# Real roots of $1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots +\frac{x^n}{n!}$ [duplicate]

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.

• hint: derivative of $Q_n(x)$ is $Q_{n-1}(x)$ – kingW3 Aug 15 '19 at 23:17
• @kingW3 That's enough! Thank you! – Bach Aug 15 '19 at 23:25
• Duplicate of math.stackexchange.com/q/1397428? That question is only about the case $n=6$, but the answer is for all $n$. – Martin R Aug 16 '19 at 7:40
• You can state a stronger conjecture: for $x>0$, $0<Q_0< Q_1<\cdots Q_n<e^x$ and for $x<0$, $Q_{2n-1}<e^x$, $Q_{2n}>e^x$. – Yves Daoust Aug 16 '19 at 15:29

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 at has one real root at must. 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 $$Q_{n+1}(x_0)=Q_n(x_0)+\frac{{x_0}^{n+1}}{(n+1)!}=\frac{{x_0}^{n+1}}{(n+1)!}>0$$, since $$n+1$$ is even and $$x_0\neq0$$ (since $$Q_n(0)=1\neq0$$).

• 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
• 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 7:31
• A non-induction proof is here: math.stackexchange.com/a/1397473. – Martin R Aug 16 '19 at 7:38
• Thank you for your answer. – NoChance Aug 16 '19 at 8:49

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.

• 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. isn't a mathematician Aug 16 '19 at 0:30
• @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