cosh x inequality

Question

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.

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. — Gautam Shenoy, Apr 26 '16 at 14:59
The inequality is in Achim Klenke's "Probability theory" (Exercise 9.2.4), but without proof. — Elias Strehle, Oct 22 '17 at 10:17

score 32 · Accepted Answer · answered Mar 15 '13 at 17:23

32

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

answered Mar 15 '13 at 17:23

3

These infinite product representations of functions are handy just like Taylor series. – Mhenni Benghorbal Mar 15 '13 at 19:42
1

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. – Mar 15 '13 at 19:44
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. – Gautam Shenoy Mar 16 '13 at 18:07
1

How do you find the infinite product representation? – melatonin15 Apr 27 '20 at 17:30

score 19 · Answer 2 · edited Apr 17 '19 at 11:57

19

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 $f$.

edited Apr 17 '19 at 11:57

Davide Giraudo

172,925

answered Mar 15 '13 at 17:31

M.H

11,498
3
30
66

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. – Gautam Shenoy Mar 16 '13 at 18:03

score 9 · Answer 3 · answered Apr 26 '16 at 11:46

Noticing that $\frac{d}{dx}\tanh x = 1 -\tanh^2x$, note that:

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

score 0 · Answer 4 · answered Apr 28 '23 at 08:12

0

Also worth noting the alternating series for $\log\cosh(x)$:

$$ \log\cosh(x) = \frac{x^2}{2} - \frac{x^4}{12} + \frac{x^6}{45} + \mathcal{O}(x^8) $$

Since the series is alternating you have the inequalities:

$$\frac{x^2}{2} - \frac{x^4}{12} \le \log\cosh(x) \le \frac{x^2}{2}$$

Hence

$$ e^{\frac{x^2}{2} - \frac{x^4}{12}} \le \cosh(x) \le e^{\frac{x^2}{2}} $$

Keeping more and more terms of the series you can get increasingly tighter inequalities.

answered Apr 28 '23 at 08:12

a06e

6,665

Related: https://math.stackexchange.com/questions/4688080/closed-form-maclaurin-series-of-log-coshx/4688115 – a06e Apr 28 '23 at 10:25

cosh x inequality

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