ignored.
> > You have already gotten TWO counterexamples.
> >
> > f(x)=x³, a=0 and b=1. Then f'(x)=3x², f(b)=1, f(a)=0.
> > (b-a)f'(c)=f(b)-f(a)=1.
> >
> > f'(c)=3c²=1.
> > We end up with something we can't solve only using rational numbers.
> I have already answered this question. The square root of 1/3 is not described by any number. It is a CONSTANT which is a numeric approximation but not the actual value.
>
> The ratio square root of 1/3 is realised from a right angled triangle with one leg equal to leg (p) of original isosceles right-angled triangle and the other leg (q) to hypotenuse of original isosceles right-angled triangle..
>
> Thus the ratio is p:q and this ratio has no measure. If we attempt to measure it, we end up with the same failed measure that is the constant 0.5773 (which is an approximation or failed measure of p:q).
>
> You fail to understand these things because you don't understand ratio and number.
What you call "magnitudes" are in fact just real numbers.

We need something more than just rationals for the mean value theorem. That *something* is real numbers.

On Saturday, 5 August 2023 at 06:46:52 UTC-4, markus...
> > > > > > > > > > > > > > > > > > > https://www.academia.edu/45154026/Teaching_the_fundamental_theorem_of_calculus
> > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > I was the FIRST HUMAN to fully understand the mean value theorem and to prove constructively.
> > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > NONE of the stupid FUCKS who came before me where up to the task:
> > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > https://www.academia.edu/81300370/Mainstream_mathematics_academics_are_arrogant_and_incorrigible_ignoramuses_The_mean_value_theorem_IS_the_fundamental_theorem_of_calculus
> > > > > > > > > > > > > > > > > Your proof is invalid.
> > > > > > > > > > > > > > > > You've said that about a lot of things and every time you have been wrong.
> > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > You're wrong here yet again.
> > > > > > > > > > > > > > > > > The real mean value theorem requires real numbers.
> > > > > > > > > > > > > > > > It doesn't have anything to do with an ill-formed object you think of as a "real number".
> > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > The mean value theorem is about a level magnitude (what you erroneously call an "arithmetic mean"). It's the reason calculus works at all.
> > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > You're right - I don't understand the "limits" of your ignorance and stupidity.
> > > > > > > > > > > > > > > The mean value theorem doesn't work without real numbers. A very elementary counterexample is f(x)=0 if x²>2 and f(x)=1 if x²<2. It is continuous and differentiable everywhere on the rational number line with f'=0. If we consider [a, b]=[0, 2], we have a counterexample to the MVT (mean value theorem). From (b-a)f'(c)=f(b)-f(a) we have
> > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > f'(c)=0-1=-1.
> > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > But there is no such c. If we attempt to explicitly find c, we find that it doesn't exist.
> > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > The mean value theorem requires a real number line.
> > > > > > > > > > > > > > > See a psychiatrist soon.
> > > > > > > > > > > > > >
> > > > > > > > > > > > > > You most definitely do need to see a psychiatrist. You are mentally ill.
> > > > > > > > > > > > > > > I don't need to.
> > > > > > > > > > > > > > > But you can consider f(x)=x³ over [0, 1] and see why the mean value theorem requires real numbers.
> > > > > > > > > > > > > >
> > > > > > > > > > > > > > It doesn't show at all what you claim. In fact, the mvt doesn't give a shit about anything but magnitudes.
> > > > > > > > > > > > > >
> > > > > > > > > > > > > > You obviously don't understand the mvt which is about a level magnitude (aka arithmetic mean in your bullshit mathematics)..
> > > > > > > > > > > > > Then try it.
> > > > > > > > > > > > >
> > > > > > > > > > > > > f(x)=x³, a=0 and b=0. Then f'(x)=3x², f(b)=1, f(a)=0.
> > > > > > > > > > > > What?! f(b)=f(0)=0, not 1, you imbecile!
> > > > > > > > > > > >
> > > > > > > > > > > > <scheißen>
> > > > > > > > > > > b=0 was a typo. b=1, so f(b) is indeed one. You will get f'(c)=3c³=1, and this can't be solved with rational numbers only.
> > > > > > > > > > It still is wrong, whether a typo or not. Incoherent gibberish out of your juvenile brain.
> > > > > > > > > Can you solve f'(c)=3c²=1 without real numbers?
> > > > > > > > Yes, idiot! It is the CONSTANT known as the square root of 1/3 and there is NO number that decribes its measure.
> > > > > > > The square root of 1/3 isn't rational.
> > > > > > And so? Stating the obvious incorrectly. There is no number that describes the measure of the constant root(3).
> > > > > > > Thus the mean value theorem fails without real numbers.
> > > > > > It doesn't because there is no such thing as "real number".
> > > > > >
> > > > > > <END OF DISCUSSION>
> > > > > If there is no such c, then the mean value theorem is false. The theorem states there is a such c.
> > > > You'll have to tell me again what is your example. I stop after the first mistake I find and you get ignored.
> > > You have already gotten TWO counterexamples.
> > >
> > > f(x)=x³, a=0 and b=1. Then f'(x)=3x², f(b)=1, f(a)=0.
> > > (b-a)f'(c)=f(b)-f(a)=1.
> > >
> > > f'(c)=3c²=1.
> > > We end up with something we can't solve only using rational numbers.
> > I have already answered this question. The square root of 1/3 is not described by any number. It is a CONSTANT which is a numeric approximation but not the actual value.
> >
> > The ratio square root of 1/3 is realised from a right angled triangle with one leg equal to leg (p) of original isosceles right-angled triangle and the other leg (q) to hypotenuse of original isosceles right-angled triangle.
> >
> > Thus the ratio is p:q and this ratio has no measure. If we attempt to measure it, we end up with the same failed measure that is the constant 0.5773 (which is an approximation or failed measure of p:q).
> >
> > You fail to understand these things because you don't understand ratio and number.
> What you call "magnitudes" are in fact just real numbers.

You are just too stupid.

Here is the proof of the mean value theorem using the 100% rigorous New Calculus:

The New Calculus derivative:

Given any interval (c-m,c+n), the New Calculus (henceforth NC) derivative is given by:

f'(c)= (f(c+n)-f(c-m))/(m+n)

f'(c) represents the slope of a secant line with end points (c-m, f(c-m)) and ( c+n, f(c+n)) that is parallel to the tangent line at x=c.
m and n are horizontal distances from c.

The interval (c-m,c+n) can be partitioned into equal sub-intervals of (m+n)/k.

Each sub-interval has μ_s as the abscissa of f' so that

f'(μ_s) = [ f(c-m + ((m+n)(s+1))/k) - f(c-m + ((m+n)s)/k) ] / ((m+n)/k)

The level magnitude (aka arithmetic mean in mainstream mathematics) is given by:

f'(c) = (1/k) \sum_{s=1}^k f'(μ_s)

From the above statement, we want to show that

f'(c) = [ f'(μ_s1) + f'(μ_s2) + f'(μ_s3) + ... + f'(μ_k-1) + f'(μ_k) ]/k

By replacing each of the means with a derivative, we have:

f'(c) = (1/k) { [ f(c-m + (m+n)/k)-f(c-m) ]/((m+n)/k)
+ [ f(c-m + 2(m+n)/k)-f(c-m + (m+n)/k) ]/((m+n)/k)]
+ [ f(c-m + 3(m+n)/k)-f(c-m + 2(m+n)/k) ]/((m+n)/k)]
+ ...
+ [ f(c-m + ((k-1)(m+n))/k)-f(c-m + ((k-2)(m+n))/k) ]/((m+n)/k)]
+ [ f(c+n)-f(c-m + ((k-1)(m+n))/k) ]/((m+n)/k)]
}
The above reduces to f'(c) = (f(c+n)-f(c-m))/(m+n)

(1/(m+n)) \int_{c-m}^{c+n} f'(x) dx = (f(c+n)-f(c-m))/(m+n)

So, what we have proved above is that f'(c) is the level magnitude of all the y-ordinates of f'(x) in the interval (c-m, c+n) and we can find the product of f'(c) and m+n to give us the area under f'(x) from c-m to c+n where c-m < c < c+n.

lördag 5 augusti 2023 kl. 18:11:13 UTC+2 skrev Eram semper recta:
> On Saturday, 5 August 2023 at 06:46:52 UTC-4, markus...
> > > > > > > > > > > > > > > > > > > > https://www.academia.edu/45154026/Teaching_the_fundamental_theorem_of_calculus
> > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > I was the FIRST HUMAN to fully understand the mean value theorem and to prove constructively.
> > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > NONE of the stupid FUCKS who came before me where up to the task:
> > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > https://www.academia.edu/81300370/Mainstream_mathematics_academics_are_arrogant_and_incorrigible_ignoramuses_The_mean_value_theorem_IS_the_fundamental_theorem_of_calculus
> > > > > > > > > > > > > > > > > > Your proof is invalid.
> > > > > > > > > > > > > > > > > You've said that about a lot of things and every time you have been wrong.
> > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > You're wrong here yet again.
> > > > > > > > > > > > > > > > > > The real mean value theorem requires real numbers.
> > > > > > > > > > > > > > > > > It doesn't have anything to do with an ill-formed object you think of as a "real number".
> > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > The mean value theorem is about a level magnitude (what you erroneously call an "arithmetic mean"). It's the reason calculus works at all.
> > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > You're right - I don't understand the "limits" of your ignorance and stupidity.
> > > > > > > > > > > > > > > > The mean value theorem doesn't work without real numbers. A very elementary counterexample is f(x)=0 if x²>2 and f(x)=1 if x²<2. It is continuous and differentiable everywhere on the rational number line with f'=0. If we consider [a, b]=[0, 2], we have a counterexample to the MVT (mean value theorem). From (b-a)f'(c)=f(b)-f(a) we have
> > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > f'(c)=0-1=-1.
> > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > But there is no such c. If we attempt to explicitly find c, we find that it doesn't exist.
> > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > The mean value theorem requires a real number line.
> > > > > > > > > > > > > > > > See a psychiatrist soon.
> > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > You most definitely do need to see a psychiatrist. You are mentally ill.
> > > > > > > > > > > > > > > > I don't need to.
> > > > > > > > > > > > > > > > But you can consider f(x)=x³ over [0, 1] and see why the mean value theorem requires real numbers.
> > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > It doesn't show at all what you claim. In fact, the mvt doesn't give a shit about anything but magnitudes.
> > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > You obviously don't understand the mvt which is about a level magnitude (aka arithmetic mean in your bullshit mathematics).
> > > > > > > > > > > > > > Then try it.
> > > > > > > > > > > > > >
> > > > > > > > > > > > > > f(x)=x³, a=0 and b=0. Then f'(x)=3x², f(b)=1, f(a)=0.
> > > > > > > > > > > > > What?! f(b)=f(0)=0, not 1, you imbecile!
> > > > > > > > > > > > >
> > > > > > > > > > > > > <scheißen>
> > > > > > > > > > > > b=0 was a typo. b=1, so f(b) is indeed one. You will get f'(c)=3c³=1, and this can't be solved with rational numbers only.
> > > > > > > > > > > It still is wrong, whether a typo or not. Incoherent gibberish out of your juvenile brain.
> > > > > > > > > > Can you solve f'(c)=3c²=1 without real numbers?
> > > > > > > > > Yes, idiot! It is the CONSTANT known as the square root of 1/3 and there is NO number that decribes its measure.
> > > > > > > > The square root of 1/3 isn't rational.
> > > > > > > And so? Stating the obvious incorrectly. There is no number that describes the measure of the constant root(3).
> > > > > > > > Thus the mean value theorem fails without real numbers.
> > > > > > > It doesn't because there is no such thing as "real number".
> > > > > > >
> > > > > > > <END OF DISCUSSION>
> > > > > > If there is no such c, then the mean value theorem is false. The theorem states there is a such c.
> > > > > You'll have to tell me again what is your example. I stop after the first mistake I find and you get ignored.
> > > > You have already gotten TWO counterexamples.
> > > >
> > > > f(x)=x³, a=0 and b=1. Then f'(x)=3x², f(b)=1, f(a)=0.
> > > > (b-a)f'(c)=f(b)-f(a)=1.
> > > >
> > > > f'(c)=3c²=1.
> > > > We end up with something we can't solve only using rational numbers..
> > > I have already answered this question. The square root of 1/3 is not described by any number. It is a CONSTANT which is a numeric approximation but not the actual value.
> > >
> > > The ratio square root of 1/3 is realised from a right angled triangle with one leg equal to leg (p) of original isosceles right-angled triangle and the other leg (q) to hypotenuse of original isosceles right-angled triangle.
> > >
> > > Thus the ratio is p:q and this ratio has no measure. If we attempt to measure it, we end up with the same failed measure that is the constant 0.5773 (which is an approximation or failed measure of p:q).
> > >
> > > You fail to understand these things because you don't understand ratio and number.
> > What you call "magnitudes" are in fact just real numbers.
> You are just too stupid.
>
> Here is the proof of the mean value theorem using the 100% rigorous New Calculus:
>
> The New Calculus derivative:
>
> Given any interval (c-m,c+n), the New Calculus (henceforth NC) derivative is given by:
>
> f'(c)= (f(c+n)-f(c-m))/(m+n)
>
> f'(c) represents the slope of a secant line with end points (c-m, f(c-m)) and ( c+n, f(c+n)) that is parallel to the tangent line at x=c.
> m and n are horizontal distances from c.
>
> The interval (c-m,c+n) can be partitioned into equal sub-intervals of (m+n)/k.
>
> Each sub-interval has μ_s as the abscissa of f' so that
>
> f'(μ_s) = [ f(c-m + ((m+n)(s+1))/k) - f(c-m + ((m+n)s)/k) ] / ((m+n)/k)
>
> The level magnitude (aka arithmetic mean in mainstream mathematics) is given by:
>
> f'(c) = (1/k) \sum_{s=1}^k f'(μ_s)
>
> From the above statement, we want to show that
>
> f'(c) = [ f'(μ_s1) + f'(μ_s2) + f'(μ_s3) + ... + f'(μ_k-1) + f'(μ_k) ]/k
>
> By replacing each of the means with a derivative, we have:
>
> f'(c) = (1/k) { [ f(c-m + (m+n)/k)-f(c-m) ]/((m+n)/k)
> + [ f(c-m + 2(m+n)/k)-f(c-m + (m+n)/k) ]/((m+n)/k)]
> + [ f(c-m + 3(m+n)/k)-f(c-m + 2(m+n)/k) ]/((m+n)/k)]
> + ...
> + [ f(c-m + ((k-1)(m+n))/k)-f(c-m + ((k-2)(m+n))/k) ]/((m+n)/k)]
> + [ f(c+n)-f(c-m + ((k-1)(m+n))/k) ]/((m+n)/k)]
> }
> The above reduces to f'(c) = (f(c+n)-f(c-m))/(m+n)
>
> (1/(m+n)) \int_{c-m}^{c+n} f'(x) dx = (f(c+n)-f(c-m))/(m+n)
>
> So, what we have proved above is that f'(c) is the level magnitude of all the y-ordinates of f'(x) in the interval (c-m, c+n) and we can find the product of f'(c) and m+n to give us the area under f'(x) from c-m to c+n where c-m < c < c+n.
>
> Did you see anything there about "real numbers", you imbecile?
> >
> > We need something more than just rationals for the mean value theorem. That *something* is real numbers.
> No. What you need is understanding. You have NONE.
How do you know those μ_s exist?

On Saturday, 5 August 2023 at 12:24:51 UTC-4, markus...
> lördag 5 augusti 2023 kl. 18:11:13 UTC+2 skrev Eram semper recta:
> > On Saturday, 5 August 2023 at 06:46:52 UTC-4, markus...
> > > > > > > > > > > > > > > > > > > > > https://www.academia.edu/45154026/Teaching_the_fundamental_theorem_of_calculus
> > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > I was the FIRST HUMAN to fully understand the mean value theorem and to prove constructively.
> > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > NONE of the stupid FUCKS who came before me where up to the task:
> > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > https://www.academia.edu/81300370/Mainstream_mathematics_academics_are_arrogant_and_incorrigible_ignoramuses_The_mean_value_theorem_IS_the_fundamental_theorem_of_calculus
> > > > > > > > > > > > > > > > > > > Your proof is invalid.
> > > > > > > > > > > > > > > > > > You've said that about a lot of things and every time you have been wrong.
> > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > You're wrong here yet again.
> > > > > > > > > > > > > > > > > > > The real mean value theorem requires real numbers.
> > > > > > > > > > > > > > > > > > It doesn't have anything to do with an ill-formed object you think of as a "real number".
> > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > The mean value theorem is about a level magnitude (what you erroneously call an "arithmetic mean"). It's the reason calculus works at all.
> > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > You're right - I don't understand the "limits" of your ignorance and stupidity.
> > > > > > > > > > > > > > > > > The mean value theorem doesn't work without real numbers. A very elementary counterexample is f(x)=0 if x²>2 and f(x)=1 if x²<2. It is continuous and differentiable everywhere on the rational number line with f'=0. If we consider [a, b]=[0, 2], we have a counterexample to the MVT (mean value theorem). From (b-a)f'(c)=f(b)-f(a) we have
> > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > f'(c)=0-1=-1.
> > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > But there is no such c. If we attempt to explicitly find c, we find that it doesn't exist.
> > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > The mean value theorem requires a real number line.
> > > > > > > > > > > > > > > > > See a psychiatrist soon.
> > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > You most definitely do need to see a psychiatrist. You are mentally ill.
> > > > > > > > > > > > > > > > > I don't need to.
> > > > > > > > > > > > > > > > > But you can consider f(x)=x³ over [0, 1] and see why the mean value theorem requires real numbers.
> > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > It doesn't show at all what you claim. In fact, the mvt doesn't give a shit about anything but magnitudes.
> > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > You obviously don't understand the mvt which is about a level magnitude (aka arithmetic mean in your bullshit mathematics).
> > > > > > > > > > > > > > > Then try it.
> > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > f(x)=x³, a=0 and b=0. Then f'(x)=3x², f(b)=1, f(a)=0.
> > > > > > > > > > > > > > What?! f(b)=f(0)=0, not 1, you imbecile!
> > > > > > > > > > > > > >
> > > > > > > > > > > > > > <scheißen>
> > > > > > > > > > > > > b=0 was a typo. b=1, so f(b) is indeed one. You will get f'(c)=3c³=1, and this can't be solved with rational numbers only.
> > > > > > > > > > > > It still is wrong, whether a typo or not. Incoherent gibberish out of your juvenile brain.
> > > > > > > > > > > Can you solve f'(c)=3c²=1 without real numbers?
> > > > > > > > > > Yes, idiot! It is the CONSTANT known as the square root of 1/3 and there is NO number that decribes its measure.
> > > > > > > > > The square root of 1/3 isn't rational.
> > > > > > > > And so? Stating the obvious incorrectly. There is no number that describes the measure of the constant root(3).
> > > > > > > > > Thus the mean value theorem fails without real numbers.
> > > > > > > > It doesn't because there is no such thing as "real number".
> > > > > > > >
> > > > > > > > <END OF DISCUSSION>
> > > > > > > If there is no such c, then the mean value theorem is false. The theorem states there is a such c.
> > > > > > You'll have to tell me again what is your example. I stop after the first mistake I find and you get ignored.
> > > > > You have already gotten TWO counterexamples.
> > > > >
> > > > > f(x)=x³, a=0 and b=1. Then f'(x)=3x², f(b)=1, f(a)=0.
> > > > > (b-a)f'(c)=f(b)-f(a)=1.
> > > > >
> > > > > f'(c)=3c²=1.
> > > > > We end up with something we can't solve only using rational numbers.
> > > > I have already answered this question. The square root of 1/3 is not described by any number. It is a CONSTANT which is a numeric approximation but not the actual value.
> > > >
> > > > The ratio square root of 1/3 is realised from a right angled triangle with one leg equal to leg (p) of original isosceles right-angled triangle and the other leg (q) to hypotenuse of original isosceles right-angled triangle.
> > > >
> > > > Thus the ratio is p:q and this ratio has no measure. If we attempt to measure it, we end up with the same failed measure that is the constant 0.5773 (which is an approximation or failed measure of p:q).
> > > >
> > > > You fail to understand these things because you don't understand ratio and number.
> > > What you call "magnitudes" are in fact just real numbers.
> > You are just too stupid.
> >
> > Here is the proof of the mean value theorem using the 100% rigorous New Calculus:
> >
> > The New Calculus derivative:
> >
> > Given any interval (c-m,c+n), the New Calculus (henceforth NC) derivative is given by:
> >
> > f'(c)= (f(c+n)-f(c-m))/(m+n)
> >
> > f'(c) represents the slope of a secant line with end points (c-m, f(c-m)) and ( c+n, f(c+n)) that is parallel to the tangent line at x=c.
> > m and n are horizontal distances from c.
> >
> > The interval (c-m,c+n) can be partitioned into equal sub-intervals of (m+n)/k.
> >
> > Each sub-interval has μ_s as the abscissa of f' so that
> >
> > f'(μ_s) = [ f(c-m + ((m+n)(s+1))/k) - f(c-m + ((m+n)s)/k) ] / ((m+n)/k)
> >
> > The level magnitude (aka arithmetic mean in mainstream mathematics) is given by:
> >
> > f'(c) = (1/k) \sum_{s=1}^k f'(μ_s)
> >
> > From the above statement, we want to show that
> >
> > f'(c) = [ f'(μ_s1) + f'(μ_s2) + f'(μ_s3) + ... + f'(μ_k-1) + f'(μ_k) ]/k
> >
> > By replacing each of the means with a derivative, we have:
> >
> > f'(c) = (1/k) { [ f(c-m + (m+n)/k)-f(c-m) ]/((m+n)/k)
> > + [ f(c-m + 2(m+n)/k)-f(c-m + (m+n)/k) ]/((m+n)/k)]
> > + [ f(c-m + 3(m+n)/k)-f(c-m + 2(m+n)/k) ]/((m+n)/k)]
> > + ...
> > + [ f(c-m + ((k-1)(m+n))/k)-f(c-m + ((k-2)(m+n))/k) ]/((m+n)/k)]
> > + [ f(c+n)-f(c-m + ((k-1)(m+n))/k) ]/((m+n)/k)]
> > }
> > The above reduces to f'(c) = (f(c+n)-f(c-m))/(m+n)
> >
> > (1/(m+n)) \int_{c-m}^{c+n} f'(x) dx = (f(c+n)-f(c-m))/(m+n)
> >
> > So, what we have proved above is that f'(c) is the level magnitude of all the y-ordinates of f'(x) in the interval (c-m, c+n) and we can find the product of f'(c) and m+n to give us the area under f'(x) from c-m to c+n where c-m < c < c+n.
> >
> > Did you see anything there about "real numbers", you imbecile?
> > >
> > > We need something more than just rationals for the mean value theorem.. That *something* is real numbers.
> > No. What you need is understanding. You have NONE.

