Root of a Power Series

  • #1
I have to find the "second smallest root" of the following equation :

[itex]1-x+(x^2)/(2!)^2-(x^3)/(3!)^2+(x^4)/(4!)^2+...=0[/itex]

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
 

Answers and Replies

  • #2
HallsofIvy
Science Advisor
Homework Helper
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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.
 
  • #3
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 [itex]e^x[/itex] is [itex]1+x+(x^2)/(2!)+(x^3)/(3!)+...[/itex]

The problem with this series is the square of the nth factorial is involved.
 
  • #4
D H
Staff Emeritus
Science Advisor
Insights Author
15,393
685
Why would matlab not give a correct answer.
Two reasons: (1) you're not feeding it the right problem, and (2) the companion matrix is quite ill-formed for a truncated series of order 6 or more.

Try graphing the series. You will need to be a bit creative in how you perform the summation if you want to see behavior for anything but small values of x.
 

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