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Power series help

  • Thread starter mkienbau
  • Start date
  • #1
12
0
How do I multiply power series?

Homework Statement


Find the power series:
[tex] e^x arctan(x) [/tex]

Homework Equations



[tex] e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} [/tex]

[tex] arctan(x) = 0 + x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7}[/tex]

The Attempt at a Solution



So do I multiply 1 by 0, x by x and so forth? Or do I go 1 by 0, 1 by x? Or is there another way?
 
Last edited:

Answers and Replies

  • #2
You have to multiply 1 by the whole acrtan series, x by the whole arctan series, and so on. There might be a way to simplify it though. Wikipedia has this under "power series"

[tex]f(x)g(x) = \left(\sum_{n=0}^\infty a_n (x-c)^n\right)\left(\sum_{n=0}^\infty b_n (x-c)^n\right)[/tex]

[tex] = \sum_{i=0}^\infty \sum_{j=0}^\infty a_i b_j (x-c)^{i+j}[/tex]

[tex] = \sum_{n=0}^\infty \left(\sum_{i=0}^n a_i b_{n-i}\right) (x-c)^n [/tex]
 
  • #3
12
0
So I kind of treat it like F.O.I.L.?
 
  • #4
Tom Mattson
Staff Emeritus
Science Advisor
Gold Member
5,500
7
Sort of. FOIL is the distrubutive law for (binomial)X(binomial). Here, you've got two infinitely long "polynomials". Obviously, you won't be able to write out all of the terms. :tongue:
 
  • #5
12
0
Awesome, I think I got it, I only had to take it out to the [tex] x^5 [/tex] term.
 

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