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

[dacruick]

Homework Statement

By definition, e^{z} = \sum\frac{1}{n!}z^{n}

Use this to show the relationship in the question title

The Attempt at a Solution

Well, what I've tried to do is as follows:
e^{z_{1}}e^{z_{2}} = \sum(z_{1}z_{2})^{n}/{n!^{2}}

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

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

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

 

[Poopsilon]

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:

(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

 

[dacruick]

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:

(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

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.

 

[Poopsilon]

Starbucks is far too successful already, I want all of it.

 

[dacruick]

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.