On 4/26/2022 6:26 AM, WM wrote:
> Eram semper recta schrieb am Dienstag, 26. April 2022 um 11:13:43 UTC+2:
>> On Sunday, 24 April 2022 at 14:06:59 UTC+3, WM wrote:
>
>>>> One is forced to accept that such an "ultimate ratio" exists based on faith. In actual fact, none of the n results in [f(x+h_n)-f(x)]/h_n ever being f'(x).
>>> That is clear because the limit is not reached by any definable natural argument.
>> My point exactly. The definition f'(x)=lim_{h->0} [f(x+h)-f(x)]/h is spurious.
>
> It is like the definition of the limit 0 of the sequence (1/n).
>
> Regards, WM

[ crank to crank broadcast ]

On 4/26/2022 6:20 AM, WM wrote:
> zelos...@gmail.com schrieb am Dienstag, 26. April 2022 um 06:53:26 UTC+2:
>> måndag 25 april 2022 kl. 12:23:23 UTC+2 skrev WM:
>>> zelos...@gmail.com schrieb am Montag, 25. April 2022 um 07:30:59 UTC+2:
>>>> torsdag 21 april 2022 kl. 16:16:32 UTC+2 skrev WM:
>>>>> William schrieb am Mittwoch, 20. April 2022 um 19:02:38 UTC+2:
>>>>>> On Wednesday, April 20, 2022 at 9:38:20 AM UTC-3, WM wrote:
>>>>>>> William schrieb am Dienstag, 19. April 2022 um 18:04:38 UTC+2:
>>>>>>>
>>>>>>>> A "potentially infinite set" is nonsense.
>>>>>>> Alas it is the only set that we can apply in mathematics.
>>>>>> Piffle. A result you do not like is not a contradiction,
>>>>> By induction we prove that every definable natural number, i.e., every natural number which is subject to induction, belongs to a FISON but has ℵo successors which cannot be removed whatever natural number you consider. That means they cannot be used as individuals.
>>>> This is a pure non-sequitor
>>> Try logic:
>> I am better at logic than you.
>>> Peano naturals: ∀n ∈ ℕ_def: {1, 2, 3, ..., omega} \ {1, 2, 3, ..., n} = {n+1, n+2, ..., omega} .
>> That applies to all of N so N_def=N, try using logic.
>>> Dark naturals: {1, 2, 3, ..., omega} \ {1, 2, 3, ...} = {omega}
>> That does not define dark in any meaningful way.
>
> You are not good enough at logic.
>
> Regards, WM

where in the equations are your Dark Numbers ? which term ?

On Tuesday, April 26, 2022 at 8:19:32 AM UTC-3, WM wrote:
> William schrieb am Montag, 25. April 2022 um 17:19:18 UTC+2:

> > The natural numbers are a Peano set. Thus a "dark element" cannot be a natural number.
> Fact is in |N there are infinitely many natnumbers which cannot be treated individually:

Piffle,

> We can reach omega by collectively treating the natural numbers
> {1, 2, 3, ..., omega} \ {1, 2, 3, ...} = {omega} ,

A statement about the *set* {1, 2, 3,...}

> but not by individually treating them
> ∀n ∈ ℕ_[p]: {1, 2, 3, ..., omega} \ {1, 2, 3, ..., n} = {n+1, n+2, ..., omega} .

Statements about the *elements* of {1, 2, 3,...} . If all *elements* of a set have some property, the *set* may or may not have this property.

--
William Hughes

William schrieb am Dienstag, 26. April 2022 um 18:08:31 UTC+2:

> Statements about the *elements* of {1, 2, 3,...} . If all *elements* of a set have some property, the *set* may or may not have this property.

Here we are concerned with the elements.
All definable elements obey
n ∈ ℕ_[p]: {1, 2, 3, ..., omega} \ {1, 2, 3, ..., n} = {n+1, n+2, ..., omega} .
All elements allow
{1, 2, 3, ..., omega} \ {1, 2, 3, ...} = {omega}

That's a difference.

Regards, WM

An unsavory Idiot WM drivels:

> That's a difference.

LOL

You are simply imbecile.

Piss off already.

