# Why the length of the zigzag curve approximating the circle does not approach the length of the circle?

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

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:

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?

• what derivatives? – Jorge Fernández Hidalgo Jul 19 '15 at 5:27
• @dREaM If you have a curve that is given by $f(x)$, its length is computed by $\int \sqrt{f'(x)+1} dx$ – Omar Nagib Jul 19 '15 at 5:28
• there are continuous functions that are nowhere differentiable. – Jorge Fernández Hidalgo Jul 19 '15 at 5:28
• Let $F_n(x)=\frac1n\sin(nx)$. Then clearly $F_n(x)\to0$ but $F_n'(x)=\cos(nx)\not\to0$. – Rahul Jul 19 '15 at 5:40
• 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. – Did Jul 19 '15 at 6:14

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.
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.