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I want to prove that the function:

$f(x) = \displaystyle \lim_{y\rightarrow x} \dfrac{\text{tanh}^{-1}(y)}{y}$

is convex in $(-1,1)$, where $\text{tanh}^{-1}()$ is the inverse hyperbolic tangent function. Any help would be appreciated.

Thanks!

HK Lee
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Vokram8
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  • It's a bit "tedious" but maybe you can try from calculating the 2nd derivative of the function – Evan William Chandra May 14 '19 at 03:14
  • Isn't $f(x)$ simply $\dfrac{\tanh^{-1}(x)}{x}$ except when $x=0 $ ? have you considered Taylor series? – J. W. Tanner May 14 '19 at 03:16
  • I computed the second derivative but did not manage to show it is positive. Yes, the limit is just there in order to handle the case $x=0$. I have not used Taylor series, I will try that thanks. – Vokram8 May 14 '19 at 03:32

1 Answers1

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$\operatorname{arctanh} x= \sum_{k=1,3,5,...} {1 \over k} x^k$, so $f(x) = \sum_{k=1,3,5,...} {1 \over k} x^{k-1}$.

Hence $f''(x) = \sum_{k=3,5,7,...} {1 \over k} (k-1)(k-2) x^{k-3}$ and it is straightforward to observe that $f''(x) \ge 0$.

copper.hat
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