On 4/26/2022 11:31 AM, WM wrote:
> William schrieb am Dienstag, 26. April 2022 um 18:08:31 UTC+2:
>
>> Statements about the *elements* of {1, 2, 3,...} . If all *elements* of a set have some property, the *set* may or may not have this property.
>
> Here we are concerned with the elements.
> All definable elements obey

Wrong. Elements are members of a set, they do not "obey".

> n ∈ ℕ_[p]: {1, 2, 3, ..., omega} \ {1, 2, 3, ..., n} = {n+1, n+2, ..., omega} .
> All elements allow

Wrong. Elements are members of a set, they do not "allow", they are not an operator.

> {1, 2, 3, ..., omega} \ {1, 2, 3, ...} = {omega}
>
> That's a difference.

what does this relationship between these two DIFFERENT sets mean ?

{1, 2, 3, ...} = {omega}

>
> Regards, WM

go read a math book.

On Tuesday, April 26, 2022 at 1:31:33 PM UTC-3, WM wrote:

> All

Piffle.

--
William Hughes

tisdag 26 april 2022 kl. 18:31:33 UTC+2 skrev WM:
> William schrieb am Dienstag, 26. April 2022 um 18:08:31 UTC+2:
>
> > Statements about the *elements* of {1, 2, 3,...} . If all *elements* of a set have some property, the *set* may or may not have this property.
> Here we are concerned with the elements.
> All definable elements obey
> n ∈ ℕ_[p]: {1, 2, 3, ..., omega} \ {1, 2, 3, ..., n} = {n+1, n+2, ..., omega} .
> All elements allow
> {1, 2, 3, ..., omega} \ {1, 2, 3, ...} = {omega}
> That's a difference.
>
> Regards, WM
your "definable" is still just N

On Tuesday, 26 April 2022 at 14:27:01 UTC+3, WM wrote:
> Eram semper recta schrieb am Dienstag, 26. April 2022 um 11:13:43 UTC+2:
> > On Sunday, 24 April 2022 at 14:06:59 UTC+3, WM wrote:
>
> > > > One is forced to accept that such an "ultimate ratio" exists based on faith. In actual fact, none of the n results in [f(x+h_n)-f(x)]/h_n ever being f'(x).
> > > That is clear because the limit is not reached by any definable natural argument.
> > My point exactly. The definition f'(x)=lim_{h->0} [f(x+h)-f(x)]/h is spurious.
> It is like the definition of the limit 0 of the sequence (1/n).

Not at all. In the case of the sequence 1/n we know that 0 is the least value not obtainable. In the case of the derivative, the limit hangs between two "points" somewhere in the ether.

>
> Regards, WM

On Wednesday, 27 April 2022 at 09:18:50 UTC+3, Eram semper recta wrote:
> On Tuesday, 26 April 2022 at 14:27:01 UTC+3, WM wrote:
> > Eram semper recta schrieb am Dienstag, 26. April 2022 um 11:13:43 UTC+2:
> > > On Sunday, 24 April 2022 at 14:06:59 UTC+3, WM wrote:
> >
> > > > > One is forced to accept that such an "ultimate ratio" exists based on faith. In actual fact, none of the n results in [f(x+h_n)-f(x)]/h_n ever being f'(x).
> > > > That is clear because the limit is not reached by any definable natural argument.
> > > My point exactly. The definition f'(x)=lim_{h->0} [f(x+h)-f(x)]/h is spurious.
> > It is like the definition of the limit 0 of the sequence (1/n).
> Not at all. In the case of the sequence 1/n we know that 0 is the least value not obtainable. In the case of the derivative, the limit hangs between two "points" somewhere in the ether.

In any case, the definition is clearly wrong because as my historic geometric theorem states:

[f(x+h)-f(x)]/h = f'(x) + Q(x,h) {GI}

all the above are constant slopes, meaning they do not change. Therefore,

lim_{h->0} [f(x+h)-f(x)]/h = lim_{h->0} f'(x) + lim_{h->0} Q(x,h)

The only way the LHS can be equal to f'(x) is if Q(x,h)=0 meaning that h=0, the very thing that terrifies you!

f'(x)=f'(x) + lim_{h->0} Q(x,h)

But Q(x,h) is a *constant* for any given non-parallel secant line. There is exactly one (x,h) pair that satisfies the geometric identity {GI}.

