Fritz Feldhase schrieb am Sonntag, 1. Mai 2022 um 16:02:35 UTC+2:
> 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.
>
Here is the disproof
∀n ∈ ℕ_P: |ℕ \ {1, 2, 3, ..., n}| = ℵo
|ℕ \ {1, 2, 3, ...}| = 0 .
Unfortunately you cannot read it. What a pity.

Regards, WM

On Sunday, 1 May 2022 at 17:18:13 UTC-3, WM wrote:
[...]
> ∀n ∈ ℕ_P: |ℕ \ {1, 2, 3, ..., n}| = ℵo
> |ℕ \ {1, 2, 3, ...}| = 0 .

Quantifier shift. How original!

On 5/1/2022 3:18 PM, WM wrote:
> Fritz Feldhase schrieb am Sonntag, 1. Mai 2022 um 16:02:35 UTC+2:
>> 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.
>>
> Here is the disproof
> ∀n ∈ ℕ_P: |ℕ \ {1, 2, 3, ..., n}| = ℵo

that is an Endsegment, which are infinite.

> |ℕ \ {1, 2, 3, ...}| = 0 .

that is the set of natural numbers less the set of natural numbers.

So how do these two trivial statements prove ℕ is not a Peano set, again ?

>
> Regards, WM
>

söndag 1 maj 2022 kl. 22:18:13 UTC+2 skrev WM:
> Fritz Feldhase schrieb am Sonntag, 1. Mai 2022 um 16:02:35 UTC+2:
> > 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.
> >
> Here is the disproof
> ∀n ∈ ℕ_P: |ℕ \ {1, 2, 3, ..., n}| = ℵo
> |ℕ \ {1, 2, 3, ...}| = 0 .
> Unfortunately you cannot read it. What a pity.
>
> Regards, WM
THat proves nothing of what you want. That still means natural numbers are in a Peano set

On Sunday, 1 May 2022 at 23:13:42 UTC+3, WM wrote:
> Eram semper recta schrieb am Sonntag, 1. Mai 2022 um 07:56:54 UTC+2:
> > 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.
> >
> The derivative of x^n = nx^(n-1), proved by limit method.

Limit theory is not a proof of any kind - it merely helps you validate your *guess* that nx^(n-1) is the correct derivative.

> Similary
> sinx' = cos x

The sine series was discovered long before the bogus mainstream calculus.

> e^x' = e^x

e^x = f(x,0) where f(x,n)= (1+xn)^(1/n)

So once again, nothing but the binomial theorem here. No limit theory whatsoever.

Any other methods of your claim that there are "so many ways"?

>
> Regards, WM

On Monday, 2 May 2022 at 11:16:31 UTC+3, Eram semper recta wrote:
> On Sunday, 1 May 2022 at 23:13:42 UTC+3, WM wrote:
> > Eram semper recta schrieb am Sonntag, 1. Mai 2022 um 07:56:54 UTC+2:
> > > 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.
> > >
> > The derivative of x^n = nx^(n-1), proved by limit method.
> Limit theory is not a proof of any kind - it merely helps you validate your *guess* that nx^(n-1) is the correct derivative.
> > Similary
> > sinx' = cos x
> The sine series was discovered long before the bogus mainstream calculus.
>
> > e^x' = e^x
>
> e^x = f(x,0) where f(x,n)= (1+xn)^(1/n)

Discarding every term with n in it (in the expanded binomial) has nothing to do with limits. Similarly,
sin(x)/x = 1 - x^2/3! + x^4/5! - ... evaluates to 1 when x=0.

Nothing about limits here also. Limititis is a disease of the dysfunctional mainstream brain.

>
> So once again, nothing but the binomial theorem here. No limit theory whatsoever.
>
> Any other methods of your claim that there are "so many ways"?
>
> >
> > Regards, WM

