Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question

2 Answers 2

up vote 6 down vote accepted

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

share|improve this answer
    
These infinite product representations of functions are handy just like Taylor series. –  Mhenni Benghorbal Mar 15 '13 at 19:42
    
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. –  Byron Schmuland 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

Hint: $\ln(\cosh x) \leq \ln(e^{x^2/2}) $ then $$\ln(\cosh x) \leq {x^2/2}$$ now define this function $f(x)=\ln(\cosh x) - {x^2/2}$ and $f'(x)=0$ and find maximum of $f$.

edit: $f'(x)=\tanh(x)-x$

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.