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Probably a simple question, but I wonder about the following:

To prove that $\exp(z_1+z_2) = \exp(z_1)\exp(z_2)$, I use : $$\exp(z_1+z_2) = \sum_{n=0}^{\infty}\sum_{k=0}^n\frac{1}{k!(n-k)!}z_1^kz_2^{n-k} $$ by using the binomial expansion. Now the property has been proved if this sum equals: $$\sum_{k=0}^{\infty}\sum_{m=0}^{\infty}\frac{z_1^{k}z_2^{m}}{k!m!}$$

Strange enough, I don't see exactly why this is true (although they use this without explanation in many books). I see that both sums contain "all terms" formally, but I would be glad if someone could show rigorously that both sums converge to the same complex number. Probably it is just a property of series that I'm missing here.


marked as duplicate by Martin Sleziak, J. W. Perry, user127.0.0.1, Claude Leibovici, Stefan Hansen Feb 20 '14 at 8:43

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What you want to use is the Cauchy product or convolution.

Note that $$\sum_{k\geqslant 0} a_k\cdot \sum_{k\geqslant 0} b_k=a_0b_0+(a_0b_1+a_1b_0)+(a_0b_2+a_1b_1+a_2b_0)+\cdots$$

and this is true whenever either series converges absolutely and the other converges. In your case, both powerseries converge absolutely, so you're more than safe.

This can be written as

$$\sum_{k\geqslant 0} a_k\cdot \sum_{k\geqslant 0} b_k=\sum_{k\geqslant 0}\left(\sum_{j+i=k}a_ib_j\right)$$

But we can also write this as

$$\sum_{k\geqslant 0} a_k\cdot \sum_{k\geqslant 0} b_k=\sum_{k\geqslant 0}\sum_{n=0}^k a_n b_{k-n}$$

Now see what happens when $$a_k=\frac {z_1^k}{k!}$$ $$b_k=\frac{z_2^k}{k!}$$

  • $\begingroup$ Thank you! exactly what I was looking for. $\endgroup$ – yarnamc Jun 15 '13 at 20:01

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