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I've read the argument in Rudin, but I think I need a little clarification

\begin{align} e^a e^b &= \sum_{k=0}^{\infty} \frac{a^k}{k!} \sum_{m=0}^{\infty} \frac{b^m}{m!}\\ &= \sum_{n=0}^{\infty} \frac{ n!}{n!} \sum_{k=0}^{n} \frac{a^k}{k!} \frac{b^{n-k}}{(n-k)!} \\ &= \sum_{n=0}^{\infty} \frac{(a+b)^n}{n!}\\ &=e^{a+b}\\ \end{align}

I'd like to understand how we get from the first line to the second. Of course $m =n-k$, but how do we ensure that we haven't missed any addends in the limit changing process?

Changing limits: $m=0 \rightarrow n-k=0 \rightarrow k=n$ and $m=\infty \rightarrow n-k=\infty \rightarrow k=0$, assuming $n \to \infty$.

\begin{align} \frac{b^m}{m!} \sum_{k=0}^{\infty} \frac{a^k}{k!} &= \sum_{k=0}^{\infty} \frac{a^k}{k!} \frac{b^{m}}{m!} \\ &=\sum_{k=0}^{?} \frac{a^k}{k!} \frac{b^{n-k}}{(n-k)!} \\ \end{align}

how to justify that "?"$=n$ ?

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This is an application of the Cauchy product. – Javier May 25 '13 at 21:39
up vote 2 down vote accepted

Suppose we have two series

$$p(x)=\sum_{k=1}^\infty a_kx^k$$

$$q(x)=\sum_{k=1}^\infty b_kx^k$$

Then multiplying and collecting terms gives $$(p\cdot q)(x)=a_0b_0+(a_1b_0+b_1a_0)x+(a_2b_0+a_1b_1+a_0b_2)x^2+\dots \\=\sum_{k=0}^\infty \left(\sum_{j+i=k}a_jb_i\right)x^k\\=\sum_{k=0}^\infty \left(\sum_{j=0}^k a_jb_{k-j}\right)x^k$$

This informal approach actually works whenever either one of the series converges and the other converges absolutely. Try to see what happens when we choose the series to be those of $e^a,e^b$.

ADD The case when the series are not power series, but just plain series gives $$\sum\limits_{k = 0}^\infty {\left( {\sum\limits_{j = 0}^k {{a_j}} {b_{k - j}}} \right)} = \sum\limits_{k = 0}^\infty {{a_k}} \cdot \sum\limits_{k = 0}^\infty {{b_k}} $$

This gives when $$\eqalign{ & {a_k} = \frac{{{a^k}}}{{k!}} \cr & {b_k} = \frac{{{b^k}}}{{k!}} \cr} $$ that

$$\displaylines{ {e^a}{e^b} = \sum\limits_{k = 0}^\infty {\left( {\sum\limits_{j = 0}^k {\frac{{{a^j}}}{{j!}}} \frac{{{b^{k - j}}}}{{\left( {k - j} \right)!}}} \right)} \cr = \sum\limits_{k = 0}^\infty {\frac{1}{{k!}}\left( {\sum\limits_{j = 0}^k {\frac{{k!}}{{j!\left( {k - j} \right)!}}{a^j}} {b^{k - j}}} \right)} \cr = \sum\limits_{k = 0}^\infty {\frac{1}{{k!}}{{\left( {a + b} \right)}^k}} \cr = {e^{a + b}} \cr} $$

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Yeah, you're not thinking of this right. Think about taking the first quadrant in $mk$-space and filling it up with (anti-)diagonal lines $n=k+m=\text{constant}$. As you let $n$ vary from $1$ to $\infty$, you hit every $(m,k)$.

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