# Find new generating function, given an arbitrary generating function

In a discrete mathematics past paper, I am asked to find the generating function for the sequence $$\langle a_0, 0, a_2, 0, a_4, 0, \ldots \rangle,$$ given that $A(x)$ is the generating function for the sequence $$\langle a_0, a_1, a_2, a_3, \ldots \rangle.$$

I have been struggling with this for a while, with no success.
I think that I am missing something simple and would appreciate a hint to help me solve the problem.

You want to kill the terms of odd degree, which can be done by projecting onto the space of even power series parallel to the space of odd power series (the two spaces are complementary inside the space of all power series). If your generating series is given by an expression $A(x)$ (and unless you are working over a field of characteristic$~2$, which I will suppose) you can build this "even image" of $A$ as a linear combination of $A(x)$ and of the result $A(-x)$ of substituting $-x$ for $x$; the coefficients are easily found.
• I find it difficult to understand the technical language used in your answer. I am not familiar with the operation of 'projecting onto [a] space of ... power series'. Neither am I familiar with the phrase 'working over a field of characteristic $2$'. – Caleb Owusu-Yianoma May 21 '15 at 12:12
• @CKKOY: Do you know what even and odd function of (a real value) $x$ are, and that every function of $x$ can be written uniquely as the sum of an even and an odd function? Then projection to the space of even functions is simply mapping $f$ the its even component in this sum (forgetting the odd component). Things are entirely similar for power series, with just a very slight difference in the meaning of "even" and "odd" (for functions "even" means having the same value in any $x$ as in $-x$, for power series it means not changing when $-x$ is substituted for $x$ in the series). – Marc van Leeuwen May 21 '15 at 13:17
• As for fields of characteristic$~2$, if you don't know what they are then obviously you are not working over one, and just forget the remark. I said so just because you did not say what kind of coefficients your power series have, and in the special case of values in such a field it would not be possible to divide them by$~2$, and my hint would not apply (but again, you need not worry about this eventuality since it is not your case). – Marc van Leeuwen May 21 '15 at 13:22