lördag 5 augusti 2023 kl. 21:02:56 UTC+2 skrev Eram semper recta:
> On Saturday, 5 August 2023 at 12:24:51 UTC-4, markus...
> > lördag 5 augusti 2023 kl. 18:11:13 UTC+2 skrev Eram semper recta:
> > > On Saturday, 5 August 2023 at 06:46:52 UTC-4, markus...
> > > > > > > > > > > > > > > > > > > > > > https://www.academia.edu/45154026/Teaching_the_fundamental_theorem_of_calculus
> > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > I was the FIRST HUMAN to fully understand the mean value theorem and to prove constructively.
> > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > NONE of the stupid FUCKS who came before me where up to the task:
> > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > https://www.academia.edu/81300370/Mainstream_mathematics_academics_are_arrogant_and_incorrigible_ignoramuses_The_mean_value_theorem_IS_the_fundamental_theorem_of_calculus
> > > > > > > > > > > > > > > > > > > > Your proof is invalid.
> > > > > > > > > > > > > > > > > > > You've said that about a lot of things and every time you have been wrong.
> > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > You're wrong here yet again.
> > > > > > > > > > > > > > > > > > > > The real mean value theorem requires real numbers.
> > > > > > > > > > > > > > > > > > > It doesn't have anything to do with an ill-formed object you think of as a "real number".
> > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > The mean value theorem is about a level magnitude (what you erroneously call an "arithmetic mean"). It's the reason calculus works at all.
> > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > You're right - I don't understand the "limits" of your ignorance and stupidity.
> > > > > > > > > > > > > > > > > > The mean value theorem doesn't work without real numbers. A very elementary counterexample is f(x)=0 if x²>2 and f(x)=1 if x²<2. It is continuous and differentiable everywhere on the rational number line with f'=0. If we consider [a, b]=[0, 2], we have a counterexample to the MVT (mean value theorem). From (b-a)f'(c)=f(b)-f(a) we have
> > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > f'(c)=0-1=-1.
> > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > But there is no such c. If we attempt to explicitly find c, we find that it doesn't exist.
> > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > The mean value theorem requires a real number line.
> > > > > > > > > > > > > > > > > > See a psychiatrist soon.
> > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > You most definitely do need to see a psychiatrist. You are mentally ill.
> > > > > > > > > > > > > > > > > > I don't need to.
> > > > > > > > > > > > > > > > > > But you can consider f(x)=x³ over [0, 1] and see why the mean value theorem requires real numbers.
> > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > It doesn't show at all what you claim. In fact, the mvt doesn't give a shit about anything but magnitudes.
> > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > You obviously don't understand the mvt which is about a level magnitude (aka arithmetic mean in your bullshit mathematics).
> > > > > > > > > > > > > > > > Then try it.
> > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > f(x)=x³, a=0 and b=0. Then f'(x)=3x², f(b)=1, f(a)=0.
> > > > > > > > > > > > > > > What?! f(b)=f(0)=0, not 1, you imbecile!
> > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > <scheißen>
> > > > > > > > > > > > > > b=0 was a typo. b=1, so f(b) is indeed one. You will get f'(c)=3c³=1, and this can't be solved with rational numbers only.
> > > > > > > > > > > > > It still is wrong, whether a typo or not. Incoherent gibberish out of your juvenile brain.
> > > > > > > > > > > > Can you solve f'(c)=3c²=1 without real numbers?
> > > > > > > > > > > Yes, idiot! It is the CONSTANT known as the square root of 1/3 and there is NO number that decribes its measure.
> > > > > > > > > > The square root of 1/3 isn't rational.
> > > > > > > > > And so? Stating the obvious incorrectly. There is no number that describes the measure of the constant root(3).
> > > > > > > > > > Thus the mean value theorem fails without real numbers.
> > > > > > > > > It doesn't because there is no such thing as "real number".
> > > > > > > > >
> > > > > > > > > <END OF DISCUSSION>
> > > > > > > > If there is no such c, then the mean value theorem is false.. The theorem states there is a such c.
> > > > > > > You'll have to tell me again what is your example. I stop after the first mistake I find and you get ignored.
> > > > > > You have already gotten TWO counterexamples.
> > > > > >
> > > > > > f(x)=x³, a=0 and b=1. Then f'(x)=3x², f(b)=1, f(a)=0.
> > > > > > (b-a)f'(c)=f(b)-f(a)=1.
> > > > > >
> > > > > > f'(c)=3c²=1.
> > > > > > We end up with something we can't solve only using rational numbers.
> > > > > I have already answered this question. The square root of 1/3 is not described by any number. It is a CONSTANT which is a numeric approximation but not the actual value.
> > > > >
> > > > > The ratio square root of 1/3 is realised from a right angled triangle with one leg equal to leg (p) of original isosceles right-angled triangle and the other leg (q) to hypotenuse of original isosceles right-angled triangle.
> > > > >
> > > > > Thus the ratio is p:q and this ratio has no measure. If we attempt to measure it, we end up with the same failed measure that is the constant 0.5773 (which is an approximation or failed measure of p:q).
> > > > >
> > > > > You fail to understand these things because you don't understand ratio and number.
> > > > What you call "magnitudes" are in fact just real numbers.
> > > You are just too stupid.
> > >
> > > Here is the proof of the mean value theorem using the 100% rigorous New Calculus:
> > >
> > > The New Calculus derivative:
> > >
> > > Given any interval (c-m,c+n), the New Calculus (henceforth NC) derivative is given by:
> > >
> > > f'(c)= (f(c+n)-f(c-m))/(m+n)
> > >
> > > f'(c) represents the slope of a secant line with end points (c-m, f(c-m)) and ( c+n, f(c+n)) that is parallel to the tangent line at x=c.
> > > m and n are horizontal distances from c.
> > >
> > > The interval (c-m,c+n) can be partitioned into equal sub-intervals of (m+n)/k.
> > >
> > > Each sub-interval has μ_s as the abscissa of f' so that
> > >
> > > f'(μ_s) = [ f(c-m + ((m+n)(s+1))/k) - f(c-m + ((m+n)s)/k) ] / ((m+n)/k)
> > >
> > > The level magnitude (aka arithmetic mean in mainstream mathematics) is given by:
> > >
> > > f'(c) = (1/k) \sum_{s=1}^k f'(μ_s)
> > >
> > > From the above statement, we want to show that
> > >
> > > f'(c) = [ f'(μ_s1) + f'(μ_s2) + f'(μ_s3) + ... + f'(μ_k-1) + f'(μ_k) ]/k
> > >
> > > By replacing each of the means with a derivative, we have:
> > >
> > > f'(c) = (1/k) { [ f(c-m + (m+n)/k)-f(c-m) ]/((m+n)/k)
> > > + [ f(c-m + 2(m+n)/k)-f(c-m + (m+n)/k) ]/((m+n)/k)]
> > > + [ f(c-m + 3(m+n)/k)-f(c-m + 2(m+n)/k) ]/((m+n)/k)]
> > > + ...
> > > + [ f(c-m + ((k-1)(m+n))/k)-f(c-m + ((k-2)(m+n))/k) ]/((m+n)/k)]
> > > + [ f(c+n)-f(c-m + ((k-1)(m+n))/k) ]/((m+n)/k)]
> > > }
> > > The above reduces to f'(c) = (f(c+n)-f(c-m))/(m+n)
> > >
> > > (1/(m+n)) \int_{c-m}^{c+n} f'(x) dx = (f(c+n)-f(c-m))/(m+n)
> > >
> > > So, what we have proved above is that f'(c) is the level magnitude of all the y-ordinates of f'(x) in the interval (c-m, c+n) and we can find the product of f'(c) and m+n to give us the area under f'(x) from c-m to c+n where c-m < c < c+n.
> > >
> > > Did you see anything there about "real numbers", you imbecile?
> > > >
> > > > We need something more than just rationals for the mean value theorem. That *something* is real numbers.
> > > No. What you need is understanding. You have NONE.
>
> > How do you know those μ_s exist?
> How can you know that you are a moron?
Prove those mu_s. How do you know they exist?

On Saturday, 5 August 2023 at 17:27:46 UTC-4, markus...
> > > > > > > > > > > > > > > > > > > > > > > https://www.academia.edu/45154026/Teaching_the_fundamental_theorem_of_calculus
> > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > I was the FIRST HUMAN to fully understand the mean value theorem and to prove constructively.
> > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > NONE of the stupid FUCKS who came before me where up to the task:
> > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > https://www.academia.edu/81300370/Mainstream_mathematics_academics_are_arrogant_and_incorrigible_ignoramuses_The_mean_value_theorem_IS_the_fundamental_theorem_of_calculus
> > > > > > > > > > > > > > > > > > > > > Your proof is invalid.
> > > > > > > > > > > > > > > > > > > > You've said that about a lot of things and every time you have been wrong.
> > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > You're wrong here yet again.
> > > > > > > > > > > > > > > > > > > > > The real mean value theorem requires real numbers.
> > > > > > > > > > > > > > > > > > > > It doesn't have anything to do with an ill-formed object you think of as a "real number".
> > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > The mean value theorem is about a level magnitude (what you erroneously call an "arithmetic mean"). It's the reason calculus works at all.
> > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > You're right - I don't understand the "limits" of your ignorance and stupidity.
> > > > > > > > > > > > > > > > > > > The mean value theorem doesn't work without real numbers. A very elementary counterexample is f(x)=0 if x²>2 and f(x)=1 if x²<2. It is continuous and differentiable everywhere on the rational number line with f'=0. If we consider [a, b]=[0, 2], we have a counterexample to the MVT (mean value theorem). From (b-a)f'(c)=f(b)-f(a) we have
> > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > f'(c)=0-1=-1.
> > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > But there is no such c. If we attempt to explicitly find c, we find that it doesn't exist.
> > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > The mean value theorem requires a real number line.
> > > > > > > > > > > > > > > > > > > See a psychiatrist soon.
> > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > You most definitely do need to see a psychiatrist. You are mentally ill.
> > > > > > > > > > > > > > > > > > > I don't need to.
> > > > > > > > > > > > > > > > > > > But you can consider f(x)=x³ over [0, 1] and see why the mean value theorem requires real numbers.
> > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > It doesn't show at all what you claim. In fact, the mvt doesn't give a shit about anything but magnitudes.
> > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > You obviously don't understand the mvt which is about a level magnitude (aka arithmetic mean in your bullshit mathematics).
> > > > > > > > > > > > > > > > > Then try it.
> > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > f(x)=x³, a=0 and b=0. Then f'(x)=3x², f(b)=1, f(a)=0.
> > > > > > > > > > > > > > > > What?! f(b)=f(0)=0, not 1, you imbecile!
> > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > <scheißen>
> > > > > > > > > > > > > > > b=0 was a typo. b=1, so f(b) is indeed one. You will get f'(c)=3c³=1, and this can't be solved with rational numbers only.
> > > > > > > > > > > > > > It still is wrong, whether a typo or not. Incoherent gibberish out of your juvenile brain.
> > > > > > > > > > > > > Can you solve f'(c)=3c²=1 without real numbers?
> > > > > > > > > > > > Yes, idiot! It is the CONSTANT known as the square root of 1/3 and there is NO number that decribes its measure.
> > > > > > > > > > > The square root of 1/3 isn't rational.
> > > > > > > > > > And so? Stating the obvious incorrectly. There is no number that describes the measure of the constant root(3).
> > > > > > > > > > > Thus the mean value theorem fails without real numbers.
> > > > > > > > > > It doesn't because there is no such thing as "real number".
> > > > > > > > > >
> > > > > > > > > > <END OF DISCUSSION>
> > > > > > > > > If there is no such c, then the mean value theorem is false. The theorem states there is a such c.
> > > > > > > > You'll have to tell me again what is your example. I stop after the first mistake I find and you get ignored.
> > > > > > > You have already gotten TWO counterexamples.
> > > > > > >
> > > > > > > f(x)=x³, a=0 and b=1. Then f'(x)=3x², f(b)=1, f(a)=0.
> > > > > > > (b-a)f'(c)=f(b)-f(a)=1.
> > > > > > >
> > > > > > > f'(c)=3c²=1.
> > > > > > > We end up with something we can't solve only using rational numbers.
> > > > > > I have already answered this question. The square root of 1/3 is not described by any number. It is a CONSTANT which is a numeric approximation but not the actual value.
> > > > > >
> > > > > > The ratio square root of 1/3 is realised from a right angled triangle with one leg equal to leg (p) of original isosceles right-angled triangle and the other leg (q) to hypotenuse of original isosceles right-angled triangle.
> > > > > >
> > > > > > Thus the ratio is p:q and this ratio has no measure. If we attempt to measure it, we end up with the same failed measure that is the constant 0.5773 (which is an approximation or failed measure of p:q).
> > > > > >
> > > > > > You fail to understand these things because you don't understand ratio and number.
> > > > > What you call "magnitudes" are in fact just real numbers.
> > > > You are just too stupid.
> > > >
> > > > Here is the proof of the mean value theorem using the 100% rigorous New Calculus:
> > > >
> > > > The New Calculus derivative:
> > > >
> > > > Given any interval (c-m,c+n), the New Calculus (henceforth NC) derivative is given by:
> > > >
> > > > f'(c)= (f(c+n)-f(c-m))/(m+n)
> > > >
> > > > f'(c) represents the slope of a secant line with end points (c-m, f(c-m)) and ( c+n, f(c+n)) that is parallel to the tangent line at x=c.
> > > > m and n are horizontal distances from c.
> > > >
> > > > The interval (c-m,c+n) can be partitioned into equal sub-intervals of (m+n)/k.
> > > >
> > > > Each sub-interval has μ_s as the abscissa of f' so that
> > > >
> > > > f'(μ_s) = [ f(c-m + ((m+n)(s+1))/k) - f(c-m + ((m+n)s)/k) ] / ((m+n)/k)
> > > >
> > > > The level magnitude (aka arithmetic mean in mainstream mathematics) is given by:
> > > >
> > > > f'(c) = (1/k) \sum_{s=1}^k f'(μ_s)
> > > >
> > > > From the above statement, we want to show that
> > > >
> > > > f'(c) = [ f'(μ_s1) + f'(μ_s2) + f'(μ_s3) + ... + f'(μ_k-1) + f'(μ_k) ]/k
> > > >
> > > > By replacing each of the means with a derivative, we have:
> > > >
> > > > f'(c) = (1/k) { [ f(c-m + (m+n)/k)-f(c-m) ]/((m+n)/k)
> > > > + [ f(c-m + 2(m+n)/k)-f(c-m + (m+n)/k) ]/((m+n)/k)]
> > > > + [ f(c-m + 3(m+n)/k)-f(c-m + 2(m+n)/k) ]/((m+n)/k)]
> > > > + ...
> > > > + [ f(c-m + ((k-1)(m+n))/k)-f(c-m + ((k-2)(m+n))/k) ]/((m+n)/k)]
> > > > + [ f(c+n)-f(c-m + ((k-1)(m+n))/k) ]/((m+n)/k)]
> > > > }
> > > > The above reduces to f'(c) = (f(c+n)-f(c-m))/(m+n)
> > > >
> > > > (1/(m+n)) \int_{c-m}^{c+n} f'(x) dx = (f(c+n)-f(c-m))/(m+n)
> > > >
> > > > So, what we have proved above is that f'(c) is the level magnitude of all the y-ordinates of f'(x) in the interval (c-m, c+n) and we can find the product of f'(c) and m+n to give us the area under f'(x) from c-m to c+n where c-m < c < c+n.
> > > >
> > > > Did you see anything there about "real numbers", you imbecile?
> > > > >
> > > > > We need something more than just rationals for the mean value theorem. That *something* is real numbers.
> > > > No. What you need is understanding. You have NONE.
> >
> > > How do you know those μ_s exist?
> > How can you know that you are a moron?
> Prove those mu_s. How do you know they exist?

söndag 6 augusti 2023 kl. 02:03:23 UTC+2 skrev Eram semper recta:
> On Saturday, 5 August 2023 at 17:27:46 UTC-4, markus...
> > > > > > > > > > > > > > > > > > > > > > > > https://www.academia.edu/45154026/Teaching_the_fundamental_theorem_of_calculus
> > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > I was the FIRST HUMAN to fully understand the mean value theorem and to prove constructively.
> > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > NONE of the stupid FUCKS who came before me where up to the task:
> > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > https://www.academia.edu/81300370/Mainstream_mathematics_academics_are_arrogant_and_incorrigible_ignoramuses_The_mean_value_theorem_IS_the_fundamental_theorem_of_calculus
> > > > > > > > > > > > > > > > > > > > > > Your proof is invalid.
> > > > > > > > > > > > > > > > > > > > > You've said that about a lot of things and every time you have been wrong.
> > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > You're wrong here yet again.
> > > > > > > > > > > > > > > > > > > > > > The real mean value theorem requires real numbers.
> > > > > > > > > > > > > > > > > > > > > It doesn't have anything to do with an ill-formed object you think of as a "real number".
> > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > The mean value theorem is about a level magnitude (what you erroneously call an "arithmetic mean"). It's the reason calculus works at all.
> > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > You're right - I don't understand the "limits" of your ignorance and stupidity.
> > > > > > > > > > > > > > > > > > > > The mean value theorem doesn't work without real numbers. A very elementary counterexample is f(x)=0 if x²>2 and f(x)=1 if x²<2. It is continuous and differentiable everywhere on the rational number line with f'=0. If we consider [a, b]=[0, 2], we have a counterexample to the MVT (mean value theorem). From (b-a)f'(c)=f(b)-f(a) we have
> > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > f'(c)=0-1=-1.
> > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > But there is no such c. If we attempt to explicitly find c, we find that it doesn't exist.
> > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > The mean value theorem requires a real number line.
> > > > > > > > > > > > > > > > > > > > See a psychiatrist soon.
> > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > You most definitely do need to see a psychiatrist. You are mentally ill.
> > > > > > > > > > > > > > > > > > > > I don't need to.
> > > > > > > > > > > > > > > > > > > > But you can consider f(x)=x³ over [0, 1] and see why the mean value theorem requires real numbers.
> > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > It doesn't show at all what you claim.. In fact, the mvt doesn't give a shit about anything but magnitudes.
> > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > You obviously don't understand the mvt which is about a level magnitude (aka arithmetic mean in your bullshit mathematics).
> > > > > > > > > > > > > > > > > > Then try it.
> > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > f(x)=x³, a=0 and b=0. Then f'(x)=3x², f(b)=1, f(a)=0.
> > > > > > > > > > > > > > > > > What?! f(b)=f(0)=0, not 1, you imbecile!
> > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > <scheißen>
> > > > > > > > > > > > > > > > b=0 was a typo. b=1, so f(b) is indeed one. You will get f'(c)=3c³=1, and this can't be solved with rational numbers only.
> > > > > > > > > > > > > > > It still is wrong, whether a typo or not. Incoherent gibberish out of your juvenile brain.
> > > > > > > > > > > > > > Can you solve f'(c)=3c²=1 without real numbers?
> > > > > > > > > > > > > Yes, idiot! It is the CONSTANT known as the square root of 1/3 and there is NO number that decribes its measure.
> > > > > > > > > > > > The square root of 1/3 isn't rational.
> > > > > > > > > > > And so? Stating the obvious incorrectly. There is no number that describes the measure of the constant root(3).
> > > > > > > > > > > > Thus the mean value theorem fails without real numbers.
> > > > > > > > > > > It doesn't because there is no such thing as "real number".
> > > > > > > > > > >
> > > > > > > > > > > <END OF DISCUSSION>
> > > > > > > > > > If there is no such c, then the mean value theorem is false. The theorem states there is a such c.
> > > > > > > > > You'll have to tell me again what is your example. I stop after the first mistake I find and you get ignored.
> > > > > > > > You have already gotten TWO counterexamples.
> > > > > > > >
> > > > > > > > f(x)=x³, a=0 and b=1. Then f'(x)=3x², f(b)=1, f(a)=0.
> > > > > > > > (b-a)f'(c)=f(b)-f(a)=1.
> > > > > > > >
> > > > > > > > f'(c)=3c²=1.
> > > > > > > > We end up with something we can't solve only using rational numbers.
> > > > > > > I have already answered this question. The square root of 1/3 is not described by any number. It is a CONSTANT which is a numeric approximation but not the actual value.
> > > > > > >
> > > > > > > The ratio square root of 1/3 is realised from a right angled triangle with one leg equal to leg (p) of original isosceles right-angled triangle and the other leg (q) to hypotenuse of original isosceles right-angled triangle.
> > > > > > >
> > > > > > > Thus the ratio is p:q and this ratio has no measure. If we attempt to measure it, we end up with the same failed measure that is the constant 0.5773 (which is an approximation or failed measure of p:q).
> > > > > > >
> > > > > > > You fail to understand these things because you don't understand ratio and number.
> > > > > > What you call "magnitudes" are in fact just real numbers.
> > > > > You are just too stupid.
> > > > >
> > > > > Here is the proof of the mean value theorem using the 100% rigorous New Calculus:
> > > > >
> > > > > The New Calculus derivative:
> > > > >
> > > > > Given any interval (c-m,c+n), the New Calculus (henceforth NC) derivative is given by:
> > > > >
> > > > > f'(c)= (f(c+n)-f(c-m))/(m+n)
> > > > >
> > > > > f'(c) represents the slope of a secant line with end points (c-m, f(c-m)) and ( c+n, f(c+n)) that is parallel to the tangent line at x=c.
> > > > > m and n are horizontal distances from c.
> > > > >
> > > > > The interval (c-m,c+n) can be partitioned into equal sub-intervals of (m+n)/k.
> > > > >
> > > > > Each sub-interval has μ_s as the abscissa of f' so that
> > > > >
> > > > > f'(μ_s) = [ f(c-m + ((m+n)(s+1))/k) - f(c-m + ((m+n)s)/k) ] / ((m+n)/k)
> > > > >
> > > > > The level magnitude (aka arithmetic mean in mainstream mathematics) is given by:
> > > > >
> > > > > f'(c) = (1/k) \sum_{s=1}^k f'(μ_s)
> > > > >
> > > > > From the above statement, we want to show that
> > > > >
> > > > > f'(c) = [ f'(μ_s1) + f'(μ_s2) + f'(μ_s3) + ... + f'(μ_k-1) + f'(μ_k) ]/k
> > > > >
> > > > > By replacing each of the means with a derivative, we have:
> > > > >
> > > > > f'(c) = (1/k) { [ f(c-m + (m+n)/k)-f(c-m) ]/((m+n)/k)
> > > > > + [ f(c-m + 2(m+n)/k)-f(c-m + (m+n)/k) ]/((m+n)/k)]
> > > > > + [ f(c-m + 3(m+n)/k)-f(c-m + 2(m+n)/k) ]/((m+n)/k)]
> > > > > + ...
> > > > > + [ f(c-m + ((k-1)(m+n))/k)-f(c-m + ((k-2)(m+n))/k) ]/((m+n)/k)]
> > > > > + [ f(c+n)-f(c-m + ((k-1)(m+n))/k) ]/((m+n)/k)]
> > > > > }
> > > > > The above reduces to f'(c) = (f(c+n)-f(c-m))/(m+n)
> > > > >
> > > > > (1/(m+n)) \int_{c-m}^{c+n} f'(x) dx = (f(c+n)-f(c-m))/(m+n)
> > > > >
> > > > > So, what we have proved above is that f'(c) is the level magnitude of all the y-ordinates of f'(x) in the interval (c-m, c+n) and we can find the product of f'(c) and m+n to give us the area under f'(x) from c-m to c+n where c-m < c < c+n.
> > > > >
> > > > > Did you see anything there about "real numbers", you imbecile?
> > > > > >
> > > > > > We need something more than just rationals for the mean value theorem. That *something* is real numbers.
> > > > > No. What you need is understanding. You have NONE.
> > >
> > > > How do you know those μ_s exist?
> > > How can you know that you are a moron?
> > Prove those mu_s. How do you know they exist?
> It's simple but probably too complicated for you. I will not give you the answer but try to lead you to it.
>
> If f is defined everywhere in (a,b) then is it true or false that at least one y ordinate represents the level magnitude (aka arithmetic mean)?
Yes, it's true by the mean value theorem. Which you seek to prove. That's circular.

On Sunday, 6 August 2023 at 14:19:23 UTC-4, markus...@gmail.com wrote:
> söndag 6 augusti 2023 kl. 02:03:23 UTC+2 skrev Eram semper recta:
> > On Saturday, 5 August 2023 at 17:27:46 UTC-4, markus...
> > > > > > > > > > > > > > > > > > > > > > > > > https://www.academia.edu/45154026/Teaching_the_fundamental_theorem_of_calculus
> > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > I was the FIRST HUMAN to fully understand the mean value theorem and to prove constructively.
> > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > NONE of the stupid FUCKS who came before me where up to the task:
> > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > https://www.academia.edu/81300370/Mainstream_mathematics_academics_are_arrogant_and_incorrigible_ignoramuses_The_mean_value_theorem_IS_the_fundamental_theorem_of_calculus
> > > > > > > > > > > > > > > > > > > > > > > Your proof is invalid.
> > > > > > > > > > > > > > > > > > > > > > You've said that about a lot of things and every time you have been wrong.
> > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > You're wrong here yet again.
> > > > > > > > > > > > > > > > > > > > > > > The real mean value theorem requires real numbers.
> > > > > > > > > > > > > > > > > > > > > > It doesn't have anything to do with an ill-formed object you think of as a "real number".
> > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > The mean value theorem is about a level magnitude (what you erroneously call an "arithmetic mean"). It's the reason calculus works at all.
> > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > You're right - I don't understand the "limits" of your ignorance and stupidity.
> > > > > > > > > > > > > > > > > > > > > The mean value theorem doesn't work without real numbers. A very elementary counterexample is f(x)=0 if x²>2 and f(x)=1 if x²<2. It is continuous and differentiable everywhere on the rational number line with f'=0. If we consider [a, b]=[0, 2], we have a counterexample to the MVT (mean value theorem). From (b-a)f'(c)=f(b)-f(a) we have
> > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > f'(c)=0-1=-1.
> > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > But there is no such c. If we attempt to explicitly find c, we find that it doesn't exist.
> > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > The mean value theorem requires a real number line.
> > > > > > > > > > > > > > > > > > > > > See a psychiatrist soon.
> > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > You most definitely do need to see a psychiatrist. You are mentally ill.
> > > > > > > > > > > > > > > > > > > > > I don't need to.
> > > > > > > > > > > > > > > > > > > > > But you can consider f(x)=x³ over [0, 1] and see why the mean value theorem requires real numbers.
> > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > It doesn't show at all what you claim. In fact, the mvt doesn't give a shit about anything but magnitudes.
> > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > You obviously don't understand the mvt which is about a level magnitude (aka arithmetic mean in your bullshit mathematics).
> > > > > > > > > > > > > > > > > > > Then try it.
> > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > f(x)=x³, a=0 and b=0. Then f'(x)=3x², f(b)=1, f(a)=0.
> > > > > > > > > > > > > > > > > > What?! f(b)=f(0)=0, not 1, you imbecile!
> > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > <scheißen>
> > > > > > > > > > > > > > > > > b=0 was a typo. b=1, so f(b) is indeed one. You will get f'(c)=3c³=1, and this can't be solved with rational numbers only.
> > > > > > > > > > > > > > > > It still is wrong, whether a typo or not. Incoherent gibberish out of your juvenile brain.
> > > > > > > > > > > > > > > Can you solve f'(c)=3c²=1 without real numbers?
> > > > > > > > > > > > > > Yes, idiot! It is the CONSTANT known as the square root of 1/3 and there is NO number that decribes its measure.
> > > > > > > > > > > > > The square root of 1/3 isn't rational.
> > > > > > > > > > > > And so? Stating the obvious incorrectly. There is no number that describes the measure of the constant root(3).
> > > > > > > > > > > > > Thus the mean value theorem fails without real numbers.
> > > > > > > > > > > > It doesn't because there is no such thing as "real number".
> > > > > > > > > > > >
> > > > > > > > > > > > <END OF DISCUSSION>
> > > > > > > > > > > If there is no such c, then the mean value theorem is false. The theorem states there is a such c.
> > > > > > > > > > You'll have to tell me again what is your example. I stop after the first mistake I find and you get ignored.
> > > > > > > > > You have already gotten TWO counterexamples.
> > > > > > > > >
> > > > > > > > > f(x)=x³, a=0 and b=1. Then f'(x)=3x², f(b)=1, f(a)=0.
> > > > > > > > > (b-a)f'(c)=f(b)-f(a)=1.
> > > > > > > > >
> > > > > > > > > f'(c)=3c²=1.
> > > > > > > > > We end up with something we can't solve only using rational numbers.
> > > > > > > > I have already answered this question. The square root of 1/3 is not described by any number. It is a CONSTANT which is a numeric approximation but not the actual value.
> > > > > > > >
> > > > > > > > The ratio square root of 1/3 is realised from a right angled triangle with one leg equal to leg (p) of original isosceles right-angled triangle and the other leg (q) to hypotenuse of original isosceles right-angled triangle.
> > > > > > > >
> > > > > > > > Thus the ratio is p:q and this ratio has no measure. If we attempt to measure it, we end up with the same failed measure that is the constant 0.5773 (which is an approximation or failed measure of p:q).
> > > > > > > >
> > > > > > > > You fail to understand these things because you don't understand ratio and number.
> > > > > > > What you call "magnitudes" are in fact just real numbers.
> > > > > > You are just too stupid.
> > > > > >
> > > > > > Here is the proof of the mean value theorem using the 100% rigorous New Calculus:
> > > > > >
> > > > > > The New Calculus derivative:
> > > > > >
> > > > > > Given any interval (c-m,c+n), the New Calculus (henceforth NC) derivative is given by:
> > > > > >
> > > > > > f'(c)= (f(c+n)-f(c-m))/(m+n)
> > > > > >
> > > > > > f'(c) represents the slope of a secant line with end points (c-m, f(c-m)) and ( c+n, f(c+n)) that is parallel to the tangent line at x=c..
> > > > > > m and n are horizontal distances from c.
> > > > > >
> > > > > > The interval (c-m,c+n) can be partitioned into equal sub-intervals of (m+n)/k.
> > > > > >
> > > > > > Each sub-interval has μ_s as the abscissa of f' so that
> > > > > >
> > > > > > f'(μ_s) = [ f(c-m + ((m+n)(s+1))/k) - f(c-m + ((m+n)s)/k) ] / ((m+n)/k)
> > > > > >
> > > > > > The level magnitude (aka arithmetic mean in mainstream mathematics) is given by:
> > > > > >
> > > > > > f'(c) = (1/k) \sum_{s=1}^k f'(μ_s)
> > > > > >
> > > > > > From the above statement, we want to show that
> > > > > >
> > > > > > f'(c) = [ f'(μ_s1) + f'(μ_s2) + f'(μ_s3) + ... + f'(μ_k-1) + f'(μ_k) ]/k
> > > > > >
> > > > > > By replacing each of the means with a derivative, we have:
> > > > > >
> > > > > > f'(c) = (1/k) { [ f(c-m + (m+n)/k)-f(c-m) ]/((m+n)/k)
> > > > > > + [ f(c-m + 2(m+n)/k)-f(c-m + (m+n)/k) ]/((m+n)/k)]
> > > > > > + [ f(c-m + 3(m+n)/k)-f(c-m + 2(m+n)/k) ]/((m+n)/k)]
> > > > > > + ...
> > > > > > + [ f(c-m + ((k-1)(m+n))/k)-f(c-m + ((k-2)(m+n))/k) ]/((m+n)/k)]
> > > > > > + [ f(c+n)-f(c-m + ((k-1)(m+n))/k) ]/((m+n)/k)]
> > > > > > }
> > > > > > The above reduces to f'(c) = (f(c+n)-f(c-m))/(m+n)
> > > > > >
> > > > > > (1/(m+n)) \int_{c-m}^{c+n} f'(x) dx = (f(c+n)-f(c-m))/(m+n)
> > > > > >
> > > > > > So, what we have proved above is that f'(c) is the level magnitude of all the y-ordinates of f'(x) in the interval (c-m, c+n) and we can find the product of f'(c) and m+n to give us the area under f'(x) from c-m to c+n where c-m < c < c+n.
> > > > > >
> > > > > > Did you see anything there about "real numbers", you imbecile?
> > > > > > >
> > > > > > > We need something more than just rationals for the mean value theorem. That *something* is real numbers.
> > > > > > No. What you need is understanding. You have NONE.
> > > >
> > > > > How do you know those μ_s exist?
> > > > How can you know that you are a moron?
> > > Prove those mu_s. How do you know they exist?
> > It's simple but probably too complicated for you. I will not give you the answer but try to lead you to it.
> >
> > If f is defined everywhere in (a,b) then is it true or false that at least one y ordinate represents the level magnitude (aka arithmetic mean)?
> Yes, it's true by the mean value theorem.

