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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.

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    $\begingroup$ hint: derivative of $Q_n(x)$ is $Q_{n-1}(x)$ $\endgroup$ – kingW3 Aug 15 '19 at 23:17
  • $\begingroup$ @kingW3 That's enough! Thank you! $\endgroup$ – Bach Aug 15 '19 at 23:25
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    $\begingroup$ Duplicate of math.stackexchange.com/q/1397428? That question is only about the case $n=6$, but the answer is for all $n$. $\endgroup$ – Martin R Aug 16 '19 at 7:40
  • $\begingroup$ You can state a stronger conjecture: for $x>0$, $0<Q_0< Q_1<\cdots Q_n<e^x$ and for $x<0$, $Q_{2n-1}<e^x$, $Q_{2n}>e^x$. $\endgroup$ – Yves Daoust Aug 16 '19 at 15:29
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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 $Q_n(0)=1\neq0$).

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  • $\begingroup$ 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. $\endgroup$ – NoChance Aug 15 '19 at 23:32
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    $\begingroup$ 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. $\endgroup$ – José Carlos Santos Aug 16 '19 at 7:31
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    $\begingroup$ A non-induction proof is here: math.stackexchange.com/a/1397473. $\endgroup$ – Martin R Aug 16 '19 at 7:38
  • $\begingroup$ Thank you for your answer. $\endgroup$ – NoChance Aug 16 '19 at 8:49
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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.

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  • $\begingroup$ This is not much of a hint, but I will add that the Szegő curve has been previously discussed in answers on this site. $\endgroup$ – J. M. isn't a mathematician Aug 16 '19 at 0:30
  • $\begingroup$ @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 $\endgroup$ – G Cab Aug 16 '19 at 14:08

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