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I have come across a generating function that is similar to another generating function.

First, some preliminaries. I call a closed form a function like $\frac{1}{1-x}$. In other words, a closed form does not involve a summation or integral, but rather a bunch of arithmetic that describes a generating function.

Now, I have come across a closed form of the generating function:

$$\sum_{k=0}^\infty{X_k y^k} = X_0 + X_1 y^1 + X_2 y^2 + \dots$$

where the $X_k$'s are themselves generating functions:

$$X_k = \frac{1}{1-(z_k)} = 1 + (z_k) + (z_k)^2 + \dots$$

I'd like to convert the $X_k$'s into $z_k$'s. In other words, each coefficient of the original series is $\frac{1}{1-(z_k)}$. I'd like to convert from that to just $z_k$. I'm wondering if there is some sort of transformation that can do this all at once.

I would greatly appreciate any help on this.

IMPORTANT EXPLANATION

I know how to get the $z$ values in the formula. What I really want is to change the formula into something that is not equal to the original formula. In other words, say I have:

$$y\frac{1}{1-z_1}+y^2\frac{1}{1-z_2}$$

Then I want:

$$y z_1 + y^2 z_2$$

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Why not just substitute the formula you have for $X_k$ into the sum? You will get something like $$\sum_{k, l} z_k^l y^k$$ This is fairly messy, but I think it has to be that way, unless $z_k$ is something nice. –  Feanor Jan 26 '13 at 23:05
    
@Feanor: The problem is that I have a closed form. I start with something like $\displaystyle \frac{1}{1-y(\frac{1}{1-z})}$. This isn't the best example, but I really can't modify the individual $X_k$'s because the $z$'s are mixed together and are not in the form of a summation. I have everything in closed form, and that's my real problem. I'd like to modify the closed forms, if possible. –  Matt Groff Jan 26 '13 at 23:20
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1 Answer

Not all at once - don't worry about convergence, just substitute:

$\sum_{k=0}^\infty{X_k y^k} =\sum_{k=0}^\infty{\frac{ y^k}{1-z_k}} =\sum_{k=0}^\infty y^k\sum_{j=0}^\infty{z_k^j} = \sum_{j=0}^\infty \sum_{k=0}^\infty y^k {z_k^j} $.

From here on, it's up to you. If there is a more explicit form for the $z_k$, substitute it here and proceed.

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I'm not sure that I can proceed like this. I'm actually trying to change the formula. I don't want $\sum{z_k}$ in the result. I want to transform $\sum{z_k} \to z_k$. –  Matt Groff Jan 26 '13 at 23:26
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