måndag 7 augusti 2023 kl. 00:35:54 UTC+2 skrev Eram semper recta:
> On Sunday, 6 August 2023 at 14:19:23 UTC-4, markus...@gmail.com wrote:
> > söndag 6 augusti 2023 kl. 02:03:23 UTC+2 skrev Eram semper recta:
> > > On Saturday, 5 August 2023 at 17:27:46 UTC-4, markus...
> > > > > > > > > > > > > > > > > > > > > > > > > > https://www.academia.edu/45154026/Teaching_the_fundamental_theorem_of_calculus
> > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > I was the FIRST HUMAN to fully understand the mean value theorem and to prove constructively.
> > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > NONE of the stupid FUCKS who came before me where up to the task:
> > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > https://www.academia.edu/81300370/Mainstream_mathematics_academics_are_arrogant_and_incorrigible_ignoramuses_The_mean_value_theorem_IS_the_fundamental_theorem_of_calculus
> > > > > > > > > > > > > > > > > > > > > > > > Your proof is invalid.
> > > > > > > > > > > > > > > > > > > > > > > You've said that about a lot of things and every time you have been wrong.
> > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > You're wrong here yet again.
> > > > > > > > > > > > > > > > > > > > > > > > The real mean value theorem requires real numbers.
> > > > > > > > > > > > > > > > > > > > > > > It doesn't have anything to do with an ill-formed object you think of as a "real number".
> > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > The mean value theorem is about a level magnitude (what you erroneously call an "arithmetic mean"). It's the reason calculus works at all.
> > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > You're right - I don't understand the "limits" of your ignorance and stupidity.
> > > > > > > > > > > > > > > > > > > > > > The mean value theorem doesn't work without real numbers. A very elementary counterexample is f(x)=0 if x²>2 and f(x)=1 if x²<2. It is continuous and differentiable everywhere on the rational number line with f'=0. If we consider [a, b]=[0, 2], we have a counterexample to the MVT (mean value theorem). From (b-a)f'(c)=f(b)-f(a) we have
> > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > f'(c)=0-1=-1.
> > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > But there is no such c. If we attempt to explicitly find c, we find that it doesn't exist.
> > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > The mean value theorem requires a real number line.
> > > > > > > > > > > > > > > > > > > > > > See a psychiatrist soon.
> > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > You most definitely do need to see a psychiatrist. You are mentally ill.
> > > > > > > > > > > > > > > > > > > > > > I don't need to.
> > > > > > > > > > > > > > > > > > > > > > But you can consider f(x)=x³ over [0, 1] and see why the mean value theorem requires real numbers..
> > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > It doesn't show at all what you claim. In fact, the mvt doesn't give a shit about anything but magnitudes.
> > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > You obviously don't understand the mvt which is about a level magnitude (aka arithmetic mean in your bullshit mathematics).
> > > > > > > > > > > > > > > > > > > > Then try it.
> > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > f(x)=x³, a=0 and b=0. Then f'(x)=3x², f(b)=1, f(a)=0.
> > > > > > > > > > > > > > > > > > > What?! f(b)=f(0)=0, not 1, you imbecile!
> > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > <scheißen>
> > > > > > > > > > > > > > > > > > b=0 was a typo. b=1, so f(b) is indeed one. You will get f'(c)=3c³=1, and this can't be solved with rational numbers only.
> > > > > > > > > > > > > > > > > It still is wrong, whether a typo or not. Incoherent gibberish out of your juvenile brain.
> > > > > > > > > > > > > > > > Can you solve f'(c)=3c²=1 without real numbers?
> > > > > > > > > > > > > > > Yes, idiot! It is the CONSTANT known as the square root of 1/3 and there is NO number that decribes its measure.
> > > > > > > > > > > > > > The square root of 1/3 isn't rational.
> > > > > > > > > > > > > And so? Stating the obvious incorrectly. There is no number that describes the measure of the constant root(3).
> > > > > > > > > > > > > > Thus the mean value theorem fails without real numbers.
> > > > > > > > > > > > > It doesn't because there is no such thing as "real number".
> > > > > > > > > > > > >
> > > > > > > > > > > > > <END OF DISCUSSION>
> > > > > > > > > > > > If there is no such c, then the mean value theorem is false. The theorem states there is a such c.
> > > > > > > > > > > You'll have to tell me again what is your example. I stop after the first mistake I find and you get ignored.
> > > > > > > > > > You have already gotten TWO counterexamples.
> > > > > > > > > >
> > > > > > > > > > f(x)=x³, a=0 and b=1. Then f'(x)=3x², f(b)=1, f(a)=0.
> > > > > > > > > > (b-a)f'(c)=f(b)-f(a)=1.
> > > > > > > > > >
> > > > > > > > > > f'(c)=3c²=1.
> > > > > > > > > > We end up with something we can't solve only using rational numbers.
> > > > > > > > > I have already answered this question. The square root of 1/3 is not described by any number. It is a CONSTANT which is a numeric approximation but not the actual value.
> > > > > > > > >
> > > > > > > > > The ratio square root of 1/3 is realised from a right angled triangle with one leg equal to leg (p) of original isosceles right-angled triangle and the other leg (q) to hypotenuse of original isosceles right-angled triangle.
> > > > > > > > >
> > > > > > > > > Thus the ratio is p:q and this ratio has no measure. If we attempt to measure it, we end up with the same failed measure that is the constant 0.5773 (which is an approximation or failed measure of p:q).
> > > > > > > > >
> > > > > > > > > You fail to understand these things because you don't understand ratio and number.
> > > > > > > > What you call "magnitudes" are in fact just real numbers.
> > > > > > > You are just too stupid.
> > > > > > >
> > > > > > > Here is the proof of the mean value theorem using the 100% rigorous New Calculus:
> > > > > > >
> > > > > > > The New Calculus derivative:
> > > > > > >
> > > > > > > Given any interval (c-m,c+n), the New Calculus (henceforth NC) derivative is given by:
> > > > > > >
> > > > > > > f'(c)= (f(c+n)-f(c-m))/(m+n)
> > > > > > >
> > > > > > > f'(c) represents the slope of a secant line with end points (c-m, f(c-m)) and ( c+n, f(c+n)) that is parallel to the tangent line at x=c.
> > > > > > > m and n are horizontal distances from c.
> > > > > > >
> > > > > > > The interval (c-m,c+n) can be partitioned into equal sub-intervals of (m+n)/k.
> > > > > > >
> > > > > > > Each sub-interval has μ_s as the abscissa of f' so that
> > > > > > >
> > > > > > > f'(μ_s) = [ f(c-m + ((m+n)(s+1))/k) - f(c-m + ((m+n)s)/k) ] / ((m+n)/k)
> > > > > > >
> > > > > > > The level magnitude (aka arithmetic mean in mainstream mathematics) is given by:
> > > > > > >
> > > > > > > f'(c) = (1/k) \sum_{s=1}^k f'(μ_s)
> > > > > > >
> > > > > > > From the above statement, we want to show that
> > > > > > >
> > > > > > > f'(c) = [ f'(μ_s1) + f'(μ_s2) + f'(μ_s3) + .... + f'(μ_k-1) + f'(μ_k) ]/k
> > > > > > >
> > > > > > > By replacing each of the means with a derivative, we have:
> > > > > > >
> > > > > > > f'(c) = (1/k) { [ f(c-m + (m+n)/k)-f(c-m) ]/((m+n)/k)
> > > > > > > + [ f(c-m + 2(m+n)/k)-f(c-m + (m+n)/k) ]/((m+n)/k)]
> > > > > > > + [ f(c-m + 3(m+n)/k)-f(c-m + 2(m+n)/k) ]/((m+n)/k)]
> > > > > > > + ...
> > > > > > > + [ f(c-m + ((k-1)(m+n))/k)-f(c-m + ((k-2)(m+n))/k) ]/((m+n)/k)]
> > > > > > > + [ f(c+n)-f(c-m + ((k-1)(m+n))/k) ]/((m+n)/k)]
> > > > > > > }
> > > > > > > The above reduces to f'(c) = (f(c+n)-f(c-m))/(m+n)
> > > > > > >
> > > > > > > (1/(m+n)) \int_{c-m}^{c+n} f'(x) dx = (f(c+n)-f(c-m))/(m+n)
> > > > > > >
> > > > > > > So, what we have proved above is that f'(c) is the level magnitude of all the y-ordinates of f'(x) in the interval (c-m, c+n) and we can find the product of f'(c) and m+n to give us the area under f'(x) from c-m to c+n where c-m < c < c+n.
> > > > > > >
> > > > > > > Did you see anything there about "real numbers", you imbecile?
> > > > > > > >
> > > > > > > > We need something more than just rationals for the mean value theorem. That *something* is real numbers.
> > > > > > > No. What you need is understanding. You have NONE.
> > > > >
> > > > > > How do you know those μ_s exist?
> > > > > How can you know that you are a moron?
> > > > Prove those mu_s. How do you know they exist?
> > > It's simple but probably too complicated for you. I will not give you the answer but try to lead you to it.
> > >
> > > If f is defined everywhere in (a,b) then is it true or false that at least one y ordinate represents the level magnitude (aka arithmetic mean)?
> > Yes, it's true by the mean value theorem.
> Not by the mean value theorem because one will find a level magnitude even if the curve is not smooth. The mvt applies only to smooth curves.
> f need not be smooth anywhere in the given interval and there will still be an arithmetic mean. Do you disagree still?
> > Which you seek to prove. That's circular.
> No. That's wrong.
How do you get those mu_s then?

On Monday, 7 August 2023 at 05:02:12 UTC-4, markus...@gmail.com wrote:
> måndag 7 augusti 2023 kl. 00:35:54 UTC+2 skrev Eram semper recta:
> > On Sunday, 6 August 2023 at 14:19:23 UTC-4, markus...@gmail.com wrote:
> > > söndag 6 augusti 2023 kl. 02:03:23 UTC+2 skrev Eram semper recta:
> > > > On Saturday, 5 August 2023 at 17:27:46 UTC-4, markus...
> > > > > > > > > > > > > > > > > > > > > > > > > > > https://www.academia.edu/45154026/Teaching_the_fundamental_theorem_of_calculus
> > > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > > I was the FIRST HUMAN to fully understand the mean value theorem and to prove constructively.
> > > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > > NONE of the stupid FUCKS who came before me where up to the task:
> > > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > > https://www.academia.edu/81300370/Mainstream_mathematics_academics_are_arrogant_and_incorrigible_ignoramuses_The_mean_value_theorem_IS_the_fundamental_theorem_of_calculus
> > > > > > > > > > > > > > > > > > > > > > > > > Your proof is invalid.
> > > > > > > > > > > > > > > > > > > > > > > > You've said that about a lot of things and every time you have been wrong.
> > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > You're wrong here yet again..
> > > > > > > > > > > > > > > > > > > > > > > > > The real mean value theorem requires real numbers.
> > > > > > > > > > > > > > > > > > > > > > > > It doesn't have anything to do with an ill-formed object you think of as a "real number".
> > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > The mean value theorem is about a level magnitude (what you erroneously call an "arithmetic mean"). It's the reason calculus works at all.
> > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > You're right - I don't understand the "limits" of your ignorance and stupidity.
> > > > > > > > > > > > > > > > > > > > > > > The mean value theorem doesn't work without real numbers. A very elementary counterexample is f(x)=0 if x²>2 and f(x)=1 if x²<2. It is continuous and differentiable everywhere on the rational number line with f'=0. If we consider [a, b]=[0, 2], we have a counterexample to the MVT (mean value theorem). From (b-a)f'(c)=f(b)-f(a) we have
> > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > f'(c)=0-1=-1.
> > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > But there is no such c. If we attempt to explicitly find c, we find that it doesn't exist.
> > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > The mean value theorem requires a real number line.
> > > > > > > > > > > > > > > > > > > > > > > See a psychiatrist soon.
> > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > You most definitely do need to see a psychiatrist. You are mentally ill.
> > > > > > > > > > > > > > > > > > > > > > > I don't need to.
> > > > > > > > > > > > > > > > > > > > > > > But you can consider f(x)=x³ over [0, 1] and see why the mean value theorem requires real numbers..
> > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > It doesn't show at all what you claim. In fact, the mvt doesn't give a shit about anything but magnitudes.
> > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > You obviously don't understand the mvt which is about a level magnitude (aka arithmetic mean in your bullshit mathematics).
> > > > > > > > > > > > > > > > > > > > > Then try it.
> > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > f(x)=x³, a=0 and b=0. Then f'(x)=3x², f(b)=1, f(a)=0.
> > > > > > > > > > > > > > > > > > > > What?! f(b)=f(0)=0, not 1, you imbecile!
> > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > <scheißen>
> > > > > > > > > > > > > > > > > > > b=0 was a typo. b=1, so f(b) is indeed one. You will get f'(c)=3c³=1, and this can't be solved with rational numbers only.
> > > > > > > > > > > > > > > > > > It still is wrong, whether a typo or not. Incoherent gibberish out of your juvenile brain.
> > > > > > > > > > > > > > > > > Can you solve f'(c)=3c²=1 without real numbers?
> > > > > > > > > > > > > > > > Yes, idiot! It is the CONSTANT known as the square root of 1/3 and there is NO number that decribes its measure.
> > > > > > > > > > > > > > > The square root of 1/3 isn't rational.
> > > > > > > > > > > > > > And so? Stating the obvious incorrectly. There is no number that describes the measure of the constant root(3).
> > > > > > > > > > > > > > > Thus the mean value theorem fails without real numbers.
> > > > > > > > > > > > > > It doesn't because there is no such thing as "real number".
> > > > > > > > > > > > > >
> > > > > > > > > > > > > > <END OF DISCUSSION>
> > > > > > > > > > > > > If there is no such c, then the mean value theorem is false. The theorem states there is a such c.
> > > > > > > > > > > > You'll have to tell me again what is your example. I stop after the first mistake I find and you get ignored.
> > > > > > > > > > > You have already gotten TWO counterexamples.
> > > > > > > > > > >
> > > > > > > > > > > f(x)=x³, a=0 and b=1. Then f'(x)=3x², f(b)=1, f(a)=0.
> > > > > > > > > > > (b-a)f'(c)=f(b)-f(a)=1.
> > > > > > > > > > >
> > > > > > > > > > > f'(c)=3c²=1.
> > > > > > > > > > > We end up with something we can't solve only using rational numbers.
> > > > > > > > > > I have already answered this question. The square root of 1/3 is not described by any number. It is a CONSTANT which is a numeric approximation but not the actual value.
> > > > > > > > > >
> > > > > > > > > > The ratio square root of 1/3 is realised from a right angled triangle with one leg equal to leg (p) of original isosceles right-angled triangle and the other leg (q) to hypotenuse of original isosceles right-angled triangle.
> > > > > > > > > >
> > > > > > > > > > Thus the ratio is p:q and this ratio has no measure. If we attempt to measure it, we end up with the same failed measure that is the constant 0.5773 (which is an approximation or failed measure of p:q).
> > > > > > > > > >
> > > > > > > > > > You fail to understand these things because you don't understand ratio and number.
> > > > > > > > > What you call "magnitudes" are in fact just real numbers.
> > > > > > > > You are just too stupid.
> > > > > > > >
> > > > > > > > Here is the proof of the mean value theorem using the 100% rigorous New Calculus:
> > > > > > > >
> > > > > > > > The New Calculus derivative:
> > > > > > > >
> > > > > > > > Given any interval (c-m,c+n), the New Calculus (henceforth NC) derivative is given by:
> > > > > > > >
> > > > > > > > f'(c)= (f(c+n)-f(c-m))/(m+n)
> > > > > > > >
> > > > > > > > f'(c) represents the slope of a secant line with end points (c-m, f(c-m)) and ( c+n, f(c+n)) that is parallel to the tangent line at x=c.
> > > > > > > > m and n are horizontal distances from c.
> > > > > > > >
> > > > > > > > The interval (c-m,c+n) can be partitioned into equal sub-intervals of (m+n)/k.
> > > > > > > >
> > > > > > > > Each sub-interval has μ_s as the abscissa of f' so that
> > > > > > > >
> > > > > > > > f'(μ_s) = [ f(c-m + ((m+n)(s+1))/k) - f(c-m + ((m+n)s)/k) ] / ((m+n)/k)
> > > > > > > >
> > > > > > > > The level magnitude (aka arithmetic mean in mainstream mathematics) is given by:
> > > > > > > >
> > > > > > > > f'(c) = (1/k) \sum_{s=1}^k f'(μ_s)
> > > > > > > >
> > > > > > > > From the above statement, we want to show that
> > > > > > > >
> > > > > > > > f'(c) = [ f'(μ_s1) + f'(μ_s2) + f'(μ_s3) + ... + f'(μ_k-1) + f'(μ_k) ]/k
> > > > > > > >
> > > > > > > > By replacing each of the means with a derivative, we have:
> > > > > > > >
> > > > > > > > f'(c) = (1/k) { [ f(c-m + (m+n)/k)-f(c-m) ]/((m+n)/k)
> > > > > > > > + [ f(c-m + 2(m+n)/k)-f(c-m + (m+n)/k) ]/((m+n)/k)]
> > > > > > > > + [ f(c-m + 3(m+n)/k)-f(c-m + 2(m+n)/k) ]/((m+n)/k)]
> > > > > > > > + ...
> > > > > > > > + [ f(c-m + ((k-1)(m+n))/k)-f(c-m + ((k-2)(m+n))/k) ]/((m+n)/k)]
> > > > > > > > + [ f(c+n)-f(c-m + ((k-1)(m+n))/k) ]/((m+n)/k)]
> > > > > > > > }
> > > > > > > > The above reduces to f'(c) = (f(c+n)-f(c-m))/(m+n)
> > > > > > > >
> > > > > > > > (1/(m+n)) \int_{c-m}^{c+n} f'(x) dx = (f(c+n)-f(c-m))/(m+n)
> > > > > > > >
> > > > > > > > So, what we have proved above is that f'(c) is the level magnitude of all the y-ordinates of f'(x) in the interval (c-m, c+n) and we can find the product of f'(c) and m+n to give us the area under f'(x) from c-m to c+n where c-m < c < c+n.
> > > > > > > >
> > > > > > > > Did you see anything there about "real numbers", you imbecile?
> > > > > > > > >
> > > > > > > > > We need something more than just rationals for the mean value theorem. That *something* is real numbers.
> > > > > > > > No. What you need is understanding. You have NONE.
> > > > > >
> > > > > > > How do you know those μ_s exist?
> > > > > > How can you know that you are a moron?
> > > > > Prove those mu_s. How do you know they exist?
> > > > It's simple but probably too complicated for you. I will not give you the answer but try to lead you to it.
> > > >
> > > > If f is defined everywhere in (a,b) then is it true or false that at least one y ordinate represents the level magnitude (aka arithmetic mean)?
> > > Yes, it's true by the mean value theorem.
> > Not by the mean value theorem because one will find a level magnitude even if the curve is not smooth. The mvt applies only to smooth curves.
> > f need not be smooth anywhere in the given interval and there will still be an arithmetic mean. Do you disagree still?
> > > Which you seek to prove. That's circular.
> > No. That's wrong.
> How do you get those mu_s then?

måndag 7 augusti 2023 kl. 14:27:42 UTC+2 skrev Eram semper recta:
> On Monday, 7 August 2023 at 05:02:12 UTC-4, markus...@gmail.com wrote:
> > måndag 7 augusti 2023 kl. 00:35:54 UTC+2 skrev Eram semper recta:
> > > On Sunday, 6 August 2023 at 14:19:23 UTC-4, markus...@gmail.com wrote:
> > > > söndag 6 augusti 2023 kl. 02:03:23 UTC+2 skrev Eram semper recta:
> > > > > On Saturday, 5 August 2023 at 17:27:46 UTC-4, markus...
> > > > > > > > > > > > > > > > > > > > > > > > > > > > https://www.academia.edu/45154026/Teaching_the_fundamental_theorem_of_calculus
> > > > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > > > I was the FIRST HUMAN to fully understand the mean value theorem and to prove constructively.
> > > > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > > > NONE of the stupid FUCKS who came before me where up to the task:
> > > > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > > > https://www.academia.edu/81300370/Mainstream_mathematics_academics_are_arrogant_and_incorrigible_ignoramuses_The_mean_value_theorem_IS_the_fundamental_theorem_of_calculus
> > > > > > > > > > > > > > > > > > > > > > > > > > Your proof is invalid.
> > > > > > > > > > > > > > > > > > > > > > > > > You've said that about a lot of things and every time you have been wrong.
> > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > You're wrong here yet again.
> > > > > > > > > > > > > > > > > > > > > > > > > > The real mean value theorem requires real numbers.
> > > > > > > > > > > > > > > > > > > > > > > > > It doesn't have anything to do with an ill-formed object you think of as a "real number".
> > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > The mean value theorem is about a level magnitude (what you erroneously call an "arithmetic mean"). It's the reason calculus works at all.
> > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > You're right - I don't understand the "limits" of your ignorance and stupidity.
> > > > > > > > > > > > > > > > > > > > > > > > The mean value theorem doesn't work without real numbers. A very elementary counterexample is f(x)=0 if x²>2 and f(x)=1 if x²<2. It is continuous and differentiable everywhere on the rational number line with f'=0. If we consider [a, b]=[0, 2], we have a counterexample to the MVT (mean value theorem). From (b-a)f'(c)=f(b)-f(a) we have
> > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > f'(c)=0-1=-1.
> > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > But there is no such c. If we attempt to explicitly find c, we find that it doesn't exist.
> > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > The mean value theorem requires a real number line.
> > > > > > > > > > > > > > > > > > > > > > > > See a psychiatrist soon.
> > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > You most definitely do need to see a psychiatrist. You are mentally ill.
> > > > > > > > > > > > > > > > > > > > > > > > I don't need to.
> > > > > > > > > > > > > > > > > > > > > > > > But you can consider f(x)=x³ over [0, 1] and see why the mean value theorem requires real numbers.
> > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > It doesn't show at all what you claim. In fact, the mvt doesn't give a shit about anything but magnitudes.
> > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > You obviously don't understand the mvt which is about a level magnitude (aka arithmetic mean in your bullshit mathematics).
> > > > > > > > > > > > > > > > > > > > > > Then try it.
> > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > f(x)=x³, a=0 and b=0.. Then f'(x)=3x², f(b)=1, f(a)=0.
> > > > > > > > > > > > > > > > > > > > > What?! f(b)=f(0)=0, not 1, you imbecile!
> > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > <scheißen>
> > > > > > > > > > > > > > > > > > > > b=0 was a typo. b=1, so f(b) is indeed one. You will get f'(c)=3c³=1, and this can't be solved with rational numbers only.
> > > > > > > > > > > > > > > > > > > It still is wrong, whether a typo or not. Incoherent gibberish out of your juvenile brain.
> > > > > > > > > > > > > > > > > > Can you solve f'(c)=3c²=1 without real numbers?
> > > > > > > > > > > > > > > > > Yes, idiot! It is the CONSTANT known as the square root of 1/3 and there is NO number that decribes its measure.
> > > > > > > > > > > > > > > > The square root of 1/3 isn't rational.
> > > > > > > > > > > > > > > And so? Stating the obvious incorrectly. There is no number that describes the measure of the constant root(3).
> > > > > > > > > > > > > > > > Thus the mean value theorem fails without real numbers.
> > > > > > > > > > > > > > > It doesn't because there is no such thing as "real number".
> > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > <END OF DISCUSSION>
> > > > > > > > > > > > > > If there is no such c, then the mean value theorem is false. The theorem states there is a such c.
> > > > > > > > > > > > > You'll have to tell me again what is your example.. I stop after the first mistake I find and you get ignored.
> > > > > > > > > > > > You have already gotten TWO counterexamples.
> > > > > > > > > > > >
> > > > > > > > > > > > f(x)=x³, a=0 and b=1. Then f'(x)=3x², f(b)=1, f(a)=0.
> > > > > > > > > > > > (b-a)f'(c)=f(b)-f(a)=1.
> > > > > > > > > > > >
> > > > > > > > > > > > f'(c)=3c²=1.
> > > > > > > > > > > > We end up with something we can't solve only using rational numbers.
> > > > > > > > > > > I have already answered this question. The square root of 1/3 is not described by any number. It is a CONSTANT which is a numeric approximation but not the actual value.
> > > > > > > > > > >
> > > > > > > > > > > The ratio square root of 1/3 is realised from a right angled triangle with one leg equal to leg (p) of original isosceles right-angled triangle and the other leg (q) to hypotenuse of original isosceles right-angled triangle.
> > > > > > > > > > >
> > > > > > > > > > > Thus the ratio is p:q and this ratio has no measure. If we attempt to measure it, we end up with the same failed measure that is the constant 0.5773 (which is an approximation or failed measure of p:q).
> > > > > > > > > > >
> > > > > > > > > > > You fail to understand these things because you don't understand ratio and number.
> > > > > > > > > > What you call "magnitudes" are in fact just real numbers.
> > > > > > > > > You are just too stupid.
> > > > > > > > >
> > > > > > > > > Here is the proof of the mean value theorem using the 100% rigorous New Calculus:
> > > > > > > > >
> > > > > > > > > The New Calculus derivative:
> > > > > > > > >
> > > > > > > > > Given any interval (c-m,c+n), the New Calculus (henceforth NC) derivative is given by:
> > > > > > > > >
> > > > > > > > > f'(c)= (f(c+n)-f(c-m))/(m+n)
> > > > > > > > >
> > > > > > > > > f'(c) represents the slope of a secant line with end points (c-m, f(c-m)) and ( c+n, f(c+n)) that is parallel to the tangent line at x=c.
> > > > > > > > > m and n are horizontal distances from c.
> > > > > > > > >
> > > > > > > > > The interval (c-m,c+n) can be partitioned into equal sub-intervals of (m+n)/k.
> > > > > > > > >
> > > > > > > > > Each sub-interval has μ_s as the abscissa of f' so that
> > > > > > > > >
> > > > > > > > > f'(μ_s) = [ f(c-m + ((m+n)(s+1))/k) - f(c-m + ((m+n)s)/k) ] / ((m+n)/k)
> > > > > > > > >
> > > > > > > > > The level magnitude (aka arithmetic mean in mainstream mathematics) is given by:
> > > > > > > > >
> > > > > > > > > f'(c) = (1/k) \sum_{s=1}^k f'(μ_s)
> > > > > > > > >
> > > > > > > > > From the above statement, we want to show that
> > > > > > > > >
> > > > > > > > > f'(c) = [ f'(μ_s1) + f'(μ_s2) + f'(μ_s3) + ... + f'(μ_k-1) + f'(μ_k) ]/k
> > > > > > > > >
> > > > > > > > > By replacing each of the means with a derivative, we have:
> > > > > > > > >
> > > > > > > > > f'(c) = (1/k) { [ f(c-m + (m+n)/k)-f(c-m) ]/((m+n)/k)
> > > > > > > > > + [ f(c-m + 2(m+n)/k)-f(c-m + (m+n)/k) ]/((m+n)/k)]
> > > > > > > > > + [ f(c-m + 3(m+n)/k)-f(c-m + 2(m+n)/k) ]/((m+n)/k)]
> > > > > > > > > + ...
> > > > > > > > > + [ f(c-m + ((k-1)(m+n))/k)-f(c-m + ((k-2)(m+n))/k) ]/((m+n)/k)]
> > > > > > > > > + [ f(c+n)-f(c-m + ((k-1)(m+n))/k) ]/((m+n)/k)]
> > > > > > > > > }
> > > > > > > > > The above reduces to f'(c) = (f(c+n)-f(c-m))/(m+n)
> > > > > > > > >
> > > > > > > > > (1/(m+n)) \int_{c-m}^{c+n} f'(x) dx = (f(c+n)-f(c-m))/(m+n)
> > > > > > > > >
> > > > > > > > > So, what we have proved above is that f'(c) is the level magnitude of all the y-ordinates of f'(x) in the interval (c-m, c+n) and we can find the product of f'(c) and m+n to give us the area under f'(x) from c-m to c+n where c-m < c < c+n.
> > > > > > > > >
> > > > > > > > > Did you see anything there about "real numbers", you imbecile?
> > > > > > > > > >
> > > > > > > > > > We need something more than just rationals for the mean value theorem. That *something* is real numbers.
> > > > > > > > > No. What you need is understanding. You have NONE.
> > > > > > >
> > > > > > > > How do you know those μ_s exist?
> > > > > > > How can you know that you are a moron?
> > > > > > Prove those mu_s. How do you know they exist?
> > > > > It's simple but probably too complicated for you. I will not give you the answer but try to lead you to it.
> > > > >
> > > > > If f is defined everywhere in (a,b) then is it true or false that at least one y ordinate represents the level magnitude (aka arithmetic mean)?
> > > > Yes, it's true by the mean value theorem.
> > > Not by the mean value theorem because one will find a level magnitude even if the curve is not smooth. The mvt applies only to smooth curves.
> > > f need not be smooth anywhere in the given interval and there will still be an arithmetic mean. Do you disagree still?
> > > > Which you seek to prove. That's circular.
> > > No. That's wrong.
> > How do you get those mu_s then?
> Answer the question dumbo!
> Not by the mean value theorem because one will find a level magnitude even if the curve is not smooth. The mvt applies only to smooth curves.
> f need not be smooth anywhere in the given interval and there will still be an arithmetic mean.
> ------> Do you disagree still?
Then how do you find it?