Taking the limit of Q(x,h) is EQUIVALENT to setting Q(x,h)=0 implying that h=0.

>
> >
> > Regards, WM

On Wednesday, 27 April 2022 at 09:25:35 UTC+3, Eram semper recta wrote:
> On Wednesday, 27 April 2022 at 09:18:50 UTC+3, Eram semper recta wrote:
> > On Tuesday, 26 April 2022 at 14:27:01 UTC+3, WM wrote:
> > > Eram semper recta schrieb am Dienstag, 26. April 2022 um 11:13:43 UTC+2:
> > > > On Sunday, 24 April 2022 at 14:06:59 UTC+3, WM wrote:
> > >
> > > > > > One is forced to accept that such an "ultimate ratio" exists based on faith. In actual fact, none of the n results in [f(x+h_n)-f(x)]/h_n ever being f'(x).
> > > > > That is clear because the limit is not reached by any definable natural argument.
> > > > My point exactly. The definition f'(x)=lim_{h->0} [f(x+h)-f(x)]/h is spurious.
> > > It is like the definition of the limit 0 of the sequence (1/n).
> > Not at all. In the case of the sequence 1/n we know that 0 is the least value not obtainable. In the case of the derivative, the limit hangs between two "points" somewhere in the ether.
> In any case, the definition is clearly wrong because as my historic geometric theorem states:
>
> [f(x+h)-f(x)]/h = f'(x) + Q(x,h) {GI}
>
> all the above are constant slopes, meaning they do not change. Therefore,
>
> lim_{h->0} [f(x+h)-f(x)]/h = lim_{h->0} f'(x) + lim_{h->0} Q(x,h)
>
> The only way the LHS can be equal to f'(x) is if Q(x,h)=0 meaning that h=0, the very thing that terrifies you!
>
> f'(x)=f'(x) + lim_{h->0} Q(x,h)
>
> But Q(x,h) is a *constant* for any given non-parallel secant line. There is exactly one (x,h) pair that satisfies the geometric identity {GI}.
>
> Taking the limit of Q(x,h) is EQUIVALENT to setting Q(x,h)=0 implying that h=0.

You have no valid systematic way of finding the derivative in mainstream calculus. You are forced to guess it using flawed arithmetic of the first principles method which treats h as non-zero and then treats it as 0.

>
> >
> > >
> > > Regards, WM

onsdag 27 april 2022 kl. 08:18:50 UTC+2 skrev Eram semper recta:
> On Tuesday, 26 April 2022 at 14:27:01 UTC+3, WM wrote:
> > Eram semper recta schrieb am Dienstag, 26. April 2022 um 11:13:43 UTC+2:
> > > On Sunday, 24 April 2022 at 14:06:59 UTC+3, WM wrote:
> >
> > > > > One is forced to accept that such an "ultimate ratio" exists based on faith. In actual fact, none of the n results in [f(x+h_n)-f(x)]/h_n ever being f'(x).
> > > > That is clear because the limit is not reached by any definable natural argument.
> > > My point exactly. The definition f'(x)=lim_{h->0} [f(x+h)-f(x)]/h is spurious.
> > It is like the definition of the limit 0 of the sequence (1/n).
> Not at all. In the case of the sequence 1/n we know that 0 is the least value not obtainable. In the case of the derivative, the limit hangs between two "points" somewhere in the ether.
>
> >
> > Regards, WM
making it the limit you imbecile.

