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Find all real solutions of:

$1$ + $x$ + $\dfrac{x^2}{2!}$ + $\dfrac{x^3}{3!}$ +...+ $\dfrac{x^{2n}}{(2n)!}$ = $0$

This is math club question I haven't made much headway on.

JMP
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T. Jones
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    @JohnMa Of course, I had complex roots in mind, it's much easier with real roots :) – Jean-Claude Arbaut Nov 05 '15 at 13:16
  • is this really a duplicate it asks about the exact solutions, not about the existence of them. – tired Nov 05 '15 at 13:17
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    @tired : In the duplicate it shows there is no solution. (for $n$ even) –  Nov 05 '15 at 13:18
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    The distinction seems moot if there are none. @tired – Macavity Nov 05 '15 at 13:18
  • ok i didn't read the answer properly, sry – tired Nov 05 '15 at 13:19
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    For the genereal question about the root location see the nice graphic in https://math.stackexchange.com/a/109605/115115, "Gábor Szegő (1924) and Jean Dieudonné (1935) both showed that the roots of the scaled exponential sum function $e_n(nz)$ approach the portion of the curve $|z·exp(1−z)|=1$ … within the unit disk." – Lutz Lehmann Nov 05 '15 at 13:22
  • Thanks @JohnMa this was what I was looking for; I tried searching for similar questions before I posted, but I didn't find any. – T. Jones Nov 05 '15 at 13:33
  • I think that's it's extremely hard to search for duplicate here. @T.Jones Sometimes I can't even find my own answer..... –  Nov 05 '15 at 13:36
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    Here is something about the complex roots, in french (Centrale-Supelec 2001 MP Math-1). – Jean-Claude Arbaut Nov 05 '15 at 13:41

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