I re-write entirely the previous posted (and wrong) proof. I think that this new proof is correct but I need some confirmation.
I would like to see too some alternative proof easier than what I did here, because my attempted proof seems overcomplicated (with probably some gaps somewhere). Thank you in advance.
Let $f_n(x):=\sum_{k=0}^n\frac{x^k}{k!}$ for $n\in\Bbb N_{>0}$. Show that the equation $f_n(x)=0$ have only one real solution for $n$ odd and no real solutions for $n$ even.
Proof
Some previous facts:
- $\partial f_n=f_{n-1}$ for all $n\in\Bbb N_{>0}$.
- $f_n(0)=1$ for all $n\in\Bbb N_{>0}$.
- $f_n(x)>0$ whenever $x\ge 0$.
For this exercise I will use the next theorem proved here:
Let $p(x)=\sum_{k=0}^n a_k x^k$ a polynomial function over $\Bbb K$ with $a_n=1$ and $n\ge2$. Let $R:=\sum_{k=0}^n |a_k|$. Then $|p(x)|>R$ when $|x|>R$.
For our $f_n$ this can be written as
$$|f_n(x)|>\sum_{k=0}^n\frac1{k!}\quad\text{ whenever }\quad|x|>n!\sum_{k=0}^n\frac1{k!}$$
Then for our polynomial functions $f_n$ for some $x_1<0$ we have that $|f_n(x_1)|>1$.
(Alternatively to this theorem we can use the bound $|s-s_n|<x_{n+1}$ for an alternating series $s:=\sum_{k=0}^\infty(-1)^k x_k$ with $(|x_k|)\downarrow 0$ with equivalent results).
Now Im ready to start a proof by strong induction:
Base case: we have that $f_1(x)=x+1$ is strictly increasing and have a unique real solution to $f_1(x)=0$ in $x=-1$.
We have too that $f_2(x)=x^2/2+x+1>0$ for all $x\in\Bbb R$, so it dont have any real solution to $f_2(x)=0$.
Induction hypothesis: assume that for all $2n+1\le m$ its true that $f_{2n+1}$ is strictly increasing and have a unique real solution to $f_{2n+1}(x)=0$.
We assume too that for all $2n\le m$ its true that $f_{2n}(x)>0$ for $x\in\Bbb R$.
Induction step: first we assume that $m$ is even, then because $\partial f_{m+1}=f_m$ we have that $f_{m+1}$ is strictly increasing, so we need to prove that $f_{m+1}$ have a unique real solution to $f_{m+1}(x_0)=0$ for some $x_0\in\Bbb R$.
Because $f_{m+1}$ is continuous and $f_{m+1}(x)>0$ for $x\ge 0$ it is enough to show that $f_{m+1}$ take negative values somewhere when $x<0$. From the theorem stated above we know that exists some $x_1<0$ such that $|f_{m+1}(x_1)|>1$. Because $f_{m+1}(0)=1$ and $f_{m+1}$ is strictly increasing we conclude that $f_{m+1}(x_1)<-1$, and then by the continuity of $f_{m+1}$ exists an unique $x_0$ such that $f_{m+1}(x_0)=0$.
The above implies that $f_{m+2}$ have an absolute extremum or an inflection point in $x_0$ because $\partial f_{m+2}=f_{m+1}$. Now observe that $f_{m+2}(x_0)=x_0^{m+2}/(m+2)!>0$. Because $f_m(x)>0$ for all $x\in\Bbb R$ then $f_{m+2}$ is strictly convex, what imply that $x_0$ cannot be an inflection point, hence it is an absolute minimum (because $f$ is not bounded above and $x_0$ is the unique zero of $f_{m+1}$).
Now to conclude this part because $x_0$ is an absolute extremum, with $f_{m+2}(0)=1$ and $f_{m+2}(x_0)>0$ then $f_{m+2}(x)>0$ for all $x\in\Bbb R$, as desired.
If $m$ is odd then we can repeat the same analysis with same results.$\Box$