1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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!

  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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook