# Given a sequence ${a_k}, k=1,2,3...$, find the generating function for ${a_2k}$

This is a problem on an exam review sheet for a discrete mathematics course.

Consider a generating function

$$F(x)=a_0+a_1x+a_2x^2+...$$

Using operations on generating functions, obtain a generating function for the sequence $a_0, a_2, a_4, a_6, ..., a_k$ .

I guess I'm just kind of confused as to what exactly I'm supposed to find. Would I need to find a way to subtract the odd power terms in $F(x)$, or do I find a function of the form

$$G(x)=a_0+a_2x+a_4x^2+a_6x^3+...+a_{2k}x^k?$$

If I already wrote out this function is it given or do I need to somehow put it in terms of $F(x)$? Sorry if this is a silly question, I'm sure if I just figure out what I'm supposed to be finding I can figure out the rest on my own.

• You can use the fact that every function can be written as a sum of an odd and an even function and that this decomposition is unique. Feb 26, 2015 at 19:12
• Thanks! So like Ron Gordon's answer below, I basically just have to decompose G in terms of F? I was making this problem way more confusing for myself than I needed to... Feb 26, 2015 at 19:22

$$G(x) = \frac12 \left [F \left ( \sqrt{x} \right ) + F \left (- \sqrt{x} \right ) \right ]$$
I feel the answer is self-explanatory. As an additional exercise, try computing the G.F. for the sequence $a_1, a_3,\ldots$.
• My work is a little stream-of-consciousness right now, so please bear with me: Sort of segue-ing off what you did, $$G(x)=a_0+a_3x+a_6x^2+... = a_0+a_3(x^{1/3})^3+a_6(x^{1/3})^6...$$ so I need to manipulate $F(x)$ a bit. So now have $$F(x^{1/3})=a_0+a_1(x^{1/3})+a_2(x^{1/3})^2+a_3(x^{1/3})^3+...$$ I need to find inputs of $x$ so that I can somehow get rid of the $a_{3k+1}$ and $a_{3k+2}$ terms, correct? Feb 26, 2015 at 20:10
Hint: We have $$e^x=\cosh(x)+\sinh(x)$$ so $$\cosh(x)=e^x-\sinh(x)$$