On Monday, 7 August 2023 at 18:55:07 UTC-4, markus...@gmail.com wrote:
> måndag 7 augusti 2023 kl. 14:27:42 UTC+2 skrev Eram semper recta:
> > On Monday, 7 August 2023 at 05:02:12 UTC-4, markus...@gmail.com wrote:
> > > måndag 7 augusti 2023 kl. 00:35:54 UTC+2 skrev Eram semper recta:
> > > > On Sunday, 6 August 2023 at 14:19:23 UTC-4, markus...@gmail.com wrote:
> > > > > söndag 6 augusti 2023 kl. 02:03:23 UTC+2 skrev Eram semper recta:
> > > > > > On Saturday, 5 August 2023 at 17:27:46 UTC-4, markus...
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > https://www.academia.edu/45154026/Teaching_the_fundamental_theorem_of_calculus
> > > > > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > > > > I was the FIRST HUMAN to fully understand the mean value theorem and to prove constructively.
> > > > > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > > > > NONE of the stupid FUCKS who came before me where up to the task:
> > > > > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > > > > https://www.academia.edu/81300370/Mainstream_mathematics_academics_are_arrogant_and_incorrigible_ignoramuses_The_mean_value_theorem_IS_the_fundamental_theorem_of_calculus
> > > > > > > > > > > > > > > > > > > > > > > > > > > Your proof is invalid..
> > > > > > > > > > > > > > > > > > > > > > > > > > You've said that about a lot of things and every time you have been wrong.
> > > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > > You're wrong here yet again.
> > > > > > > > > > > > > > > > > > > > > > > > > > > The real mean value theorem requires real numbers.
> > > > > > > > > > > > > > > > > > > > > > > > > > It doesn't have anything to do with an ill-formed object you think of as a "real number".
> > > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > > The mean value theorem is about a level magnitude (what you erroneously call an "arithmetic mean").. It's the reason calculus works at all.
> > > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > > You're right - I don't understand the "limits" of your ignorance and stupidity.
> > > > > > > > > > > > > > > > > > > > > > > > > The mean value theorem doesn't work without real numbers. A very elementary counterexample is f(x)=0 if x²>2 and f(x)=1 if x²<2. It is continuous and differentiable everywhere on the rational number line with f'=0. If we consider [a, b]=[0, 2], we have a counterexample to the MVT (mean value theorem). From (b-a)f'(c)=f(b)-f(a) we have
> > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > f'(c)=0-1=-1.
> > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > But there is no such c. If we attempt to explicitly find c, we find that it doesn't exist.
> > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > The mean value theorem requires a real number line.
> > > > > > > > > > > > > > > > > > > > > > > > > See a psychiatrist soon.
> > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > You most definitely do need to see a psychiatrist. You are mentally ill.
> > > > > > > > > > > > > > > > > > > > > > > > > I don't need to.
> > > > > > > > > > > > > > > > > > > > > > > > > But you can consider f(x)=x³ over [0, 1] and see why the mean value theorem requires real numbers.
> > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > It doesn't show at all what you claim. In fact, the mvt doesn't give a shit about anything but magnitudes.
> > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > You obviously don't understand the mvt which is about a level magnitude (aka arithmetic mean in your bullshit mathematics).
> > > > > > > > > > > > > > > > > > > > > > > Then try it.
> > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > f(x)=x³, a=0 and b=0. Then f'(x)=3x², f(b)=1, f(a)=0.
> > > > > > > > > > > > > > > > > > > > > > What?! f(b)=f(0)=0, not 1, you imbecile!
> > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > <scheißen>
> > > > > > > > > > > > > > > > > > > > > b=0 was a typo. b=1, so f(b) is indeed one. You will get f'(c)=3c³=1, and this can't be solved with rational numbers only.
> > > > > > > > > > > > > > > > > > > > It still is wrong, whether a typo or not. Incoherent gibberish out of your juvenile brain.
> > > > > > > > > > > > > > > > > > > Can you solve f'(c)=3c²=1 without real numbers?
> > > > > > > > > > > > > > > > > > Yes, idiot! It is the CONSTANT known as the square root of 1/3 and there is NO number that decribes its measure.
> > > > > > > > > > > > > > > > > The square root of 1/3 isn't rational.
> > > > > > > > > > > > > > > > And so? Stating the obvious incorrectly. There is no number that describes the measure of the constant root(3).
> > > > > > > > > > > > > > > > > Thus the mean value theorem fails without real numbers.
> > > > > > > > > > > > > > > > It doesn't because there is no such thing as "real number".
> > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > <END OF DISCUSSION>
> > > > > > > > > > > > > > > If there is no such c, then the mean value theorem is false. The theorem states there is a such c.
> > > > > > > > > > > > > > You'll have to tell me again what is your example. I stop after the first mistake I find and you get ignored.
> > > > > > > > > > > > > You have already gotten TWO counterexamples.
> > > > > > > > > > > > >
> > > > > > > > > > > > > f(x)=x³, a=0 and b=1. Then f'(x)=3x², f(b)=1, f(a)=0.
> > > > > > > > > > > > > (b-a)f'(c)=f(b)-f(a)=1.
> > > > > > > > > > > > >
> > > > > > > > > > > > > f'(c)=3c²=1.
> > > > > > > > > > > > > We end up with something we can't solve only using rational numbers.
> > > > > > > > > > > > I have already answered this question. The square root of 1/3 is not described by any number. It is a CONSTANT which is a numeric approximation but not the actual value.
> > > > > > > > > > > >
> > > > > > > > > > > > The ratio square root of 1/3 is realised from a right angled triangle with one leg equal to leg (p) of original isosceles right-angled triangle and the other leg (q) to hypotenuse of original isosceles right-angled triangle.
> > > > > > > > > > > >
> > > > > > > > > > > > Thus the ratio is p:q and this ratio has no measure.. If we attempt to measure it, we end up with the same failed measure that is the constant 0.5773 (which is an approximation or failed measure of p:q)..
> > > > > > > > > > > >
> > > > > > > > > > > > You fail to understand these things because you don't understand ratio and number.
> > > > > > > > > > > What you call "magnitudes" are in fact just real numbers.
> > > > > > > > > > You are just too stupid.
> > > > > > > > > >
> > > > > > > > > > Here is the proof of the mean value theorem using the 100% rigorous New Calculus:
> > > > > > > > > >
> > > > > > > > > > The New Calculus derivative:
> > > > > > > > > >
> > > > > > > > > > Given any interval (c-m,c+n), the New Calculus (henceforth NC) derivative is given by:
> > > > > > > > > >
> > > > > > > > > > f'(c)= (f(c+n)-f(c-m))/(m+n)
> > > > > > > > > >
> > > > > > > > > > f'(c) represents the slope of a secant line with end points (c-m, f(c-m)) and ( c+n, f(c+n)) that is parallel to the tangent line at x=c.
> > > > > > > > > > m and n are horizontal distances from c.
> > > > > > > > > >
> > > > > > > > > > The interval (c-m,c+n) can be partitioned into equal sub-intervals of (m+n)/k.
> > > > > > > > > >
> > > > > > > > > > Each sub-interval has μ_s as the abscissa of f' so that
> > > > > > > > > >
> > > > > > > > > > f'(μ_s) = [ f(c-m + ((m+n)(s+1))/k) - f(c-m + ((m+n)s)/k) ] / ((m+n)/k)
> > > > > > > > > >
> > > > > > > > > > The level magnitude (aka arithmetic mean in mainstream mathematics) is given by:
> > > > > > > > > >
> > > > > > > > > > f'(c) = (1/k) \sum_{s=1}^k f'(μ_s)
> > > > > > > > > >
> > > > > > > > > > From the above statement, we want to show that
> > > > > > > > > >
> > > > > > > > > > f'(c) = [ f'(μ_s1) + f'(μ_s2) + f'(μ_s3) + ... + f'(μ_k-1) + f'(μ_k) ]/k
> > > > > > > > > >
> > > > > > > > > > By replacing each of the means with a derivative, we have:
> > > > > > > > > >
> > > > > > > > > > f'(c) = (1/k) { [ f(c-m + (m+n)/k)-f(c-m) ]/((m+n)/k)
> > > > > > > > > > + [ f(c-m + 2(m+n)/k)-f(c-m + (m+n)/k) ]/((m+n)/k)]
> > > > > > > > > > + [ f(c-m + 3(m+n)/k)-f(c-m + 2(m+n)/k) ]/((m+n)/k)]
> > > > > > > > > > + ...
> > > > > > > > > > + [ f(c-m + ((k-1)(m+n))/k)-f(c-m + ((k-2)(m+n))/k) ]/((m+n)/k)]
> > > > > > > > > > + [ f(c+n)-f(c-m + ((k-1)(m+n))/k) ]/((m+n)/k)]
> > > > > > > > > > }
> > > > > > > > > > The above reduces to f'(c) = (f(c+n)-f(c-m))/(m+n)
> > > > > > > > > >
> > > > > > > > > > (1/(m+n)) \int_{c-m}^{c+n} f'(x) dx = (f(c+n)-f(c-m))/(m+n)
> > > > > > > > > >
> > > > > > > > > > So, what we have proved above is that f'(c) is the level magnitude of all the y-ordinates of f'(x) in the interval (c-m, c+n) and we can find the product of f'(c) and m+n to give us the area under f'(x) from c-m to c+n where c-m < c < c+n.
> > > > > > > > > >
> > > > > > > > > > Did you see anything there about "real numbers", you imbecile?
> > > > > > > > > > >
> > > > > > > > > > > We need something more than just rationals for the mean value theorem. That *something* is real numbers.
> > > > > > > > > > No. What you need is understanding. You have NONE.
> > > > > > > >
> > > > > > > > > How do you know those μ_s exist?
> > > > > > > > How can you know that you are a moron?
> > > > > > > Prove those mu_s. How do you know they exist?
> > > > > > It's simple but probably too complicated for you. I will not give you the answer but try to lead you to it.
> > > > > >
> > > > > > If f is defined everywhere in (a,b) then is it true or false that at least one y ordinate represents the level magnitude (aka arithmetic mean)?
> > > > > Yes, it's true by the mean value theorem.
> > > > Not by the mean value theorem because one will find a level magnitude even if the curve is not smooth. The mvt applies only to smooth curves.
> > > > f need not be smooth anywhere in the given interval and there will still be an arithmetic mean. Do you disagree still?
> > > > > Which you seek to prove. That's circular.
> > > > No. That's wrong.
> > > How do you get those mu_s then?
> > Answer the question dumbo!
> > Not by the mean value theorem because one will find a level magnitude even if the curve is not smooth. The mvt applies only to smooth curves.
> > f need not be smooth anywhere in the given interval and there will still be an arithmetic mean.
> > ------> Do you disagree still?
> Then how do you find it?

tisdag 8 augusti 2023 kl. 03:08:27 UTC+2 skrev Eram semper recta:
> On Monday, 7 August 2023 at 18:55:07 UTC-4, markus...@gmail.com wrote:
> > måndag 7 augusti 2023 kl. 14:27:42 UTC+2 skrev Eram semper recta:
> > > On Monday, 7 August 2023 at 05:02:12 UTC-4, markus...@gmail.com wrote:
> > > > måndag 7 augusti 2023 kl. 00:35:54 UTC+2 skrev Eram semper recta:
> > > > > On Sunday, 6 August 2023 at 14:19:23 UTC-4, markus...@gmail.com wrote:
> > > > > > söndag 6 augusti 2023 kl. 02:03:23 UTC+2 skrev Eram semper recta:
> > > > > > > On Saturday, 5 August 2023 at 17:27:46 UTC-4, markus...
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > https://www.academia.edu/45154026/Teaching_the_fundamental_theorem_of_calculus
> > > > > > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > I was the FIRST HUMAN to fully understand the mean value theorem and to prove constructively..
> > > > > > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > NONE of the stupid FUCKS who came before me where up to the task:
> > > > > > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > https://www.academia.edu/81300370/Mainstream_mathematics_academics_are_arrogant_and_incorrigible_ignoramuses_The_mean_value_theorem_IS_the_fundamental_theorem_of_calculus
> > > > > > > > > > > > > > > > > > > > > > > > > > > > Your proof is invalid.
> > > > > > > > > > > > > > > > > > > > > > > > > > > You've said that about a lot of things and every time you have been wrong.
> > > > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > > > You're wrong here yet again.
> > > > > > > > > > > > > > > > > > > > > > > > > > > > The real mean value theorem requires real numbers.
> > > > > > > > > > > > > > > > > > > > > > > > > > > It doesn't have anything to do with an ill-formed object you think of as a "real number".
> > > > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > > > The mean value theorem is about a level magnitude (what you erroneously call an "arithmetic mean"). It's the reason calculus works at all.
> > > > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > > > You're right - I don't understand the "limits" of your ignorance and stupidity.
> > > > > > > > > > > > > > > > > > > > > > > > > > The mean value theorem doesn't work without real numbers. A very elementary counterexample is f(x)=0 if x²>2 and f(x)=1 if x²<2. It is continuous and differentiable everywhere on the rational number line with f'=0. If we consider [a, b]=[0, 2], we have a counterexample to the MVT (mean value theorem). From (b-a)f'(c)=f(b)-f(a) we have
> > > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > > f'(c)=0-1=-1.
> > > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > > But there is no such c. If we attempt to explicitly find c, we find that it doesn't exist.
> > > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > > The mean value theorem requires a real number line.
> > > > > > > > > > > > > > > > > > > > > > > > > > See a psychiatrist soon..
> > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > You most definitely do need to see a psychiatrist. You are mentally ill.
> > > > > > > > > > > > > > > > > > > > > > > > > > I don't need to.
> > > > > > > > > > > > > > > > > > > > > > > > > > But you can consider f(x)=x³ over [0, 1] and see why the mean value theorem requires real numbers.
> > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > It doesn't show at all what you claim. In fact, the mvt doesn't give a shit about anything but magnitudes.
> > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > You obviously don't understand the mvt which is about a level magnitude (aka arithmetic mean in your bullshit mathematics).
> > > > > > > > > > > > > > > > > > > > > > > > Then try it.
> > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > f(x)=x³, a=0 and b=0. Then f'(x)=3x², f(b)=1, f(a)=0.
> > > > > > > > > > > > > > > > > > > > > > > What?! f(b)=f(0)=0, not 1, you imbecile!
> > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > <scheißen>
> > > > > > > > > > > > > > > > > > > > > > b=0 was a typo. b=1, so f(b) is indeed one. You will get f'(c)=3c³=1, and this can't be solved with rational numbers only.
> > > > > > > > > > > > > > > > > > > > > It still is wrong, whether a typo or not. Incoherent gibberish out of your juvenile brain.
> > > > > > > > > > > > > > > > > > > > Can you solve f'(c)=3c²=1 without real numbers?
> > > > > > > > > > > > > > > > > > > Yes, idiot! It is the CONSTANT known as the square root of 1/3 and there is NO number that decribes its measure.
> > > > > > > > > > > > > > > > > > The square root of 1/3 isn't rational.
> > > > > > > > > > > > > > > > > And so? Stating the obvious incorrectly. There is no number that describes the measure of the constant root(3).
> > > > > > > > > > > > > > > > > > Thus the mean value theorem fails without real numbers.
> > > > > > > > > > > > > > > > > It doesn't because there is no such thing as "real number".
> > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > <END OF DISCUSSION>
> > > > > > > > > > > > > > > > If there is no such c, then the mean value theorem is false. The theorem states there is a such c.
> > > > > > > > > > > > > > > You'll have to tell me again what is your example. I stop after the first mistake I find and you get ignored.
> > > > > > > > > > > > > > You have already gotten TWO counterexamples.
> > > > > > > > > > > > > >
> > > > > > > > > > > > > > f(x)=x³, a=0 and b=1. Then f'(x)=3x², f(b)=1, f(a)=0.
> > > > > > > > > > > > > > (b-a)f'(c)=f(b)-f(a)=1.
> > > > > > > > > > > > > >
> > > > > > > > > > > > > > f'(c)=3c²=1.
> > > > > > > > > > > > > > We end up with something we can't solve only using rational numbers.
> > > > > > > > > > > > > I have already answered this question. The square root of 1/3 is not described by any number. It is a CONSTANT which is a numeric approximation but not the actual value.
> > > > > > > > > > > > >
> > > > > > > > > > > > > The ratio square root of 1/3 is realised from a right angled triangle with one leg equal to leg (p) of original isosceles right-angled triangle and the other leg (q) to hypotenuse of original isosceles right-angled triangle.
> > > > > > > > > > > > >
> > > > > > > > > > > > > Thus the ratio is p:q and this ratio has no measure. If we attempt to measure it, we end up with the same failed measure that is the constant 0.5773 (which is an approximation or failed measure of p:q).
> > > > > > > > > > > > >
> > > > > > > > > > > > > You fail to understand these things because you don't understand ratio and number.
> > > > > > > > > > > > What you call "magnitudes" are in fact just real numbers.
> > > > > > > > > > > You are just too stupid.
> > > > > > > > > > >
> > > > > > > > > > > Here is the proof of the mean value theorem using the 100% rigorous New Calculus:
> > > > > > > > > > >
> > > > > > > > > > > The New Calculus derivative:
> > > > > > > > > > >
> > > > > > > > > > > Given any interval (c-m,c+n), the New Calculus (henceforth NC) derivative is given by:
> > > > > > > > > > >
> > > > > > > > > > > f'(c)= (f(c+n)-f(c-m))/(m+n)
> > > > > > > > > > >
> > > > > > > > > > > f'(c) represents the slope of a secant line with end points (c-m, f(c-m)) and ( c+n, f(c+n)) that is parallel to the tangent line at x=c.
> > > > > > > > > > > m and n are horizontal distances from c.
> > > > > > > > > > >
> > > > > > > > > > > The interval (c-m,c+n) can be partitioned into equal sub-intervals of (m+n)/k.
> > > > > > > > > > >
> > > > > > > > > > > Each sub-interval has μ_s as the abscissa of f' so that
> > > > > > > > > > >
> > > > > > > > > > > f'(μ_s) = [ f(c-m + ((m+n)(s+1))/k) - f(c-m + ((m+n)s)/k) ] / ((m+n)/k)
> > > > > > > > > > >
> > > > > > > > > > > The level magnitude (aka arithmetic mean in mainstream mathematics) is given by:
> > > > > > > > > > >
> > > > > > > > > > > f'(c) = (1/k) \sum_{s=1}^k f'(μ_s)
> > > > > > > > > > >
> > > > > > > > > > > From the above statement, we want to show that
> > > > > > > > > > >
> > > > > > > > > > > f'(c) = [ f'(μ_s1) + f'(μ_s2) + f'(μ_s3) + ... + f'(μ_k-1) + f'(μ_k) ]/k
> > > > > > > > > > >
> > > > > > > > > > > By replacing each of the means with a derivative, we have:
> > > > > > > > > > >
> > > > > > > > > > > f'(c) = (1/k) { [ f(c-m + (m+n)/k)-f(c-m) ]/((m+n)/k)
> > > > > > > > > > > + [ f(c-m + 2(m+n)/k)-f(c-m + (m+n)/k) ]/((m+n)/k)]
> > > > > > > > > > > + [ f(c-m + 3(m+n)/k)-f(c-m + 2(m+n)/k) ]/((m+n)/k)]
> > > > > > > > > > > + ...
> > > > > > > > > > > + [ f(c-m + ((k-1)(m+n))/k)-f(c-m + ((k-2)(m+n))/k) ]/((m+n)/k)]
> > > > > > > > > > > + [ f(c+n)-f(c-m + ((k-1)(m+n))/k) ]/((m+n)/k)]
> > > > > > > > > > > }
> > > > > > > > > > > The above reduces to f'(c) = (f(c+n)-f(c-m))/(m+n)
> > > > > > > > > > >
> > > > > > > > > > > (1/(m+n)) \int_{c-m}^{c+n} f'(x) dx = (f(c+n)-f(c-m))/(m+n)
> > > > > > > > > > >
> > > > > > > > > > > So, what we have proved above is that f'(c) is the level magnitude of all the y-ordinates of f'(x) in the interval (c-m, c+n) and we can find the product of f'(c) and m+n to give us the area under f'(x) from c-m to c+n where c-m < c < c+n.
> > > > > > > > > > >
> > > > > > > > > > > Did you see anything there about "real numbers", you imbecile?
> > > > > > > > > > > >
> > > > > > > > > > > > We need something more than just rationals for the mean value theorem. That *something* is real numbers.
> > > > > > > > > > > No. What you need is understanding. You have NONE.
> > > > > > > > >
> > > > > > > > > > How do you know those μ_s exist?
> > > > > > > > > How can you know that you are a moron?
> > > > > > > > Prove those mu_s. How do you know they exist?
> > > > > > > It's simple but probably too complicated for you. I will not give you the answer but try to lead you to it.
> > > > > > >
> > > > > > > If f is defined everywhere in (a,b) then is it true or false that at least one y ordinate represents the level magnitude (aka arithmetic mean)?
> > > > > > Yes, it's true by the mean value theorem.
> > > > > Not by the mean value theorem because one will find a level magnitude even if the curve is not smooth. The mvt applies only to smooth curves..
> > > > > f need not be smooth anywhere in the given interval and there will still be an arithmetic mean. Do you disagree still?
> > > > > > Which you seek to prove. That's circular.
> > > > > No. That's wrong.
> > > > How do you get those mu_s then?
> > > Answer the question dumbo!
> > > Not by the mean value theorem because one will find a level magnitude even if the curve is not smooth. The mvt applies only to smooth curves.
> > > f need not be smooth anywhere in the given interval and there will still be an arithmetic mean.
> > > ------> Do you disagree still?
> > Then how do you find it?
> Answer the question, retard!
>
> Do you disagree still?
I disagree that you can find the mu_s without using the MVT in the first place

