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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.

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  • $\begingroup$ Note that $P_n'(x)= P_{n-1}(x)$. I guess you can do by induction. $\endgroup$ – user99914 Nov 22 '14 at 11:53
  • $\begingroup$ @John: Induction on what ? $n$ ? I am not sure ... $\endgroup$ – user123733 Nov 22 '14 at 11:57
  • $\begingroup$ Its obvious that there are no positive real roots. This is an $n$-degree polynomial with positive coefficients. By Descartes Rule of Signs there are no positive real roots. This is true regardless of the value of $n$. $\endgroup$ – CogitoErgoCogitoSum Nov 26 '14 at 6:47
  • $\begingroup$ I dont know if its helpful but $P_n$ is also the $n$th partial sum for the Taylor series, centered at $x=0$, for the exponential function $e^x$. Just pointing that out if it helps. The exponential function is positive everywhere and thus never crosses the x-axis. $\endgroup$ – CogitoErgoCogitoSum Nov 26 '14 at 6:49
  • $\begingroup$ @CogitoErgoCogitoSum: You don't have to use Descarte's rule of signs , just simply note that $P_n(x)>o$ whenever $x>0$ $\endgroup$ – user123733 Nov 26 '14 at 14:24
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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 $n$.

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  • $\begingroup$ form $P_k(y)=0$ how do you conclude that $P_k(x)<0$ for $x<y$ ? It might happen that $P_k$ is +ve everywhere else , looking like an U curve passing through the origin . $\endgroup$ – user123733 Nov 22 '14 at 12:35
  • $\begingroup$ I use implicitly the fact that $P_k$ tends to $-\infty$ as $x \to -\infty$ and $P_k(x) \to\infty $ as $x \to \infty$. @user123733 $\endgroup$ – user99914 Nov 22 '14 at 12:37
  • $\begingroup$ The inequalities follow only from that ? $\endgroup$ – user123733 Nov 22 '14 at 12:44
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    $\begingroup$ Yes. If there is $x<y$ so that $P_k(x)>0$, then as $P_k(x')<0$ for some $x'<x$, there is $x''$ between $x'$ and $x$ so that $P_k(x'')=0$ (IVT). @user123733 $\endgroup$ – user99914 Nov 22 '14 at 12:46
  • $\begingroup$ @user99914 What did you mean by "need to check the asymptotics"? $\endgroup$ – user439126 Feb 7 at 16:29

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