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Exponential Power Series Expansion

  1. Sep 16, 2011 #1
    I want to show that [itex] e^x e^x = e^{2x} [/itex] using a power series expansion. So I start with

    [tex] \sum_{n=0}^\infty \frac{x^n}{n!} \sum_{m=0}^\infty \frac{x^m}{m!} [/tex]
    [tex] \sum_{n=0}^\infty \sum_{m=0}^\infty \frac{x^n}{n!} \frac{x^m}{m!} [/tex]
    [tex] \sum_{n=0}^\infty \sum_{m=0}^\infty \frac{x^{m+n}}{m!n!} [/tex]

    I am at a loss of where to go from here. I want to reduce the last expression down to [itex] \sum_{n=0}^\infty \frac{(2x)^n}{n!} [/itex] but I am not sure of how to get rid of one of the summations.
     
  2. jcsd
  3. Sep 16, 2011 #2
    I think I got it...in case anyone was interested.

    I can use a substitution on the indicies of the double series as follows. Let [itex] m=q [/itex] and [itex] n=p-q [/itex], with the condition that [itex] q \le p [/itex]. This gives

    [tex] \sum_{p=0}^\infty \sum_{q=0}^p \frac{x^q}{q!}\frac{x^{p-q}}{(p-q)!} [/tex]
    Which one recognizes as a binomial expansion. Where [itex] \frac{(x+x)^n}{n!} = \sum_{k=0}^n \frac{1}{k!(n-k)!} x^k x^{n-k} [/itex]. Thus our double series reduces down to

    [tex] \sum_{p=0}^\infty \frac{(x+x)^p}{p!} = e^{2x} [/tex]
     
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