On Tuesday, 8 August 2023 at 04:36:57 UTC-4, markus...@gmail.com wrote:
> tisdag 8 augusti 2023 kl. 03:08:27 UTC+2 skrev Eram semper recta:
> > On Monday, 7 August 2023 at 18:55:07 UTC-4, markus...@gmail.com wrote:
> > > måndag 7 augusti 2023 kl. 14:27:42 UTC+2 skrev Eram semper recta:
> > > > On Monday, 7 August 2023 at 05:02:12 UTC-4, markus...@gmail.com wrote:
> > > > > måndag 7 augusti 2023 kl. 00:35:54 UTC+2 skrev Eram semper recta:
> > > > > > On Sunday, 6 August 2023 at 14:19:23 UTC-4, markus...@gmail.com wrote:
> > > > > > > söndag 6 augusti 2023 kl. 02:03:23 UTC+2 skrev Eram semper recta:
> > > > > > > > On Saturday, 5 August 2023 at 17:27:46 UTC-4, markus...
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > https://www.academia.edu/45154026/Teaching_the_fundamental_theorem_of_calculus
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > I was the FIRST HUMAN to fully understand the mean value theorem and to prove constructively.
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > NONE of the stupid FUCKS who came before me where up to the task:
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > https://www.academia.edu/81300370/Mainstream_mathematics_academics_are_arrogant_and_incorrigible_ignoramuses_The_mean_value_theorem_IS_the_fundamental_theorem_of_calculus
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > Your proof is invalid.
> > > > > > > > > > > > > > > > > > > > > > > > > > > > You've said that about a lot of things and every time you have been wrong.
> > > > > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > > > > You're wrong here yet again.
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > The real mean value theorem requires real numbers.
> > > > > > > > > > > > > > > > > > > > > > > > > > > > It doesn't have anything to do with an ill-formed object you think of as a "real number".
> > > > > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > > > > The mean value theorem is about a level magnitude (what you erroneously call an "arithmetic mean"). It's the reason calculus works at all.
> > > > > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > > > > You're right - I don't understand the "limits" of your ignorance and stupidity.
> > > > > > > > > > > > > > > > > > > > > > > > > > > The mean value theorem doesn't work without real numbers. A very elementary counterexample is f(x)=0 if x²>2 and f(x)=1 if x²<2. It is continuous and differentiable everywhere on the rational number line with f'=0. If we consider [a, b]=[0, 2], we have a counterexample to the MVT (mean value theorem). From (b-a)f'(c)=f(b)-f(a) we have
> > > > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > > > f'(c)=0-1=-1.
> > > > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > > > But there is no such c. If we attempt to explicitly find c, we find that it doesn't exist.
> > > > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > > > The mean value theorem requires a real number line.
> > > > > > > > > > > > > > > > > > > > > > > > > > > See a psychiatrist soon.
> > > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > > You most definitely do need to see a psychiatrist. You are mentally ill.
> > > > > > > > > > > > > > > > > > > > > > > > > > > I don't need to.
> > > > > > > > > > > > > > > > > > > > > > > > > > > But you can consider f(x)=x³ over [0, 1] and see why the mean value theorem requires real numbers.
> > > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > > It doesn't show at all what you claim. In fact, the mvt doesn't give a shit about anything but magnitudes.
> > > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > > You obviously don't understand the mvt which is about a level magnitude (aka arithmetic mean in your bullshit mathematics).
> > > > > > > > > > > > > > > > > > > > > > > > > Then try it.
> > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > f(x)=x³, a=0 and b=0. Then f'(x)=3x², f(b)=1, f(a)=0.
> > > > > > > > > > > > > > > > > > > > > > > > What?! f(b)=f(0)=0, not 1, you imbecile!
> > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > <scheißen>
> > > > > > > > > > > > > > > > > > > > > > > b=0 was a typo. b=1, so f(b) is indeed one. You will get f'(c)=3c³=1, and this can't be solved with rational numbers only.
> > > > > > > > > > > > > > > > > > > > > > It still is wrong, whether a typo or not. Incoherent gibberish out of your juvenile brain.
> > > > > > > > > > > > > > > > > > > > > Can you solve f'(c)=3c²=1 without real numbers?
> > > > > > > > > > > > > > > > > > > > Yes, idiot! It is the CONSTANT known as the square root of 1/3 and there is NO number that decribes its measure.
> > > > > > > > > > > > > > > > > > > The square root of 1/3 isn't rational..
> > > > > > > > > > > > > > > > > > And so? Stating the obvious incorrectly.. There is no number that describes the measure of the constant root(3).
> > > > > > > > > > > > > > > > > > > Thus the mean value theorem fails without real numbers.
> > > > > > > > > > > > > > > > > > It doesn't because there is no such thing as "real number".
> > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > <END OF DISCUSSION>
> > > > > > > > > > > > > > > > > If there is no such c, then the mean value theorem is false. The theorem states there is a such c.
> > > > > > > > > > > > > > > > You'll have to tell me again what is your example. I stop after the first mistake I find and you get ignored.
> > > > > > > > > > > > > > > You have already gotten TWO counterexamples.
> > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > f(x)=x³, a=0 and b=1. Then f'(x)=3x², f(b)=1, f(a)=0.
> > > > > > > > > > > > > > > (b-a)f'(c)=f(b)-f(a)=1.
> > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > f'(c)=3c²=1.
> > > > > > > > > > > > > > > We end up with something we can't solve only using rational numbers.
> > > > > > > > > > > > > > I have already answered this question. The square root of 1/3 is not described by any number. It is a CONSTANT which is a numeric approximation but not the actual value.
> > > > > > > > > > > > > >
> > > > > > > > > > > > > > The ratio square root of 1/3 is realised from a right angled triangle with one leg equal to leg (p) of original isosceles right-angled triangle and the other leg (q) to hypotenuse of original isosceles right-angled triangle.
> > > > > > > > > > > > > >
> > > > > > > > > > > > > > Thus the ratio is p:q and this ratio has no measure. If we attempt to measure it, we end up with the same failed measure that is the constant 0.5773 (which is an approximation or failed measure of p:q).
> > > > > > > > > > > > > >
> > > > > > > > > > > > > > You fail to understand these things because you don't understand ratio and number.
> > > > > > > > > > > > > What you call "magnitudes" are in fact just real numbers.
> > > > > > > > > > > > You are just too stupid.
> > > > > > > > > > > >
> > > > > > > > > > > > Here is the proof of the mean value theorem using the 100% rigorous New Calculus:
> > > > > > > > > > > >
> > > > > > > > > > > > The New Calculus derivative:
> > > > > > > > > > > >
> > > > > > > > > > > > Given any interval (c-m,c+n), the New Calculus (henceforth NC) derivative is given by:
> > > > > > > > > > > >
> > > > > > > > > > > > f'(c)= (f(c+n)-f(c-m))/(m+n)
> > > > > > > > > > > >
> > > > > > > > > > > > f'(c) represents the slope of a secant line with end points (c-m, f(c-m)) and ( c+n, f(c+n)) that is parallel to the tangent line at x=c.
> > > > > > > > > > > > m and n are horizontal distances from c.
> > > > > > > > > > > >
> > > > > > > > > > > > The interval (c-m,c+n) can be partitioned into equal sub-intervals of (m+n)/k.
> > > > > > > > > > > >
> > > > > > > > > > > > Each sub-interval has μ_s as the abscissa of f' so that
> > > > > > > > > > > >
> > > > > > > > > > > > f'(μ_s) = [ f(c-m + ((m+n)(s+1))/k) - f(c-m + ((m+n)s)/k) ] / ((m+n)/k)
> > > > > > > > > > > >
> > > > > > > > > > > > The level magnitude (aka arithmetic mean in mainstream mathematics) is given by:
> > > > > > > > > > > >
> > > > > > > > > > > > f'(c) = (1/k) \sum_{s=1}^k f'(μ_s)
> > > > > > > > > > > >
> > > > > > > > > > > > From the above statement, we want to show that
> > > > > > > > > > > >
> > > > > > > > > > > > f'(c) = [ f'(μ_s1) + f'(μ_s2) + f'(μ_s3) + ... + f'(μ_k-1) + f'(μ_k) ]/k
> > > > > > > > > > > >
> > > > > > > > > > > > By replacing each of the means with a derivative, we have:
> > > > > > > > > > > >
> > > > > > > > > > > > f'(c) = (1/k) { [ f(c-m + (m+n)/k)-f(c-m) ]/((m+n)/k)
> > > > > > > > > > > > + [ f(c-m + 2(m+n)/k)-f(c-m + (m+n)/k) ]/((m+n)/k)]
> > > > > > > > > > > > + [ f(c-m + 3(m+n)/k)-f(c-m + 2(m+n)/k) ]/((m+n)/k)]
> > > > > > > > > > > > + ...
> > > > > > > > > > > > + [ f(c-m + ((k-1)(m+n))/k)-f(c-m + ((k-2)(m+n))/k) ]/((m+n)/k)]
> > > > > > > > > > > > + [ f(c+n)-f(c-m + ((k-1)(m+n))/k) ]/((m+n)/k)]
> > > > > > > > > > > > }
> > > > > > > > > > > > The above reduces to f'(c) = (f(c+n)-f(c-m))/(m+n)
> > > > > > > > > > > >
> > > > > > > > > > > > (1/(m+n)) \int_{c-m}^{c+n} f'(x) dx = (f(c+n)-f(c-m))/(m+n)
> > > > > > > > > > > >
> > > > > > > > > > > > So, what we have proved above is that f'(c) is the level magnitude of all the y-ordinates of f'(x) in the interval (c-m, c+n) and we can find the product of f'(c) and m+n to give us the area under f'(x) from c-m to c+n where c-m < c < c+n.
> > > > > > > > > > > >
> > > > > > > > > > > > Did you see anything there about "real numbers", you imbecile?
> > > > > > > > > > > > >
> > > > > > > > > > > > > We need something more than just rationals for the mean value theorem. That *something* is real numbers.
> > > > > > > > > > > > No. What you need is understanding. You have NONE.
> > > > > > > > > >
> > > > > > > > > > > How do you know those μ_s exist?
> > > > > > > > > > How can you know that you are a moron?
> > > > > > > > > Prove those mu_s. How do you know they exist?
> > > > > > > > It's simple but probably too complicated for you. I will not give you the answer but try to lead you to it.
> > > > > > > >
> > > > > > > > If f is defined everywhere in (a,b) then is it true or false that at least one y ordinate represents the level magnitude (aka arithmetic mean)?
> > > > > > > Yes, it's true by the mean value theorem.
> > > > > > Not by the mean value theorem because one will find a level magnitude even if the curve is not smooth. The mvt applies only to smooth curves.
> > > > > > f need not be smooth anywhere in the given interval and there will still be an arithmetic mean. Do you disagree still?
> > > > > > > Which you seek to prove. That's circular.
> > > > > > No. That's wrong.
> > > > > How do you get those mu_s then?
> > > > Answer the question dumbo!
> > > > Not by the mean value theorem because one will find a level magnitude even if the curve is not smooth. The mvt applies only to smooth curves.
> > > > f need not be smooth anywhere in the given interval and there will still be an arithmetic mean.
> > > > ------> Do you disagree still?
> > > Then how do you find it?
> > Answer the question, retard!
> >
> > Do you disagree still?
> I disagree that you can find the mu_s without using the MVT in the first place

tisdag 8 augusti 2023 kl. 18:40:26 UTC+2 skrev Eram semper recta:
> On Tuesday, 8 August 2023 at 04:36:57 UTC-4, markus...@gmail.com wrote:
> > tisdag 8 augusti 2023 kl. 03:08:27 UTC+2 skrev Eram semper recta:
> > > On Monday, 7 August 2023 at 18:55:07 UTC-4, markus...@gmail.com wrote:
> > > > måndag 7 augusti 2023 kl. 14:27:42 UTC+2 skrev Eram semper recta:
> > > > > On Monday, 7 August 2023 at 05:02:12 UTC-4, markus...@gmail.com wrote:
> > > > > > måndag 7 augusti 2023 kl. 00:35:54 UTC+2 skrev Eram semper recta:
> > > > > > > On Sunday, 6 August 2023 at 14:19:23 UTC-4, markus...@gmail.com wrote:
> > > > > > > > söndag 6 augusti 2023 kl. 02:03:23 UTC+2 skrev Eram semper recta:
> > > > > > > > > On Saturday, 5 August 2023 at 17:27:46 UTC-4, markus...
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > https://www..academia.edu/45154026/Teaching_the_fundamental_theorem_of_calculus
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > I was the FIRST HUMAN to fully understand the mean value theorem and to prove constructively.
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > NONE of the stupid FUCKS who came before me where up to the task:
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > https://www.academia.edu/81300370/Mainstream_mathematics_academics_are_arrogant_and_incorrigible_ignoramuses_The_mean_value_theorem_IS_the_fundamental_theorem_of_calculus
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > Your proof is invalid.
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > You've said that about a lot of things and every time you have been wrong.
> > > > > > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > You're wrong here yet again.
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > The real mean value theorem requires real numbers.
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > It doesn't have anything to do with an ill-formed object you think of as a "real number".
> > > > > > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > The mean value theorem is about a level magnitude (what you erroneously call an "arithmetic mean"). It's the reason calculus works at all.
> > > > > > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > You're right - I don't understand the "limits" of your ignorance and stupidity.
> > > > > > > > > > > > > > > > > > > > > > > > > > > > The mean value theorem doesn't work without real numbers. A very elementary counterexample is f(x)=0 if x²>2 and f(x)=1 if x²<2. It is continuous and differentiable everywhere on the rational number line with f'=0. If we consider [a, b]=[0, 2], we have a counterexample to the MVT (mean value theorem).. From (b-a)f'(c)=f(b)-f(a) we have
> > > > > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > > > > f'(c)=0-1=-1.
> > > > > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > > > > But there is no such c. If we attempt to explicitly find c, we find that it doesn't exist.
> > > > > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > > > > The mean value theorem requires a real number line.
> > > > > > > > > > > > > > > > > > > > > > > > > > > > See a psychiatrist soon.
> > > > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > > > You most definitely do need to see a psychiatrist. You are mentally ill.
> > > > > > > > > > > > > > > > > > > > > > > > > > > > I don't need to.
> > > > > > > > > > > > > > > > > > > > > > > > > > > > But you can consider f(x)=x³ over [0, 1] and see why the mean value theorem requires real numbers.
> > > > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > > > It doesn't show at all what you claim. In fact, the mvt doesn't give a shit about anything but magnitudes.
> > > > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > > > You obviously don't understand the mvt which is about a level magnitude (aka arithmetic mean in your bullshit mathematics).
> > > > > > > > > > > > > > > > > > > > > > > > > > Then try it.
> > > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > > f(x)=x³, a=0 and b=0. Then f'(x)=3x², f(b)=1, f(a)=0.
> > > > > > > > > > > > > > > > > > > > > > > > > What?! f(b)=f(0)=0, not 1, you imbecile!
> > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > <scheißen>
> > > > > > > > > > > > > > > > > > > > > > > > b=0 was a typo. b=1, so f(b) is indeed one. You will get f'(c)=3c³=1, and this can't be solved with rational numbers only.
> > > > > > > > > > > > > > > > > > > > > > > It still is wrong, whether a typo or not. Incoherent gibberish out of your juvenile brain.
> > > > > > > > > > > > > > > > > > > > > > Can you solve f'(c)=3c²=1 without real numbers?
> > > > > > > > > > > > > > > > > > > > > Yes, idiot! It is the CONSTANT known as the square root of 1/3 and there is NO number that decribes its measure.
> > > > > > > > > > > > > > > > > > > > The square root of 1/3 isn't rational.
> > > > > > > > > > > > > > > > > > > And so? Stating the obvious incorrectly. There is no number that describes the measure of the constant root(3).
> > > > > > > > > > > > > > > > > > > > Thus the mean value theorem fails without real numbers.
> > > > > > > > > > > > > > > > > > > It doesn't because there is no such thing as "real number".
> > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > <END OF DISCUSSION>
> > > > > > > > > > > > > > > > > > If there is no such c, then the mean value theorem is false. The theorem states there is a such c.
> > > > > > > > > > > > > > > > > You'll have to tell me again what is your example. I stop after the first mistake I find and you get ignored.
> > > > > > > > > > > > > > > > You have already gotten TWO counterexamples..
> > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > f(x)=x³, a=0 and b=1. Then f'(x)=3x², f(b)=1, f(a)=0.
> > > > > > > > > > > > > > > > (b-a)f'(c)=f(b)-f(a)=1.
> > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > f'(c)=3c²=1.
> > > > > > > > > > > > > > > > We end up with something we can't solve only using rational numbers.
> > > > > > > > > > > > > > > I have already answered this question. The square root of 1/3 is not described by any number. It is a CONSTANT which is a numeric approximation but not the actual value.
> > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > The ratio square root of 1/3 is realised from a right angled triangle with one leg equal to leg (p) of original isosceles right-angled triangle and the other leg (q) to hypotenuse of original isosceles right-angled triangle.
> > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > Thus the ratio is p:q and this ratio has no measure. If we attempt to measure it, we end up with the same failed measure that is the constant 0.5773 (which is an approximation or failed measure of p:q).
> > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > You fail to understand these things because you don't understand ratio and number.
> > > > > > > > > > > > > > What you call "magnitudes" are in fact just real numbers.
> > > > > > > > > > > > > You are just too stupid.
> > > > > > > > > > > > >
> > > > > > > > > > > > > Here is the proof of the mean value theorem using the 100% rigorous New Calculus:
> > > > > > > > > > > > >
> > > > > > > > > > > > > The New Calculus derivative:
> > > > > > > > > > > > >
> > > > > > > > > > > > > Given any interval (c-m,c+n), the New Calculus (henceforth NC) derivative is given by:
> > > > > > > > > > > > >
> > > > > > > > > > > > > f'(c)= (f(c+n)-f(c-m))/(m+n)
> > > > > > > > > > > > >
> > > > > > > > > > > > > f'(c) represents the slope of a secant line with end points (c-m, f(c-m)) and ( c+n, f(c+n)) that is parallel to the tangent line at x=c.
> > > > > > > > > > > > > m and n are horizontal distances from c.
> > > > > > > > > > > > >
> > > > > > > > > > > > > The interval (c-m,c+n) can be partitioned into equal sub-intervals of (m+n)/k.
> > > > > > > > > > > > >
> > > > > > > > > > > > > Each sub-interval has μ_s as the abscissa of f' so that
> > > > > > > > > > > > >
> > > > > > > > > > > > > f'(μ_s) = [ f(c-m + ((m+n)(s+1))/k) - f(c-m + ((m+n)s)/k) ] / ((m+n)/k)
> > > > > > > > > > > > >
> > > > > > > > > > > > > The level magnitude (aka arithmetic mean in mainstream mathematics) is given by:
> > > > > > > > > > > > >
> > > > > > > > > > > > > f'(c) = (1/k) \sum_{s=1}^k f'(μ_s)
> > > > > > > > > > > > >
> > > > > > > > > > > > > From the above statement, we want to show that
> > > > > > > > > > > > >
> > > > > > > > > > > > > f'(c) = [ f'(μ_s1) + f'(μ_s2) + f'(μ_s3) + ... + f'(μ_k-1) + f'(μ_k) ]/k
> > > > > > > > > > > > >
> > > > > > > > > > > > > By replacing each of the means with a derivative, we have:
> > > > > > > > > > > > >
> > > > > > > > > > > > > f'(c) = (1/k) { [ f(c-m + (m+n)/k)-f(c-m) ]/((m+n)/k)
> > > > > > > > > > > > > + [ f(c-m + 2(m+n)/k)-f(c-m + (m+n)/k) ]/((m+n)/k)]
> > > > > > > > > > > > > + [ f(c-m + 3(m+n)/k)-f(c-m + 2(m+n)/k) ]/((m+n)/k)]
> > > > > > > > > > > > > + ...
> > > > > > > > > > > > > + [ f(c-m + ((k-1)(m+n))/k)-f(c-m + ((k-2)(m+n))/k) ]/((m+n)/k)]
> > > > > > > > > > > > > + [ f(c+n)-f(c-m + ((k-1)(m+n))/k) ]/((m+n)/k)]
> > > > > > > > > > > > > }
> > > > > > > > > > > > > The above reduces to f'(c) = (f(c+n)-f(c-m))/(m+n)
> > > > > > > > > > > > >
> > > > > > > > > > > > > (1/(m+n)) \int_{c-m}^{c+n} f'(x) dx = (f(c+n)-f(c-m))/(m+n)
> > > > > > > > > > > > >
> > > > > > > > > > > > > So, what we have proved above is that f'(c) is the level magnitude of all the y-ordinates of f'(x) in the interval (c-m, c+n) and we can find the product of f'(c) and m+n to give us the area under f'(x) from c-m to c+n where c-m < c < c+n.
> > > > > > > > > > > > >
> > > > > > > > > > > > > Did you see anything there about "real numbers", you imbecile?
> > > > > > > > > > > > > >
> > > > > > > > > > > > > > We need something more than just rationals for the mean value theorem. That *something* is real numbers.
> > > > > > > > > > > > > No. What you need is understanding. You have NONE..
> > > > > > > > > > >
> > > > > > > > > > > > How do you know those μ_s exist?
> > > > > > > > > > > How can you know that you are a moron?
> > > > > > > > > > Prove those mu_s. How do you know they exist?
> > > > > > > > > It's simple but probably too complicated for you. I will not give you the answer but try to lead you to it.
> > > > > > > > >
> > > > > > > > > If f is defined everywhere in (a,b) then is it true or false that at least one y ordinate represents the level magnitude (aka arithmetic mean)?
> > > > > > > > Yes, it's true by the mean value theorem.
> > > > > > > Not by the mean value theorem because one will find a level magnitude even if the curve is not smooth. The mvt applies only to smooth curves.
> > > > > > > f need not be smooth anywhere in the given interval and there will still be an arithmetic mean. Do you disagree still?
> > > > > > > > Which you seek to prove. That's circular.
> > > > > > > No. That's wrong.
> > > > > > How do you get those mu_s then?
> > > > > Answer the question dumbo!
> > > > > Not by the mean value theorem because one will find a level magnitude even if the curve is not smooth. The mvt applies only to smooth curves..
> > > > > f need not be smooth anywhere in the given interval and there will still be an arithmetic mean.
> > > > > ------> Do you disagree still?
> > > > Then how do you find it?
> > > Answer the question, retard!
> > >
> > > Do you disagree still?
> > I disagree that you can find the mu_s without using the MVT in the first place
> Well, you are wrong because if a curve is not smooth but everywhere defined means that an arithmetic mean is possible even though the mvt does not apply in the case of non-smooth curves.
>
> Explain what your understanding of arithmetic mean is. Don't tell me how to construct it because every fool knows how to do that. In words, explain the meaning of arithmetic mean. Remember: Adding up set of values and dividing by total, does not explain arithmetic mean, only how to determined it.

