Proving positivity of the exponential function

Question

Question. Without using the semigroup property ($\mathrm{e}^{x}\mathrm{e}^{y}=\mathrm{e}^{x+y}$), how can we show that $\mathrm{e}^{x}>0$ for all $x\in\mathbb{R}$ only by using the series expansion?

Explanation. From the series expansion of $\mathrm{e}^{x}=\sum_{k=0}^{\infty}\frac{x^{k}}{k!}$ for $x\in\mathbb{R}$, we see that $\mathrm{e}^{x}>0$ for $x\geq0$. Thus, if the series becomes negative, this can only happen for negative values of $x$. So proving $\mathrm{e}^{-x}$ for $x>0$ will complete the proof. As the series converges uniformly on any compact interval $I\subset\mathrm{R}$, we can rearrange the terms of the series and write $\mathrm{e}^{-x}=\lim_{n\to\infty}g_{n}(x)$ for $x\geq0$, where $g_{n}(x):=1+\sum_{k=1}^{n}\Big(\frac{x^{2k}}{(2k)!}-\frac{x^{2k-1}}{(2k-1)!}\Big)$ for $x\geq0$ and $n\in\mathbb{N}$. Obviously, $g_{n}$ is decreasing on $[0,1]$ and $g_{n}(1)>\frac{1}{\mathrm{e}}$.

I need to prove the following.

Claim. There exists an increasing divergent sequence $\{\xi_{n}\}\subset(0,\infty)$ such that $g_{n}$ is decreasing on $[0,\xi_{n}]$ with $g_{n}(\xi_{n})>0$ for $n\in\mathbb{N}$.

Strengthened Claim. $\xi_{n}:=\sum_{k=1}^{n}\frac{1}{k}$ for $k\in\mathbb{N}$.

Answer 1

The series expansion is $$ e^x=\sum_{n=0}^\infty\frac{x^n}{n!}=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\cdots $$ For $x\ge0$ we have $1$ and a bunch of nonnegative numbers, so the result is clearly positive.

For $x<0$ notice that: $$ \frac1{e^x}=e^{-x} $$ So positivity of $e^x$ clearly implies that $e^{-x}$ is positive.

Answer 2

Using termwise differentiation one finds that $\exp$ satisfies the linear differential equation $y'=y$, which obvioulsy satisfies the assumptions of the existence and uniqueness theorem. The function $y_0(x):\equiv0$ is a solution, and no other solution can cross the graph of $y_0$. It follows that $x\mapsto e^x$, which is positive when $x=0$, is positive on its full domain ${\mathbb R}$.

Answer 3

A hyperbolic trigonometry approach. Set $$ C(x)=\sum_{k=0}^\infty\frac{x^{2k}}{(2k)!}\quad\text{and}\quad S(x)=\sum_{k=1}^\infty\frac{x^{2k-1}}{(2k-1)!} $$ It suffices to show that $C(x)>S(x)$, for every $x\in\mathbb R$.

First observe that: $C'(x)=S(x)$ and $S'(x)=C(x)$. Then observe that $$ \big(C^2(x)-S^2(x)\big)'=2\big(C(x)C'(x)-S(x)S'(x)\big)=2\big(C(x)S(x)-S(x)C(x)\big)=0, $$ and hence $$ C^2(x)-S^2(x)=C^2(0)-S^2(0)=1. $$ Thus, for every $x\in\mathbb R$, $$ C(x)=\sqrt{S^2(x)+1}>S(x). $$