Proving $1+x+\frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!}>0$.

Question

I've been asked to prove that $1+x+\frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!}>0$. Proving that $1+x+\frac{x^2}{2!}>0$ is simple as you just need to consider the discriminant and show that it's less that 0. Showing $1+x+\frac{x^2}{2!} + \frac{x^3}{3!}=0$ only has 1 solution is to just consider the derivative and using the previous proof that $1+x+\frac{x^2}{2!}>0$ has no zero (so it is always increasing).

With this information, how would you be able to prove $1+x+\frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!}>0$? I've tried to show that the global minimum (the lone solution to $1+x+\frac{x^2}{2!} + \frac{x^3}{3!}=0$) gives a positive answer, however I have not been successful. How can I show this?

What's the domain of $x$? Taking the derivative (a cubic) shows that it cannot be positive for all real $x.$ — Allawonder, May 01 '20 at 09:36
Are you allowed to know that this is the first five terms of the maclaurin expansion of $e^x$? — Andrew Chin, May 01 '20 at 09:38
Also a consequence of https://math.stackexchange.com/q/225063/42969. — Martin R, May 01 '20 at 09:42

score 2 · Accepted Answer · answered May 01 '20 at 09:34

2

As you say, $1+x+x^2/2+x^3/6=0$ has a unique solution, (with $x<0$). For this $x$, $$1+x+\frac{x^2}2+\frac{x^3}{3!}+\frac{x^4}{4!}=\frac{x^4}{4!}>0.$$

answered May 01 '20 at 09:34

Angina Seng

158,341

Proving $1+x+\frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!}>0$.

1 Answers1