onsdag 27 april 2022 kl. 08:25:35 UTC+2 skrev Eram semper recta:
> On Wednesday, 27 April 2022 at 09:18:50 UTC+3, Eram semper recta wrote:
> > On Tuesday, 26 April 2022 at 14:27:01 UTC+3, WM wrote:
> > > Eram semper recta schrieb am Dienstag, 26. April 2022 um 11:13:43 UTC+2:
> > > > On Sunday, 24 April 2022 at 14:06:59 UTC+3, WM wrote:
> > >
> > > > > > One is forced to accept that such an "ultimate ratio" exists based on faith. In actual fact, none of the n results in [f(x+h_n)-f(x)]/h_n ever being f'(x).
> > > > > That is clear because the limit is not reached by any definable natural argument.
> > > > My point exactly. The definition f'(x)=lim_{h->0} [f(x+h)-f(x)]/h is spurious.
> > > It is like the definition of the limit 0 of the sequence (1/n).
> > Not at all. In the case of the sequence 1/n we know that 0 is the least value not obtainable. In the case of the derivative, the limit hangs between two "points" somewhere in the ether.
> In any case, the definition is clearly wrong because as my historic geometric theorem states:
>
> [f(x+h)-f(x)]/h = f'(x) + Q(x,h) {GI}
>
> all the above are constant slopes, meaning they do not change. Therefore,
>
> lim_{h->0} [f(x+h)-f(x)]/h = lim_{h->0} f'(x) + lim_{h->0} Q(x,h)
>
> The only way the LHS can be equal to f'(x) is if Q(x,h)=0 meaning that h=0, the very thing that terrifies you!
>
> f'(x)=f'(x) + lim_{h->0} Q(x,h)
>
> But Q(x,h) is a *constant* for any given non-parallel secant line. There is exactly one (x,h) pair that satisfies the geometric identity {GI}.
>
> Taking the limit of Q(x,h) is EQUIVALENT to setting Q(x,h)=0 implying that h=0.
>
> >
> > >
> > > Regards, WM

Yet h is never zero :) You still do not understand limits

onsdag 27 april 2022 kl. 08:35:30 UTC+2 skrev Eram semper recta:
> On Wednesday, 27 April 2022 at 09:25:35 UTC+3, Eram semper recta wrote:
> > On Wednesday, 27 April 2022 at 09:18:50 UTC+3, Eram semper recta wrote:
> > > On Tuesday, 26 April 2022 at 14:27:01 UTC+3, WM wrote:
> > > > Eram semper recta schrieb am Dienstag, 26. April 2022 um 11:13:43 UTC+2:
> > > > > On Sunday, 24 April 2022 at 14:06:59 UTC+3, WM wrote:
> > > >
> > > > > > > One is forced to accept that such an "ultimate ratio" exists based on faith. In actual fact, none of the n results in [f(x+h_n)-f(x)]/h_n ever being f'(x).
> > > > > > That is clear because the limit is not reached by any definable natural argument.
> > > > > My point exactly. The definition f'(x)=lim_{h->0} [f(x+h)-f(x)]/h is spurious.
> > > > It is like the definition of the limit 0 of the sequence (1/n).
> > > Not at all. In the case of the sequence 1/n we know that 0 is the least value not obtainable. In the case of the derivative, the limit hangs between two "points" somewhere in the ether.
> > In any case, the definition is clearly wrong because as my historic geometric theorem states:
> >
> > [f(x+h)-f(x)]/h = f'(x) + Q(x,h) {GI}
> >
> > all the above are constant slopes, meaning they do not change. Therefore,
> >
> > lim_{h->0} [f(x+h)-f(x)]/h = lim_{h->0} f'(x) + lim_{h->0} Q(x,h)
> >
> > The only way the LHS can be equal to f'(x) is if Q(x,h)=0 meaning that h=0, the very thing that terrifies you!
> >
> > f'(x)=f'(x) + lim_{h->0} Q(x,h)
> >
> > But Q(x,h) is a *constant* for any given non-parallel secant line. There is exactly one (x,h) pair that satisfies the geometric identity {GI}.
> >
> > Taking the limit of Q(x,h) is EQUIVALENT to setting Q(x,h)=0 implying that h=0.
> You have no valid systematic way of finding the derivative in mainstream calculus. You are forced to guess it using flawed arithmetic of the first principles method which treats h as non-zero and then treats it as 0.
>
> >
> > >
> > > >
> > > > Regards, WM

False, all the derivation rules can be hsown to be valid given definitions and systematic :)

There is no guessing going on!

