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I see this question being answered multiple times on this website including here but it still seems inconsistent with the mean value theorem.

If we have $\sin(x)$ defined on $[0,1]$ and we have $[x,y]\subset [0,1]$, then according to the mean value theorem, there exists $c\in (x,y)$ such that $|\sin(x)-\sin(y)|=\cos(c)|x-y|$.

Since $\cos(c)\in (0,1)$ for $c\in (0,1)$ doesn't this imply that $\sin(x)$ is indeed a contraction?

Stylel
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  • The order of quantifiers matters. – Marek Kryspin Nov 26 '23 at 12:55
  • @MarekKryspin Oh I finally see thanks. This seems to be quite a widespread miscalculation since online some people assumed $\sin(x)$ was a contraction with the above method while others said it wasn't because $\lim_{x\to 0}\frac{\sin(x)}{x}=1$ ( this paper that can be found with a simple google search which jumped on the same conclusion...)

    I have a follow-up question: What is the correct way to show that $x_{n+1}=\sin(x_n)$ converges to 0?

     – Stylel Nov 26 '23 at 13:10

1 Answers1

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Everything you said in the question is correct, except for your conclusion that "$\sin(x)$ is indeed a contraction".

It is true (by MVT, like you say), that for every $0\leq x<y\leq 1,$ there is such a $c\in (0,1)$ such that $|\sin(x)-\sin(y)|=\cos(c)|x-y|$. But for $x$ and $y$ both approaching $0,$ this forces $\cos (c)$ to get very (arbitrarily) close to $1$ (from below $1$), and so there is no universal $c<1$ (or in wikipedia's notation $k<1$) that satisfies the contraction definition, as we require $c<1$ (or in wikipedia's notation $k<1$).

A simple way to think of a contraction is that for all $x\in D,\vert f'(x)\vert <t$ for some $t\in (0,1).$ That is, assuming $f$ is differentiable...

Adam Rubinson
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