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I recently bumped into this question which asks why $\pi=4$ is wrong.

enter image description here

And some answers(see the answer of user TCL, for example) stated that this has to do with functions and their derivatives.

Their answers were something like this:

Let $F_n(x)$ be a sequence of curves (the zigzag curves in the above picture) that approach $g(x)$ as n tends to $\infty$.

And $g(x)$ the curve that represents the circle.

$$\lim_{n \to \infty} F_n(x) = g(x).$$

Does not imply

$$\lim_{n \to \infty} F'_n(x) = g'(x).$$

And lengths has to do with derivatives($\int \sqrt{f'(x)+1} \, dx$), therefore convergence of two curves does not mean convergence of their length.

And based upon my understanding of this:

enter image description here

the condition that the function should satisfy for the above implication to hold is that it has to be a continuous function,but the curve(see the linked question) approximating the circle it is not continuous.

Is my understanding correct?

Does convergence of two continuous functions implies the convergence of their derivatives?

If this is the case, then why the function that approximates the circles is not continuous?

I saw this question, But I don't know if its related to my question about continuity?

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  • $\begingroup$ what derivatives? $\endgroup$ – Jorge Fernández Hidalgo Jul 19 '15 at 5:27
  • $\begingroup$ @dREaM If you have a curve that is given by $f(x)$, its length is computed by $\int \sqrt{f'(x)+1} dx$ $\endgroup$ – Omar Nagib Jul 19 '15 at 5:28
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    $\begingroup$ there are continuous functions that are nowhere differentiable. $\endgroup$ – Jorge Fernández Hidalgo Jul 19 '15 at 5:28
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    $\begingroup$ Let $F_n(x)=\frac1n\sin(nx)$. Then clearly $F_n(x)\to0$ but $F_n'(x)=\cos(nx)\not\to0$. $\endgroup$ – Rahul Jul 19 '15 at 5:40
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    $\begingroup$ For one, you are confusing the fact that the functional $\gamma\mapsto\ell(\gamma)$ is not continuous (for the uniform norm) wih the fact that one applies $\ell$ to continuous functions. $\endgroup$ – Did Jul 19 '15 at 6:14
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No, you misunderstand Emanuele Paolini's answer. He notes the problem that $\ell$ is not a continuous function. But his $\ell$ does not refer to one of the curves, which you label $F_n$ and $g$. Instead, $\ell$ is the function that inputs a curve and outputs its length. What it means for such a "higher-order" function to be continuous or discontinuous takes some work to explain... You can try researching the topology of uniform convergence.

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  • $\begingroup$ Alright, that's regarding Emanuele's answer, what about TCL's answer(the second Answer in the linked question)? He states that when two functions converges, this does not imply thier derivatives do. When he's talking about functions, is he talking about the functions that represent the curves or the function which "takes a curve as an input and output its lenght"? $\endgroup$ – Omar Nagib Jul 19 '15 at 6:29
  • $\begingroup$ TCL is talking about the curves. Note the sentence "let x=a(t) ... be the parameterizations of the two curves". $\endgroup$ – Chris Culter Jul 19 '15 at 6:34
  • $\begingroup$ So based on the answer of TCL, which says that convergence of functions (those representing the curves) does not imply convergence of their derivatives(which are used in computing the length), And this is the reason why the fact that the zigzag approaches the circle does not imply that their length converge. Is my understanding correct? $\endgroup$ – Omar Nagib Jul 19 '15 at 6:44
  • $\begingroup$ That's correct! Still, you should be careful not to over-generalize the idea. I'm sure there are other sequences of curves where the derivatives behave very badly yet the lengths still miraculously converge. $\endgroup$ – Chris Culter Jul 19 '15 at 6:59
  • $\begingroup$ So what is the neccessary condition that must be satisfied for the implication to holds true? $\endgroup$ – Omar Nagib Jul 19 '15 at 7:04
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Let $f(x)=\frac1x\sin x^2$, $g(x)\equiv0$. Then $$ f'(x)=-\frac{1}{x^2}\sin x^2+\frac1x\cdot 2x\cos x^2, $$ which has no limit at infinity, while derivative of $g$ is 0.

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  • $\begingroup$ I have edit my question, So if you wish to edit in response. $\endgroup$ – Omar Nagib Jul 19 '15 at 5:57

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