tisdag 15 augusti 2023 kl. 00:57:04 UTC+2 skrev Eram semper recta:
> 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.
You prove the MVT using the MVT. Not a good start, pal. John, oh dear.

On Monday, July 31, 2023 at 4:13:59 PM UTC-4, Eram semper recta wrote:
> On Monday, 31 July 2023 at 12:56:09 UTC-4, markus... wrote:
> > måndag 31 juli 2023 kl. 11:18:09 UTC+2 skrev Eram semper recta:
> > > https://www.academia.edu/45154026/Teaching_the_fundamental_theorem_of_calculus
> > The mean value theorem is not a trivial fact, nor an "unremarkable statement". It is the basis for most theorems in real analysis.
> >
> > How do you know a such c exists? If you only allow for rational numbers, it's easy to construct a function that fails the mean value theorem.
> SHUT UP you ignorant little queer!
>
> Eat my SHIT and DIE!!!!!

Such lucid argument. Who can top this high level of maths???

On Tuesday, August 1, 2023 at 6:39:02 PM UTC-4, Eram semper recta wrote:
> On Tuesday, 1 August 2023 at 17:34:32 UTC-4, markus...@gmail.com wrote:
> > tisdag 1 augusti 2023 kl. 14:20:30 UTC+2 skrev Eram semper recta:
> > > On Tuesday, 1 August 2023 at 06:35:56 UTC-4, markus...@gmail.com wrote:
> > > > tisdag 1 augusti 2023 kl. 11:27:09 UTC+2 skrev Eram semper recta:
> > > > > On Monday, 31 July 2023 at 16:54:53 UTC-4, markus...@gmail.com wrote:
> > > > > > måndag 31 juli 2023 kl. 22:22:18 UTC+2 skrev Eram semper recta:
> > > > > > > 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
> > > > > > >
> > > > > > > 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.
> > > Not a valid or coherent example because the mvt works only on ONE function, not multiple functions. Like I have often instructed you, the methods of calculus apply ONLY to SMOOTH curves.
> > >
> > > The rest of your drivel is not even relevant. Try again, troll!
> > > > 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.

Not even Andrew Wiles can top this with his proof of FLT!

> > No. You're <shit>
> 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On Friday, 25 August 2023 at 09:58:13 UTC-4, markus.
> tisdag 15 augusti 2023 kl. 00:57:04 UTC+2 skrev Eram semper recta:
> > 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.
> You prove the MVT using the MVT. Not a good start, pal. John, oh dear.

No. It's proved using the historic identity. Your failure to understand is well-known.

söndag 27 augusti 2023 kl. 16:41:45 UTC+2 skrev Eram semper recta:
> On Friday, 25 August 2023 at 09:58:13 UTC-4, markus.
> > tisdag 15 augusti 2023 kl. 00:57:04 UTC+2 skrev Eram semper recta:
> > > 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.
> > You prove the MVT using the MVT. Not a good start, pal. John, oh dear.
> No. It's proved using the historic identity. Your failure to understand is well-known.
How do you get the mu_s then?

On Sunday, 27 August 2023 at 19:07:53 UTC-4, markus...wrote:
> söndag 27 augusti 2023 kl. 16:41:45 UTC+2 skrev Eram semper recta:
> > On Friday, 25 August 2023 at 09:58:13 UTC-4, markus.
> > > tisdag 15 augusti 2023 kl. 00:57:04 UTC+2 skrev Eram semper recta:
> > > > 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.
> > > You prove the MVT using the MVT. Not a good start, pal. John, oh dear..
> > No. It's proved using the historic identity. Your failure to understand is well-known.
> How do you get the mu_s then?

We state the New Calculus derivative as follows:

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)

NOWHERE IN THE ABOVE IS THE MVT USED, ONLY THE DEFINITION OF NEW CALCULUS DERIVATIVE.

(f(+n)-f(x-m))/(m+n) = f'(x) + Q(x,m,n)

where

(f(+n)-f(x-m))/(m+n) is ALWAYS the slope a parallel secant line to the tangent line, hence it is the slope of the tangent line at x.

f'(x) denotes the tangent line slope and Q(x,m,n) the difference in slopes which is ALWAYS zero, unlike in my historic geometric theorem where the only case when Q(x,h) is zero is when f is a linear function.

NO CIRCULARITY WHATSOEVER. The mvt is not used anywhere, but these facts help to prove it.

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.

måndag 28 augusti 2023 kl. 17:11:00 UTC+2 skrev Eram semper recta:
> On Sunday, 27 August 2023 at 19:07:53 UTC-4, markus...wrote:
> > söndag 27 augusti 2023 kl. 16:41:45 UTC+2 skrev Eram semper recta:
> > > On Friday, 25 August 2023 at 09:58:13 UTC-4, markus.
> > > > tisdag 15 augusti 2023 kl. 00:57:04 UTC+2 skrev Eram semper recta:
> > > > > 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.
> > > > You prove the MVT using the MVT. Not a good start, pal. John, oh dear.
> > > No. It's proved using the historic identity. Your failure to understand is well-known.
> > How do you get the mu_s then?
> We state the New Calculus derivative as follows:
> 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)
> NOWHERE IN THE ABOVE IS THE MVT USED, ONLY THE DEFINITION OF NEW CALCULUS DERIVATIVE.
>
> (f(+n)-f(x-m))/(m+n) = f'(x) + Q(x,m,n)
>
> where
>
> (f(+n)-f(x-m))/(m+n) is ALWAYS the slope a parallel secant line to the tangent line, hence it is the slope of the tangent line at x.
>
> f'(x) denotes the tangent line slope and Q(x,m,n) the difference in slopes which is ALWAYS zero, unlike in my historic geometric theorem where the only case when Q(x,h) is zero is when f is a linear function.
>
> NO CIRCULARITY WHATSOEVER. The mvt is not used anywhere, but these facts help to prove it.
> 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.
"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)"

And how do you know that, John?

