Exponentiation a power series

Question

I find formula to calculate $f(z)=(\sum_{i=0}^{\infty} {x^{2i+1}})^n.$ I know that $\sum_{i=0}^{\infty} {x^{2i+1}}=\frac {x}{1-x^2}.$ But I need function $f(x)$ as a power series. This is my attempt so far. $$f(z)=(\lim_{m \to \infty} \sum_{i=0}^{\infty} {x^{2i+1}})^n=x^n \lim_{m \to \infty}(\sum_{i=0}^{\infty} {x^{2i}})^n$$ I tried to use Multinomial theorem http://en.wikipedia.org/wiki/Multinomial_theorem. But I got tripped up on it. Help me please. Sorry for my English.

Answer 1

If by "I need the function as a power series", you mean to get an explicit formula for the coefficients of the terms of the series, then, yes, there's a nice solution of this problem. In fact, the coefficients of the series for your function turn out to be just binomial coefficients.

The main step in this calculation uses the repeated derivative of the geometric series, namely $\frac1{(1-x)^{m+1}}=\sum_{i=0}^\infty\binom{m+i}m x^i$. Also, let's use the standard notation for a coefficient of a power series, $[x^m]\sum_{i=0}^\infty a_i x^i=a_m$. Then, for your function $f(x)=\frac {x^n}{(1-x^2)^n}$, the coefficient of $x^m$ is $$ \begin{aligned} \big[ x^m\big] f(x)&=[x^{m-n}]\frac1{(1-x^2)^n}\\ &=[x^{m-n}]\sum_{i=0}^\infty\binom{n-1+i}{n-1}(x^2)^i\\ &=\begin{cases} \binom{({m+n-2})/2}{n-1} &\mbox{if } m-n \mbox{ is even} \\ 0& \mbox{otherwise.}\end{cases} \end{aligned}$$ The first line accounts for the power $x^n$ in the numerator of your function. The second line uses the derivative of the geometric function and substitutes $x^2$ for $x$. The last line is a slight algebraic simplification of the second. So, your function has only even terms if $n$ is even or only odd terms if $n$ is odd, with the coefficients listed in the last line. For instance, when $n$ is even, this shows $$f(x)=\sum_{i=0}^\infty \binom{i-1+n/2}{n-1} x^{2i}.$$