måndag 2 maj 2022 kl. 10:22:07 UTC+2 skrev Eram semper recta:
> On Monday, 2 May 2022 at 11:16:31 UTC+3, Eram semper recta wrote:
> > On Sunday, 1 May 2022 at 23:13:42 UTC+3, WM wrote:
> > > Eram semper recta schrieb am Sonntag, 1. Mai 2022 um 07:56:54 UTC+2:
> > > > 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.
> > > >
> > > The derivative of x^n = nx^(n-1), proved by limit method.
> > Limit theory is not a proof of any kind - it merely helps you validate your *guess* that nx^(n-1) is the correct derivative.
> > > Similary
> > > sinx' = cos x
> > The sine series was discovered long before the bogus mainstream calculus.
> >
> > > e^x' = e^x
> >
> > e^x = f(x,0) where f(x,n)= (1+xn)^(1/n)
> Discarding every term with n in it (in the expanded binomial) has nothing to do with limits. Similarly,
> sin(x)/x = 1 - x^2/3! + x^4/5! - ... evaluates to 1 when x=0.
>
> Nothing about limits here also. Limititis is a disease of the dysfunctional mainstream brain.
> >
> > So once again, nothing but the binomial theorem here. No limit theory whatsoever.
> >
> > Any other methods of your claim that there are "so many ways"?
> >
> > >
> > > Regards, WM

To know that sin(x)/x = 1 - x^2/3! + x^4/5! - ...
You need to know the derivative first, which requires limits
So thsi is very circular.

horand....@gmail.com schrieb am Sonntag, 1. Mai 2022 um 23:01:07 UTC+2:
> On Sunday, 1 May 2022 at 17:18:13 UTC-3, WM wrote:
> [...]
> > ∀n ∈ ℕ_P: |ℕ \ {1, 2, 3, ..., n}| = ℵo
> > |ℕ \ {1, 2, 3, ...}| = 0 .
> Quantifier shift. How original!

My statements are simply mathematical facts - if there is the complete set ℕ. Do you deny it?

Regards, W

zelos...@gmail.com schrieb am Montag, 2. Mai 2022 um 06:42:03 UTC+2:
> söndag 1 maj 2022 kl. 22:18:13 UTC+2 skrev WM:
> > Fritz Feldhase schrieb am Sonntag, 1. Mai 2022 um 16:02:35 UTC+2:
> > > 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.
> > >
> > Here is the disproof
> > ∀n ∈ ℕ_P: |ℕ \ {1, 2, 3, ..., n}| = ℵo
> > |ℕ \ {1, 2, 3, ...}| = 0 .
> > Unfortunately you cannot read it. What a pity.
> >
> THat proves nothing of what you want. That still means natural numbers are in a Peano set

Unfortunately you cannot read it. What a pity.

Regards, WM

On Monday, 2 May 2022 at 08:30:58 UTC-3, WM wrote:
> horand....@gmail.com schrieb am Sonntag, 1. Mai 2022 um 23:01:07 UTC+2:
> > On Sunday, 1 May 2022 at 17:18:13 UTC-3, WM wrote:
> > [...]
> > > ∀n ∈ ℕ_P: |ℕ \ {1, 2, 3, ..., n}| = ℵo
> > > |ℕ \ {1, 2, 3, ...}| = 0 .
> > Quantifier shift. How original!
> My statements are simply mathematical facts - if there is the complete set ℕ. Do you deny it?

Nonetheless, whether you understand or accept it or not, in those two correct statements the quantifiers are used differently. Are you trying to deny *that*?

horand....@gmail.com schrieb am Montag, 2. Mai 2022 um 13:46:28 UTC+2:
> On Monday, 2 May 2022 at 08:30:58 UTC-3, WM wrote:
> > horand....@gmail.com schrieb am Sonntag, 1. Mai 2022 um 23:01:07 UTC+2:
> > > On Sunday, 1 May 2022 at 17:18:13 UTC-3, WM wrote:
> > > [...]
> > > > ∀n ∈ ℕ_P: |ℕ \ {1, 2, 3, ..., n}| = ℵo
> > > > |ℕ \ {1, 2, 3, ...}| = 0 .
> > > Quantifier shift. How original!
> > My statements are simply mathematical facts - if there is the complete set ℕ. Do you deny it?
> Nonetheless, whether you understand or accept it or not, in those two correct statements the quantifiers are used differently.

Of course. The first statement says that every FISON will not exhaust the set ℕ. The second statement says that subtracting all natural numbers will exhaust the set ℕ. This proves that not all natural numbers have FISONs.

