Talk:Fundamental increment lemma

The assertion is FALSE because ...

"The lemma asserts that the existence of this derivative implies the existence of a function phi ..."

Let's take as an example f(x)= (x-1)^(1/3) + 1 for which f'(x) does not exist. This neither implies nor rejects the fact that a function phi(h) exists in any way whatsoever.

Q(x,h) = ( (x+h-1)^(1/3) - (x-1)^(1/3) )/h - ( (x-1)^(-2/3) )/3

It's not just that Wikipedia is a load of rubbish. It's that it is a dangerous load of rubbish.

"The lemma asserts that the existence of this derivative implies the existence of a function phi ..."

FALSE in its entirety. Whether derivative exists or not is immaterial to the function Q(x,h) or phi(h) as you incorrectly call it in this article because the function depends on both x and h.

The function Q(x,h) is made up of the terms containing a factor of h and may also contain x. However, Q(x,h) contains no terms without a factor of h. — Preceding unsigned comment added by RealMathematician (talk • contribs) 15:38, 9 September 2023 (UTC)

RealMathematician (talk) 14:28, 9 September 2023 (UTC)

What's wrong with the assertion "that the existence of this derivative implies the existence of a function phi such that" so-and-so? It's a rather simple fact. Actually, proven in the article.

Concerning your "example". It seems that you do not understand the logical significance of an "implication": The claim is "If P then Q". Considering a case with not-P is completely irrelevant concerning the validity of the original claim ("If P then Q"). 84.155.152.129 (talk) 15:03, 9 September 2023 (UTC)

So, this IP is a clown none other than Markus Klyver.

You can't reason with this type. He is an idiot.

The assertion is WRONG, you idiot! I am not here to teach you propositional logic. RealMathematician (talk) 15:36, 9 September 2023 (UTC)

John Gabriel usually calls himself "the greatest mathematician since Archimedes" but he doesn't understand how implications work. Clearly "P implies Q" isn't the same as "not P implies not Q". But John Gabriel doesnt understand this, unfortunately. This could explain why he fails to understand the limit definition or why he thinks "3 ⇔ 2+1" makes sense. Or why he thinks we can't say that 3≤4.

Of course, f'(a) must exist for the formula to make sense.

The lemma states that difference quotient = derivative at x=a + error, and this is a special case of the limit definition. 81.225.32.185 (talk) 15:37, 9 September 2023 (UTC)

There is no error whatsoever. Q(x,h) does not care whether a derivative exists or not and does not depend on it in any way.

The "lemma" incorrectly ASSERTS that the "existence of the derivative" implies Q(x,h) exists. It DOES NOT. Simple as this and I provided a "counterexample" if you know what that means. RealMathematician (talk) 15:41, 9 September 2023 (UTC)

You haven't given a counterexample. The function you have given is differentiable, just not everywhere (it isn't even defined everywhere).

You can check the limit yourself and see that Q(x, h) converges to zero when h goes to zero. 81.225.32.185 (talk) 15:45, 9 September 2023 (UTC)

"and I provided a 'counterexample'". Nope. All your "example" shows is that you don't understand logic. Hint: You can't "disprove" the assertion "P implies Q" by considering a special case with non-P. 84.155.152.129 (talk) 15:55, 9 September 2023 (UTC)

"The 'lemma' incorrectly ASSERTS that the 'existence of the derivative' implies Q(x,h) exists. It DOES NOT."

The lemma *correctly* asserts "that the existence of this derivative implies the existence of a function phi such that" so-and-so (that's why it's called a lemma). See the short proof in the article. 84.155.152.129 (talk) 16:09, 9 September 2023 (UTC)

1. It's not a lemma. It's a false assertion and I *correctly* gave a counterexample. Whether the derivative exists or not has nothing to do with the function phi.

2. There is no proof in the article, only incoherent gibberish. RealMathematician (talk) 18:48, 9 September 2023 (UTC)

Your counterexample is not correct.

The lemma is of the form "if P, then Q".

Your counterexample assumes not-P, so it isn't a counterexample. A counterexample would assume P and show not-Q.

The proof given is correct. 81.225.32.185 (talk) 19:17, 9 September 2023 (UTC)

Exactly. 84.155.152.129 (talk) 20:17, 9 September 2023 (UTC)

Yes idiot. I still think you cannot say 3 <= 4 and I am correct. RealMathematician (talk) 18:51, 9 September 2023 (UTC)

"Real mathematician" 🙄 81.225.32.185 (talk) 19:15, 9 September 2023 (UTC)

Sounds like a joke. :-) 84.155.152.129 (talk) 20:13, 9 September 2023 (UTC)

No, you bumbling idiot!

