# Proving that $\frac{e^x + e^{-x}}2 \le e^{x^2/2}$

Prove the following inequality: $$\dfrac{e^x + e^{-x}}2 \le e^{\frac{x^2}{2}}$$

This should be solved using Taylor series.

I tried expanding the left to the 5th degree and the right site to the 3rd degree, but it didn't help.

Any tips?

• Why not use the full series? – Antonio Vargas Jan 28 '14 at 18:09

$$\frac{e^x+e^{-x}}{2} = \frac{\left(1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots \right) + \left(1 - x + \frac{x^2}{2!} - \frac{x^3}{3!} + \cdots \right)}{2} = \frac{2 + 2 \frac{x^2}{2!} + 2 \frac{x^4}{4!} + \cdots}{2}$$

$$= 1 + \frac{x^2}{2} + \frac{x^4}{24} + \frac{x^6}{720}+ \cdots$$

$$e^{\frac{x^2}{2}} = 1 + \frac{x^2}{2} + \frac{\left(\frac{x^2}{2}\right)^2}{2!} + \frac{\left(\frac{x^2}{2}\right)^3}{3!} + \cdots = 1 + \frac{x^2}{2} + \frac{x^4}{8} + \frac{x^6}{48} + \cdots$$

Basically, it comes down to comparing $\displaystyle \frac{1}{(2n)!}$ and $\displaystyle \frac{1}{2^n n!}$, because they are the coefficients for $\displaystyle \frac{e^x + e^{-x}}{2}$ and $e^{\frac{x^2}{2}}$, respectively. We note that $$\frac{1}{(2n)!} = \frac{1}{1 \cdot 2 \cdot 3 \cdots (2n-1) \cdot (2n)}$$ while $$\frac{1}{2^n n!} = \frac{1}{2 \cdot 4 \cdot 6 \cdots (2n-2) \cdot (2n)}$$ Because the denominator for the second one is obviously smaller than that of the first, the second fraction is bigger than the first. This applies to every $n$, and therefore $$\frac{e^x+e^{-x}}{2} \le e^{\frac{x^2}{2}}$$

• I did those three dots (...) at the end once in homework and they wrote on it that "three dots aren't magic" and I can't write the polynomial like that. – GinKin Feb 26 '14 at 18:16
• Then you can either write it as $$1 + \frac{x^2}{2} + \frac{x^4}{24} + \frac{x^6}{720} + O(x^8)$$ or as $$\sum_{n=0}^{\infty} \frac{x^{2n}}{(2n)!}$$ I personally prefer the latter. Similarly for the other polynomial, either as $$1 + \frac{x^2}{2} + \frac{x^4}{8} + \frac{x^6}{48} + O(x^8)$$ or as $$\sum_{n=0}^{\infty} \frac{x^{2n}}{2^n n!}$$ – 2012ssohn Feb 27 '14 at 0:53

Using $e^x = \sum_{k=0}^\infty \frac{1}{k!} x^k$, we get $$\frac{1}{2}\left( e^x + e^{-x}\right) = \frac{1}{2} \sum_{k=0}^\infty \frac{1}{k!} [ x^k + (-x)^k] = \sum_{i=0}^\infty ??? x^{2i},$$ and $$e^\frac{x^2}{2} = \sum_{k=0}^\infty ??? x^{2k}.$$

Fill in ???, and show that each term in the first sum is smaller than the corresponding term in the latter sum.

Obscene overkill: since $$\cosh(z)=\prod_{n\geq 0}\left(1+\frac{4z^2}{(2n+1)^2\pi^2}\right)$$ we have $$\log\cosh(z)\leq \sum_{n\geq 0}\frac{4z^2}{(2n+1)^2\pi^2} =\frac{z^2}{2}$$ and the claim follows by exponentiating both sides. In other terms, it is sufficient to integrate both sides of $$\tanh(z)\leq z$$ over $$[0,x\in\mathbb{R}^+]$$.

• Mind providing an elementary proof of the first identity? – Dwagg Jan 21 '19 at 21:43
• @Dwagg: a proof through Chebyshev polynomials can be found in my notes. It is the Weierstrass product for the cosine function. – Jack D'Aurizio Jan 21 '19 at 21:47