Determining the coefficient of a generating function

Question

I've been doing some practice problems and I have encountered some questions. How would I determine the coefficient of $x^{25}$ in the generating function $F(x)$ with closed form expression? I can't factor the numerator which is why I did for the previous ones.

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

Answer 1

First simplify the rational function as follows: $$ \frac{1-2x+2x^2}{(1-x)^2} =\frac{(1-x)^2+x^2}{(1-x)^2} =1+\frac{x^2}{(1-x)^2}.\tag{1} $$ Then use the identity $$ \frac{1}{(1-x)^k}=\sum_{n=0}^\infty \binom{k+n-1}{k-1}x^n\tag{2} $$ (which can be obtained by repeatedly differentiating the geometric series) where $k\geq 1$ to get that $$ [x^n]\left(\frac{1}{(1-x)^2}\right)=n+1\tag{3} $$ and hence $$ [x^n]\left(\frac{x^2}{(1-x)^2}\right)=n-1.\tag{4} $$ Finally, to answer the question note that $$ [x^{25}]\left(\frac{1-2x+2x^2}{(1-x)^2}\right)= [x^{25}]\left(1+\frac{x^2}{(1-x)^2}\right)=[x^{25}]\left(\frac{x^2}{(1-x)^2}\right)=25-1=\color{blue}{24} $$ by (1) and (4).

Answer 2

HINT

Note that if $G(x) = (1-x)^{-1}$ then $G'(x) = (1-x)^{-2}$, and you have $$ G(x) = \sum_{k=0}^\infty x^k\\ G'(x) = \sum_{k=1}^\infty kx^{k-1}\\ $$ and $$ F(x) = G'(x) (1-2x+2x^2) $$

Can you finish this?

UPDATE

$$ \begin{split} \left[x^{25}\right] F(x) &= \left[x^{25}\right] (1-2x+2x^2) \sum_{k=1}^\infty kx^{k-1} \\ &= \left[x^{25}\right] \sum_{k=1}^\infty kx^{k-1} -2 \left[x^{24}\right] \sum_{k=1}^\infty kx^{k-1} +2 \left[x^{23}\right] \sum_{k=1}^\infty kx^{k-1}\\ \end{split} $$

Can you finish this?

Answer 3

It is convenient to use the coefficient of operator $[x^n]$ to denote the coefficient of $x^n$ in a series.

We obtain \begin{align*} \color{blue}{[x^{25}]\frac{1-2x+2x^2}{(1-x)^2}}&=[x^{25}](1-2x+2x^2)\sum_{j=0}^\infty\binom{-2}{j}(-x)^j\tag{1}\\ &=\left([x^{25}]-2[x^{24}]+2[x^{23}]\right)\sum_{j=0}^\infty\binom{j+1}{1}x^j\tag{2}\\ &=26-2\cdot 25+2\cdot 24\tag{3}\\ &=\color{blue}{24} \end{align*}

Comment:

• In (1) we apply the binomial series expansion.

• In (2) we use the linearity of the coefficient of operator and apply the rule \begin{align*} [x^{p-q}]A(x)=[x^p]x^qA(x) \end{align*} We also use the binomial identity \begin{align*} \binom{-p}{q}=\binom{p+q-1}{p-1}(-1)^q \end{align*}

• In (3) we select the coefficients accordingly.