Regards, WM

On 5/2/2022 6:55 AM, WM wrote:
> horand....@gmail.com schrieb am Montag, 2. Mai 2022 um 13:46:28 UTC+2:
>> On Monday, 2 May 2022 at 08:30:58 UTC-3, WM wrote:
>>> horand....@gmail.com schrieb am Sonntag, 1. Mai 2022 um 23:01:07 UTC+2:
>>>> On Sunday, 1 May 2022 at 17:18:13 UTC-3, WM wrote:
>>>> [...]
>>>>> ∀n ∈ ℕ_P: |ℕ \ {1, 2, 3, ..., n}| = ℵo
>>>>> |ℕ \ {1, 2, 3, ...}| = 0 .
>>>> Quantifier shift. How original!
>>> My statements are simply mathematical facts - if there is the complete set ℕ. Do you deny it?
>> Nonetheless, whether you understand or accept it or not, in those two correct statements the quantifiers are used differently.
>
> Of course. The first statement says that every FISON will not exhaust the set ℕ. The second statement says that subtracting all natural numbers will exhaust the set ℕ. This proves that not all natural numbers have FISONs.
>
> Regards, WM

wrong again.

What it does prove is that you do not know math, or you do not know how to form correct conclusions.

Please post just 1 post where you get everything right, and correct. Can you do try to do that ?

On 5/2/2022 6:34 AM, WM wrote:
> zelos...@gmail.com schrieb am Montag, 2. Mai 2022 um 06:42:03 UTC+2:
>> söndag 1 maj 2022 kl. 22:18:13 UTC+2 skrev WM:
>>> Fritz Feldhase schrieb am Sonntag, 1. Mai 2022 um 16:02:35 UTC+2:
>>>> 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.
>>>>
>>> Here is the disproof
>>> ∀n ∈ ℕ_P: |ℕ \ {1, 2, 3, ..., n}| = ℵo
>>> |ℕ \ {1, 2, 3, ...}| = 0 .
>>> Unfortunately you cannot read it. What a pity.
>>>
>> THat proves nothing of what you want. That still means natural numbers are in a Peano set
>
> Unfortunately you cannot read it. What a pity.
>
> Regards, WM
>

your two equations do not lead to your conclusion at all.

On Monday, 2 May 2022 at 08:55:27 UTC-3, WM wrote:
> horand....@gmail.com schrieb am Montag, 2. Mai 2022 um 13:46:28 UTC+2:
> > On Monday, 2 May 2022 at 08:30:58 UTC-3, WM wrote:
> > > horand....@gmail.com schrieb am Sonntag, 1. Mai 2022 um 23:01:07 UTC+2:
> > > > On Sunday, 1 May 2022 at 17:18:13 UTC-3, WM wrote:
> > > > [...]
> > > > > ∀n ∈ ℕ_P: |ℕ \ {1, 2, 3, ..., n}| = ℵo
> > > > > |ℕ \ {1, 2, 3, ...}| = 0 .
> > > > Quantifier shift. How original!
> > > My statements are simply mathematical facts - if there is the complete set ℕ. Do you deny it?
> > Nonetheless, whether you understand or accept it or not, in those two correct statements the quantifiers are used differently.
> Of course. The first statement says that every FISON will not exhaust the set ℕ. The second statement says that subtracting all natural numbers will exhaust the set ℕ. This proves that not all natural numbers have FISONs.

In Summary: Quantifier shift! Why the hell do you think that is news?

> On Monday, 2 May 2022 at 08:55:27 UTC-3, WM wrote:
> > horand....@gmail.com schrieb am Montag, 2. Mai 2022 um 13:46:28 UTC+2:
> > > On Monday, 2 May 2022 at 08:30:58 UTC-3, WM wrote:
> > > > horand....@gmail.com schrieb am Sonntag, 1. Mai 2022 um 23:01:07 UTC+2:
> > > > > On Sunday, 1 May 2022 at 17:18:13 UTC-3, WM wrote:
> > > > > [...]
> > > > > > ∀n ∈ ℕ_P: |ℕ \ {1, 2, 3, ..., n}| = ℵo
> > > > > > |ℕ \ {1, 2, 3, ...}| = 0 .
> > > > > Quantifier shift. How original!
> > > > My statements are simply mathematical facts - if there is the complete set ℕ. Do you deny it?
> > > Nonetheless, whether you understand or accept it or not, in those two correct statements the quantifiers are used differently.
> > Of course. The first statement says that every FISON will not exhaust the set ℕ. The second statement says that subtracting all natural numbers will exhaust the set ℕ. This proves that not all natural numbers have FISONs.
> In Summary: Quantifier shift!

