# Root of a Power Series

I have to find the "second smallest root" of the following equation :

$1-x+(x^2)/(2!)^2-(x^3)/(3!)^2+(x^4)/(4!)^2+...=0$

Matlab returns quite a satisfactory answer. >> p=[1/518400 -1/14400 1/1576 -1/36 1/4 -1 1]

p =

0.0000 -0.0001 0.0006 -0.0278 0.2500 -1.0000 1.0000

>> roots(p)

ans =

35.5690
-4.6796 +18.5352i
-4.6796 -18.5352i
4.1776 + 3.2154i
4.1776 - 3.2154i
1.4350

But I have been asked to identify this series as well, which I am unable to do. Can anybody help me identify this series as a function or a product of functions? Thanking anybody who answers before hand

HallsofIvy
Homework Helper
But Matlab does NOT give a solution to that equation! What you have done is enter the first seven terms of the series as if it were a 6th degree polynomial. If I am understanding your question correctly, that is an infinite power sum. In fact, it looks to me like the power series for e-x and that is never 0.

Why would matlab not give a correct answer. From what I can infer, as the nth terms rises the factorial rises dramatically, therefore the fraction becomes very minuscule in value. Ten terms would be sufficient to give an answer correct to the third or fourth decimal place.

The power series of $e^x$ is $1+x+(x^2)/(2!)+(x^3)/(3!)+...$

The problem with this series is the square of the nth factorial is involved.

D H
Staff Emeritus