On Tuesday, 29 August 2023 at 02:45:52 UTC-4, markus...
> > > > > > 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.
> > > > > You prove the MVT using the MVT. Not a good start, pal. John, oh dear.
> > > > No. It's proved using the historic identity. Your failure to understand is well-known.
> > > How do you get the mu_s then?
> > We state the New Calculus derivative as follows:
> > 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)
> > NOWHERE IN THE ABOVE IS THE MVT USED, ONLY THE DEFINITION OF NEW CALCULUS DERIVATIVE.
> >
> > (f(+n)-f(x-m))/(m+n) = f'(x) + Q(x,m,n)
> >
> > where
> >
> > (f(+n)-f(x-m))/(m+n) is ALWAYS the slope a parallel secant line to the tangent line, hence it is the slope of the tangent line at x.
> >
> > f'(x) denotes the tangent line slope and Q(x,m,n) the difference in slopes which is ALWAYS zero, unlike in my historic geometric theorem where the only case when Q(x,h) is zero is when f is a linear function.
> >
> > NO CIRCULARITY WHATSOEVER. The mvt is not used anywhere, but these facts help to prove it.
> > 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.
> "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)"
> And how do you know that, John?

Because f'(x) = (f(x+n)-f(x-m))/(m+n) and this is used in the proof to arrive at f'(c). You would have known if you weren't such a moron.

DO NOT address me by my first name, you little shit! I am old enough to be your grandfather.

tisdag 29 augusti 2023 kl. 21:05:12 UTC+2 skrev Eram semper recta:
> On Tuesday, 29 August 2023 at 02:45:52 UTC-4, markus...
> > > > > > > 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.
> > > > > > You prove the MVT using the MVT. Not a good start, pal. John, oh dear.
> > > > > No. It's proved using the historic identity. Your failure to understand is well-known.
> > > > How do you get the mu_s then?
> > > We state the New Calculus derivative as follows:
> > > 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)
> > > NOWHERE IN THE ABOVE IS THE MVT USED, ONLY THE DEFINITION OF NEW CALCULUS DERIVATIVE.
> > >
> > > (f(+n)-f(x-m))/(m+n) = f'(x) + Q(x,m,n)
> > >
> > > where
> > >
> > > (f(+n)-f(x-m))/(m+n) is ALWAYS the slope a parallel secant line to the tangent line, hence it is the slope of the tangent line at x.
> > >
> > > f'(x) denotes the tangent line slope and Q(x,m,n) the difference in slopes which is ALWAYS zero, unlike in my historic geometric theorem where the only case when Q(x,h) is zero is when f is a linear function.
> > >
> > > NO CIRCULARITY WHATSOEVER. The mvt is not used anywhere, but these facts help to prove it.
> > > 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.
> > "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)"
> > And how do you know that, John?
> Because f'(x) = (f(x+n)-f(x-m))/(m+n) and this is used in the proof to arrive at f'(c). You would have known if you weren't such a moron.
>
> DO NOT address me by my first name, you little shit! I am old enough to be your grandfather.
You still haven't explained how you know there is a much mu_s, John.

On Tuesday, 29 August 2023 at 15:05:12 UTC-4, Eram semper recta wrote:
> On Tuesday, 29 August 2023 at 02:45:52 UTC-4, markus...
> > > > > > > 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.
> > > > > > You prove the MVT using the MVT. Not a good start, pal. John, oh dear.
> > > > > No. It's proved using the historic identity. Your failure to understand is well-known.
> > > > How do you get the mu_s then?
> > > We state the New Calculus derivative as follows:
> > > 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)
> > > NOWHERE IN THE ABOVE IS THE MVT USED, ONLY THE DEFINITION OF NEW CALCULUS DERIVATIVE.
> > >
> > > (f(+n)-f(x-m))/(m+n) = f'(x) + Q(x,m,n)
> > >
> > > where
> > >
> > > (f(+n)-f(x-m))/(m+n) is ALWAYS the slope a parallel secant line to the tangent line, hence it is the slope of the tangent line at x.
> > >
> > > f'(x) denotes the tangent line slope and Q(x,m,n) the difference in slopes which is ALWAYS zero, unlike in my historic geometric theorem where the only case when Q(x,h) is zero is when f is a linear function.
> > >
> > > NO CIRCULARITY WHATSOEVER. The mvt is not used anywhere, but these facts help to prove it.
> > > 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.
> > "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)"
> > And how do you know that, John?
> Because f'(x) = (f(x+n)-f(x-m))/(m+n) and this is used in the proof to arrive at f'(c). You would have known if you weren't such a moron.
>
> DO NOT address me by my first name, you little shit! I am old enough to be your grandfather.

Defiant idiot Klyver doesn't understand that the x in the above New Calculus derivative definition is the same as a mu.

You can't fix stupid!