Prove the following, without induction. Is it possible?
The equation $\sum_{k=0}^{n}\frac{x^{k}}{k!}=0$ has no real root if $n$ is even.
And if $n$ is odd, it has only one real root.
I also tried searching the proofs many times with search key words "number of real root" or "exponential function" or "$\sum_{k=0}^{n}\frac{x^{k}}{k!}$". But failed. What is the way to search the related topics?
Note that, $f'(x)+\frac{x^k}{k!}=f(x).$ If $k$ is even and $x_0$ is the smallest root of $f,$ then $f'(x_0)<0.$ So immediately after $x_0$ $f(x)<0$ and the function decreases. Thus, in order for the function $f$ to have its second zero which it must have as a polynomial of even degree (the multiplicity of each root is clearly one) it has to increase. However, if the function $f$ increases at a point then $f(x)>0$ (follows from our identity). This shows that our assumption is wrong. So, $f$ does not have a root. Once even case is established, odd case follows immediately.
Consider the function $f(x)=\sum_{k=0}^{2n}\frac{x^{k}}{k!}$
Note that $f'(x)+\frac{x^{2n}}{(2n)!}=f(x) \large \star$
Suppose there exist a minimum value for the function(at $x=c$) then $f'(c)=0$.
From $\star$ as $2n$ is even $f(c)=\frac{c^{2n}}{(2n)!} \ge 0 \quad\forall c \in \mathbb{R}$ (at $c=0$ $f(0)=1$ and the value of function is positive )
So $ \large {\sum_{k=0}^{2n}\frac{x^{k}}{k!}}>0 \quad \forall x\in \mathbb{R}$
Now consider the case of $f(x)=\sum_{k=0}^{2n+1}\frac{x^{k}}{k!}$
Noting that $f'(x)=\sum_{k=0}^{2n}\frac{x^{k}}{k!} >0$ the function is monotonous over real numbers and as the degree is odd the function has exactly one root.
Let $f_n(x) = \sum_{k=0}^{n} \frac{x^k}{k!}$. Then we know that:
$f_0(x) = 1\\f_1(x)=1+x\\f_2(x)=1+x+\frac{x^2}{2}$
etc.
and for $n>0$ then $f_{n+1}$ is related to $f_n$ in two ways:
(1) $f_{n+1}(x) = f_n(x) + \frac{x^{n+1}}{{n+1}!}$
(2) $f_n(x) = \frac{df_{n+1}(x)}{dx}$
Now we can use induction in two ways.
First, if $f_n(x)$ has no real roots and is positive for all $x$ (as $f_0$ is, for example) then (2) tell us that $f_{n+1}(x)$ is monotonically increasing and has exactly one real root. Suppose this root is at $x=x_0$.
Secondly, if $f_{n+1}(x)$ has one real root at $x=x_0$ then (1) tells us that $f_{n+2}(x_0) = \frac{x^{n+2}}{n+2!}$ because $f_{n+1}(x_0)=0$. So $f_{n+2}(x)$ has a global minimum or a global maximum at $x=x_0$. And if we also know that $f_{n+1}(x)$ is monotonically increasing them this must be a global minimum. And if $n$ is even then $x_0^{n+2} > 0$ so $f_{n+2}(x_0) > 0$ and so $f_{n+2}(x)$ is positive for all $x$.
You can build this into a formal proof by induction to prove that $f_n(x)$ has no real roots for even $n$ and exactly one real root for odd $n$.
Look up "roots of truncated exponential series" in Google.
In particular look at http://demonstrations.wolfram.com/SzegoeCurve/.