On Wednesday, 27 April 2022 at 09:35:30 UTC+3, Eram semper recta wrote:
> On Wednesday, 27 April 2022 at 09:25:35 UTC+3, Eram semper recta wrote:
> > On Wednesday, 27 April 2022 at 09:18:50 UTC+3, Eram semper recta wrote:
> > > On Tuesday, 26 April 2022 at 14:27:01 UTC+3, WM wrote:
> > > > Eram semper recta schrieb am Dienstag, 26. April 2022 um 11:13:43 UTC+2:
> > > > > On Sunday, 24 April 2022 at 14:06:59 UTC+3, WM wrote:
> > > >
> > > > > > > One is forced to accept that such an "ultimate ratio" exists based on faith. In actual fact, none of the n results in [f(x+h_n)-f(x)]/h_n ever being f'(x).
> > > > > > That is clear because the limit is not reached by any definable natural argument.
> > > > > My point exactly. The definition f'(x)=lim_{h->0} [f(x+h)-f(x)]/h is spurious.
> > > > It is like the definition of the limit 0 of the sequence (1/n).
> > > Not at all. In the case of the sequence 1/n we know that 0 is the least value not obtainable. In the case of the derivative, the limit hangs between two "points" somewhere in the ether.
> > In any case, the definition is clearly wrong because as my historic geometric theorem states:
> >
> > [f(x+h)-f(x)]/h = f'(x) + Q(x,h) {GI}
> >
> > all the above are constant slopes, meaning they do not change. Therefore,
> >
> > lim_{h->0} [f(x+h)-f(x)]/h = lim_{h->0} f'(x) + lim_{h->0} Q(x,h)
> >
> > The only way the LHS can be equal to f'(x) is if Q(x,h)=0 meaning that h=0, the very thing that terrifies you!
> >
> > f'(x)=f'(x) + lim_{h->0} Q(x,h)
> >
> > But Q(x,h) is a *constant* for any given non-parallel secant line. There is exactly one (x,h) pair that satisfies the geometric identity {GI}.
> >
> > Taking the limit of Q(x,h) is EQUIVALENT to setting Q(x,h)=0 implying that h=0.
> You have no valid systematic way of finding the derivative in mainstream calculus. You are forced to guess it using flawed arithmetic of the first principles method which treats h as non-zero and then treats it as 0.

You can't claim that 0<|x-c|<𝛿 => |f(x)-L|<ε somehow rigorises your mainstream derivative f'(x)=lim_{h->0} [f(x+h)-f(x)]/h because f'(x)=lim_{h->0} [f(x+h)-f(x)]/h means:

For every ε>0, 𝛿>0, if 0<|x-c|<𝛿 then |f(x)-L|<ε.

For starters, it's pretty circular because L = f'(x) and you cannot use the derivative in its own definition.

If you retort that the epsilonics statement is a verification, well then you still haven't explained how you guessed L.

You have no valid systematic way of finding the derivative in the flawed mainstream formulation.

