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Find coefficient of generating function.

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

MY WAY OF SOLVING SIMILAR PROBLEM:

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

2) partial fraction $g(x) = \frac{A}{1-x} + \frac{B}{1+x} $

3) $ g(x) = \sum\limits_{n=0}^\infty Ax^{n} + \sum\limits_{n=0}^\infty B (-1)^nx^{n} = \sum\limits_{n=0}^\infty (A+(-1)^nB)x^{n} $ -solution

But what can I do with $f(x)$? I can't use my method because:

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

$\frac{-x^3 +3x}{1-x^2} = \frac{A}{1-x} + \frac{B}{1+x}$

$ -x^3+3x = A(1+x) + B(1-x) $

$-x^3 = 0 \cdot x^3 $

$ -1 =0 $

  • 2
    Do the same, just don't mix the standalone $x$ into the fraction. – 2017-01-25
  • 0
    Partial fractions – that's the hard way to do $g$. It's just a geometric series. And for $f$, partial fractions only works when degree of numerator less than degree of denominator. – 2017-01-25

1 Answers 1

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There is no reason to add up the right hand terms, since the term $x$ is simple and convenient. The other term can be expanded using the geometric series expansion \begin{align*} \frac{1}{1-y}=\sum_{n=0}^\infty y^n\qquad\qquad |y|<1 \end{align*} with $y=x^2$.

We obtain \begin{align*} f(x)&=\frac{2x}{1-x^2}+x\\ &=2x\sum_{n=0}^\infty x^{2n}+x\\ &=3x+2\sum_{n=1}^\infty x^{2n+1} \end{align*}

We conclude the coefficient $[x^n]$ of $f(x)$ is \begin{align*} [x^n]f(x)= \begin{cases} 3&n=1\\ 2&n>1, odd\\ 0&n\geq 0, even \end{cases} \end{align*}