On Tuesday, 8 August 2023 at 17:50:58 UTC-4, markus...@gmail.com wrote:
> tisdag 8 augusti 2023 kl. 18:40:26 UTC+2 skrev Eram semper recta:
> > On Tuesday, 8 August 2023 at 04:36:57 UTC-4, markus...@gmail.com wrote:
> > > tisdag 8 augusti 2023 kl. 03:08:27 UTC+2 skrev Eram semper recta:
> > > > On Monday, 7 August 2023 at 18:55:07 UTC-4, markus...@gmail.com wrote:
> > > > > måndag 7 augusti 2023 kl. 14:27:42 UTC+2 skrev Eram semper recta:
> > > > > > On Monday, 7 August 2023 at 05:02:12 UTC-4, markus...@gmail.com wrote:
> > > > > > > måndag 7 augusti 2023 kl. 00:35:54 UTC+2 skrev Eram semper recta:
> > > > > > > > On Sunday, 6 August 2023 at 14:19:23 UTC-4, markus...@gmail..com wrote:
> > > > > > > > > söndag 6 augusti 2023 kl. 02:03:23 UTC+2 skrev Eram semper recta:
> > > > > > > > > > On Saturday, 5 August 2023 at 17:27:46 UTC-4, markus...
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > https://www.academia.edu/45154026/Teaching_the_fundamental_theorem_of_calculus
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > I was the FIRST HUMAN to fully understand the mean value theorem and to prove constructively.
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > NONE of the stupid FUCKS who came before me where up to the task:
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > https://www..academia.edu/81300370/Mainstream_mathematics_academics_are_arrogant_and_incorrigible_ignoramuses_The_mean_value_theorem_IS_the_fundamental_theorem_of_calculus
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > Your proof is invalid.
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > You've said that about a lot of things and every time you have been wrong.
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > You're wrong here yet again.
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > The real mean value theorem requires real numbers.
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > It doesn't have anything to do with an ill-formed object you think of as a "real number".
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > The mean value theorem is about a level magnitude (what you erroneously call an "arithmetic mean"). It's the reason calculus works at all.
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > You're right - I don't understand the "limits" of your ignorance and stupidity.
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > The mean value theorem doesn't work without real numbers. A very elementary counterexample is f(x)=0 if x²>2 and f(x)=1 if x²<2. It is continuous and differentiable everywhere on the rational number line with f'=0. If we consider [a, b]=[0, 2], we have a counterexample to the MVT (mean value theorem). From (b-a)f'(c)=f(b)-f(a) we have
> > > > > > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > f'(c)=0-1=-1.
> > > > > > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > But there is no such c. If we attempt to explicitly find c, we find that it doesn't exist.
> > > > > > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > The mean value theorem requires a real number line.
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > See a psychiatrist soon.
> > > > > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > > > > You most definitely do need to see a psychiatrist. You are mentally ill.
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > I don't need to.
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > But you can consider f(x)=x³ over [0, 1] and see why the mean value theorem requires real numbers.
> > > > > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > > > > It doesn't show at all what you claim. In fact, the mvt doesn't give a shit about anything but magnitudes.
> > > > > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > > > > You obviously don't understand the mvt which is about a level magnitude (aka arithmetic mean in your bullshit mathematics).
> > > > > > > > > > > > > > > > > > > > > > > > > > > Then try it.
> > > > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > > > f(x)=x³, a=0 and b=0. Then f'(x)=3x², f(b)=1, f(a)=0.
> > > > > > > > > > > > > > > > > > > > > > > > > > What?! f(b)=f(0)=0, not 1, you imbecile!
> > > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > > <scheißen>
> > > > > > > > > > > > > > > > > > > > > > > > > b=0 was a typo. b=1, so f(b) is indeed one. You will get f'(c)=3c³=1, and this can't be solved with rational numbers only.
> > > > > > > > > > > > > > > > > > > > > > > > It still is wrong, whether a typo or not. Incoherent gibberish out of your juvenile brain.
> > > > > > > > > > > > > > > > > > > > > > > Can you solve f'(c)=3c²=1 without real numbers?
> > > > > > > > > > > > > > > > > > > > > > Yes, idiot! It is the CONSTANT known as the square root of 1/3 and there is NO number that decribes its measure.
> > > > > > > > > > > > > > > > > > > > > The square root of 1/3 isn't rational.
> > > > > > > > > > > > > > > > > > > > And so? Stating the obvious incorrectly. There is no number that describes the measure of the constant root(3)..
> > > > > > > > > > > > > > > > > > > > > Thus the mean value theorem fails without real numbers.
> > > > > > > > > > > > > > > > > > > > It doesn't because there is no such thing as "real number".
> > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > <END OF DISCUSSION>
> > > > > > > > > > > > > > > > > > > If there is no such c, then the mean value theorem is false. The theorem states there is a such c.
> > > > > > > > > > > > > > > > > > You'll have to tell me again what is your example. I stop after the first mistake I find and you get ignored.
> > > > > > > > > > > > > > > > > You have already gotten TWO counterexamples.
> > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > f(x)=x³, a=0 and b=1. Then f'(x)=3x², f(b)=1, f(a)=0.
> > > > > > > > > > > > > > > > > (b-a)f'(c)=f(b)-f(a)=1.
> > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > f'(c)=3c²=1.
> > > > > > > > > > > > > > > > > We end up with something we can't solve only using rational numbers.
> > > > > > > > > > > > > > > > I have already answered this question. The square root of 1/3 is not described by any number. It is a CONSTANT which is a numeric approximation but not the actual value.
> > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > The ratio square root of 1/3 is realised from a right angled triangle with one leg equal to leg (p) of original isosceles right-angled triangle and the other leg (q) to hypotenuse of original isosceles right-angled triangle.
> > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > Thus the ratio is p:q and this ratio has no measure. If we attempt to measure it, we end up with the same failed measure that is the constant 0.5773 (which is an approximation or failed measure of p:q).
> > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > You fail to understand these things because you don't understand ratio and number.
> > > > > > > > > > > > > > > What you call "magnitudes" are in fact just real numbers.
> > > > > > > > > > > > > > You are just too stupid.
> > > > > > > > > > > > > >
> > > > > > > > > > > > > > Here is the proof of the mean value theorem using the 100% rigorous New Calculus:
> > > > > > > > > > > > > >
> > > > > > > > > > > > > > The New Calculus derivative:
> > > > > > > > > > > > > >
> > > > > > > > > > > > > > Given any interval (c-m,c+n), the New Calculus (henceforth NC) derivative is given by:
> > > > > > > > > > > > > >
> > > > > > > > > > > > > > f'(c)= (f(c+n)-f(c-m))/(m+n)
> > > > > > > > > > > > > >
> > > > > > > > > > > > > > f'(c) represents the slope of a secant line with end points (c-m, f(c-m)) and ( c+n, f(c+n)) that is parallel to the tangent line at x=c.
> > > > > > > > > > > > > > m and n are horizontal distances from c.
> > > > > > > > > > > > > >
> > > > > > > > > > > > > > The interval (c-m,c+n) can be partitioned into equal sub-intervals of (m+n)/k.
> > > > > > > > > > > > > >
> > > > > > > > > > > > > > Each sub-interval has μ_s as the abscissa of f' so that
> > > > > > > > > > > > > >
> > > > > > > > > > > > > > f'(μ_s) = [ f(c-m + ((m+n)(s+1))/k) - f(c-m + ((m+n)s)/k) ] / ((m+n)/k)
> > > > > > > > > > > > > >
> > > > > > > > > > > > > > The level magnitude (aka arithmetic mean in mainstream mathematics) is given by:
> > > > > > > > > > > > > >
> > > > > > > > > > > > > > f'(c) = (1/k) \sum_{s=1}^k f'(μ_s)
> > > > > > > > > > > > > >
> > > > > > > > > > > > > > From the above statement, we want to show that
> > > > > > > > > > > > > >
> > > > > > > > > > > > > > f'(c) = [ f'(μ_s1) + f'(μ_s2) + f'(μ_s3) + ... + f'(μ_k-1) + f'(μ_k) ]/k
> > > > > > > > > > > > > >
> > > > > > > > > > > > > > By replacing each of the means with a derivative, we have:
> > > > > > > > > > > > > >
> > > > > > > > > > > > > > f'(c) = (1/k) { [ f(c-m + (m+n)/k)-f(c-m) ]/((m+n)/k)
> > > > > > > > > > > > > > + [ f(c-m + 2(m+n)/k)-f(c-m + (m+n)/k) ]/((m+n)/k)]
> > > > > > > > > > > > > > + [ f(c-m + 3(m+n)/k)-f(c-m + 2(m+n)/k) ]/((m+n)/k)]
> > > > > > > > > > > > > > + ...
> > > > > > > > > > > > > > + [ f(c-m + ((k-1)(m+n))/k)-f(c-m + ((k-2)(m+n))/k) ]/((m+n)/k)]
> > > > > > > > > > > > > > + [ f(c+n)-f(c-m + ((k-1)(m+n))/k) ]/((m+n)/k)]
> > > > > > > > > > > > > > }
> > > > > > > > > > > > > > The above reduces to f'(c) = (f(c+n)-f(c-m))/(m+n)
> > > > > > > > > > > > > >
> > > > > > > > > > > > > > (1/(m+n)) \int_{c-m}^{c+n} f'(x) dx = (f(c+n)-f(c-m))/(m+n)
> > > > > > > > > > > > > >
> > > > > > > > > > > > > > So, what we have proved above is that f'(c) is the level magnitude of all the y-ordinates of f'(x) in the interval (c-m, c+n) and we can find the product of f'(c) and m+n to give us the area under f'(x) from c-m to c+n where c-m < c < c+n.
> > > > > > > > > > > > > >
> > > > > > > > > > > > > > Did you see anything there about "real numbers", you imbecile?
> > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > We need something more than just rationals for the mean value theorem. That *something* is real numbers.
> > > > > > > > > > > > > > No. What you need is understanding. You have NONE.
> > > > > > > > > > > >
> > > > > > > > > > > > > How do you know those μ_s exist?
> > > > > > > > > > > > How can you know that you are a moron?
> > > > > > > > > > > Prove those mu_s. How do you know they exist?
> > > > > > > > > > It's simple but probably too complicated for you. I will not give you the answer but try to lead you to it.
> > > > > > > > > >
> > > > > > > > > > If f is defined everywhere in (a,b) then is it true or false that at least one y ordinate represents the level magnitude (aka arithmetic mean)?
> > > > > > > > > Yes, it's true by the mean value theorem.
> > > > > > > > Not by the mean value theorem because one will find a level magnitude even if the curve is not smooth. The mvt applies only to smooth curves.
> > > > > > > > f need not be smooth anywhere in the given interval and there will still be an arithmetic mean. Do you disagree still?
> > > > > > > > > Which you seek to prove. That's circular.
> > > > > > > > No. That's wrong.
> > > > > > > How do you get those mu_s then?
> > > > > > Answer the question dumbo!
> > > > > > Not by the mean value theorem because one will find a level magnitude even if the curve is not smooth. The mvt applies only to smooth curves.
> > > > > > f need not be smooth anywhere in the given interval and there will still be an arithmetic mean.
> > > > > > ------> Do you disagree still?
> > > > > Then how do you find it?
> > > > Answer the question, retard!
> > > >
> > > > Do you disagree still?
> > > I disagree that you can find the mu_s without using the MVT in the first place
> > Well, you are wrong because if a curve is not smooth but everywhere defined means that an arithmetic mean is possible even though the mvt does not apply in the case of non-smooth curves.
> >
> > Explain what your understanding of arithmetic mean is. Don't tell me how to construct it because every fool knows how to do that. In words, explain the meaning of arithmetic mean. Remember: Adding up set of values and dividing by total, does not explain arithmetic mean, only how to determined it..
> How can you take a mean of infinitely many values?

onsdag 9 augusti 2023 kl. 02:43:12 UTC+2 skrev Eram semper recta:
> On Tuesday, 8 August 2023 at 17:50:58 UTC-4, markus...@gmail.com wrote:
> > tisdag 8 augusti 2023 kl. 18:40:26 UTC+2 skrev Eram semper recta:
> > > On Tuesday, 8 August 2023 at 04:36:57 UTC-4, markus...@gmail.com wrote:
> > > > tisdag 8 augusti 2023 kl. 03:08:27 UTC+2 skrev Eram semper recta:
> > > > > On Monday, 7 August 2023 at 18:55:07 UTC-4, markus...@gmail.com wrote:
> > > > > > måndag 7 augusti 2023 kl. 14:27:42 UTC+2 skrev Eram semper recta:
> > > > > > > On Monday, 7 August 2023 at 05:02:12 UTC-4, markus...@gmail.com wrote:
> > > > > > > > måndag 7 augusti 2023 kl. 00:35:54 UTC+2 skrev Eram semper recta:
> > > > > > > > > On Sunday, 6 August 2023 at 14:19:23 UTC-4, markus...@gmail.com wrote:
> > > > > > > > > > söndag 6 augusti 2023 kl. 02:03:23 UTC+2 skrev Eram semper recta:
> > > > > > > > > > > On Saturday, 5 August 2023 at 17:27:46 UTC-4, markus....
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > https://www.academia.edu/45154026/Teaching_the_fundamental_theorem_of_calculus
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > I was the FIRST HUMAN to fully understand the mean value theorem and to prove constructively.
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > NONE of the stupid FUCKS who came before me where up to the task:
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > https://www.academia.edu/81300370/Mainstream_mathematics_academics_are_arrogant_and_incorrigible_ignoramuses_The_mean_value_theorem_IS_the_fundamental_theorem_of_calculus
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > Your proof is invalid.
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > You've said that about a lot of things and every time you have been wrong.
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > You're wrong here yet again.
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > The real mean value theorem requires real numbers.
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > It doesn't have anything to do with an ill-formed object you think of as a "real number"..
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > The mean value theorem is about a level magnitude (what you erroneously call an "arithmetic mean"). It's the reason calculus works at all.
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > You're right - I don't understand the "limits" of your ignorance and stupidity.
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > The mean value theorem doesn't work without real numbers. A very elementary counterexample is f(x)=0 if x²>2 and f(x)=1 if x²<2. It is continuous and differentiable everywhere on the rational number line with f'=0. If we consider [a, b]=[0, 2], we have a counterexample to the MVT (mean value theorem). From (b-a)f'(c)=f(b)-f(a) we have
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > f'(c)=0-1=-1.
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > But there is no such c. If we attempt to explicitly find c, we find that it doesn't exist.
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > The mean value theorem requires a real number line.
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > See a psychiatrist soon.
> > > > > > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > You most definitely do need to see a psychiatrist. You are mentally ill.
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > I don't need to..
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > But you can consider f(x)=x³ over [0, 1] and see why the mean value theorem requires real numbers.
> > > > > > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > It doesn't show at all what you claim. In fact, the mvt doesn't give a shit about anything but magnitudes.
> > > > > > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > You obviously don't understand the mvt which is about a level magnitude (aka arithmetic mean in your bullshit mathematics).
> > > > > > > > > > > > > > > > > > > > > > > > > > > > Then try it.
> > > > > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > > > > f(x)=x³, a=0 and b=0. Then f'(x)=3x², f(b)=1, f(a)=0.
> > > > > > > > > > > > > > > > > > > > > > > > > > > What?! f(b)=f(0)=0, not 1, you imbecile!
> > > > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > > > <scheißen>
> > > > > > > > > > > > > > > > > > > > > > > > > > b=0 was a typo. b=1, so f(b) is indeed one. You will get f'(c)=3c³=1, and this can't be solved with rational numbers only.
> > > > > > > > > > > > > > > > > > > > > > > > > It still is wrong, whether a typo or not. Incoherent gibberish out of your juvenile brain.
> > > > > > > > > > > > > > > > > > > > > > > > Can you solve f'(c)=3c²=1 without real numbers?
> > > > > > > > > > > > > > > > > > > > > > > Yes, idiot! It is the CONSTANT known as the square root of 1/3 and there is NO number that decribes its measure.
> > > > > > > > > > > > > > > > > > > > > > The square root of 1/3 isn't rational.
> > > > > > > > > > > > > > > > > > > > > And so? Stating the obvious incorrectly. There is no number that describes the measure of the constant root(3).
> > > > > > > > > > > > > > > > > > > > > > Thus the mean value theorem fails without real numbers.
> > > > > > > > > > > > > > > > > > > > > It doesn't because there is no such thing as "real number".
> > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > <END OF DISCUSSION>
> > > > > > > > > > > > > > > > > > > > If there is no such c, then the mean value theorem is false. The theorem states there is a such c.
> > > > > > > > > > > > > > > > > > > You'll have to tell me again what is your example. I stop after the first mistake I find and you get ignored.
> > > > > > > > > > > > > > > > > > You have already gotten TWO counterexamples.
> > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > f(x)=x³, a=0 and b=1. Then f'(x)=3x², f(b)=1, f(a)=0.
> > > > > > > > > > > > > > > > > > (b-a)f'(c)=f(b)-f(a)=1.
> > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > f'(c)=3c²=1.
> > > > > > > > > > > > > > > > > > We end up with something we can't solve only using rational numbers.
> > > > > > > > > > > > > > > > > I have already answered this question. The square root of 1/3 is not described by any number. It is a CONSTANT which is a numeric approximation but not the actual value.
> > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > The ratio square root of 1/3 is realised from a right angled triangle with one leg equal to leg (p) of original isosceles right-angled triangle and the other leg (q) to hypotenuse of original isosceles right-angled triangle.
> > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > Thus the ratio is p:q and this ratio has no measure. If we attempt to measure it, we end up with the same failed measure that is the constant 0.5773 (which is an approximation or failed measure of p:q).
> > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > You fail to understand these things because you don't understand ratio and number.
> > > > > > > > > > > > > > > > What you call "magnitudes" are in fact just real numbers.
> > > > > > > > > > > > > > > You are just too stupid.
> > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > Here is the proof of the mean value theorem using the 100% rigorous New Calculus:
> > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > The New Calculus derivative:
> > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > Given any interval (c-m,c+n), the New Calculus (henceforth NC) derivative is given by:
> > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > f'(c)= (f(c+n)-f(c-m))/(m+n)
> > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > f'(c) represents the slope of a secant line with end points (c-m, f(c-m)) and ( c+n, f(c+n)) that is parallel to the tangent line at x=c.
> > > > > > > > > > > > > > > m and n are horizontal distances from c.
> > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > The interval (c-m,c+n) can be partitioned into equal sub-intervals of (m+n)/k.
> > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > Each sub-interval has μ_s as the abscissa of f' so that
> > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > f'(μ_s) = [ f(c-m + ((m+n)(s+1))/k) - f(c-m + ((m+n)s)/k) ] / ((m+n)/k)
> > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > The level magnitude (aka arithmetic mean in mainstream mathematics) is given by:
> > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > f'(c) = (1/k) \sum_{s=1}^k f'(μ_s)
> > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > From the above statement, we want to show that
> > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > f'(c) = [ f'(μ_s1) + f'(μ_s2) + f'(μ_s3) + ... + f'(μ_k-1) + f'(μ_k) ]/k
> > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > By replacing each of the means with a derivative, we have:
> > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > f'(c) = (1/k) { [ f(c-m + (m+n)/k)-f(c-m) ]/((m+n)/k)
> > > > > > > > > > > > > > > + [ f(c-m + 2(m+n)/k)-f(c-m + (m+n)/k) ]/((m+n)/k)]
> > > > > > > > > > > > > > > + [ f(c-m + 3(m+n)/k)-f(c-m + 2(m+n)/k) ]/((m+n)/k)]
> > > > > > > > > > > > > > > + ...
> > > > > > > > > > > > > > > + [ f(c-m + ((k-1)(m+n))/k)-f(c-m + ((k-2)(m+n))/k) ]/((m+n)/k)]
> > > > > > > > > > > > > > > + [ f(c+n)-f(c-m + ((k-1)(m+n))/k) ]/((m+n)/k)]
> > > > > > > > > > > > > > > }
> > > > > > > > > > > > > > > The above reduces to f'(c) = (f(c+n)-f(c-m))/(m+n)
> > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > (1/(m+n)) \int_{c-m}^{c+n} f'(x) dx = (f(c+n)-f(c-m))/(m+n)
> > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > So, what we have proved above is that f'(c) is the level magnitude of all the y-ordinates of f'(x) in the interval (c-m, c+n) and we can find the product of f'(c) and m+n to give us the area under f'(x) from c-m to c+n where c-m < c < c+n.
> > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > Did you see anything there about "real numbers", you imbecile?
> > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > We need something more than just rationals for the mean value theorem. That *something* is real numbers.
> > > > > > > > > > > > > > > No. What you need is understanding. You have NONE.
> > > > > > > > > > > > >
> > > > > > > > > > > > > > How do you know those μ_s exist?
> > > > > > > > > > > > > How can you know that you are a moron?
> > > > > > > > > > > > Prove those mu_s. How do you know they exist?
> > > > > > > > > > > It's simple but probably too complicated for you. I will not give you the answer but try to lead you to it.
> > > > > > > > > > >
> > > > > > > > > > > If f is defined everywhere in (a,b) then is it true or false that at least one y ordinate represents the level magnitude (aka arithmetic mean)?
> > > > > > > > > > Yes, it's true by the mean value theorem.
> > > > > > > > > Not by the mean value theorem because one will find a level magnitude even if the curve is not smooth. The mvt applies only to smooth curves.
> > > > > > > > > f need not be smooth anywhere in the given interval and there will still be an arithmetic mean. Do you disagree still?
> > > > > > > > > > Which you seek to prove. That's circular.
> > > > > > > > > No. That's wrong.
> > > > > > > > How do you get those mu_s then?
> > > > > > > Answer the question dumbo!
> > > > > > > Not by the mean value theorem because one will find a level magnitude even if the curve is not smooth. The mvt applies only to smooth curves.
> > > > > > > f need not be smooth anywhere in the given interval and there will still be an arithmetic mean.
> > > > > > > ------> Do you disagree still?
> > > > > > Then how do you find it?
> > > > > Answer the question, retard!
> > > > >
> > > > > Do you disagree still?
> > > > I disagree that you can find the mu_s without using the MVT in the first place
> > > Well, you are wrong because if a curve is not smooth but everywhere defined means that an arithmetic mean is possible even though the mvt does not apply in the case of non-smooth curves.
> > >
> > > Explain what your understanding of arithmetic mean is. Don't tell me how to construct it because every fool knows how to do that. In words, explain the meaning of arithmetic mean. Remember: Adding up set of values and dividing by total, does not explain arithmetic mean, only how to determined it.
> > How can you take a mean of infinitely many values?
>
> Explain what your understanding of arithmetic mean is.
> <irrelevant rubbish>
Fine, I'll give you the answers. Then you are going to answer me.

On Wednesday, 9 August 2023 at 05:21:48 UTC-4, markus...@gmail.com wrote:
> onsdag 9 augusti 2023 kl. 02:43:12 UTC+2 skrev Eram semper recta:
> > On Tuesday, 8 August 2023 at 17:50:58 UTC-4, markus...@gmail.com wrote:
> > > tisdag 8 augusti 2023 kl. 18:40:26 UTC+2 skrev Eram semper recta:
> > > > On Tuesday, 8 August 2023 at 04:36:57 UTC-4, markus...@gmail.com wrote:
> > > > > tisdag 8 augusti 2023 kl. 03:08:27 UTC+2 skrev Eram semper recta:
> > > > > > On Monday, 7 August 2023 at 18:55:07 UTC-4, markus...@gmail.com wrote:
> > > > > > > måndag 7 augusti 2023 kl. 14:27:42 UTC+2 skrev Eram semper recta:
> > > > > > > > On Monday, 7 August 2023 at 05:02:12 UTC-4, markus...@gmail..com wrote:
> > > > > > > > > måndag 7 augusti 2023 kl. 00:35:54 UTC+2 skrev Eram semper recta:
> > > > > > > > > > On Sunday, 6 August 2023 at 14:19:23 UTC-4, markus...@gmail.com wrote:
> > > > > > > > > > > söndag 6 augusti 2023 kl. 02:03:23 UTC+2 skrev Eram semper recta:
> > > > > > > > > > > > On Saturday, 5 August 2023 at 17:27:46 UTC-4, markus...
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > https://www.academia.edu/45154026/Teaching_the_fundamental_theorem_of_calculus
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > I was the FIRST HUMAN to fully understand the mean value theorem and to prove constructively.
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > NONE of the stupid FUCKS who came before me where up to the task:
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > https://www.academia.edu/81300370/Mainstream_mathematics_academics_are_arrogant_and_incorrigible_ignoramuses_The_mean_value_theorem_IS_the_fundamental_theorem_of_calculus
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > Your proof is invalid.
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > You've said that about a lot of things and every time you have been wrong.
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > You're wrong here yet again.
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > The real mean value theorem requires real numbers.
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > It doesn't have anything to do with an ill-formed object you think of as a "real number".
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > The mean value theorem is about a level magnitude (what you erroneously call an "arithmetic mean"). It's the reason calculus works at all.
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > You're right - I don't understand the "limits" of your ignorance and stupidity.
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > The mean value theorem doesn't work without real numbers. A very elementary counterexample is f(x)=0 if x²>2 and f(x)=1 if x²<2. It is continuous and differentiable everywhere on the rational number line with f'=0. If we consider [a, b]=[0, 2], we have a counterexample to the MVT (mean value theorem). From (b-a)f'(c)=f(b)-f(a) we have
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > f'(c)=0-1=-1.
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > But there is no such c. If we attempt to explicitly find c, we find that it doesn't exist.
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > The mean value theorem requires a real number line.
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > See a psychiatrist soon.
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > You most definitely do need to see a psychiatrist. You are mentally ill.
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > I don't need to.
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > But you can consider f(x)=x³ over [0, 1] and see why the mean value theorem requires real numbers.
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > It doesn't show at all what you claim. In fact, the mvt doesn't give a shit about anything but magnitudes.
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > You obviously don't understand the mvt which is about a level magnitude (aka arithmetic mean in your bullshit mathematics).
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > Then try it.
> > > > > > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > f(x)=x³, a=0 and b=0. Then f'(x)=3x², f(b)=1, f(a)=0.
> > > > > > > > > > > > > > > > > > > > > > > > > > > > What?! f(b)=f(0)=0, not 1, you imbecile!
> > > > > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > > > > <scheißen>
> > > > > > > > > > > > > > > > > > > > > > > > > > > b=0 was a typo. b=1, so f(b) is indeed one. You will get f'(c)=3c³=1, and this can't be solved with rational numbers only.
> > > > > > > > > > > > > > > > > > > > > > > > > > It still is wrong, whether a typo or not. Incoherent gibberish out of your juvenile brain.
> > > > > > > > > > > > > > > > > > > > > > > > > Can you solve f'(c)=3c²=1 without real numbers?
> > > > > > > > > > > > > > > > > > > > > > > > Yes, idiot! It is the CONSTANT known as the square root of 1/3 and there is NO number that decribes its measure.
> > > > > > > > > > > > > > > > > > > > > > > The square root of 1/3 isn't rational.
> > > > > > > > > > > > > > > > > > > > > > And so? Stating the obvious incorrectly. There is no number that describes the measure of the constant root(3).
> > > > > > > > > > > > > > > > > > > > > > > Thus the mean value theorem fails without real numbers.
> > > > > > > > > > > > > > > > > > > > > > It doesn't because there is no such thing as "real number".
> > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > <END OF DISCUSSION>
> > > > > > > > > > > > > > > > > > > > > If there is no such c, then the mean value theorem is false. The theorem states there is a such c.
> > > > > > > > > > > > > > > > > > > > You'll have to tell me again what is your example. I stop after the first mistake I find and you get ignored.
> > > > > > > > > > > > > > > > > > > You have already gotten TWO counterexamples.
> > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > f(x)=x³, a=0 and b=1. Then f'(x)=3x², f(b)=1, f(a)=0.
> > > > > > > > > > > > > > > > > > > (b-a)f'(c)=f(b)-f(a)=1.
> > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > f'(c)=3c²=1.
> > > > > > > > > > > > > > > > > > > We end up with something we can't solve only using rational numbers.
> > > > > > > > > > > > > > > > > > I have already answered this question. The square root of 1/3 is not described by any number. It is a CONSTANT which is a numeric approximation but not the actual value.
> > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > The ratio square root of 1/3 is realised from a right angled triangle with one leg equal to leg (p) of original isosceles right-angled triangle and the other leg (q) to hypotenuse of original isosceles right-angled triangle.
> > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > Thus the ratio is p:q and this ratio has no measure. If we attempt to measure it, we end up with the same failed measure that is the constant 0.5773 (which is an approximation or failed measure of p:q).
> > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > You fail to understand these things because you don't understand ratio and number.
> > > > > > > > > > > > > > > > > What you call "magnitudes" are in fact just real numbers.
> > > > > > > > > > > > > > > > You are just too stupid.
> > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > Here is the proof of the mean value theorem using the 100% rigorous New Calculus:
> > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > The New Calculus derivative:
> > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > Given any interval (c-m,c+n), the New Calculus (henceforth NC) derivative is given by:
> > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > f'(c)= (f(c+n)-f(c-m))/(m+n)
> > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > f'(c) represents the slope of a secant line with end points (c-m, f(c-m)) and ( c+n, f(c+n)) that is parallel to the tangent line at x=c.
> > > > > > > > > > > > > > > > m and n are horizontal distances from c.
> > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > The interval (c-m,c+n) can be partitioned into equal sub-intervals of (m+n)/k.
> > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > Each sub-interval has μ_s as the abscissa of f' so that
> > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > f'(μ_s) = [ f(c-m + ((m+n)(s+1))/k) - f(c-m + ((m+n)s)/k) ] / ((m+n)/k)
> > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > The level magnitude (aka arithmetic mean in mainstream mathematics) is given by:
> > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > f'(c) = (1/k) \sum_{s=1}^k f'(μ_s)
> > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > From the above statement, we want to show that
> > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > f'(c) = [ f'(μ_s1) + f'(μ_s2) + f'(μ_s3) + ... + f'(μ_k-1) + f'(μ_k) ]/k
> > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > By replacing each of the means with a derivative, we have:
> > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > f'(c) = (1/k) { [ f(c-m + (m+n)/k)-f(c-m) ]/((m+n)/k)
> > > > > > > > > > > > > > > > + [ f(c-m + 2(m+n)/k)-f(c-m + (m+n)/k) ]/((m+n)/k)]
> > > > > > > > > > > > > > > > + [ f(c-m + 3(m+n)/k)-f(c-m + 2(m+n)/k) ]/((m+n)/k)]
> > > > > > > > > > > > > > > > + ...
> > > > > > > > > > > > > > > > + [ f(c-m + ((k-1)(m+n))/k)-f(c-m + ((k-2)(m+n))/k) ]/((m+n)/k)]
> > > > > > > > > > > > > > > > + [ f(c+n)-f(c-m + ((k-1)(m+n))/k) ]/((m+n)/k)]
> > > > > > > > > > > > > > > > }
> > > > > > > > > > > > > > > > The above reduces to f'(c) = (f(c+n)-f(c-m))/(m+n)
> > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > (1/(m+n)) \int_{c-m}^{c+n} f'(x) dx = (f(c+n)-f(c-m))/(m+n)
> > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > So, what we have proved above is that f'(c) is the level magnitude of all the y-ordinates of f'(x) in the interval (c-m, c+n) and we can find the product of f'(c) and m+n to give us the area under f'(x) from c-m to c+n where c-m < c < c+n.
> > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > Did you see anything there about "real numbers", you imbecile?
> > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > We need something more than just rationals for the mean value theorem. That *something* is real numbers.
> > > > > > > > > > > > > > > > No. What you need is understanding. You have NONE.
> > > > > > > > > > > > > >
> > > > > > > > > > > > > > > How do you know those μ_s exist?
> > > > > > > > > > > > > > How can you know that you are a moron?
> > > > > > > > > > > > > Prove those mu_s. How do you know they exist?
> > > > > > > > > > > > It's simple but probably too complicated for you. I will not give you the answer but try to lead you to it.
> > > > > > > > > > > >
> > > > > > > > > > > > If f is defined everywhere in (a,b) then is it true or false that at least one y ordinate represents the level magnitude (aka arithmetic mean)?
> > > > > > > > > > > Yes, it's true by the mean value theorem.
> > > > > > > > > > Not by the mean value theorem because one will find a level magnitude even if the curve is not smooth. The mvt applies only to smooth curves.
> > > > > > > > > > f need not be smooth anywhere in the given interval and there will still be an arithmetic mean. Do you disagree still?
> > > > > > > > > > > Which you seek to prove. That's circular.
> > > > > > > > > > No. That's wrong.
> > > > > > > > > How do you get those mu_s then?
> > > > > > > > Answer the question dumbo!
> > > > > > > > Not by the mean value theorem because one will find a level magnitude even if the curve is not smooth. The mvt applies only to smooth curves.
> > > > > > > > f need not be smooth anywhere in the given interval and there will still be an arithmetic mean.
> > > > > > > > ------> Do you disagree still?
> > > > > > > Then how do you find it?
> > > > > > Answer the question, retard!
> > > > > >
> > > > > > Do you disagree still?
> > > > > I disagree that you can find the mu_s without using the MVT in the first place
> > > > Well, you are wrong because if a curve is not smooth but everywhere defined means that an arithmetic mean is possible even though the mvt does not apply in the case of non-smooth curves.
> > > >
> > > > Explain what your understanding of arithmetic mean is. Don't tell me how to construct it because every fool knows how to do that. In words, explain the meaning of arithmetic mean. Remember: Adding up set of values and dividing by total, does not explain arithmetic mean, only how to determined it.
> > > How can you take a mean of infinitely many values?
> >
> > Explain what your understanding of arithmetic mean is.
> > <irrelevant rubbish>
> Fine, I'll give you the answers. Then you are going to answer me.
>
> The mean of a finite set can be thought of intuitively as adding everything together in a big pile and then distributing it equally in n piles.