"Of course, f'(a) must exist for the formula to make sense."

That's my point exactly, twit! f'(a) existence is irrelevant. What don't you get about it, you baboon? RealMathematician (talk) 18:52, 9 September 2023 (UTC)

The equation involves f'(a). This assumes that f'(a) is defined and exists. 81.225.32.185 (talk) 19:27, 9 September 2023 (UTC)

Obviously. (One might think.) 84.155.152.129 (talk) 20:13, 9 September 2023 (UTC)

"function Q(x,h) or phi(h) as you incorrectly call it in this article because the function depends on both x and h."

To say that a function depends on x and h or just h isn't a huge difference.

It entirely depends on what you consider fixed and what's not fixed. In the derivative definition, x is usually fixed and h is allowed to vary around the origin. 81.225.32.185 (talk) 15:41, 9 September 2023 (UTC)

Who are you to claim anything is small or huge? Do you have a name, you cowardly idiot? No, again, you are WRONG. It is sufficient to say that Q(x,h) is a function that depends on x and h. Hold your 2 cents, moron! RealMathematician (talk) 15:43, 9 September 2023 (UTC)

Right. In the context of the article x is fixed: x = a here. 84.155.152.129 (talk) 15:46, 9 September 2023 (UTC)

You don't get to fix anything for a given non-parallel secant line, you utter moron!

If one is finding the slope of a non-parallel secant line whose endpoints lie on f(x)=x^2, the procedure is:

Slope = (f(x+h)+f(x))/h= 2x+h

Now choose any x and h, say x=1 and h=2 then the slope of the non-parallel secant line whose endpoints on f are (1,1) and (3,9), we have: slope=2(1)+2 = 4 and this agrees with (f(3)-f(1))/2 = (9-1)/2 = 4

If you use another value for h, then you are no longer talking about the same secant line. Get it, imbecile?

slope of non-parallel secant line = slope of tangent line + difference in slopes (NOT error because there is no error.)

The only error is YOU.

Elementary details that even a high school student can understand, but you can't! Tsk, tsk. RealMathematician (talk) 18:11, 9 September 2023 (UTC)

If you choose a different h, yes that's a different secant line. What's the problem with this? We let h approach zero, and that means f'(a) is the limit of many secant lines, all of which have a different h.

Taking a limit when h goes to zero requires the difference quotient to be defined around h=0. 81.225.32.185 (talk) 19:19, 9 September 2023 (UTC)

Plagiarism by Talman.

The identity (f(x+h)-f(x))/h = f'(x) + Q(x,h) is my (John Gabriel) identity that is proved for all smooth functions in the Historic Geometric Theorem that was realised from the New Calculus.

The slope difference function is Q(x,h) and not Q(h).

www.academia.edu/62358358/My_historic_geometric_theorem_of_January_2020

From this identity, both the derivative and definite integral can be defined:

www.academia.edu/105576431/The_Holy_Grail_of_Calculus

Also, it seems strange that he is allowed to create this Wikipedia entry given that the only supporting source is a short article by Talman. I always thought that one could not create any entry without the source being a physical book or journal?

64.99.242.121 (talk) 22:31, 30 August 2023 (UTC)

You're a funny dude John Gabriel. No, Wikipedia won't delete it because of "plagiarism". But I agree that the lemma doesn't say much and is an immediate consequence of the limit definition. 81.225.32.185 (talk) 07:14, 5 September 2023 (UTC)

I don't see any humour but can understand that feeble minds are easily amused.

Wikipedia should delete it - period because it is NONSENSE.

You're wrong about it being a consequence of the limit definition. It's the other way around. RealMathematician (talk) 21:43, 5 September 2023 (UTC)

Explain why it's 'nonsense'. 81.225.32.185 (talk) 01:25, 7 September 2023 (UTC)

I have done so. See below. RealMathematician (talk) 10:05, 7 September 2023 (UTC)

I dont't buy the "plagiarism" accusation either – this Wikipedia article was added in 2012 and the earliest reference is from 2008. Whatever your theorem says, you claim to have found in 2020.

