Question

Separating a single infinite series into a product of two infinite series

Answer 1

\[\sum_{n=0}^{\infty}a _{n}x ^{n} = \left( \sum_{n=0}^{\infty}x ^{n} \right)\left( \sum_{n=0}^{\infty}x ^{2n} \right)\]

find a_n

Answer 2

what have you tried so far ?

Answer 3

This is going to resemble the binomial expansion, I think. Recall,
\[\sum_{k=0}^n(x+y)^n=\binom nk x^ky^{n-k}\]
In this case, there are \(\dbinom nk\) ways of getting the \(k\)th power of \(x\).
\[\sum_{n=0}^\infty x^n=1+x+x^2+x^3+x^4+\cdots\]
\[\sum_{n=0}^\infty x^{2n}=1+x^2+x^4+\cdots\]
Expanding gives the following first few terms:
\[\begin{array}{c|c|c|c}
n&\text{power of }x&\text{ways of getting}&\text{coefficient}\\
\hline
0&x^0&x^0\times x^0&1\\
1&x^1&x^1\times x^0&1\\
2&x^2&x^0\times x^2,~x^2\times x^0&2\\
3&x^3&x^1\times x^2,~x^3\times x^0&2\\
4&x^4&x^0\times x^4,~x^2\times x^2,~x^4\times x^0&3\\
5&x^5&x^1\times x^4,~x^3\times x^2,~x^5\times x^0&3\\
6&x^6&x^0\times x^6,~x^2\times x^4,~x^4\times x^2,~x^6\times x^0&4
\end{array}\]
Perhaps you see the pattern by now? It's a matter of counting the ways you can add up exponents to obtain the \(n\)th power; however, you have to account for the fact that the first series contributes both odd and even powers, whereas the second contributes exclusively even powers.

Answer 4

Oh and forget what I'd said about the binomial expansion. I was thinking ahead (mistakenly) and forgot I'd typed that :/

Answer 5

nice :) i was thinking of using convolution or generating functions earlier

Answer 6

straight foward pattern

Answer 7

very close to convolution coming up