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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!

  • 0
    Nice question.+12012-12-30

4 Answers 4

3

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 $(2)$.

3

$ (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}

  • 2
    Nice demonstration without words. (+1)2012-12-12
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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 $n$.

  • 0
    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),..$2012-12-11
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Given the series can be written

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

  • 0
    Insig$h$t, rather than a proof, is what is sought.2012-12-11