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Showing that e^a * e^b = e^(a+b)

  1. Jan 12, 2012 #1
    1. The problem statement, all variables and given/known data
    By definition, [itex]e^{z}[/itex] = [itex]\sum[/itex][itex]\frac{1}{n!}[/itex][itex]z^{n}[/itex]

    Use this to show the relationship in the question title


    3. The attempt at a solution
    Well, what I've tried to do is as follows:
    [itex]e^{z_{1}}[/itex][itex]e^{z_{2}}[/itex] = [itex]\sum[/itex][itex](z_{1}z_{2})^{n}[/itex]/[itex]{n!^{2}}[/itex]

    And set that equal to
    [itex]e^{(z_{1}+z_{2})}[/itex] = [itex]\sum[/itex][itex]\frac{1}{n!}[/itex][itex](z_{1}+z_{2})^{n}[/itex]

    What I'm left with is this expression that
    [itex](z_{1}+z_{2})^{n}[/itex] = [itex](z_{1}z_{2})^{n}/n![/itex]

    those are all sums from n=0 to n=∞ still btw. Sorry for my LameTex inexperience.

    My question is, is there any way that I can show that those two are equal? Or have I gone about this all wrong. Thanks!

    dacruick
     
  2. jcsd
  3. Jan 12, 2012 #2
    You can't multiply infinite series like you've done. In fact you can't even multiply finite polynomials like that. For example consider the multiplication of the following quadratics:

    [tex](x^2 + x + 1)(y^2 + y + 1) = (xy)^2 + x^2y + xy^2 + x^2 + y^2 + x + y + 1 \neq (xy)^2 + xy + 1[/tex]
     
  4. Jan 12, 2012 #3
    I'm so stupid...Haha, not just because of the foolishness you stated but because as soon as you said that I'm pretty sure I figured out how to do it. Maybe it had something to do with that cup of coffee I just made. You'll have to tussle with Starbucks for partial credit of my pending success.
     
  5. Jan 12, 2012 #4
    Starbucks is far too successful already, I want all of it.
     
  6. Jan 12, 2012 #5
    Haha get your lawyer ready then :P.

    So I have the answer, I've expanded both of the series for n = 0, 1, 2, and 3 and I can sufficiently see that the terms will cancel out on both sides.

    However, I question the completeness of my answer. I assume that the only way to get marks for this badboy is to expand both of the series' in terms of 'n'. I don't readily see how to do this. Would it work if I just multiplied each of the first 4 values of 'n' by the last 2 terms of each series, and then see if those cancelled?

    I have next to no pure mathematical background, and I'm unclear as to what constitutes "showing" that these two series are equal.
     
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