These are two different statements. They are not new. They prove the existence of dark numbers.

Regards, WM

On Monday, 2 May 2022 at 17:04:06 UTC-3, WM wrote:
> > On Monday, 2 May 2022 at 08:55:27 UTC-3, WM wrote:
> > > horand....@gmail.com schrieb am Montag, 2. Mai 2022 um 13:46:28 UTC+2:
> > > > On Monday, 2 May 2022 at 08:30:58 UTC-3, WM wrote:
> > > > > horand....@gmail.com schrieb am Sonntag, 1. Mai 2022 um 23:01:07 UTC+2:
> > > > > > On Sunday, 1 May 2022 at 17:18:13 UTC-3, WM wrote:
> > > > > > [...]
> > > > > > > ∀n ∈ ℕ_P: |ℕ \ {1, 2, 3, ..., n}| = ℵo
> > > > > > > |ℕ \ {1, 2, 3, ...}| = 0 .
> > > > > > Quantifier shift. How original!
> > > > > My statements are simply mathematical facts - if there is the complete set ℕ. Do you deny it?
> > > > Nonetheless, whether you understand or accept it or not, in those two correct statements the quantifiers are used differently.
> > > Of course. The first statement says that every FISON will not exhaust the set ℕ. The second statement says that subtracting all natural numbers will exhaust the set ℕ. This proves that not all natural numbers have FISONs.
> > In Summary: Quantifier shift!
> These are two different statements. They are not new. They prove the existence of dark numbers.

Once a moron, always a moron, Muckenheim. These two statements prove nothing of the sort. And now, PFO.

On 5/2/2022 3:04 PM, WM wrote:
>> On Monday, 2 May 2022 at 08:55:27 UTC-3, WM wrote:
>>> horand....@gmail.com schrieb am Montag, 2. Mai 2022 um 13:46:28 UTC+2:
>>>> On Monday, 2 May 2022 at 08:30:58 UTC-3, WM wrote:
>>>>> horand....@gmail.com schrieb am Sonntag, 1. Mai 2022 um 23:01:07 UTC+2:
>>>>>> On Sunday, 1 May 2022 at 17:18:13 UTC-3, WM wrote:
>>>>>> [...]
>>>>>>> ∀n ∈ ℕ_P: |ℕ \ {1, 2, 3, ..., n}| = ℵo
>>>>>>> |ℕ \ {1, 2, 3, ...}| = 0 .
>>>>>> Quantifier shift. How original!
>>>>> My statements are simply mathematical facts - if there is the complete set ℕ. Do you deny it?
>>>> Nonetheless, whether you understand or accept it or not, in those two correct statements the quantifiers are used differently.
>>> Of course. The first statement says that every FISON will not exhaust the set ℕ. The second statement says that subtracting all natural numbers will exhaust the set ℕ. This proves that not all natural numbers have FISONs.
>> In Summary: Quantifier shift!
>
> These are two different statements. They are not new. They prove the existence of dark numbers.

your proof is spoof;

∀n ∈ ℕ_P: |ℕ \ {1, 2, 3, ..., n}| = ℵo says endsegments are infinite.

|ℕ \ {1, 2, 3, ...}| = 0 says if you delete all the elements you have nothing left.

Dark numbers are not involved in any way.

>
> Regards, WM

måndag 2 maj 2022 kl. 13:30:58 UTC+2 skrev WM:
> horand....@gmail.com schrieb am Sonntag, 1. Mai 2022 um 23:01:07 UTC+2:
> > On Sunday, 1 May 2022 at 17:18:13 UTC-3, WM wrote:
> > [...]
> > > ∀n ∈ ℕ_P: |ℕ \ {1, 2, 3, ..., n}| = ℵo
> > > |ℕ \ {1, 2, 3, ...}| = 0 .
> > Quantifier shift. How original!
> My statements are simply mathematical facts - if there is the complete set ℕ. Do you deny it?
>
> Regards, W
Nope, you make huge leaps in logic, complete non-sequitors

