4

Let $Q_n(x)$ be the degree $n$ polynomial $$ 1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots +\frac{x^n}{n!} $$ How many real roots does the equation $Q_n(x)=0$ have?

My attempt:

It is obvious that $Q_n(x)$ will have all its real roots in the negative part of the real line if there is any. Also, we notice that if $n$ is odd, then there is at least one real root by the complex conjugate root theorem. So I conjecture that there is exactly one root for $n$ odd and there is no root for $n$ even.

However, I don't know how to analyze $Q_n(x)$. All I can do is to take derivative and this does not provide more useful information. Any hint is appreciated! Thanks.

José Carlos Santos
  • 427,504
Bach
  • 5,730
  • 2
  • 20
  • 41

2 Answers2

13

Your conjecture is correct and it can be proved by induction.

The statement is trivially true if $n=1$. Assume that it is true for a certain $n$. If $n$ is even, then $(\forall x\in\mathbb R):Q_n(x)>0$. So, $Q_{n+1}$ is strictly increasing (note that $Q_{n+1}'(x)=Q_n(x)$) and therefore has at most one real root. But every polynomial whose degree is odd has at least one root. So, it has exactly one root.

And if $n$ is odd, then $Q_{n+1}$ first decreases and then increases. So, it has an absolute minimum, which is attained at the point $x_0$ such that $Q_n(x_0)=0$. But\begin{align}Q_{n+1}(x_0)&=Q_n(x_0)+\frac{{x_0}^{n+1}}{(n+1)!}\\&=\frac{{x_0}^{n+1}}{(n+1)!}\\&>0,\end{align}since $n+1$ is even and $x_0\neq0$ (since $Q_n(0)=1\neq0$).

Gary
  • 31,845
José Carlos Santos
  • 427,504
  • I learned from you great answer that $Q_n(x)$ has exactly 1 root if n is even. I am not sure I get how many roots if n is odd? Thanks. – NoChance Aug 15 '19 at 23:32
  • 1
    As I wrote in my answer, if $n$ is odd then $Q_n$ is strictly increasing and therefore it cannot have more than one root. But a polynomial with even degree always has at least one root. So, $Q_n$ has exactly one root. – José Carlos Santos Aug 16 '19 at 07:31
  • 1
    A non-induction proof is here: https://math.stackexchange.com/a/1397473. – Martin R Aug 16 '19 at 07:38
  • Thank you for your answer. – NoChance Aug 16 '19 at 08:49
0

Hint:

if you multiply your polynomial by $\Gamma(n+1) e^{-x}$, which does not have any zeros, you get the Incomplete Upper Gamma function $\Gamma(n+1,x)$.

Then refer to this related post.

G Cab
  • 35,272
  • This is not much of a hint, but I will add that the Szegő curve has been previously discussed in answers on this site. – J. M. ain't a mathematician Aug 16 '19 at 00:30
  • @J.M.isapoormathematician: thanks for signalling : I was actually mumbling where the entireness of $\Gamma(n+1,x)$ goes lost. Can you please detail some links to Szego curve? what is it ? thanks – G Cab Aug 16 '19 at 14:08