onsdag 9 augusti 2023 kl. 13:29:23 UTC+2 skrev Eram semper recta:
> On Wednesday, 9 August 2023 at 05:21:48 UTC-4, markus...@gmail.com wrote:
> > onsdag 9 augusti 2023 kl. 02:43:12 UTC+2 skrev Eram semper recta:
> > > On Tuesday, 8 August 2023 at 17:50:58 UTC-4, markus...@gmail.com wrote:
> > > > tisdag 8 augusti 2023 kl. 18:40:26 UTC+2 skrev Eram semper recta:
> > > > > On Tuesday, 8 August 2023 at 04:36:57 UTC-4, markus...@gmail.com wrote:
> > > > > > tisdag 8 augusti 2023 kl. 03:08:27 UTC+2 skrev Eram semper recta:
> > > > > > > On Monday, 7 August 2023 at 18:55:07 UTC-4, markus...@gmail.com wrote:
> > > > > > > > måndag 7 augusti 2023 kl. 14:27:42 UTC+2 skrev Eram semper recta:
> > > > > > > > > On Monday, 7 August 2023 at 05:02:12 UTC-4, markus...@gmail.com wrote:
> > > > > > > > > > måndag 7 augusti 2023 kl. 00:35:54 UTC+2 skrev Eram semper recta:
> > > > > > > > > > > On Sunday, 6 August 2023 at 14:19:23 UTC-4, markus...@gmail.com wrote:
> > > > > > > > > > > > söndag 6 augusti 2023 kl. 02:03:23 UTC+2 skrev Eram semper recta:
> > > > > > > > > > > > > On Saturday, 5 August 2023 at 17:27:46 UTC-4, markus...
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > https://www.academia.edu/45154026/Teaching_the_fundamental_theorem_of_calculus
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > I was the FIRST HUMAN to fully understand the mean value theorem and to prove constructively.
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > NONE of the stupid FUCKS who came before me where up to the task:
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > https://www.academia.edu/81300370/Mainstream_mathematics_academics_are_arrogant_and_incorrigible_ignoramuses_The_mean_value_theorem_IS_the_fundamental_theorem_of_calculus
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > Your proof is invalid.
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > You've said that about a lot of things and every time you have been wrong.
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > You're wrong here yet again.
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > The real mean value theorem requires real numbers.
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > It doesn't have anything to do with an ill-formed object you think of as a "real number".
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > The mean value theorem is about a level magnitude (what you erroneously call an "arithmetic mean"). It's the reason calculus works at all.
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > You're right - I don't understand the "limits" of your ignorance and stupidity.
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > The mean value theorem doesn't work without real numbers. A very elementary counterexample is f(x)=0 if x²>2 and f(x)=1 if x²<2. It is continuous and differentiable everywhere on the rational number line with f'=0. If we consider [a, b]=[0, 2], we have a counterexample to the MVT (mean value theorem). From (b-a)f'(c)=f(b)-f(a) we have
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > f'(c)=0-1=-1.
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > But there is no such c. If we attempt to explicitly find c, we find that it doesn't exist.
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > The mean value theorem requires a real number line.
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > See a psychiatrist soon.
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > You most definitely do need to see a psychiatrist. You are mentally ill.
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > I don't need to.
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > But you can consider f(x)=x³ over [0, 1] and see why the mean value theorem requires real numbers.
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > It doesn't show at all what you claim. In fact, the mvt doesn't give a shit about anything but magnitudes.
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > You obviously don't understand the mvt which is about a level magnitude (aka arithmetic mean in your bullshit mathematics).
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > Then try it.
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > f(x)=x³, a=0 and b=0. Then f'(x)=3x², f(b)=1, f(a)=0.
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > What?! f(b)=f(0)=0, not 1, you imbecile!
> > > > > > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > <scheißen>
> > > > > > > > > > > > > > > > > > > > > > > > > > > > b=0 was a typo. b=1, so f(b) is indeed one. You will get f'(c)=3c³=1, and this can't be solved with rational numbers only.
> > > > > > > > > > > > > > > > > > > > > > > > > > > It still is wrong, whether a typo or not. Incoherent gibberish out of your juvenile brain.
> > > > > > > > > > > > > > > > > > > > > > > > > > Can you solve f'(c)=3c²=1 without real numbers?
> > > > > > > > > > > > > > > > > > > > > > > > > Yes, idiot! It is the CONSTANT known as the square root of 1/3 and there is NO number that decribes its measure.
> > > > > > > > > > > > > > > > > > > > > > > > The square root of 1/3 isn't rational.
> > > > > > > > > > > > > > > > > > > > > > > And so? Stating the obvious incorrectly. There is no number that describes the measure of the constant root(3).
> > > > > > > > > > > > > > > > > > > > > > > > Thus the mean value theorem fails without real numbers.
> > > > > > > > > > > > > > > > > > > > > > > It doesn't because there is no such thing as "real number".
> > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > <END OF DISCUSSION>
> > > > > > > > > > > > > > > > > > > > > > If there is no such c, then the mean value theorem is false. The theorem states there is a such c.
> > > > > > > > > > > > > > > > > > > > > You'll have to tell me again what is your example. I stop after the first mistake I find and you get ignored..
> > > > > > > > > > > > > > > > > > > > You have already gotten TWO counterexamples.
> > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > f(x)=x³, a=0 and b=1. Then f'(x)=3x², f(b)=1, f(a)=0.
> > > > > > > > > > > > > > > > > > > > (b-a)f'(c)=f(b)-f(a)=1.
> > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > f'(c)=3c²=1.
> > > > > > > > > > > > > > > > > > > > We end up with something we can't solve only using rational numbers.
> > > > > > > > > > > > > > > > > > > I have already answered this question.. The square root of 1/3 is not described by any number. It is a CONSTANT which is a numeric approximation but not the actual value.
> > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > The ratio square root of 1/3 is realised from a right angled triangle with one leg equal to leg (p) of original isosceles right-angled triangle and the other leg (q) to hypotenuse of original isosceles right-angled triangle.
> > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > Thus the ratio is p:q and this ratio has no measure. If we attempt to measure it, we end up with the same failed measure that is the constant 0.5773 (which is an approximation or failed measure of p:q).
> > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > You fail to understand these things because you don't understand ratio and number.
> > > > > > > > > > > > > > > > > > What you call "magnitudes" are in fact just real numbers.
> > > > > > > > > > > > > > > > > You are just too stupid.
> > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > Here is the proof of the mean value theorem using the 100% rigorous New Calculus:
> > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > The New Calculus derivative:
> > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > Given any interval (c-m,c+n), the New Calculus (henceforth NC) derivative is given by:
> > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > f'(c)= (f(c+n)-f(c-m))/(m+n)
> > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > f'(c) represents the slope of a secant line with end points (c-m, f(c-m)) and ( c+n, f(c+n)) that is parallel to the tangent line at x=c.
> > > > > > > > > > > > > > > > > m and n are horizontal distances from c.
> > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > The interval (c-m,c+n) can be partitioned into equal sub-intervals of (m+n)/k.
> > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > Each sub-interval has μ_s as the abscissa of f' so that
> > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > f'(μ_s) = [ f(c-m + ((m+n)(s+1))/k) - f(c-m + ((m+n)s)/k) ] / ((m+n)/k)
> > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > The level magnitude (aka arithmetic mean in mainstream mathematics) is given by:
> > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > f'(c) = (1/k) \sum_{s=1}^k f'(μ_s)
> > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > From the above statement, we want to show that
> > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > f'(c) = [ f'(μ_s1) + f'(μ_s2) + f'(μ_s3) + ... + f'(μ_k-1) + f'(μ_k) ]/k
> > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > By replacing each of the means with a derivative, we have:
> > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > f'(c) = (1/k) { [ f(c-m + (m+n)/k)-f(c-m) ]/((m+n)/k)
> > > > > > > > > > > > > > > > > + [ f(c-m + 2(m+n)/k)-f(c-m + (m+n)/k) ]/((m+n)/k)]
> > > > > > > > > > > > > > > > > + [ f(c-m + 3(m+n)/k)-f(c-m + 2(m+n)/k) ]/((m+n)/k)]
> > > > > > > > > > > > > > > > > + ...
> > > > > > > > > > > > > > > > > + [ f(c-m + ((k-1)(m+n))/k)-f(c-m + ((k-2)(m+n))/k) ]/((m+n)/k)]
> > > > > > > > > > > > > > > > > + [ f(c+n)-f(c-m + ((k-1)(m+n))/k) ]/((m+n)/k)]
> > > > > > > > > > > > > > > > > }
> > > > > > > > > > > > > > > > > The above reduces to f'(c) = (f(c+n)-f(c-m))/(m+n)
> > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > (1/(m+n)) \int_{c-m}^{c+n} f'(x) dx = (f(c+n)-f(c-m))/(m+n)
> > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > So, what we have proved above is that f'(c) is the level magnitude of all the y-ordinates of f'(x) in the interval (c-m, c+n) and we can find the product of f'(c) and m+n to give us the area under f'(x) from c-m to c+n where c-m < c < c+n.
> > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > Did you see anything there about "real numbers", you imbecile?
> > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > We need something more than just rationals for the mean value theorem. That *something* is real numbers.
> > > > > > > > > > > > > > > > > No. What you need is understanding. You have NONE.
> > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > How do you know those μ_s exist?
> > > > > > > > > > > > > > > How can you know that you are a moron?
> > > > > > > > > > > > > > Prove those mu_s. How do you know they exist?
> > > > > > > > > > > > > It's simple but probably too complicated for you. I will not give you the answer but try to lead you to it.
> > > > > > > > > > > > >
> > > > > > > > > > > > > If f is defined everywhere in (a,b) then is it true or false that at least one y ordinate represents the level magnitude (aka arithmetic mean)?
> > > > > > > > > > > > Yes, it's true by the mean value theorem.
> > > > > > > > > > > Not by the mean value theorem because one will find a level magnitude even if the curve is not smooth. The mvt applies only to smooth curves.
> > > > > > > > > > > f need not be smooth anywhere in the given interval and there will still be an arithmetic mean. Do you disagree still?
> > > > > > > > > > > > Which you seek to prove. That's circular.
> > > > > > > > > > > No. That's wrong.
> > > > > > > > > > How do you get those mu_s then?
> > > > > > > > > Answer the question dumbo!
> > > > > > > > > Not by the mean value theorem because one will find a level magnitude even if the curve is not smooth. The mvt applies only to smooth curves.
> > > > > > > > > f need not be smooth anywhere in the given interval and there will still be an arithmetic mean.
> > > > > > > > > ------> Do you disagree still?
> > > > > > > > Then how do you find it?
> > > > > > > Answer the question, retard!
> > > > > > >
> > > > > > > Do you disagree still?
> > > > > > I disagree that you can find the mu_s without using the MVT in the first place
> > > > > Well, you are wrong because if a curve is not smooth but everywhere defined means that an arithmetic mean is possible even though the mvt does not apply in the case of non-smooth curves.
> > > > >
> > > > > Explain what your understanding of arithmetic mean is. Don't tell me how to construct it because every fool knows how to do that. In words, explain the meaning of arithmetic mean. Remember: Adding up set of values and dividing by total, does not explain arithmetic mean, only how to determined it.
> > > > How can you take a mean of infinitely many values?
> > >
> > > Explain what your understanding of arithmetic mean is.
> > > <irrelevant rubbish>
> > Fine, I'll give you the answers. Then you are going to answer me.
> >
> > The mean of a finite set can be thought of intuitively as adding everything together in a big pile and then distributing it equally in n piles.
> I did not ask about a finite or infinite set. I asked you if you there is an arithmetic mean for any given set, regardless of whether it has innumerably many elements or not. Also, I didn't state whether the values are produced by a function or otherwise.
>
> Is there an arithmetic mean for any given set (whether finite or not) of values? YES/NO
>
> <nonsense>
There are probably ways to generalize a mean to an infinite set, but the normal definition is given for a finite set.

On Wednesday, 9 August 2023 at 15:50:39 UTC-4, markus...@gmail.com wrote:
> onsdag 9 augusti 2023 kl. 13:29:23 UTC+2 skrev Eram semper recta:
> > On Wednesday, 9 August 2023 at 05:21:48 UTC-4, markus...@gmail.com wrote:
> > > onsdag 9 augusti 2023 kl. 02:43:12 UTC+2 skrev Eram semper recta:
> > > > On Tuesday, 8 August 2023 at 17:50:58 UTC-4, markus...@gmail.com wrote:
> > > > > tisdag 8 augusti 2023 kl. 18:40:26 UTC+2 skrev Eram semper recta:
> > > > > > On Tuesday, 8 August 2023 at 04:36:57 UTC-4, markus...@gmail.com wrote:
> > > > > > > tisdag 8 augusti 2023 kl. 03:08:27 UTC+2 skrev Eram semper recta:
> > > > > > > > On Monday, 7 August 2023 at 18:55:07 UTC-4, markus...@gmail..com wrote:
> > > > > > > > > måndag 7 augusti 2023 kl. 14:27:42 UTC+2 skrev Eram semper recta:
> > > > > > > > > > On Monday, 7 August 2023 at 05:02:12 UTC-4, markus...@gmail.com wrote:
> > > > > > > > > > > måndag 7 augusti 2023 kl. 00:35:54 UTC+2 skrev Eram semper recta:
> > > > > > > > > > > > On Sunday, 6 August 2023 at 14:19:23 UTC-4, markus....@gmail.com wrote:
> > > > > > > > > > > > > söndag 6 augusti 2023 kl. 02:03:23 UTC+2 skrev Eram semper recta:
> > > > > > > > > > > > > > On Saturday, 5 August 2023 at 17:27:46 UTC-4, markus...
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > https://www.academia.edu/45154026/Teaching_the_fundamental_theorem_of_calculus
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > I was the FIRST HUMAN to fully understand the mean value theorem and to prove constructively.
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > NONE of the stupid FUCKS who came before me where up to the task:
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > https://www.academia.edu/81300370/Mainstream_mathematics_academics_are_arrogant_and_incorrigible_ignoramuses_The_mean_value_theorem_IS_the_fundamental_theorem_of_calculus
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > Your proof is invalid.
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > You've said that about a lot of things and every time you have been wrong.
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > You're wrong here yet again.
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > The real mean value theorem requires real numbers.
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > It doesn't have anything to do with an ill-formed object you think of as a "real number".
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > The mean value theorem is about a level magnitude (what you erroneously call an "arithmetic mean"). It's the reason calculus works at all.
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > You're right - I don't understand the "limits" of your ignorance and stupidity.
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > The mean value theorem doesn't work without real numbers. A very elementary counterexample is f(x)=0 if x²>2 and f(x)=1 if x²<2. It is continuous and differentiable everywhere on the rational number line with f'=0. If we consider [a, b]=[0, 2], we have a counterexample to the MVT (mean value theorem). From (b-a)f'(c)=f(b)-f(a) we have
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > f'(c)=0-1=-1.
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > But there is no such c. If we attempt to explicitly find c, we find that it doesn't exist.
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > The mean value theorem requires a real number line.
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > See a psychiatrist soon.
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > You most definitely do need to see a psychiatrist. You are mentally ill.
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > I don't need to.
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > But you can consider f(x)=x³ over [0, 1] and see why the mean value theorem requires real numbers.
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > It doesn't show at all what you claim. In fact, the mvt doesn't give a shit about anything but magnitudes.
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > You obviously don't understand the mvt which is about a level magnitude (aka arithmetic mean in your bullshit mathematics).
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > Then try it.
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > f(x)=x³, a=0 and b=0. Then f'(x)=3x², f(b)=1, f(a)=0.
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > What?! f(b)=f(0)=0, not 1, you imbecile!
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > <scheißen>
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > b=0 was a typo. b=1, so f(b) is indeed one. You will get f'(c)=3c³=1, and this can't be solved with rational numbers only.
> > > > > > > > > > > > > > > > > > > > > > > > > > > > It still is wrong, whether a typo or not. Incoherent gibberish out of your juvenile brain.
> > > > > > > > > > > > > > > > > > > > > > > > > > > Can you solve f'(c)=3c²=1 without real numbers?
> > > > > > > > > > > > > > > > > > > > > > > > > > Yes, idiot! It is the CONSTANT known as the square root of 1/3 and there is NO number that decribes its measure.
> > > > > > > > > > > > > > > > > > > > > > > > > The square root of 1/3 isn't rational.
> > > > > > > > > > > > > > > > > > > > > > > > And so? Stating the obvious incorrectly. There is no number that describes the measure of the constant root(3).
> > > > > > > > > > > > > > > > > > > > > > > > > Thus the mean value theorem fails without real numbers.
> > > > > > > > > > > > > > > > > > > > > > > > It doesn't because there is no such thing as "real number".
> > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > <END OF DISCUSSION>
> > > > > > > > > > > > > > > > > > > > > > > If there is no such c, then the mean value theorem is false. The theorem states there is a such c.
> > > > > > > > > > > > > > > > > > > > > > You'll have to tell me again what is your example. I stop after the first mistake I find and you get ignored.
> > > > > > > > > > > > > > > > > > > > > You have already gotten TWO counterexamples.
> > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > f(x)=x³, a=0 and b=1. Then f'(x)=3x², f(b)=1, f(a)=0.
> > > > > > > > > > > > > > > > > > > > > (b-a)f'(c)=f(b)-f(a)=1.
> > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > f'(c)=3c²=1.
> > > > > > > > > > > > > > > > > > > > > We end up with something we can't solve only using rational numbers.
> > > > > > > > > > > > > > > > > > > > I have already answered this question. The square root of 1/3 is not described by any number. It is a CONSTANT which is a numeric approximation but not the actual value.
> > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > The ratio square root of 1/3 is realised from a right angled triangle with one leg equal to leg (p) of original isosceles right-angled triangle and the other leg (q) to hypotenuse of original isosceles right-angled triangle.
> > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > Thus the ratio is p:q and this ratio has no measure. If we attempt to measure it, we end up with the same failed measure that is the constant 0.5773 (which is an approximation or failed measure of p:q).
> > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > You fail to understand these things because you don't understand ratio and number.
> > > > > > > > > > > > > > > > > > > What you call "magnitudes" are in fact just real numbers.
> > > > > > > > > > > > > > > > > > You are just too stupid.
> > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > Here is the proof of the mean value theorem using the 100% rigorous New Calculus:
> > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > The New Calculus derivative:
> > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > Given any interval (c-m,c+n), the New Calculus (henceforth NC) derivative is given by:
> > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > f'(c)= (f(c+n)-f(c-m))/(m+n)
> > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > f'(c) represents the slope of a secant line with end points (c-m, f(c-m)) and ( c+n, f(c+n)) that is parallel to the tangent line at x=c.
> > > > > > > > > > > > > > > > > > m and n are horizontal distances from c..
> > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > The interval (c-m,c+n) can be partitioned into equal sub-intervals of (m+n)/k.
> > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > Each sub-interval has μ_s as the abscissa of f' so that
> > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > f'(μ_s) = [ f(c-m + ((m+n)(s+1))/k) - f(c-m + ((m+n)s)/k) ] / ((m+n)/k)
> > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > The level magnitude (aka arithmetic mean in mainstream mathematics) is given by:
> > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > f'(c) = (1/k) \sum_{s=1}^k f'(μ_s)
> > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > From the above statement, we want to show that
> > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > f'(c) = [ f'(μ_s1) + f'(μ_s2) + f'(μ_s3) + ... + f'(μ_k-1) + f'(μ_k) ]/k
> > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > By replacing each of the means with a derivative, we have:
> > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > f'(c) = (1/k) { [ f(c-m + (m+n)/k)-f(c-m) ]/((m+n)/k)
> > > > > > > > > > > > > > > > > > + [ f(c-m + 2(m+n)/k)-f(c-m + (m+n)/k) ]/((m+n)/k)]
> > > > > > > > > > > > > > > > > > + [ f(c-m + 3(m+n)/k)-f(c-m + 2(m+n)/k) ]/((m+n)/k)]
> > > > > > > > > > > > > > > > > > + ...
> > > > > > > > > > > > > > > > > > + [ f(c-m + ((k-1)(m+n))/k)-f(c-m + ((k-2)(m+n))/k) ]/((m+n)/k)]
> > > > > > > > > > > > > > > > > > + [ f(c+n)-f(c-m + ((k-1)(m+n))/k) ]/((m+n)/k)]
> > > > > > > > > > > > > > > > > > }
> > > > > > > > > > > > > > > > > > The above reduces to f'(c) = (f(c+n)-f(c-m))/(m+n)
> > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > (1/(m+n)) \int_{c-m}^{c+n} f'(x) dx = (f(c+n)-f(c-m))/(m+n)
> > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > So, what we have proved above is that f'(c) is the level magnitude of all the y-ordinates of f'(x) in the interval (c-m, c+n) and we can find the product of f'(c) and m+n to give us the area under f'(x) from c-m to c+n where c-m < c < c+n.
> > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > Did you see anything there about "real numbers", you imbecile?
> > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > We need something more than just rationals for the mean value theorem. That *something* is real numbers.
> > > > > > > > > > > > > > > > > > No. What you need is understanding. You have NONE.
> > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > How do you know those μ_s exist?
> > > > > > > > > > > > > > > > How can you know that you are a moron?
> > > > > > > > > > > > > > > Prove those mu_s. How do you know they exist?
> > > > > > > > > > > > > > It's simple but probably too complicated for you. I will not give you the answer but try to lead you to it.
> > > > > > > > > > > > > >
> > > > > > > > > > > > > > If f is defined everywhere in (a,b) then is it true or false that at least one y ordinate represents the level magnitude (aka arithmetic mean)?
> > > > > > > > > > > > > Yes, it's true by the mean value theorem.
> > > > > > > > > > > > Not by the mean value theorem because one will find a level magnitude even if the curve is not smooth. The mvt applies only to smooth curves.
> > > > > > > > > > > > f need not be smooth anywhere in the given interval and there will still be an arithmetic mean. Do you disagree still?
> > > > > > > > > > > > > Which you seek to prove. That's circular.
> > > > > > > > > > > > No. That's wrong.
> > > > > > > > > > > How do you get those mu_s then?
> > > > > > > > > > Answer the question dumbo!
> > > > > > > > > > Not by the mean value theorem because one will find a level magnitude even if the curve is not smooth. The mvt applies only to smooth curves.
> > > > > > > > > > f need not be smooth anywhere in the given interval and there will still be an arithmetic mean.
> > > > > > > > > > ------> Do you disagree still?
> > > > > > > > > Then how do you find it?
> > > > > > > > Answer the question, retard!
> > > > > > > >
> > > > > > > > Do you disagree still?
> > > > > > > I disagree that you can find the mu_s without using the MVT in the first place
> > > > > > Well, you are wrong because if a curve is not smooth but everywhere defined means that an arithmetic mean is possible even though the mvt does not apply in the case of non-smooth curves.
> > > > > >
> > > > > > Explain what your understanding of arithmetic mean is. Don't tell me how to construct it because every fool knows how to do that. In words, explain the meaning of arithmetic mean. Remember: Adding up set of values and dividing by total, does not explain arithmetic mean, only how to determined it.
> > > > > How can you take a mean of infinitely many values?
> > > >
> > > > Explain what your understanding of arithmetic mean is.
> > > > <irrelevant rubbish>
> > > Fine, I'll give you the answers. Then you are going to answer me.
> > >
> > > The mean of a finite set can be thought of intuitively as adding everything together in a big pile and then distributing it equally in n piles.
> > I did not ask about a finite or infinite set. I asked you if you there is an arithmetic mean for any given set, regardless of whether it has innumerably many elements or not. Also, I didn't state whether the values are produced by a function or otherwise.
> >
> > Is there an arithmetic mean for any given set (whether finite or not) of values? YES/NO
> >
> > <nonsense>
> There are probably ways to generalize a mean to an infinite set, but the normal definition is given for a finite set.