måndag 2 maj 2022 kl. 13:34:17 UTC+2 skrev WM:
> zelos...@gmail.com schrieb am Montag, 2. Mai 2022 um 06:42:03 UTC+2:
> > söndag 1 maj 2022 kl. 22:18:13 UTC+2 skrev WM:
> > > Fritz Feldhase schrieb am Sonntag, 1. Mai 2022 um 16:02:35 UTC+2:
> > > > 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.
> > > >
> > > Here is the disproof
> > > ∀n ∈ ℕ_P: |ℕ \ {1, 2, 3, ..., n}| = ℵo
> > > |ℕ \ {1, 2, 3, ...}| = 0 .
> > > Unfortunately you cannot read it. What a pity.
> > >
> > THat proves nothing of what you want. That still means natural numbers are in a Peano set
>
> Unfortunately you cannot read it. What a pity.
>
> Regards, WM
I can read mathematics way better than you

måndag 2 maj 2022 kl. 13:55:27 UTC+2 skrev WM:
> horand....@gmail.com schrieb am Montag, 2. Mai 2022 um 13:46:28 UTC+2:
> > On Monday, 2 May 2022 at 08:30:58 UTC-3, WM wrote:
> > > horand....@gmail.com schrieb am Sonntag, 1. Mai 2022 um 23:01:07 UTC+2:
> > > > On Sunday, 1 May 2022 at 17:18:13 UTC-3, WM wrote:
> > > > [...]
> > > > > ∀n ∈ ℕ_P: |ℕ \ {1, 2, 3, ..., n}| = ℵo
> > > > > |ℕ \ {1, 2, 3, ...}| = 0 .
> > > > Quantifier shift. How original!
> > > My statements are simply mathematical facts - if there is the complete set ℕ. Do you deny it?
> > Nonetheless, whether you understand or accept it or not, in those two correct statements the quantifiers are used differently.
> Of course. The first statement says that every FISON will not exhaust the set ℕ. The second statement says that subtracting all natural numbers will exhaust the set ℕ. This proves that not all natural numbers have FISONs.
>
> Regards, WM
This is a non-sequitor.
1: Exhaust means nothing in mathematics.
2: N\F_n ~ = {} and N\N={} do not entail what you want at all.

måndag 2 maj 2022 kl. 22:04:06 UTC+2 skrev WM:
> > On Monday, 2 May 2022 at 08:55:27 UTC-3, WM wrote:
> > > horand....@gmail.com schrieb am Montag, 2. Mai 2022 um 13:46:28 UTC+2:
> > > > On Monday, 2 May 2022 at 08:30:58 UTC-3, WM wrote:
> > > > > horand....@gmail.com schrieb am Sonntag, 1. Mai 2022 um 23:01:07 UTC+2:
> > > > > > On Sunday, 1 May 2022 at 17:18:13 UTC-3, WM wrote:
> > > > > > [...]
> > > > > > > ∀n ∈ ℕ_P: |ℕ \ {1, 2, 3, ..., n}| = ℵo
> > > > > > > |ℕ \ {1, 2, 3, ...}| = 0 .
> > > > > > Quantifier shift. How original!
> > > > > My statements are simply mathematical facts - if there is the complete set ℕ. Do you deny it?
> > > > Nonetheless, whether you understand or accept it or not, in those two correct statements the quantifiers are used differently.
> > > Of course. The first statement says that every FISON will not exhaust the set ℕ. The second statement says that subtracting all natural numbers will exhaust the set ℕ. This proves that not all natural numbers have FISONs.
> > In Summary: Quantifier shift!
> These are two different statements. They are not new. They prove the existence of dark numbers.
>
> Regards, WM
They prove nothing of the sort

On Monday, 2 May 2022 at 14:39:47 UTC+3, WM wrote:
> Eram semper recta schrieb am Montag, 2. Mai 2022 um 10:16:31 UTC+2:
> > On Sunday, 1 May 2022 at 23:13:42 UTC+3, WM wrote:
>
> > > The derivative of x^n = nx^(n-1), proved by limit method.
> > Limit theory is not a proof of any kind - it merely helps you validate your *guess* that nx^(n-1) is the correct derivative.
> (x^2 + 2xh + h^2 - x^2)/h = 2x + h

(x^2 + 2xh + h^2 - x^2)/h = 2x + h is the first principles method which is a *guess*.

> For h -->0 the result is precisely 2x.There is no guess.

Unfortunately no, that is not the case because

Lim{h->0} (x^2 + 2xh + h^2 - x^2)/h

means: 0<|x-c|<delta => | [f(x+h)-f(x)]/h - L|<epsilon

where L = 2x. Pretty circular indeed. :-)

The guess L used in the epsilon-delta verifinition is 2x which you obtain from the "definition" itself.

