How do I show the uniform continuity of $\tan^{-1}$ over $\mathbb{R}$ ? I am trying to use the $\epsilon - \delta$ definition. I have just started learning this topic.
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1The derivative is bounded by $1$ hence $\delta=\varepsilon$ works, that is: $$\forall\varepsilon\gt0,\ \forall (x,y),\ |x-y|\leqslant\varepsilon\implies|f(x)-f(y)|\leqslant\varepsilon.$$ – Did Sep 19 '13 at 12:16
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"derivative is bounded by 1..." the basic book i am refering to uses only the epsilon-delta definition. I do not know the use of deriatives to prove/disprove uniform continuity. can you please suggest some link where i can read about it ? – Aman Mittal Sep 19 '13 at 12:28
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@AmanMittal: I gave you the technique that uses the derivative. – Mhenni Benghorbal Sep 19 '13 at 12:30
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@MhenniBenghorbal can you pleas share some link where i can read more about it ? – Aman Mittal Sep 19 '13 at 12:34
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Yet another idea. Use $$\arctan(x) - \arctan(y) = \arctan\left(\frac{x - y}{1 + xy} \right)$$ and the uniform continuity of $\arctan$ in the compact interval $[-h,h]$, $h>0$.
Martín-Blas Pérez Pinilla
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A related problem. Hint: You can use the mean value theorem.
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$$ \Bigg|\frac{\arctan(x+h)-\arctan(x)}{(x+h)-x}\Bigg| = |\arctan(\zeta)'| \leq 1. $$
$$ \implies \Big|{\arctan(x+h)-\arctan(x)}\Big| \leq |h|< \epsilon=\delta.$$
Mhenni Benghorbal
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i haven't reached that part yet. Do you think i should finish MVT and then come back to this problem ? – Aman Mittal Sep 19 '13 at 12:16
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@AmanMittal: This is one approach to the problem which uses the boundedness of the derivative. – Mhenni Benghorbal Sep 19 '13 at 12:23
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ok, from Lipschitz theorem, a function is uniformly continuous if its derivative is bounded. which holds good here. so where did we apply MVT in this ? – Aman Mittal Sep 19 '13 at 12:41
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1That cleared it for me. Your another post here http://math.stackexchange.com/questions/322890/uniform-continuity-of-fx-x3 was very helpful too. Thanks a lot !! – Aman Mittal Sep 19 '13 at 12:55
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One side question. If we have $y=\tan x$ then we cannot use this method. can you give some pointer for this problem ? – Aman Mittal Sep 19 '13 at 13:18
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It is unbounded on $(0,\pi/2)$. Any uniformly continuous function is necessarily bounded on any bounded interval. @AmanMittal – Thomas Andrews Sep 19 '13 at 13:20
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Here $f : x \mapsto\mathbb \tan^{-1}(x)$ is continuous function and $\displaystyle\lim_{x\rightarrow \infty} f(x)$ and $\displaystyle\lim_{x\rightarrow -\infty} f(x)$ exists then $f$ is uniformly continuous.(you can prove this for any continuous function)
jim
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@AmanMittal just look at this prove http://math.stackexchange.com/questions/346292/uniform-continuity-on-0-infty?rq=1 and try to modify it. – jim Sep 19 '13 at 12:21
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Hint: $$ \frac{\mathrm{d}}{\mathrm{d}x}\tan^{-1}(x)=\frac1{1+x^2}\le1 $$
robjohn
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Hints were permitted in 2013, if that’s what the downvote is about. Otherwise, I don’t see what’s wrong. I would appreciate a comment. – robjohn Sep 30 '23 at 07:03