Proving Upper Bound on Negative Exponential

Question

I was reading a paper when I came across this lemma.

$$\forall x \geq 0,\ \exp(-x) \leq 1 - x + \frac{1}{2}x^2$$

Visually, this seems to be true:

Can someone provide a succinct proof, preferably simple or using Taylor Series? Thanks

Answer 1

We have: $f(x) = e^{-x} - 1+x -\dfrac{x^2}{2}\implies f'(x) = -e^{-x}+1-x\implies f''(x)= e^{-x} -1\le 0\implies f'(x) \le f'(0)= 0\implies f(x) \le f(0)=0$, and the result follows.

Answer 2

One way is to divide the proof into two cases: $x<4$ and $x \geq 4$. In the first case look at the Taylor series $1-x+x^{2}/2- (x^{3}/3!-x^{4}/4!)-(x^{5}/5!-x^{6}/6!)+\cdots$ and verify that the numbers $x^{3}/3!-x^{4}/4!,x^{5}/5!-x^{6}/6!$ etc are all positive. For $x\geq 4$ just note that $e^{-x}\leq 1$ and $1 \leq 1-x+x^{2}/2$.

Answer 3

We know $\exp(-x)\geq 1-x$ for all $x\in\mathbb{R}$, since $\exp$ is convex. So integrating gives, $$ \int_0^x\exp(-\xi)\,\mathrm{d}\xi\geq\int_0^x (1-\xi)\,\mathrm{d}\xi $$ which rearranges to $$ \exp(-x)\leq 1-x+\frac12 x^2. $$