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OK, so you know the geometric series, right? It goes like this:

[tex]\sum_{k=0}^{\infty} z^k = \frac{1}{1-z},\forall z\in\mathbb{C}\mbox{ such that }\left| z\right|<1[/tex]

How about this one? Call it, say, the geometric product:

[tex]\prod_{k=0}^{\infty} \left( 1+ z^{2^{k}}\right) = \frac{1}{1-z},\forall z\in\mathbb{C}\mbox{ such that }\left| z\right|<1[/tex]

Prove it in two ways.

Suppose a function [tex]f(z)[/tex] is equal to its own logarithmic derivative, namely, suppose that

[tex] \exists f(z)\mbox{ such that } \frac{d}{dz}\ln\left( f(z)\right) = \frac{f^{\prime}(z)}{f(z)} = f(z) [/tex]

Prove that [itex]f(z)=\frac{1}{C-z}[/itex] where C is a constant, is a family of solutions to this differential equation.

That being known, and taking C=1 above, we have (hand-waving the convergence details):

[tex] \frac{d}{dz}\ln\left[ \prod_{k=0}^{\infty} \left( 1+ z^{2^{k}}\right) \right] = \frac{1}{1-z},\forall z\in\mathbb{C}\mbox{ such that }\left| z\right|<1[/tex]

We have derived what new series by virtue of the fact that the log of a product is the sum of the logs?

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# A Geometric Product (a series of exercises for the curious)

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