On Sunday, 13 August 2023 at 10:52:28 UTC-4, markus...@gmail.com wrote:
> onsdag 9 augusti 2023 kl. 22:24:49 UTC+2 skrev Eram semper recta:
> > On Wednesday, 9 August 2023 at 15:50:39 UTC-4, markus...@gmail.com wrote:
> > > onsdag 9 augusti 2023 kl. 13:29:23 UTC+2 skrev Eram semper recta:
> > > > On Wednesday, 9 August 2023 at 05:21:48 UTC-4, markus...@gmail.com wrote:
> > > > > onsdag 9 augusti 2023 kl. 02:43:12 UTC+2 skrev Eram semper recta:
> > > > > > On Tuesday, 8 August 2023 at 17:50:58 UTC-4, markus...@gmail.com wrote:
> > > > > > > tisdag 8 augusti 2023 kl. 18:40:26 UTC+2 skrev Eram semper recta:
> > > > > > > > On Tuesday, 8 August 2023 at 04:36:57 UTC-4, markus...@gmail.com wrote:
> > > > > > > > > tisdag 8 augusti 2023 kl. 03:08:27 UTC+2 skrev Eram semper recta:
> > > > > > > > > > On Monday, 7 August 2023 at 18:55:07 UTC-4, markus...@gmail.com wrote:
> > > > > > > > > > > måndag 7 augusti 2023 kl. 14:27:42 UTC+2 skrev Eram semper recta:
> > > > > > > > > > > > On Monday, 7 August 2023 at 05:02:12 UTC-4, markus....@gmail.com wrote:
> > > > > > > > > > > > > måndag 7 augusti 2023 kl. 00:35:54 UTC+2 skrev Eram semper recta:
> > > > > > > > > > > > > > On Sunday, 6 August 2023 at 14:19:23 UTC-4, markus...@gmail.com wrote:
> > > > > > > > > > > > > > > söndag 6 augusti 2023 kl. 02:03:23 UTC+2 skrev Eram semper recta:
> > > > > > > > > > > > > > > > On Saturday, 5 August 2023 at 17:27:46 UTC-4, markus...
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > https://www.academia.edu/45154026/Teaching_the_fundamental_theorem_of_calculus
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > I was the FIRST HUMAN to fully understand the mean value theorem and to prove constructively.
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > NONE of the stupid FUCKS who came before me where up to the task:
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > https://www.academia.edu/81300370/Mainstream_mathematics_academics_are_arrogant_and_incorrigible_ignoramuses_The_mean_value_theorem_IS_the_fundamental_theorem_of_calculus
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > Your proof is invalid.
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > You've said that about a lot of things and every time you have been wrong.
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > You're wrong here yet again.
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > The real mean value theorem requires real numbers.
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > It doesn't have anything to do with an ill-formed object you think of as a "real number".
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > The mean value theorem is about a level magnitude (what you erroneously call an "arithmetic mean"). It's the reason calculus works at all.
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > You're right - I don't understand the "limits" of your ignorance and stupidity..
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > The mean value theorem doesn't work without real numbers. A very elementary counterexample is f(x)=0 if x²>2 and f(x)=1 if x²<2. It is continuous and differentiable everywhere on the rational number line with f'=0. If we consider [a, b]=[0, 2], we have a counterexample to the MVT (mean value theorem). From (b-a)f'(c)=f(b)-f(a) we have
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > f'(c)=0-1=-1.
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > But there is no such c. If we attempt to explicitly find c, we find that it doesn't exist.
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > The mean value theorem requires a real number line.
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > See a psychiatrist soon.
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > You most definitely do need to see a psychiatrist. You are mentally ill.
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > I don't need to.
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > But you can consider f(x)=x³ over [0, 1] and see why the mean value theorem requires real numbers.
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > It doesn't show at all what you claim. In fact, the mvt doesn't give a shit about anything but magnitudes.
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > You obviously don't understand the mvt which is about a level magnitude (aka arithmetic mean in your bullshit mathematics).
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > Then try it.
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > f(x)=x³, a=0 and b=0. Then f'(x)=3x², f(b)=1, f(a)=0.
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > What?! f(b)=f(0)=0, not 1, you imbecile!
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > <scheißen>
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > b=0 was a typo. b=1, so f(b) is indeed one. You will get f'(c)=3c³=1, and this can't be solved with rational numbers only.
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > It still is wrong, whether a typo or not. Incoherent gibberish out of your juvenile brain.
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > Can you solve f'(c)=3c²=1 without real numbers?
> > > > > > > > > > > > > > > > > > > > > > > > > > > > Yes, idiot! It is the CONSTANT known as the square root of 1/3 and there is NO number that decribes its measure.
> > > > > > > > > > > > > > > > > > > > > > > > > > > The square root of 1/3 isn't rational.
> > > > > > > > > > > > > > > > > > > > > > > > > > And so? Stating the obvious incorrectly. There is no number that describes the measure of the constant root(3).
> > > > > > > > > > > > > > > > > > > > > > > > > > > Thus the mean value theorem fails without real numbers.
> > > > > > > > > > > > > > > > > > > > > > > > > > It doesn't because there is no such thing as "real number".
> > > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > > <END OF DISCUSSION>
> > > > > > > > > > > > > > > > > > > > > > > > > If there is no such c, then the mean value theorem is false. The theorem states there is a such c.
> > > > > > > > > > > > > > > > > > > > > > > > You'll have to tell me again what is your example. I stop after the first mistake I find and you get ignored.
> > > > > > > > > > > > > > > > > > > > > > > You have already gotten TWO counterexamples.
> > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > f(x)=x³, a=0 and b=1. Then f'(x)=3x², f(b)=1, f(a)=0.
> > > > > > > > > > > > > > > > > > > > > > > (b-a)f'(c)=f(b)-f(a)=1.
> > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > f'(c)=3c²=1.
> > > > > > > > > > > > > > > > > > > > > > > We end up with something we can't solve only using rational numbers.
> > > > > > > > > > > > > > > > > > > > > > I have already answered this question. The square root of 1/3 is not described by any number. It is a CONSTANT which is a numeric approximation but not the actual value.
> > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > The ratio square root of 1/3 is realised from a right angled triangle with one leg equal to leg (p) of original isosceles right-angled triangle and the other leg (q) to hypotenuse of original isosceles right-angled triangle.
> > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > Thus the ratio is p:q and this ratio has no measure. If we attempt to measure it, we end up with the same failed measure that is the constant 0.5773 (which is an approximation or failed measure of p:q).
> > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > You fail to understand these things because you don't understand ratio and number.
> > > > > > > > > > > > > > > > > > > > > What you call "magnitudes" are in fact just real numbers.
> > > > > > > > > > > > > > > > > > > > You are just too stupid.
> > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > Here is the proof of the mean value theorem using the 100% rigorous New Calculus:
> > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > The New Calculus derivative:
> > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > Given any interval (c-m,c+n), the New Calculus (henceforth NC) derivative is given by:
> > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > f'(c)= (f(c+n)-f(c-m))/(m+n)
> > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > f'(c) represents the slope of a secant line with end points (c-m, f(c-m)) and ( c+n, f(c+n)) that is parallel to the tangent line at x=c.
> > > > > > > > > > > > > > > > > > > > m and n are horizontal distances from c.
> > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > The interval (c-m,c+n) can be partitioned into equal sub-intervals of (m+n)/k.
> > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > Each sub-interval has μ_s as the abscissa of f' so that
> > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > f'(μ_s) = [ f(c-m + ((m+n)(s+1))/k) - f(c-m + ((m+n)s)/k) ] / ((m+n)/k)
> > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > The level magnitude (aka arithmetic mean in mainstream mathematics) is given by:
> > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > f'(c) = (1/k) \sum_{s=1}^k f'(μ_s)
> > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > From the above statement, we want to show that
> > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > f'(c) = [ f'(μ_s1) + f'(μ_s2) + f'(μ_s3) + ... + f'(μ_k-1) + f'(μ_k) ]/k
> > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > By replacing each of the means with a derivative, we have:
> > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > f'(c) = (1/k) { [ f(c-m + (m+n)/k)-f(c-m) ]/((m+n)/k)
> > > > > > > > > > > > > > > > > > > > + [ f(c-m + 2(m+n)/k)-f(c-m + (m+n)/k) ]/((m+n)/k)]
> > > > > > > > > > > > > > > > > > > > + [ f(c-m + 3(m+n)/k)-f(c-m + 2(m+n)/k) ]/((m+n)/k)]
> > > > > > > > > > > > > > > > > > > > + ...
> > > > > > > > > > > > > > > > > > > > + [ f(c-m + ((k-1)(m+n))/k)-f(c-m + ((k-2)(m+n))/k) ]/((m+n)/k)]
> > > > > > > > > > > > > > > > > > > > + [ f(c+n)-f(c-m + ((k-1)(m+n))/k) ]/((m+n)/k)]
> > > > > > > > > > > > > > > > > > > > }
> > > > > > > > > > > > > > > > > > > > The above reduces to f'(c) = (f(c+n)-f(c-m))/(m+n)
> > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > (1/(m+n)) \int_{c-m}^{c+n} f'(x) dx = (f(c+n)-f(c-m))/(m+n)
> > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > So, what we have proved above is that f'(c) is the level magnitude of all the y-ordinates of f'(x) in the interval (c-m, c+n) and we can find the product of f'(c) and m+n to give us the area under f'(x) from c-m to c+n where c-m < c < c+n.
> > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > Did you see anything there about "real numbers", you imbecile?
> > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > We need something more than just rationals for the mean value theorem. That *something* is real numbers.
> > > > > > > > > > > > > > > > > > > > No. What you need is understanding. You have NONE.
> > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > How do you know those μ_s exist?
> > > > > > > > > > > > > > > > > > How can you know that you are a moron?
> > > > > > > > > > > > > > > > > Prove those mu_s. How do you know they exist?
> > > > > > > > > > > > > > > > It's simple but probably too complicated for you. I will not give you the answer but try to lead you to it.
> > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > If f is defined everywhere in (a,b) then is it true or false that at least one y ordinate represents the level magnitude (aka arithmetic mean)?
> > > > > > > > > > > > > > > Yes, it's true by the mean value theorem.
> > > > > > > > > > > > > > Not by the mean value theorem because one will find a level magnitude even if the curve is not smooth. The mvt applies only to smooth curves.
> > > > > > > > > > > > > > f need not be smooth anywhere in the given interval and there will still be an arithmetic mean. Do you disagree still?
> > > > > > > > > > > > > > > Which you seek to prove. That's circular.
> > > > > > > > > > > > > > No. That's wrong.
> > > > > > > > > > > > > How do you get those mu_s then?
> > > > > > > > > > > > Answer the question dumbo!
> > > > > > > > > > > > Not by the mean value theorem because one will find a level magnitude even if the curve is not smooth. The mvt applies only to smooth curves.
> > > > > > > > > > > > f need not be smooth anywhere in the given interval and there will still be an arithmetic mean.
> > > > > > > > > > > > ------> Do you disagree still?
> > > > > > > > > > > Then how do you find it?
> > > > > > > > > > Answer the question, retard!
> > > > > > > > > >
> > > > > > > > > > Do you disagree still?
> > > > > > > > > I disagree that you can find the mu_s without using the MVT in the first place
> > > > > > > > Well, you are wrong because if a curve is not smooth but everywhere defined means that an arithmetic mean is possible even though the mvt does not apply in the case of non-smooth curves.
> > > > > > > >
> > > > > > > > Explain what your understanding of arithmetic mean is. Don't tell me how to construct it because every fool knows how to do that. In words, explain the meaning of arithmetic mean. Remember: Adding up set of values and dividing by total, does not explain arithmetic mean, only how to determined it.
> > > > > > > How can you take a mean of infinitely many values?
> > > > > >
> > > > > > Explain what your understanding of arithmetic mean is.
> > > > > > <irrelevant rubbish>
> > > > > Fine, I'll give you the answers. Then you are going to answer me.
> > > > >
> > > > > The mean of a finite set can be thought of intuitively as adding everything together in a big pile and then distributing it equally in n piles.
> > > > I did not ask about a finite or infinite set. I asked you if you there is an arithmetic mean for any given set, regardless of whether it has innumerably many elements or not. Also, I didn't state whether the values are produced by a function or otherwise.
> > > >
> > > > Is there an arithmetic mean for any given set (whether finite or not) of values? YES/NO
> > > >
> > > > <nonsense>
> > > There are probably ways to generalize a mean to an infinite set, but the normal definition is given for a finite set.
> > Wrong. Any given set of values over a bounded interval (which you may consider infinite), there is always an arithmetic mean whether the values relate to a function or not.
> How do you define it?

måndag 14 augusti 2023 kl. 14:21:35 UTC+2 skrev Eram semper recta:
> On Sunday, 13 August 2023 at 10:52:28 UTC-4, markus...@gmail.com wrote:
> > onsdag 9 augusti 2023 kl. 22:24:49 UTC+2 skrev Eram semper recta:
> > > On Wednesday, 9 August 2023 at 15:50:39 UTC-4, markus...@gmail.com wrote:
> > > > onsdag 9 augusti 2023 kl. 13:29:23 UTC+2 skrev Eram semper recta:
> > > > > On Wednesday, 9 August 2023 at 05:21:48 UTC-4, markus...@gmail.com wrote:
> > > > > > onsdag 9 augusti 2023 kl. 02:43:12 UTC+2 skrev Eram semper recta:
> > > > > > > On Tuesday, 8 August 2023 at 17:50:58 UTC-4, markus...@gmail.com wrote:
> > > > > > > > tisdag 8 augusti 2023 kl. 18:40:26 UTC+2 skrev Eram semper recta:
> > > > > > > > > On Tuesday, 8 August 2023 at 04:36:57 UTC-4, markus...@gmail.com wrote:
> > > > > > > > > > tisdag 8 augusti 2023 kl. 03:08:27 UTC+2 skrev Eram semper recta:
> > > > > > > > > > > On Monday, 7 August 2023 at 18:55:07 UTC-4, markus...@gmail.com wrote:
> > > > > > > > > > > > måndag 7 augusti 2023 kl. 14:27:42 UTC+2 skrev Eram semper recta:
> > > > > > > > > > > > > On Monday, 7 August 2023 at 05:02:12 UTC-4, markus...@gmail.com wrote:
> > > > > > > > > > > > > > måndag 7 augusti 2023 kl. 00:35:54 UTC+2 skrev Eram semper recta:
> > > > > > > > > > > > > > > On Sunday, 6 August 2023 at 14:19:23 UTC-4, markus...@gmail.com wrote:
> > > > > > > > > > > > > > > > söndag 6 augusti 2023 kl. 02:03:23 UTC+2 skrev Eram semper recta:
> > > > > > > > > > > > > > > > > On Saturday, 5 August 2023 at 17:27:46 UTC-4, markus...
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > https://www.academia.edu/45154026/Teaching_the_fundamental_theorem_of_calculus
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > I was the FIRST HUMAN to fully understand the mean value theorem and to prove constructively.
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > NONE of the stupid FUCKS who came before me where up to the task:
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > https://www.academia.edu/81300370/Mainstream_mathematics_academics_are_arrogant_and_incorrigible_ignoramuses_The_mean_value_theorem_IS_the_fundamental_theorem_of_calculus
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > Your proof is invalid.
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > You've said that about a lot of things and every time you have been wrong.
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > You're wrong here yet again.
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > The real mean value theorem requires real numbers.
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > It doesn't have anything to do with an ill-formed object you think of as a "real number".
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > The mean value theorem is about a level magnitude (what you erroneously call an "arithmetic mean"). It's the reason calculus works at all.
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > You're right - I don't understand the "limits" of your ignorance and stupidity.
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > The mean value theorem doesn't work without real numbers. A very elementary counterexample is f(x)=0 if x²>2 and f(x)=1 if x²<2. It is continuous and differentiable everywhere on the rational number line with f'=0. If we consider [a, b]=[0, 2], we have a counterexample to the MVT (mean value theorem). From (b-a)f'(c)=f(b)-f(a) we have
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > f'(c)=0-1=-1.
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > But there is no such c. If we attempt to explicitly find c, we find that it doesn't exist.
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > The mean value theorem requires a real number line.
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > See a psychiatrist soon.
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > You most definitely do need to see a psychiatrist. You are mentally ill.
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > I don't need to.
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > But you can consider f(x)=x³ over [0, 1] and see why the mean value theorem requires real numbers.
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > It doesn't show at all what you claim. In fact, the mvt doesn't give a shit about anything but magnitudes.
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > You obviously don't understand the mvt which is about a level magnitude (aka arithmetic mean in your bullshit mathematics).
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > Then try it.
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > f(x)=x³, a=0 and b=0. Then f'(x)=3x², f(b)=1, f(a)=0.
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > What?! f(b)=f(0)=0, not 1, you imbecile!
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > <scheißen>
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > b=0 was a typo. b=1, so f(b) is indeed one. You will get f'(c)=3c³=1, and this can't be solved with rational numbers only.
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > It still is wrong, whether a typo or not. Incoherent gibberish out of your juvenile brain.
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > > Can you solve f'(c)=3c²=1 without real numbers?
> > > > > > > > > > > > > > > > > > > > > > > > > > > > > Yes, idiot! It is the CONSTANT known as the square root of 1/3 and there is NO number that decribes its measure.
> > > > > > > > > > > > > > > > > > > > > > > > > > > > The square root of 1/3 isn't rational.
> > > > > > > > > > > > > > > > > > > > > > > > > > > And so? Stating the obvious incorrectly. There is no number that describes the measure of the constant root(3).
> > > > > > > > > > > > > > > > > > > > > > > > > > > > Thus the mean value theorem fails without real numbers.
> > > > > > > > > > > > > > > > > > > > > > > > > > > It doesn't because there is no such thing as "real number".
> > > > > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > > > > <END OF DISCUSSION>
> > > > > > > > > > > > > > > > > > > > > > > > > > If there is no such c, then the mean value theorem is false. The theorem states there is a such c.
> > > > > > > > > > > > > > > > > > > > > > > > > You'll have to tell me again what is your example. I stop after the first mistake I find and you get ignored.
> > > > > > > > > > > > > > > > > > > > > > > > You have already gotten TWO counterexamples.
> > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > f(x)=x³, a=0 and b=1. Then f'(x)=3x², f(b)=1, f(a)=0.
> > > > > > > > > > > > > > > > > > > > > > > > (b-a)f'(c)=f(b)-f(a)=1.
> > > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > > f'(c)=3c²=1.
> > > > > > > > > > > > > > > > > > > > > > > > We end up with something we can't solve only using rational numbers.
> > > > > > > > > > > > > > > > > > > > > > > I have already answered this question. The square root of 1/3 is not described by any number. It is a CONSTANT which is a numeric approximation but not the actual value.
> > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > The ratio square root of 1/3 is realised from a right angled triangle with one leg equal to leg (p) of original isosceles right-angled triangle and the other leg (q) to hypotenuse of original isosceles right-angled triangle.
> > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > Thus the ratio is p:q and this ratio has no measure. If we attempt to measure it, we end up with the same failed measure that is the constant 0.5773 (which is an approximation or failed measure of p:q).
> > > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > > You fail to understand these things because you don't understand ratio and number.
> > > > > > > > > > > > > > > > > > > > > > What you call "magnitudes" are in fact just real numbers.
> > > > > > > > > > > > > > > > > > > > > You are just too stupid.
> > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > Here is the proof of the mean value theorem using the 100% rigorous New Calculus:
> > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > The New Calculus derivative:
> > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > Given any interval (c-m,c+n), the New Calculus (henceforth NC) derivative is given by:
> > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > f'(c)= (f(c+n)-f(c-m))/(m+n)
> > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > f'(c) represents the slope of a secant line with end points (c-m, f(c-m)) and ( c+n, f(c+n)) that is parallel to the tangent line at x=c.
> > > > > > > > > > > > > > > > > > > > > m and n are horizontal distances from c.
> > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > The interval (c-m,c+n) can be partitioned into equal sub-intervals of (m+n)/k.
> > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > Each sub-interval has μ_s as the abscissa of f' so that
> > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > f'(μ_s) = [ f(c-m + ((m+n)(s+1))/k) - f(c-m + ((m+n)s)/k) ] / ((m+n)/k)
> > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > The level magnitude (aka arithmetic mean in mainstream mathematics) is given by:
> > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > f'(c) = (1/k) \sum_{s=1}^k f'(μ_s)
> > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > From the above statement, we want to show that
> > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > f'(c) = [ f'(μ_s1) + f'(μ_s2) + f'(μ_s3) + ... + f'(μ_k-1) + f'(μ_k) ]/k
> > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > By replacing each of the means with a derivative, we have:
> > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > f'(c) = (1/k) { [ f(c-m + (m+n)/k)-f(c-m) ]/((m+n)/k)
> > > > > > > > > > > > > > > > > > > > > + [ f(c-m + 2(m+n)/k)-f(c-m + (m+n)/k) ]/((m+n)/k)]
> > > > > > > > > > > > > > > > > > > > > + [ f(c-m + 3(m+n)/k)-f(c-m + 2(m+n)/k) ]/((m+n)/k)]
> > > > > > > > > > > > > > > > > > > > > + ...
> > > > > > > > > > > > > > > > > > > > > + [ f(c-m + ((k-1)(m+n))/k)-f(c-m + ((k-2)(m+n))/k) ]/((m+n)/k)]
> > > > > > > > > > > > > > > > > > > > > + [ f(c+n)-f(c-m + ((k-1)(m+n))/k) ]/((m+n)/k)]
> > > > > > > > > > > > > > > > > > > > > }
> > > > > > > > > > > > > > > > > > > > > The above reduces to f'(c) = (f(c+n)-f(c-m))/(m+n)
> > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > (1/(m+n)) \int_{c-m}^{c+n} f'(x) dx = (f(c+n)-f(c-m))/(m+n)
> > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > So, what we have proved above is that f'(c) is the level magnitude of all the y-ordinates of f'(x) in the interval (c-m, c+n) and we can find the product of f'(c) and m+n to give us the area under f'(x) from c-m to c+n where c-m < c < c+n.
> > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > Did you see anything there about "real numbers", you imbecile?
> > > > > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > > > We need something more than just rationals for the mean value theorem. That *something* is real numbers.
> > > > > > > > > > > > > > > > > > > > > No. What you need is understanding. You have NONE.
> > > > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > > > > How do you know those μ_s exist?
> > > > > > > > > > > > > > > > > > > How can you know that you are a moron?
> > > > > > > > > > > > > > > > > > Prove those mu_s. How do you know they exist?
> > > > > > > > > > > > > > > > > It's simple but probably too complicated for you. I will not give you the answer but try to lead you to it.
> > > > > > > > > > > > > > > > >
> > > > > > > > > > > > > > > > > If f is defined everywhere in (a,b) then is it true or false that at least one y ordinate represents the level magnitude (aka arithmetic mean)?
> > > > > > > > > > > > > > > > Yes, it's true by the mean value theorem.
> > > > > > > > > > > > > > > Not by the mean value theorem because one will find a level magnitude even if the curve is not smooth. The mvt applies only to smooth curves.
> > > > > > > > > > > > > > > f need not be smooth anywhere in the given interval and there will still be an arithmetic mean. Do you disagree still?
> > > > > > > > > > > > > > > > Which you seek to prove. That's circular.
> > > > > > > > > > > > > > > No. That's wrong.
> > > > > > > > > > > > > > How do you get those mu_s then?
> > > > > > > > > > > > > Answer the question dumbo!
> > > > > > > > > > > > > Not by the mean value theorem because one will find a level magnitude even if the curve is not smooth. The mvt applies only to smooth curves.
> > > > > > > > > > > > > f need not be smooth anywhere in the given interval and there will still be an arithmetic mean.
> > > > > > > > > > > > > ------> Do you disagree still?
> > > > > > > > > > > > Then how do you find it?
> > > > > > > > > > > Answer the question, retard!
> > > > > > > > > > >
> > > > > > > > > > > Do you disagree still?
> > > > > > > > > > I disagree that you can find the mu_s without using the MVT in the first place
> > > > > > > > > Well, you are wrong because if a curve is not smooth but everywhere defined means that an arithmetic mean is possible even though the mvt does not apply in the case of non-smooth curves.
> > > > > > > > >
> > > > > > > > > Explain what your understanding of arithmetic mean is. Don't tell me how to construct it because every fool knows how to do that. In words, explain the meaning of arithmetic mean. Remember: Adding up set of values and dividing by total, does not explain arithmetic mean, only how to determined it.
> > > > > > > > How can you take a mean of infinitely many values?
> > > > > > >
> > > > > > > Explain what your understanding of arithmetic mean is.
> > > > > > > <irrelevant rubbish>
> > > > > > Fine, I'll give you the answers. Then you are going to answer me.
> > > > > >
> > > > > > The mean of a finite set can be thought of intuitively as adding everything together in a big pile and then distributing it equally in n piles.
> > > > > I did not ask about a finite or infinite set. I asked you if you there is an arithmetic mean for any given set, regardless of whether it has innumerably many elements or not. Also, I didn't state whether the values are produced by a function or otherwise.
> > > > >
> > > > > Is there an arithmetic mean for any given set (whether finite or not) of values? YES/NO
> > > > >
> > > > > <nonsense>
> > > > There are probably ways to generalize a mean to an infinite set, but the normal definition is given for a finite set.
> > > Wrong. Any given set of values over a bounded interval (which you may consider infinite), there is always an arithmetic mean whether the values relate to a function or not.
> > How do you define it?
> Been explained to you many times. You can't help it that you're stupid.
> >
> > What's the mean of all rational numbers?
Avoiding the question.

On Monday, 31 July 2023 at 05:18:09 UTC-4, Eram semper recta wrote:
> https://www.academia.edu/45154026/Teaching_the_fundamental_theorem_of_calculus

That article shows you what is the fundamental theorem and how to understand it, after which you might be able to teach it.