>
> >
> > >
> > > >
> > > > Regards, WM

fredag 29 april 2022 kl. 13:16:34 UTC+2 skrev Eram semper recta:
> On Wednesday, 27 April 2022 at 09:35:30 UTC+3, Eram semper recta wrote:
> > On Wednesday, 27 April 2022 at 09:25:35 UTC+3, Eram semper recta wrote:
> > > On Wednesday, 27 April 2022 at 09:18:50 UTC+3, Eram semper recta wrote:
> > > > On Tuesday, 26 April 2022 at 14:27:01 UTC+3, WM wrote:
> > > > > Eram semper recta schrieb am Dienstag, 26. April 2022 um 11:13:43 UTC+2:
> > > > > > On Sunday, 24 April 2022 at 14:06:59 UTC+3, WM wrote:
> > > > >
> > > > > > > > One is forced to accept that such an "ultimate ratio" exists based on faith. In actual fact, none of the n results in [f(x+h_n)-f(x)]/h_n ever being f'(x).
> > > > > > > That is clear because the limit is not reached by any definable natural argument.
> > > > > > My point exactly. The definition f'(x)=lim_{h->0} [f(x+h)-f(x)]/h is spurious.
> > > > > It is like the definition of the limit 0 of the sequence (1/n).
> > > > Not at all. In the case of the sequence 1/n we know that 0 is the least value not obtainable. In the case of the derivative, the limit hangs between two "points" somewhere in the ether.
> > > In any case, the definition is clearly wrong because as my historic geometric theorem states:
> > >
> > > [f(x+h)-f(x)]/h = f'(x) + Q(x,h) {GI}
> > >
> > > all the above are constant slopes, meaning they do not change. Therefore,
> > >
> > > lim_{h->0} [f(x+h)-f(x)]/h = lim_{h->0} f'(x) + lim_{h->0} Q(x,h)
> > >
> > > The only way the LHS can be equal to f'(x) is if Q(x,h)=0 meaning that h=0, the very thing that terrifies you!
> > >
> > > f'(x)=f'(x) + lim_{h->0} Q(x,h)
> > >
> > > But Q(x,h) is a *constant* for any given non-parallel secant line. There is exactly one (x,h) pair that satisfies the geometric identity {GI}.
> > >
> > > Taking the limit of Q(x,h) is EQUIVALENT to setting Q(x,h)=0 implying that h=0.
> > You have no valid systematic way of finding the derivative in mainstream calculus. You are forced to guess it using flawed arithmetic of the first principles method which treats h as non-zero and then treats it as 0.
> You can't claim that 0<|x-c|<𝛿 => |f(x)-L|<ε somehow rigorises your mainstream derivative f'(x)=lim_{h->0} [f(x+h)-f(x)]/h because f'(x)=lim_{h->0} [f(x+h)-f(x)]/h means:
>
> For every ε>0, 𝛿>0, if 0<|x-c|<𝛿 then |f(x)-L|<ε.
>
> For starters, it's pretty circular because L = f'(x) and you cannot use the derivative in its own definition.
>
> If you retort that the epsilonics statement is a verification, well then you still haven't explained how you guessed L.
>
> You have no valid systematic way of finding the derivative in the flawed mainstream formulation.
>
> >
> > >
> > > >
> > > > >
> > > > > Regards, WM

That is not the epsilon-delta definition you retard.

> For starters, it's pretty circular because L = f'(x) and you cannot use the derivative in its own definition.

You still don't get what circularity, you are conflating now the definition of limit and definition of derivative.

> You have no valid systematic way of finding the derivative in the flawed mainstream formulation.

As always, false. We have a systematic way to derive almost all functions and all traditional functions can.

On Friday, 29 April 2022 at 13:53:14 UTC+2, zelos...@gmail.com wrote:
> fredag 29 april 2022 kl. 13:16:34 UTC+2 skrev Eram semper recta:

> That is not the epsilon-delta definition you retard.

The moron and other systematic polluter of ponds is you. And I am not so
much defending him as immediately my interest for what he was saying.

> > For starters, it's pretty circular because L = f'(x) and you cannot use the derivative in its own definition.
> You still don't get what circularity, you are conflating now the definition of limit and definition of derivative.
> > You have no valid systematic way of finding the derivative in the flawed mainstream formulation.
> As always, false. We have a systematic way to derive almost all functions and all traditional functions can.

You utter idiot and stupid liar plain snip the next and substantial point which
also makes the above statements quite more reasonable and, for what
matters, understandable:

"If you retort that the epsilonics statement is a verification,
well then you still haven't explained how you guessed L."

Do you have any comment on that you fucking idiot?

No, I don't think so.

*Plonk*

Julio

zelos...@gmail.com schrieb am Mittwoch, 27. April 2022 um 07:50:07 UTC+2:
> tisdag 26 april 2022 kl. 18:31:33 UTC+2 skrev WM:
> > William schrieb am Dienstag, 26. April 2022 um 18:08:31 UTC+2:
> >
> > > Statements about the *elements* of {1, 2, 3,...} . If all *elements* of a set have some property, the *set* may or may not have this property.
> > Here we are concerned with the elements.
> > All definable elements obey
> > n ∈ ℕ_[p]: {1, 2, 3, ..., omega} \ {1, 2, 3, ..., n} = {n+1, n+2, ..., omega} .
> > All elements allow
> > {1, 2, 3, ..., omega} \ {1, 2, 3, ...} = {omega}
> > That's a difference.
> >
> your "definable" is still just N

Don't see the difference? Sorry for you.

