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I've been going through some series notes from my lecture and got stuck at this equality: $$\left(\sum_{j=0}^\infty\frac{z^j}{j!}\right)\left(\sum_{k=0}^\infty\frac{w^k}{k!}\right)=\sum_{n=0}^\infty\sum_{j=0}^n\frac{z^jw^{n-j}}{j!(n-j)!}$$ Where $z,w\in\mathbb C$.

Rather than a proof, I'm looking more for a way of understanding this equality, so next time I see something similar I go "oh, right, I know that". Right now I only have vague ideas of why that would be true.

Thanks a bunch for any help!

jimjim
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Dahn
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    Let $a_j={z^j\over j!}$ and $b_j={w^k\over k!}$. If $M$ is the infinite matrix whose $i$'th row is $[ a_0 b_i\ a_1 b_i\ a_2 b_i\ \cdots]$, the the expression on the left represents the sum of the elements of $M$. So does the expression on the right: the inner sum is the sum of the elements of the $n$'th diagonal of $M$ (the "increasing diagonals"). – David Mitra Dec 11 '12 at 22:06
  • Thanks David, together with Robert Israel's answer this makes perfect sense to me now. – Dahn Dec 11 '12 at 22:12
  • You're welcome. And, sorry for the typos in my previous comment (I should have had $b_k$, not $b_j$, in particular...). – David Mitra Dec 11 '12 at 22:15
  • Nice question.+1 – Mikasa Dec 30 '12 at 09:50

5 Answers5

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This is an example of the change of order of summation (regrouping terms). Formally (without concern for convergence), the product $$ \begin{align} \left(\sum_{i=0}^\infty a_i\right)\left(\sum_{j=0}^\infty b_j\right) &=\sum_{i=0}^\infty\sum_{j=0}^\infty a_ib_j\tag{1}\\ &=\sum_{k=0}^\infty\sum_{j=0}^k a_{k-j}b_j\tag{2} \end{align} $$ $(1)$ is the distributing multiplication over addition.

$(2)$ is a change of variables: $i+j=k$

Each product in $(1)$ appears once and once only in $(2)$.

robjohn
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$$ (1+2+3+4+\cdots)\cdot\left(\begin{array} {} & \text{one} \\[6pt] + & \text{two} \\[6pt] + & \text{three} \\[6pt] + & \text{four} \\[6pt] + & \cdots \end{array}\right) $$ $$ = \sum \left[ \begin{array}{cccc} 1\cdot\text{one}, & 2\cdot\text{one}, & 3\cdot\text{one}, & 4\cdot\text{one}, & \cdots\\[6pt] 1\cdot\text{two}, & 2\cdot\text{two}, & 3\cdot\text{two}, & 4\cdot\text{two}, & \cdots\\[6pt] 1\cdot\text{three}, & 2\cdot\text{three}, & 3\cdot\text{three}, & 4\cdot\text{three}, & \cdots\\[6pt] 1\cdot\text{four}, & 2\cdot\text{four}, & 3\cdot\text{four}, & 4\cdot\text{four}, & \cdots\\ \vdots & \vdots & \vdots & \vdots & \ddots \end{array} \right] $$ \begin{align} & = \cdots\cdots\cdots +\sum\left[ \begin{array}{cccc} \cdot & \cdot & 3\cdot\text{one}, & \cdot & \cdots\\[6pt] \cdot & 2\cdot\text{two}, & \cdot & \cdot & \cdots\\[6pt] 1\cdot\text{three}, & \cdot & \cdot & \cdot & \cdots\\[6pt] \cdot & \cdot & \cdot & \cdot & \cdots\\ \vdots & \vdots & \vdots & \vdots & \ddots \end{array} \right] \\[18pt] & {}\qquad\qquad\qquad{}+ \sum\left[ \begin{array}{cccc} \cdot & \cdot & \cdot & 4\cdot\text{one}, & \cdots\\[6pt] \cdot & \cdot & 3\cdot\text{two}, & \cdot & \cdots\\[6pt] \cdot & 2\cdot\text{three}, & \cdot & \cdot & \cdots\\[6pt] 1\cdot\text{four}, & \cdot & \cdot & \cdot & \cdots\\ \vdots & \vdots & \vdots & \vdots & \ddots \end{array} \right] + \cdots\cdots\cdots \end{align}

Michael Hardy
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Multiplying the $j$ term on the left with the $k$ term on the right gives you $\dfrac{z^j w^k}{j! k!}$. Now if $j+k=n$, this is $\dfrac{z^j w^{n-j}}{j! (n-j)!}$. Since $j$ and $k$ can be any nonnegative integers, the same is true for $n$. Given $n$, $j$ can be any integer from $0$ to $n$.

Robert Israel
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  • Thanks, I think I've got it. But let's check, is this true: So in the expression on the left, $(j,k)$ is going (a very improvised notation, hopefully it'll make sense) $(0,0),(0,1),(0,2),..,(1,0),(1,1),...$ whereas on the right we have $(0,0),(0,1),(1,0),(0,2),(1,1),(2,0),..$ – Dahn Dec 11 '12 at 22:04
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Definition: Given two series $ \sum_{n=0}^\infty a_n$ and $\sum_{n=0}^\infty b_n$, define $$c_n=\sum_{k=0}^n a_{n-k}b_k.$$ The series $\sum_{n=0}^\infty c_n$ is called Cauchy product of $\sum a_n$ and $\sum b_n$.

The Cauchy product is kinda discrete convolution. Check the following Theorem:

Mertens' theorem: If the series $\sum a_n$ and $\sum b_n$ converge to $a$ and $b$, respectively, and, at least one of them converges absolutely, then the Cauchy product of these two series converges to $ab$.

In our case, by MacLaurin, $\sum_{n=0}^\infty \frac{t^n}{n!}$ converges absolutely, for $t\in\left\{z,w\right\}$ thus $$\left(\sum_{n=0}^\infty \frac{z^n}{n!}\right)\left(\sum_{n=0}^\infty \frac{w^n}{n!}\right)=\sum_{n=0}^\infty c_n,$$ for $$c_n=\sum_{k=0}^n \frac{z^{n-k}}{(n-k)!}\frac{w^{k}}{k!}=\frac{1}{n!}\sum_{k=0}^n \frac{n!}{(n-k)!k!}z^{n-k}w^{k}=\frac{(z+w)^n}{n!}.$$ But then $$e^ze^w=\left(\sum_{n=0}^\infty \frac{z^n}{n!}\right)\left(\sum_{n=0}^\infty \frac{w^n}{n!}\right)=\sum_{n=0}^\infty \frac{(z+w)^n}{n!}=e^{z+w},$$ as desired.

For more details check:

  1. Link: https://en.wikipedia.org/wiki/Cauchy_product#Convergence_and_Mertens'_theorem
  2. Book: Tom M. Apostol, Mathematical Analysis, Second Edition, page 204, Definition 8.45 and Theorem 8.46.
Oleg Turcan
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Given the series can be written

$=e^z e^w =e^{z+w}=\sum_{n=0}^\infty =\sum \sum....$

rschwieb
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Chung. J
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