It seems to me that $\ln \cosh x\leq \frac{x^2}{2}$ for $x\in\mathbb{R}$, as suggested by graphing the difference between both functions as well as the fact that the Taylor series expansion of $\ln\cosh x$ at $x=0$ yields $\frac{x^2}{2}-\frac{x^4}{12}+\mathcal{O}(x^6)$. However, how do I prove the bound formally? Using Taylor's Theorem with the remainder seems sort of unwieldy... Any hints?
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$$\frac{d}{dx}(\ln \cosh x) = \tanh x$$ – Daniel Fischer Feb 16 '15 at 00:12
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@DanielFischer That is simple and brilliant! I'll accept your comment as an answer if you make it an answer. – HellRazor Feb 16 '15 at 00:14
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Since both sides of the inequality are even functions, we need only consider $x\geqslant 0$. Then using
$$\ln \cosh x = \int_0^x \tanh t\,dt$$
one only needs to see that $\tanh t \leqslant t$ for $t \geqslant 0$. Since this inequality is strict for $t > 0$, the original inequality is strict for $x \neq 0$.
Daniel Fischer
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One way to do this would be to compare the Maclaurin series expansions of $\cosh x$ and $e^{\frac{x^2}{2}}$:
$\displaystyle\;\;\;\cosh x=\sum_{n=0}^{\infty}\frac{x^{2n}}{(2n)!}\le\sum_{n=0}^{\infty}\frac{x^{2n}}{2^n n!}=e^{\frac{x^2}{2}}$ $\;\;$ since $(2n)!\ge2^n n!$ for all n.
user84413
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