$P_n(x):=1+ \sum_{m=1}^n\dfrac{x^m}{m!}$ has no real root for even $n$ and exactly one real root for odd $n$

Is it true that $$P_n(x):=1+ \sum_{m=1}^n\dfrac{x^m}{m!}$$ has no real root for even $n$ and exactly one real root for odd $n$? I can only prove that the polynomial cannot have any multiple roots. Am stuck , Please Help.

Solution 1:

Let $S(n)$ be the statement that

$P_n(x)$ has no real root for $n$ even, and has exactly one real roots for $n$ odd.

You can check directly that $S(1)$ and $S(2)$ are true.

Assume that $S(k)$ is true. Consider $k+1$-case:

• if $k$ is even, the induction hypothesis says that $P_k(x)$ has no real roots. As $P_k(0) = 1$, we have $P_k(x) >0$ for all $x$. As $P_{k+1}'(x) = P_k(x)$, then $P_{k+1}(x)$ is strictly increasing and so $P_{k+1}(x)$ has only one real root (Need to check the asymptotics here).

• if $k$ is odd, then there is $y\in \mathbb R$ so that $P_k(x) <0$ , $\forall x<y$, $P_k(y) = 0$ and $P_k(z) >0$ , $\forall z>y$. Then as $P_{k+1}'(x)= P_k(x)$, then $P_{k+1}(x)$ has a strict minimum at $y$. Note that $y\neq 0$. As

$$P_{k+1} (y) = P_k(y) + \dfrac{y^{k+1}}{(k+1)!} = \dfrac{y^{k+1}}{(k+1)!}>0$$

as $y\neq 0$ and $k+1$ is even. Thus $P_{k+1}$ is always positive and has no real root.

Thus $S(k+1)$ is true. By induction, the statement is true for all $n$.