Find the coefficient of $ x^{12}$ in $(1-x^2)^{-5}$ Find the coefficient of $x^{12}$ in $(1-x^2)^{-5}$
What can be said for $x^{17}$
Tried $\frac{1}{(1-x^2)^{5}}$=$\sum_{n=0}^\infty \binom{n+5-1}{n}x^n$
not sure that i can do that with $x^2$ 
 A: $$y=\frac{1}{1-x}=\sum_{n=0}^{\infty }x^n$$
the first derivative is
$$\frac{-1}{(1-x)^2}=\sum_{n=0}^{\infty }nx^{n-1}$$
the fourth derivative 
$$\frac{24}{(1-x)^5}=\sum_{n=4}^{\infty }n(n-1)(n-2)(n-3)x^{n-4}$$
let $x\rightarrow x^2$
$$\frac{24}{(1-x^2)^5}=\sum_{n=4}^{\infty }n(n-1)(n-2)(n-3)x^{2n-8}$$
to find coefficient of $x^{12}$
let
$$2n-8=12$$
$$n=10$$
so the coefficient is
$$(10)(10-1)(10-2)(10-3)/24=210$$ 
A: It's convenient to use the coefficient of operator $[x^k]$ to denote the coefficient of $x^k$ in a series. This way we can write e.g.
\begin{align*}
[x^k](1+x)^n=\binom{n}{k}
\end{align*}

We obtain
  \begin{align*}
[x^{12}]\frac{1}{(1-x^2)^5}
&=[x^{12}]\sum_{k=0}^\infty \binom{-5}{k}(-x^2)^{k}\tag{1}\\
&=[x^{12}]\sum_{k=0}^\infty \binom{k+4}{4}x^{2k}\tag{2}\\
&=\binom{10}{4}\tag{3}\\
&=210
\end{align*}

Comment:


*

*In (1) we use the binomial series expansion of $\frac{1}{(1-x^2)^5}$

*In (2) we use the binomial identity
\begin{align*}
\binom{-n}{k}=\binom{n+k-1}{k}(-1)^k
\end{align*}

*In (3) we select the coefficient of the series with $k=6$ accordingly


Note: Since the function is even, the coefficient of $x^{17}$ is zero.
A: You need to find the inverse of the power series of $(1-x^2)^5$
this one is as follows:
$[1,0,-5,0,10,0,-10,0,5,0,-1,0,0\dots$
we calculate its inverse like we would for any other power series from left to right (the algorithm is similar to that of synthetic division of a polynomial).
$[1,0,5,0,15,0,35,0,70,0,126,0,210$.
So the coefficient of $x^{12}$ is $210$ and for $x^{17}$ is clearly $0$.

I wrote a quadratic time algorithm to find the first $n+1$ terms of the inverse of a power series $P$ (given the first $n$ terms of P$).
#include <bits/stdc++.h>
using namespace std;

const int MAX=10010; // this is just the max size we work with
double P[MAX]; // here we store the coefficients of the power series
double I[MAX]; // here we store the coefficients of P^-1
int main(){
    int n; // the number of terms
    scanf("%d",&n);
    for(int i=0;i<n;i++){
        scanf("%lf",&P[i]); // here we read the first n coefficients of P
    }
    I[0]=1/P[0]; // the constant term of the inverse is the inverse of P[0]
    for(int i=1;i<=n;i++){
        double sum=0;
        for(int j=0;j<i;j++){// calculate the value of the i'th term without I[i]P[0]
            sum+=I[j]*P[i-j];
        }
        I[i]=-sum/P[0]; // then I[i] must be -sum/P[0]
    }
    for(int i=0;i<=n;i++){// this just prints the result
        printf("%g ",I[i]);
    }
    printf("\n");
    return(0);

}

A: As you noted: the negative binomial theorem, http://mathworld.wolfram.com/NegativeBinomialSeries.html, gives
$$\frac{1}{(1-y)^{5}}=\sum_{n=0}^\infty \binom{n+5-1}{n}y^n$$
Now replace $y$ with $x^2$ to get
$$\frac{1}{(1-x^2)^{5}}=\sum_{n=0}^\infty \binom{n+5-1}{n}x^{2n}$$
For the coefficient of $x^{12}$, use n=6.
For $x^{17}$ the coefficient must be zero since all terms will have even powers. 
