Prove $p_{4}(x) = 1 + x + \frac{x^2}{2!}+ \frac{x^3}{3!} + \frac{x^4}{4!} > 0 \forall x \in \mathbb{R}$

Question

Here is part of an incomplete proof that I made. First let me establish that, $p_n(x) = \sum_{k=0}^{n}=\frac{x^k}{k!}$ for $n = 1, 2, 3 \ldots$

Lemma 1: $p_2(x)$ is always greater than $0$.
Proof:
$p_2'(x) = 1 + x$ meaning $p_2$'s one and only stationary point occurs at $x = -1$. $p_2(-1) = \frac{1}{2}$. But $p_2''(x)=1$, meaning that this stationary point is a global minimum. Therefore $p_2(x) \geq \frac{1}{2}\ \forall\ x \in \mathbb{R}$. Lemma proven

Lemma 2: $p_3(x) = 1 + x + \frac{x^2}{2!}+ \frac{x^3}{3!}$ has exactly one root.
Proof:
$p_3'(x) = 1 + x + \frac{x^2}{2!} = p_2(x)$

Therefore $p_3$ is monotonically increasing. It is also differentiable and continuous across $\mathbb{R}$. $p_3(-2)=-\frac{1}{3}$ and $p_3(-1)=\frac{1}{3}$. And so by the intermediate value theorem, there exists a zero for $p_3$ between $-2$ and $-1$. This must be the only zero, since $p_3$ is monotonically increasing. Lemma proven.

Original Proof:
$p_4'(x) = p_3(x)$ and $p_4''(x)=p_2(x)$. And so it is known that there is exactly one stationary point because of Lemma 1 and that this stationary point must be a global minimum from the Lemma 2.

This global minimum will occur at $p_3(x) = 0$, which is beteween $-2$ and $-1$ as shown in Lemma 2. Incomplete. All that's needed is to show that the minimum in this interval is greater than 0.

Answer 1

An alternate approach.

Multiply by $24$ then get:

$$\begin{align} x^4+4x^3+12x^2+24x+24&=(x+1)^4+6x^2+20x+23\\&=(x+1)^4+6(x+5/3)^2+19/3 \end{align} $$

This doesn’t generalize to $p_{2n}$ easily, though.

(There is a theorem which says a real polynomial is non-negative if and only if it can be written as a sum of squares of real polynomials, I believe.)

Answer 2

You've established that $\ p_4\ $ has a global minimum $\ x_\min\ $, say, which must satisfy $\ p_3\big(x_\min\big)=0\ $ and $\ -2<x_\min<-1\ $. But $\ p_4(x)=p_3(x)+\frac{x^4}{4!}\ $. What can you say about the sign of $\ \frac{x^4}{4!}\ $ over the interval $\ -2<x<-1\ $?

Answer 3

If some calculus is allowed, then for $\,x \gt 0\,$:

$$ p_4(-x) = 1 - x + \frac{x^2}{2!} - \frac{x^3}{3!} + \frac{x^4}{4!} \,\gt\, \sum_{k \ge 0} \,(-1)^k\,\frac{x^k}{k!} \,=\, e^{-x} \,\gt\, 0 $$

This shows that $\,p_4(x)\,$ is positive on $\mathbb R^-$, and quite obviously it is positive on $\mathbb R^+$ as well. More generally, the same argument works to prove that $\,p_{2n}(x) \gt 0\,$.

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[ EDIT ] $\;$ As pointed out in Thomas Andrews' comment, the first inequality is not immmediately obvious, because the $e^{-x}$ series is alternating for $x \gt 0$, but the terms do not decrease in magnitude for small $n$, so the alternating series test cannot be applied directly.

To show that the inequality does in fact hold, it is enough to use the Lagrange form of the remainder $\,e^{-x} = p_4(-x) + \frac{(-1)^5\,e^{- t}}{5!} \,x^5\,$ for some $\,t \in (0, x)\,$, then the conclusion follows since the last term is negative. The same argument works for even-power partial sums in general:

$$ p_{2n}(-x) = \sum_{k = 0}^{2n} \,(-1)^k\,\frac{x^k}{k!} \,=\, e^{-x} \,-\, \frac{(-1)^{2n+1}\,e^{-t}}{(2n+1)!}\,x^{2n+1} \,\gt\, e^{-x} \,\gt\, 0 $$