How to show that $x^2 \ge \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \dots$?

Question

I wanted to show that for any $|x| \le 1$ we have: $e^x \le 1 + x + x^2$

So thought that using Taylor series would be a good starting point:

$$e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots$$

So all I need to do is to show that $x^2 \ge \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} +\dots$, but I've been stuck here for quite a while.

Intuitively it makes sense to me, since $x^k \le x^{k-1}$ for $k \ge 2$ and the fast increasing denominators, but I can't write it down in a formal way.

Any ideas?

EDIT: regarding the duplication, my question is specifically about showing that $x^2 \ge \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \dots$, even though I stumbled upon it when looking at exponential bounds. That question doesn't talk about my inequality.

Answer 1

If you want to do it using with Taylor series, you can think about splitting up $x^2$ into: \begin{align} x^2 &= \frac{x^2}{2} + \frac{x^2}{2^2} + \frac{x^2}{2^3} + \dots \\ e^x -1-x &= \frac{x^2}{2} + \frac{x^3}{3!}+\frac{x^4}{4!} + \dots. \end{align} You can then compare term by term, in other words, showing that $$ \frac{x^2}{2^{n-1}} \ge \frac{x^n}{n!}.$$

Answer 2

Note that if we have $e \leq 3$, we can do the following: $$\begin{align*} x^2 &= x^2(3-2) \\ &\geq x^2(e-2) \\ &= x^2\left(\frac1{2!} + \frac1{3!} + \cdots + \frac1{n!} + \cdots\right) \\ &= \frac{x^2}{2!} + \frac{x^2}{3!} + \cdots + \frac{x^2}{n!} + \cdots \\ &\geq \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots + \frac{x^n}{n!} + \cdots \\ &= e^x - 1 - x \end{align*}$$

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If you don't have $e \leq 3$, then we can show it by noting $2^{n-1} \leq n!$ for any $n \geq 1$, so $$\begin{align*}e &= \frac{1}{0!} + \frac1{1!} + \frac1{2!} + \frac1{3!} + \cdots + \frac1{n!} + \cdots \\ &\leq 1 + \left(1 + \frac1{2} + \frac1{4} + \cdots + \frac1{2^{n-1}} + \cdots\right) \\ &= 1 + 2 \\ &= 3\end{align*}$$

Answer 3

Well I don't see why you need to prove that, $$x^2 \ge \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + ...$$ You can instead work you way around it, by taking a function $f(x)$ such that, $$f(x)=1+x+x^2-e^x$$ Now all you need to do is prove,

$$f(x) \ge 0$$ for $x \in [-1,1]$

Now, $$f'(x) = 1+2x-e^x$$ $$f'(x) \le 0$$ for $x \in [-1,0]$

$$f(-1)=1-\dfrac1e \gt 0$$ and, $$f(0)=0$$

$$\therefore f(x) \ge 0;x \in [-1,0]$$ In $x \in (0,1]$, $$f'(x) \gt 0$$ $$\therefore f(x) \gt 0;x\in (0,1]$$ Hence we can conclude,

$e^x \le 1+x+x^2$ for $|x|\le 1$

Answer 4

Let

$$f=1+x+x^2-e^x$$

$$f'=1+2x-e^x, ~~f'(0)=0$$

$$f''=2-e^x,~~~f''>0\Leftrightarrow x<\ln2$$

So for $x\in [0,\ln2]$, $f'(x)$ is increasing and $f'(0)=0$, so $$f'>0,~~\forall x\in(0,\ln2]$$

where $x=\ln2$ is the maxima for $f'(x)$, next, let's see when $f'$ goes back to $0$

$$f'=1+2x-e^x=0$$

Let $$u=-x-\frac{1}2$$

We get

$$u e^u=-\frac{1}{2\sqrt e}$$

Use Lambert W-function, we can solve for $u$,

$$u=W\left(-\frac{1}{2\sqrt e}\right)$$

hence we get

$$x=-W\left(-\frac{1}{2\sqrt e}\right)-\frac{1}2\approx 1.256$$

Therefore,

$$f'>0, \forall x\in (0, 1.256)$$

which means $f$ is strictly increasing on $(0,1]\subset (0, 1.256)$. Therefore,

$$f\ge 0, \forall x\in [0,1]$$

Appendix:

For $x\in [-1,0]$

since we have shown $$f''=2-e^x,~~~f''>0\Leftrightarrow x<\ln2$$

So for $x\in (-\infty,\ln2]$, $f'(x)$ is increasing and $f'(0)=0$, so $$f'<0,~~\forall x\in(-\infty,0)$$

which means $f(x)$ is decreasing on $(-\infty,0)$, and we know $f(0)=0$, so

$$f(x)>0~~~\forall x\in (-\infty, 0)$$

Therefore,

$$f(x)\ge 0~~~\forall x\in (-\infty, 1]$$