Regards, WM

On Friday, 29 April 2022 at 14:16:34 UTC+3, Eram semper recta wrote:
> On Wednesday, 27 April 2022 at 09:35:30 UTC+3, Eram semper recta wrote:
> > On Wednesday, 27 April 2022 at 09:25:35 UTC+3, Eram semper recta wrote:
> > > On Wednesday, 27 April 2022 at 09:18:50 UTC+3, Eram semper recta wrote:
> > > > On Tuesday, 26 April 2022 at 14:27:01 UTC+3, WM wrote:
> > > > > Eram semper recta schrieb am Dienstag, 26. April 2022 um 11:13:43 UTC+2:
> > > > > > On Sunday, 24 April 2022 at 14:06:59 UTC+3, WM wrote:
> > > > >
> > > > > > > > One is forced to accept that such an "ultimate ratio" exists based on faith. In actual fact, none of the n results in [f(x+h_n)-f(x)]/h_n ever being f'(x).
> > > > > > > That is clear because the limit is not reached by any definable natural argument.
> > > > > > My point exactly. The definition f'(x)=lim_{h->0} [f(x+h)-f(x)]/h is spurious.
> > > > > It is like the definition of the limit 0 of the sequence (1/n).
> > > > Not at all. In the case of the sequence 1/n we know that 0 is the least value not obtainable. In the case of the derivative, the limit hangs between two "points" somewhere in the ether.
> > > In any case, the definition is clearly wrong because as my historic geometric theorem states:
> > >
> > > [f(x+h)-f(x)]/h = f'(x) + Q(x,h) {GI}
> > >
> > > all the above are constant slopes, meaning they do not change. Therefore,
> > >
> > > lim_{h->0} [f(x+h)-f(x)]/h = lim_{h->0} f'(x) + lim_{h->0} Q(x,h)
> > >
> > > The only way the LHS can be equal to f'(x) is if Q(x,h)=0 meaning that h=0, the very thing that terrifies you!
> > >
> > > f'(x)=f'(x) + lim_{h->0} Q(x,h)
> > >
> > > But Q(x,h) is a *constant* for any given non-parallel secant line. There is exactly one (x,h) pair that satisfies the geometric identity {GI}.
> > >
> > > Taking the limit of Q(x,h) is EQUIVALENT to setting Q(x,h)=0 implying that h=0.
> > You have no valid systematic way of finding the derivative in mainstream calculus. You are forced to guess it using flawed arithmetic of the first principles method which treats h as non-zero and then treats it as 0.
> You can't claim that 0<|x-c|<𝛿 => |f(x)-L|<ε somehow rigorises your mainstream derivative f'(x)=lim_{h->0} [f(x+h)-f(x)]/h because f'(x)=lim_{h->0} [f(x+h)-f(x)]/h means:
>
> For every ε>0, 𝛿>0, if 0<|x-c|<𝛿 then |f(x)-L|<ε.
>
> For starters, it's pretty circular because L = f'(x) and you cannot use the derivative in its own definition.
>
> If you retort that the epsilonics statement is a verification, well then you still haven't explained how you guessed L.
>
> You have no valid systematic way of finding the derivative in the flawed mainstream formulation.

No comment Wolfgang? :-) Ja, Ich weiss, Ich weiss - Du hast keine Antwort!

>
> >
> > >
> > > >
> > > > >
> > > > > Regards, WM

Eram semper recta schrieb am Samstag, 30. April 2022 um 06:43:52 UTC+2:
> On Friday, 29 April 2022 at 14:16:34 UTC+3, Eram semper recta wrote:

> > You have no valid systematic way of finding the derivative in the flawed mainstream formulation.
> No comment Wolfgang? :-) Ja, Ich weiss, Ich weiss - Du hast keine Antwort!

I show a lot of systematic ways in my book "Mathematik für die ersten Semester", 4th ed., De Gruyter, Berlin (2015), chapters 23ff.

Regards, WM

On Saturday, 30 April 2022 at 16:30:23 UTC+3, WM wrote:
> Eram semper recta schrieb am Samstag, 30. April 2022 um 06:43:52 UTC+2:
> > On Friday, 29 April 2022 at 14:16:34 UTC+3, Eram semper recta wrote:
>
> > > You have no valid systematic way of finding the derivative in the flawed mainstream formulation.
> > No comment Wolfgang? :-) Ja, Ich weiss, Ich weiss - Du hast keine Antwort!
> I show a lot of systematic ways in my book "Mathematik für die ersten Semester", 4th ed., De Gruyter, Berlin (2015), chapters 23ff.

