Why is $\frac{(e^x+e^{-x})}{2}$ less than $e^\frac{x^2}{2}$?

Question

I have read somewhere that this equality holds for all $x \in \mathbb {R}$. Is it true, and if so, why is that? $$\frac{(e^x+e^{-x})}{2} \leq e^\frac{x^2}{2}$$

Answer 1

The taylor series on the LHS is $$\sum_{n=0}^{\infty} \frac{x^{2n}}{(2n)!}$$

The taylor series on the RHS is $$\sum_{n=0}^{\infty} \frac{(x^2/2)^n}{n!} = \sum_{n=0}^{\infty} \frac{x^{2n}}{(2n)!!}$$

Where $(2n)!!$ is the double factorial $(2 \times 4 \times \ldots \times 2n)$. It is easy to then see that $(2n)!! \leq (2n)!$, namely since the latter has the extra factor of $(2n-1)!! = (1 \times 3 \times \ldots \times (2n-1))$ multiplying it.

Answer 2

The left side is $$\sum_{n=0}^{\infty} \frac{x^{2n}}{(2n)!}.$$ The right side is $$\sum_{n=0}^{\infty} \frac{x^{2n}}{2^nn!}$$

So you only need to show that $(2n)!\geq 2^nn!$, since every $x^{2n}\geq 0$.

Answer 3

Let us evaluate Taylor series of $\displaystyle e^{\frac{x^2}{2}}-\frac{e^x+e^{-x}}{2}$ near $x=0$. As we know $\displaystyle e^x=\sum_{k=0}^{\infty}\frac{1}{k!}x^k$. The radius of convergence is $\infty$, thus we can write $\displaystyle e^{\frac{x^2}{2}}=\sum_{k=0}^{\infty}\frac{1}{k!}\left(\frac{x^2}{2}\right)^k=\frac{1}{2^k k!}x^{2k}$ and $\displaystyle e^{-x}=\sum_{k=0}^{\infty}\frac{1}{k!}(-x)^k=\sum_{k=0}^{\infty}\frac{(-1)^k}{k!}x^k$. Furthermore, $$\cosh(x)=\frac{e^x+e^{-x}}{2}=\sum_{k=0}^{\infty}\frac{1+(-1)^k}{k!}x^k=\sum_{k=0}^{\infty}\frac{1}{(2k)!}x^{2k}$$

Finally $$e^{\frac{x^2}{2}}-\frac{e^x+e^{-x}}{2}=\sum_{k=0}^{\infty}\left(\frac{1}{2^k k!}-\frac{1}{(2k)!}\right)x^{2k}$$

It is easy to show by induction that $\forall \ n\in\mathbb{N}:\ 2^n\cdot{n!}<(2n)!$.

Answer 4

We may prove that: $$ \log\cosh(x) \leq \frac{x^2}{2} \tag{1}$$ by integrating the inequality: $$ \forall u\geq 0,\qquad \tanh(u)\leq u\tag{2} $$ over the interval $(0,|x|)$. $(2)$ is a simple convexity inequality.