Your brainwashing is very hard to cure.

>
> Regards, WM

On Tuesday, 3 May 2022 at 17:05:12 UTC+3, WM wrote:
> Eram semper recta schrieb am Dienstag, 3. Mai 2022 um 14:02:18 UTC+2:
> > On Monday, 2 May 2022 at 14:39:47 UTC+3, WM wrote:
> > > Eram semper recta schrieb am Montag, 2. Mai 2022 um 10:16:31 UTC+2:
> > > > On Sunday, 1 May 2022 at 23:13:42 UTC+3, WM wrote:
> > >
> > > > > The derivative of x^n = nx^(n-1), proved by limit method.
> > > > Limit theory is not a proof of any kind - it merely helps you validate your *guess* that nx^(n-1) is the correct derivative.
> > > (x^2 + 2xh + h^2 - x^2)/h = 2x + h
> > (x^2 + 2xh + h^2 - x^2)/h = 2x + h is the first principles method which is a *guess*.
> > > For h -->0 the result is precisely 2x.There is no guess.
> > Unfortunately no,
> For me it means 2x + h, and in the limit 2x.

In other words you are just as clueless as the rest. 2x+h is a guess obtained from the first principles method which is flawed because h cannot be non-zero for a time and then zero and this is precisely what taking the limit means, that is, taking the limit is EQUIVALENT to setting h=0, whether you like it or not.

Setting h=0 is absolutely not allowed, nor is using the derivative in its own "definition". Now trying to say you have a different method without setting h=0, albeit equivalent, doesn't mean it's valid.

As usual you have addressed NONE of the problems put to you. You are a hypocrite and a coward with very little difference between you and the rest of the morons you argue with on this trash heap.

>
> Regards, WM

On Tuesday, 3 May 2022 at 17:05:12 UTC+3, WM wrote:
> Eram semper recta schrieb am Dienstag, 3. Mai 2022 um 14:02:18 UTC+2:
> > On Monday, 2 May 2022 at 14:39:47 UTC+3, WM wrote:
> > > Eram semper recta schrieb am Montag, 2. Mai 2022 um 10:16:31 UTC+2:
> > > > On Sunday, 1 May 2022 at 23:13:42 UTC+3, WM wrote:
> > >
> > > > > The derivative of x^n = nx^(n-1), proved by limit method.
> > > > Limit theory is not a proof of any kind - it merely helps you validate your *guess* that nx^(n-1) is the correct derivative.
> > > (x^2 + 2xh + h^2 - x^2)/h = 2x + h
> > (x^2 + 2xh + h^2 - x^2)/h = 2x + h is the first principles method which is a *guess*.
> > > For h -->0 the result is precisely 2x.There is no guess.
> > Unfortunately no,
> For me it means 2x + h, and in the limit 2x.

2x+h _IS_ the first principles method, you ignoramus.

>
> Regards, WM

tisdag 3 maj 2022 kl. 14:02:18 UTC+2 skrev Eram semper recta:
> On Monday, 2 May 2022 at 14:39:47 UTC+3, WM wrote:
> > Eram semper recta schrieb am Montag, 2. Mai 2022 um 10:16:31 UTC+2:
> > > On Sunday, 1 May 2022 at 23:13:42 UTC+3, WM wrote:
> >
> > > > The derivative of x^n = nx^(n-1), proved by limit method.
> > > Limit theory is not a proof of any kind - it merely helps you validate your *guess* that nx^(n-1) is the correct derivative.
> > (x^2 + 2xh + h^2 - x^2)/h = 2x + h
> (x^2 + 2xh + h^2 - x^2)/h = 2x + h is the first principles method which is a *guess*.
> > For h -->0 the result is precisely 2x.There is no guess.
> Unfortunately no, that is not the case because
>
> Lim{h->0} (x^2 + 2xh + h^2 - x^2)/h
>
> means: 0<|x-c|<delta => | [f(x+h)-f(x)]/h - L|<epsilon
>
> where L = 2x. Pretty circular indeed. :-)
>
> The guess L used in the epsilon-delta verifinition is 2x which you obtain from the "definition" itself.
>
> Your brainwashing is very hard to cure.
>
> >
> > Regards, WM

YOu still have no clue what circular means.