# Generating Function & Sequence

Find the generating functions of the sequences 2, 1, 2, 1, 2, 1, . . .

I get $$\frac{1}{1+x} + \frac{1}{1-x} = \frac{2}{1-x^2}$$

But the solution ends up with $$\frac{2}{1-x^2} + \frac{x}{1-x^2} = \frac{2+x}{1-x^2}$$. The solutions starts with $$\sum_{n\ge 0} (2)x^{2n}+\sum_{n\ge 0} (1) x^{2n+1}$$

I couldn't come up with anything like that. I feel like I'm confused with something.

• you must have a typo.; $\dfrac2{1-x}+\dfrac x{1-x}=\dfrac{2+x}{1-x},$ not $\dfrac{2+x}{1-x^\color{red}2}$ – J. W. Tanner Dec 13 '19 at 1:21
• The generating function of $1,1,1,1,...$ is $\frac{1}{1-x}$, while the generating function of $1,0,1,0,1,0,...$ is $\frac{1}{1-x^2}$. Then the generating function of $2,1,2,1,2,1,...$ should be $\frac{1}{1-x}+\frac{1}{1-x^2}$. – egorovik Dec 13 '19 at 1:22
• Thanks. I fixed the typo. – user665125 Dec 13 '19 at 1:32

As $$\frac{1}{1-x}$$ is the generating series for $$1,1,1,\dots$$ and $$\frac{1}{1-x}$$ is the generating series for $$1,-1,1,-1,\dots$$, the series for which you are calculating the generating function for is $$2,0,2,0,\dots$$.

But we can get the result you want by noticing that $$2,1,2,1,\dots$$ is actually $$\frac{3}{2},\frac{3}{2},\frac{3}{2},\dots$$ plus $$\frac{1}{2},-\frac{1}{2},\frac{1}{2},-\frac{1}{2},\dots$$.

We know the first series has generating function $$\frac{3/2}{1-x}$$ and the second has generating function $$\frac{1/2}{1+x}$$. Thus, the function you are looking for is$$\frac{\frac{3}{2}}{1-x}+\frac{\frac{1}{2}}{1+x}=\frac{\frac{3}{2}(1+x)+\frac{1}{2}(1-x)}{(1-x)(1+x)}=\frac{2+x}{1-x^2}$$ which is the result you were given.

The sequence $$2,1,2,1,2,1,...$$ alternates between $$2$$ and $$1$$, being $$2$$ for even-numbered terms and $$1$$ for odd-numbered terms. The generating function is thus $$\sum\limits_{n \ge0}(2)x^{2n}+\sum\limits_{n\ge0}(1)x^{2n+1}=\sum\limits_{k\ge0}(1)x^k+\sum\limits_{k\ge0}(1)x^{2k}=\dfrac{1}{1-x}+\dfrac{1}{1-x^2}=\dfrac{2+x}{1-x^2}$$

As an alternative to the other answers, you can simply use a variant of one function, that being

$$\frac{1}{1-x^2}=\sum_{n=0}^{\infty} x^{2n}$$

If you consider that the $$2$$'s are even indexed (assuming in the sequence $$\{a_n\}$$ the first term is $$a_0$$) and the $$1$$'s are odd indexed, you have that

$$\frac{2}{1-x^2}$$

will generate the $$2$$'s and

$$\frac{x}{1-x^2}$$

will generate the $$1$$'s. This is because

$$\frac{1}{1-x^2}=1+x^2+x^4+...$$

and so

$$\frac{x}{1-x^2}=x(1+x^2+x^4+...)=x+x^3+x^5+...$$

Thus, you sequence will be generated by

$$f(x)=\frac{2+x}{1-x^2}$$

And if you want to generate $$(a_i)_{i=1}^m$$ repeatedly, the i-th term is generated by $$a_ix^i$$, and for this to repeat every $$m$$ this needs $$\dfrac{a_ix^i}{1-x^m}$$ so the final result is $$\sum_{i=1}^m\dfrac{a_ix^i}{1-x^m}$$.

In your case, with $$(1, 2)$$ repeated, this is $$\dfrac{x}{1-x^2}+\dfrac{2x^2}{1-x^2} =\dfrac{x+2x^2}{1-x^2}$$.

This starts the first term at $$x^1$$. To start it at the constant term, just reduce the exponents of the $$x^i$$ by one giving $$\sum_{i=1}^m\dfrac{a_ix^{i-1}}{1-x^m}$$, so this gives, in this case $$\dfrac{1}{1-x^2}+\dfrac{2x}{1-x^2} =\dfrac{1+2x}{1-x^2}$$.