# Evaluation of a power series

Given a power series in the form

$$f(x) = \sum_{n=0}^{\infty} \frac{p(n)}{q(n)}x^{n}$$

here $p(n)$ and $q(n)$ are Polynomials of certain degree

then can i get $f(x)$ in term of elementary functions ??

if $q(n) =1$ then i know how to get the sum but for this general case i would need some help

i believe that let be the operator $T=x \frac{d}{dx}$ then formally

$$f(x)= \frac{p(T)}{q(T)}\frac{1}{1-x}=p(T){q(T)}^{-1}\frac{1}{1-x}$$

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I don't know what it means to divide by $T$. I understand polynomials of $T$, but not the $q(T)$ – Ross Millikan Mar 13 '13 at 20:07
In terms of hypergeometric functions, but in general not in terms of elementary functions. – André Nicolas Mar 13 '13 at 20:16
how can in terms of hypergeometric function ? what would be the method ?? thanks – Jose Garcia Mar 13 '13 at 20:25
@RossMillikan you aren't dividing but $T$, because if $q(z)=z$ or more generally is divisible by $z$ then the power series doesn't make sense. Indeed, $q$ cant have any natural number roots. With no such roots, you can find the inverse of $q(T)$, which is an operator that acts on power series. If you look below, as long as $q(n)\neq 0$ for all natural $n$, which is a given, it turns out that $q(T)^{-1}$ can be easily defined for power series. (Essentially, the natural numbers are the eigenvalues of $T$...) – Thomas Andrews Mar 14 '13 at 9:51

First, my answer was overkill.

Second, it missed the point of the problem.

I'm still gonna say partial fractions.

Oh well, here's the "better" answer to a question that wasn't asked.

I'd rewrite it as $$q(T)[f(x)] = p(T)\left[\frac{1}{1-x}\right]$$ since we don't know what the meaning of $q(T)^{-1}$ is.

But in general, we can easily show that $q(T)[\sum a_nx^n] = \sum a_nq(n)x^n$. We can do this directly, since $T^k[x^n] = n^kx^n$ is easy to prove.

But since $f(x)=\sum_n \frac{p(n)}{q(n)}x^n$, that means:

$$q(T)[f(x)] = \sum_n p(n)x^n$$

On the other hand, since $\frac{1}{1-x} =\sum_n x^n$, we see that the right side is the same:

$$p(T)\left[\frac{1}{1-x}\right] =\sum_n p(n)x^n$$

So we are done.

Now, since we have proven $q(T)\left[\sum_n a_nx^n\right]=\sum q(n)a_nx^n$ and $q(n)\neq 0$ for all $n$, then $q(T)^{-1}$ is clearly defined as:

$$q(T)^{-1}\left[\sum_n a_nx^n\right] = \sum \frac{a_n}{q(n)}x^n$$

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