That's eight whole years after the article was added to Wikipedia. Clearly – not plagiarism. 84.216.129.212 (talk) 19:50, 9 September 2023 (UTC)

I also want to add that this "lemma" is incorrect because the slope of a secant line DOES NOT depend on the existence of phi(h) or f'(a). In fact, f'(a) need not exist at all. Moreover, if a derivative exists, then the identity (f(x+h)-f(x))/h = f'(x) + Q(x,h) is provable for any smooth function f. See article above. — Preceding unsigned comment added by RealMathematician (talk • contribs) 11:46, 4 September 2023 (UTC)

Hint: The lemma does not state that "the slope of a secant line depends on the existence of phi(h) or f'(a). So your complaint is moot. In addition: "if a derivative exists [at the point a], then the identity (f(a+h)-f(a))/h = f'(a) + Q(a,h) is provable [...]". That's just what the lamma says! 79.206.207.80 (talk) 00:51, 8 September 2023 (UTC)

Moderator: The previous IP is a troll called Fritz Feldhase from sci.math. RealMathematician (talk) 11:42, 8 September 2023 (UTC)

It most certainly does! "The existence of the derivative" means tangent line and non-parallel secant line are present. The main stream derivative requires both the tangent and secant line. RealMathematician (talk) 12:51, 9 September 2023 (UTC)

Delete this article as it is incorrect.

I also want to add that this "lemma" is incorrect because the slope of a secant line DOES NOT depend on the existence of phi(h) or f'(a). In fact, f'(a) need not exist at all. Moreover, if a derivative exists, then the identity (f(x+h)-f(x))/h = f'(x) + Q(x,h) is provable for any smooth function f. See article above. RealMathematician (talk) 11:55, 4 September 2023 (UTC)

What's the incorrect part? 84.216.129.226 (talk) 16:41, 7 September 2023 (UTC)

Are you a moron or what? RealMathematician (talk) 16:57, 7 September 2023 (UTC)

Nope. Please be civil, as per Wikipedia guidelines, and explain why you think the lemma is stated incorrectly. 84.216.129.226 (talk) 20:49, 7 September 2023 (UTC)

Here: "...the slope of a secant line DOES NOT depend on the existence of phi(h) or f'(a). In fact, f'(a) need not exist at all. Moreover, if a derivative exists, then the identity (f(x+h)-f(x))/h = f'(x) + Q(x,h) is provable for any smooth function f.

www.academia.edu/62358358/My_historic_geometric_theorem_of_January_2020" RealMathematician (talk) 21:26, 7 September 2023 (UTC)

Hint: The lemma does not state that "the slope of a secant line depends on the existence of phi(h) or f'(a). So your complaint is moot. In addition: "if a derivative exists [at the point a], then the identity (f(a+h)-f(a))/h = f'(a) + Q(a,h) is provable [...]". That's just what the lamma says! 79.206.207.80 (talk) 00:43, 8 September 2023 (UTC)

Just like I thought. You're am imbecile with poor reading comprehension. 64.99.242.121 (talk) 06:32, 8 September 2023 (UTC)

This kind of language does not belong on Wikipedia. 84.216.157.17 (talk) 23:08, 8 September 2023 (UTC)

Moderator: This IP is a troll from sci.math called Fritz Feldhase. RealMathematician (talk) 11:41, 8 September 2023 (UTC)

I think you misunderstand the lemma. It does not say that. It says that if f'(a) if defined, we can write f'(a) as a secant + some error that goes to zero when h goes to zero. The error term is called phi, or Q.

I think you don't quite understand what the lemma actually says. 84.216.157.17 (talk) 23:08, 8 September 2023 (UTC)

Agree. 79.206.207.80 (talk) 02:05, 9 September 2023 (UTC)

I understand well. It is you who do not understand:

"The lemma asserts that the existence of this derivative implies the existence of a function phi(h) such that."

The assertion is false. phi(x,h) exists whether or not f'(a) exists. RealMathematician (talk) 12:46, 9 September 2023 (UTC)

You are unfortunately incorrect. That's not what the lemma says. The lemma assumes f' exists at the point x=a.

The lemma makes no claims to what is true if f'(a) is not defined 84.216.129.17 (talk) 21:28, 9 September 2023 (UTC)

Moreover, Q(x,h) is not an error term of any kind. It is a difference in slope between the tangent line and the function f. One can't even begin to talk about f'(a) unless it is given that f is smooth at every point in (x, x+h). RealMathematician (talk) 12:48, 9 September 2023 (UTC)

Sorry, between the tangent line and the non-parallel secant line anchored at x. RealMathematician (talk) 12:49, 9 September 2023 (UTC)

Nope, the lemma isn't "incorrect". Get a grip, man. 84.155.152.129 (talk) 01:25, 10 September 2023 (UTC)