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I have to prove that $|\arctan x - \arctan y|\leq |x-y|$ for all real numbers $x$ and $y$. I know that it has something to do with the mean value theorem but I just don't know how to solve it. All help is appreciated

Thanks in advance :)

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The mean value theorem states that for a differntiable function $f$ and given real numbers $u$ and $v$ (where $u<v$) within the function's domain, there exists $c\in (u,v)$ such that $$f'(c)=\frac{f(v)-f(u)}{v-u}$$

Now, consider the function $f(t)=\tan{t}$ over $(-\pi/2,\pi/2)$. Therefore due to the mean value theorem there exists $c$ as explained above that $$1+\tan^2{c}=\frac{\tan v-\tan u}{v-u}$$ Obviously the LHS is always greater that one, so $$1\leq \frac{\tan v-\tan u}{v-u}$$ Or equivalently (since both sides are positive) $$1\leq \frac{|{\tan v-\tan u}|}{|v-u|}$$ Multiplying both sides by $|v-u|\ne 0$ and setting $v=\arctan x$ and $y=\arctan u$, you'll get $$|{\arctan x-\arctan y}|\leq |x-y|$$

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Let $f(t) = \arctan t$. It is differentiable everywhere. Thus, apply MVT on the interval $(x,y)$, that is, we can find some $\xi \in (x,y)$ such that

$$ | \arctan x - \arctan y| \leq | f'( \xi ) | |x- y| $$

Notice, $f'(t) = \frac{1}{1+t^2} = 1 - \frac{t^2}{1+t^2} \leq 1 $ for all $t$. Thus, in particular, for $t = \xi$, $|f'(\xi)| \leq 1$. We conclude that

$$ |\arctan x - \arctan y | \leq | x - y| $$

as desired.

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  • $\begingroup$ Shouldn't the first inequality be an equation? $\endgroup$ – Babak Oct 16 '16 at 10:16
  • $\begingroup$ I think it would be clearer if the first display reads $\arctan x-\arctan y=f'(\xi)(x-y)$, which is the standard statement of the MVT. $\endgroup$ – egreg Oct 16 '16 at 10:32

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