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I was reviewing some things I had learned on $e$ (or actually $\exp(x)$, which for $x \in \mathbb{R}$ makes sense to write as $e^x$) a few years back, and decided to try and show that $\exp(x+y) = \exp(x)\times\exp(y)$. I got really far, but don't know what to do now. This is what I have:

$$\exp(x)=\sum_{n=0}^{\infty}{\frac{x^n}{n!}}$$ $$\exp(x)\times\exp(y)=\left(\sum_{n=0}^{\infty}{\frac{x^n}{n!}}\right)\left(\sum_{m=0}^{\infty}{\frac{y^m}{m!}}\right)=\sum_{n=0}^{\infty}\sum_{m=0}^{\infty}{\frac{x^ny^m}{n!m!}}$$ $$\exp(x+y) = \sum_{n=0}^{\infty} {\frac{(x+y)^n}{n!m!}} = \sum_{n=0}^{\infty} \frac{1}{n!} \sum_{k=0}^n \binom{n}{k}x^ky^{n-k} = \sum_{n=0}^{\infty} \frac{1}{n!} \sum_{k=0}^n \frac{n!}{k!(n-k)!} \times x^ky^{n-k} = \sum_{n=0}^{\infty} \sum_{k=0}^n \frac{x^ky^{n-k}}{k!(n-k)!}$$

And that was about it. Now I'm left with showing that: $$\sum_{n=0}^{\infty}\sum_{m=0}^{\infty}{\frac{x^ny^m}{n!m!}} = \sum_{n=0}^{\infty} \sum_{k=0}^n \frac{x^ky^{n-k}}{k!(n-k)!}$$ And I don't know how to do that. I thought that doing establishing something like $j=n-k$ and rewriting would help give me some insight, but it didn't... I don't know how to handle that relationship between an infinite sum of infinite sums and an infinite sum of a simple sum, even if they both look remarkably similar. I also know that what I wrote this far is true by plugging that last sentence into WolframAlpha (unless I transcribed it into this post incorrectly).

My study of this kind of math is pretty autonomous via research, so I apologize if there was something that could have been phrased better. I'm also not really sure what tags to put in this.

EDIT: A few comments proposed posts that shared the same problem, but they both had this jump:

$$ \sum_{n=0}^{\infty}{\frac{x^n}{n!}} \sum_{m=0}^{\infty}{\frac{y^m}{m!}} = \sum_{n=0}^{\infty} \sum_{k=0}^n \frac{x^ky^{n-k}}{k!(n-k)!} $$

And didn't really explain it. I don't know why this is true... In fact, if this I understand this, then the rest of the way is self-explanatory.

The other post did something similar, but used some notation I'm not used to. I think it translates to:

$$ \sum_{n=0}^{\infty} \frac{n!}{n!} \sum_{k=0}^n \frac{x^ky^j}{k!j!} = \sum_{k=0}^{\infty} \sum_{j=0}^{\infty} \frac{x^ky^j}{k!j!} $$

And here, there are two problems: one, I'm not sure that's what's written... And if it is, then again, I don't know why that would be true.

EDIT 2: Following the suggestions in the comments, I read up on Cauchy's product and finally see why all of this happens. It is easier with the first example, I thought:

$$ \sum_{n=0}^{\infty}{\frac{x^n}{n!}} \sum_{m=0}^{\infty}{\frac{y^m}{n!}} = \sum_{n=0}^{\infty} \sum_{k=0}^n \frac{x^ky^{n-k}}{k!(n-k)!} $$

This is very much the standard form, if you consider $a_n=\frac{x^n}{n!}$ and $b_m=\frac{y^m}{m!}$:

$$ \sum_{n=0}^{\infty}a_n \sum_{m=0}^{\infty}b_m = \sum_{o=0}^{\infty}c_k = \sum_{o=0}^{\infty} \sum_{k=0}^{o} a_kb_{o-k} = \sum_{o=0}^{\infty} \sum_{k=0}^{o} \frac{x^k}{k!}\times\frac{y^{o-k}}{(o-k)!} $$

The letters are mixed up, but that's because I should have done this a bit better. The point is that I understand it know, because of the Cauchy product. Thank you all!

Érico Patto
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  • Does this help? – Joe Mar 07 '21 at 13:26
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    You should probably use the Cauchy product formula: https://en.m.wikipedia.org/wiki/Cauchy_product – Vercassivelaunos Mar 07 '21 at 13:26
  • Ok: Martín, I had actually already seen that, but got a little confused. The part that confuses me also shows up in Joe's, which I'll edit into the post. Vercassivelaunos, maybe that helps, but I'll have to spend a nice amount of time to understand it... – Érico Patto Mar 07 '21 at 13:31
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    Both series are absolutely convergent so their product is also a convergent series. That jump you mention is due to Cauchy product. – Koro Mar 07 '21 at 13:43
  • Ok, I'll take a look at it more calmly then. – Érico Patto Mar 07 '21 at 13:44
  • I am very confused by the query, and the comments that it has received. For $a,b,c \in \mathbb{R}$, with $0 < a$, the definition of $a^{(b+c)}$ is $(a^b) \times (a^c)$. At least, this is how I was taught. This means, that unless this definition is to be thrown out, the fact that $a = e = 2.718...$ is irrelevant. – user2661923 Mar 07 '21 at 13:45
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    Yes, you are absolutely correct. However, the way to prove that $\exp(x)=e^x$ is to show that it has this property. Or at least, it's a step in the right direction. – Érico Patto Mar 07 '21 at 13:47
  • @ÉricoPatto Good point, thanks for the clarification. I went back and checked my (Apostol) Calculus book, and in fact exp$(x)$ is not defined as $e^x.$ – user2661923 Mar 07 '21 at 13:53

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Theorem: Given two convergent series $X=\sum x_n$ and $Y=\sum y_n$, you define $\sum z_n$ with partial sums $Z_n=\sum_{r=0}^n x_ry_{n-r}$=coefficient of $t^n$ in $(\sum_{r=0} x_rt^r)(\sum_{s=0} y_s t^s)$, then it can be proven that if $\lim Z_n=\sum z_n$ exists (let it be equal to $Z$), then $Z=XY$. (Abel's Theorem).

In your case, $e^x=\sum\frac{x^m}{m!}, e^y=\sum\frac{y^n}{n!}$, both are absolutely convergent series that is their Cauchy product will converge. (Mertens theorem)

Now define Cauchy product, $Z_n=\sum_{r=0}^n \frac{x^ry^{n-r}}{r!(n-r)!}$

This product will converge by Mertens theorem. And by Abel's theorem, $\lim Z_n= \sum_{n=0}^\infty\sum_{r=0}^n \frac{x^ry^{n-r}}{r!(n-r)!}= (e^x)(e^y)$

$\begin{align} \implies (e^x)(e^y)&=\sum_{n=0}^\infty\sum_{r=0}^n \frac{n! x^ry^{n-r}}{r!(n-r)! n!}\\ &= \sum_{n=0}^\infty\sum_{r=0}^n \frac{^nC_r x^ry^{n-r}}{n!}\\ &=\sum_{n=0}^\infty \frac{(x+y)^n}{n!}=e^{x+y}\end{align}$.

Note that:

  1. $(x+y)^n=\sum_{r=0} ^n$ $^nC_r x^ry^{n-r}$ by Binomial Theorem.
  2. Wherever no superscript is added to $\sum$ in my answer above, please treat the superscript as $\infty$ and in case of missing subscript to $\sum$, please treat the same as index variable $=0$.
Koro
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