Maybe systematic, but none of them valid. :-)

The first principles method is systematic but it's not valid.

As for your book, it is not free and I don't have 27 euros to throw around.

Choose one of your ways and explain here and then I'll show you it's not valid.

>
>
> Regards, WM

fredag 29 april 2022 kl. 15:05:38 UTC+2 skrev ju...@diegidio.name:
> On Friday, 29 April 2022 at 13:53:14 UTC+2, zelos...@gmail.com wrote:
> > fredag 29 april 2022 kl. 13:16:34 UTC+2 skrev Eram semper recta:
>
> > That is not the epsilon-delta definition you retard.
> The moron and other systematic polluter of ponds is you. And I am not so
> much defending him as immediately my interest for what he was saying.
> > > For starters, it's pretty circular because L = f'(x) and you cannot use the derivative in its own definition.
> > You still don't get what circularity, you are conflating now the definition of limit and definition of derivative.
> > > You have no valid systematic way of finding the derivative in the flawed mainstream formulation.
> > As always, false. We have a systematic way to derive almost all functions and all traditional functions can.
> You utter idiot and stupid liar plain snip the next and substantial point which
> also makes the above statements quite more reasonable and, for what
> matters, understandable:
> "If you retort that the epsilonics statement is a verification,
> well then you still haven't explained how you guessed L."
> Do you have any comment on that you fucking idiot?
>
> No, I don't think so.
>
> *Plonk*
>
> Julio

Grow the fuck up you retard and stop defending fucking cranks.

fredag 29 april 2022 kl. 22:14:38 UTC+2 skrev WM:
> zelos...@gmail.com schrieb am Mittwoch, 27. April 2022 um 07:50:07 UTC+2:
> > tisdag 26 april 2022 kl. 18:31:33 UTC+2 skrev WM:
> > > William schrieb am Dienstag, 26. April 2022 um 18:08:31 UTC+2:
> > >
> > > > Statements about the *elements* of {1, 2, 3,...} . If all *elements* of a set have some property, the *set* may or may not have this property.
> > > Here we are concerned with the elements.
> > > All definable elements obey
> > > n ∈ ℕ_[p]: {1, 2, 3, ..., omega} \ {1, 2, 3, ..., n} = {n+1, n+2, ..., omega} .
> > > All elements allow
> > > {1, 2, 3, ..., omega} \ {1, 2, 3, ...} = {omega}
> > > That's a difference.
> > >
> > your "definable" is still just N
> Don't see the difference? Sorry for you.
>
> Regards, WM
You have provided no difference between them. They contain the exact same elements

söndag 1 maj 2022 kl. 07:56:54 UTC+2 skrev Eram semper recta:
> On Saturday, 30 April 2022 at 16:30:23 UTC+3, WM wrote:
> > Eram semper recta schrieb am Samstag, 30. April 2022 um 06:43:52 UTC+2:
> > > On Friday, 29 April 2022 at 14:16:34 UTC+3, Eram semper recta wrote:
> >
> > > > You have no valid systematic way of finding the derivative in the flawed mainstream formulation.
> > > No comment Wolfgang? :-) Ja, Ich weiss, Ich weiss - Du hast keine Antwort!
> > I show a lot of systematic ways in my book "Mathematik für die ersten Semester", 4th ed., De Gruyter, Berlin (2015), chapters 23ff.
> Maybe systematic, but none of them valid. :-)
>
> The first principles method is systematic but it's not valid.
>
> As for your book, it is not free and I don't have 27 euros to throw around.
>
> Choose one of your ways and explain here and then I'll show you it's not valid.
>
> >
> >
> > Regards, WM
you're homeless, aren't you?

On Tuesday, April 26, 2022 at 1:19:32 PM UTC+2, WM wrote:
>
> Not all natural numbers belong to a Peano set.

Since the set of (all) natural numbers *is* a Peano set, your claim is nonsensical.

You are an idiot.