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While reading an article on Hoeffding's Inequality, I came across a curious inequality. Namely

$$\cosh x \leq e^{x^2/2} \quad \forall x \in \mathbb{R}$$

I tried many ways to prove it and finally, the Taylor series approach worked:

$$e^x = 1 + x + \frac{x^2}{2!} + \cdots$$ $$e^{-x} = 1 - x + \frac{x^2}{2!} - \cdots$$ Adding the two and dividing by 2 (This operation being justified as both series converge), we get

$$\cosh x = 1 + \frac{x^2}{2!}+ \frac{x^4}{4!} + \cdots$$

Expanding $e^{x^2/2}$ yields

$$e^{x^2/2} = 1 + \frac{x^2}{2!}+ \frac{x^4}{4\times 2!} + \cdots$$

If you do a term by term comparison, you get the desired result.

My question is: Is there another more "Cute"/elegant way to get this result? If so, what is it? I tried using Jensen's Inequality but that didn't help. Also I searched for this inequality using the keywords "cosh x" and "inequality", but didn't get it.

Any ideas will be appreciated.

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  • $\begingroup$ Where do you find this beautiful inequality? $\endgroup$
    – Misa
    Apr 26, 2016 at 8:58
  • $\begingroup$ This was a long time ago, so I actually forgot where I got it from. No doubt from some good probability book. If I find the source, I will update the article. $\endgroup$ Apr 26, 2016 at 14:59
  • $\begingroup$ The inequality is in Achim Klenke's "Probability theory" (Exercise 9.2.4), but without proof. $\endgroup$ Oct 22, 2017 at 10:17

3 Answers 3

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The infinite product representation of the hyperbolic cosine function gives $$\cosh(x)=\prod_{k=1}^\infty\left(1+{4x^2\over \pi^2(2k-1)^2}\right) \leq \exp\left(\sum_{k=1}^\infty {4x^2\over \pi^2(2k-1)^2}\right) = \exp(x^2/2).$$

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    $\begingroup$ These infinite product representations of functions are handy just like Taylor series. $\endgroup$ Mar 15, 2013 at 19:42
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    $\begingroup$ Yes, I agree! It's natural for me to think of "products" when trying to bound an exponential. Though, I think Maisam's solution is better and more direct. $\endgroup$
    – user940
    Mar 15, 2013 at 19:44
  • $\begingroup$ I was unaware of such an expansion. It's pretty cute and useful. I'll wait for a few more days to see if someone posts something better. $\endgroup$ Mar 16, 2013 at 18:07
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    $\begingroup$ How do you find the infinite product representation? $\endgroup$
    – hi15
    Apr 27, 2020 at 17:30
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Hint: the wanted inequality is equivalent to $\ln(\cosh x) \leq \ln(e^{x^2/2}) $ which is in turn equivalent to $$\ln(\cosh x) \leq {x^2/2}.$$ Now define this function $f(x)=\ln(\cosh x) - {x^2/2}$ and $f'(x)=\tanh(x)-x$ and find maximum of $f$.

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  • $\begingroup$ This had also occurred to me(later), owing to the fact that all functions are differentiable. I don't suppose a Jensen inequality like proof is possible. $\endgroup$ Mar 16, 2013 at 18:03
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Noticing that $\frac{d}{dx}\tanh x = 1 -\tanh^2x$, note that:

  • $1-\tanh^2 u\le1\implies\int_0^t(1-\tanh^2 u)du\le\int_0^tdu\implies\tanh t \le t$
  • $\int_0^x \tanh t dt\le\int_0^xtdt\implies\log\cosh x \le \frac{1}{2}x^2\implies\cosh x\le e^{\frac{1}{2}x^2}$
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