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I have such $$\sum_{n\geq 0}a_nx^n=x^2(1-x^2)^{-3}$$then I have no idea how to find $a_n$

RobPratt
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xniwniw
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    Start from $(1-x^2)^3\sum a_nx_n=x^2$ and identify the coefficients (i.e. $x^2=0x^0+0x^1+1x^2+0x^3+\cdots$), you 'll find a relation between the $a_n$. Here is an example of the technique here https://math.stackexchange.com/a/4106945/399263 – zwim May 12 '22 at 21:18
  • The LHS can be thought of as the Binomial expansion of the RHS... – Adam Rubinson May 12 '22 at 21:19
  • If $|x|<1$, then $\frac1{1-x}=\sum\limits_{n\ge0}x^n$. Replace $x$ with $x^2$ and take some derivatives. – user170231 May 12 '22 at 21:23

2 Answers2

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Apply partial fraction decomposition and Newton's binomial theorem: \begin{align} \frac{x^2}{(1-x^2)^3} &= -\frac{1/16}{1-x} - \frac{1/16}{(1-x)^2} + \frac{1/8}{(1-x)^3} - \frac{1/16}{1+x} - \frac{1/16}{(1+x)^2} + \frac{1/8}{(1+x)^3} \\ &= -\frac{1}{16}\sum_{n\ge 0}x^n - \frac{1}{16}\sum_{n\ge 0}\binom{n+1}{1}x^n + \frac{1}{8}\sum_{n\ge 0}\binom{n+2}{2}x^n - \frac{1}{16}\sum_{n\ge 0}(-x)^n - \frac{1}{16}\sum_{n\ge 0}\binom{n+1}{1}(-x)^n + \frac{1}{8}\sum_{n\ge 0}\binom{n+2}{2}(-x)^n \\ &= \sum_{n\ge 0}\left(-\frac{1}{16} - \frac{n+1}{16} + \frac{(n+2)(n+1)}{16} - \frac{(-1)^n}{16} - \frac{(n+1)(-1)^n}{16} + \frac{(n+2)(n+1)(-1)^n}{16} \right)x^n\\ &= \sum_{n\ge 0}\frac{n(n+2)(1+(-1)^n)}{16} x^n, \end{align} which immediately implies that $$a_n=\frac{n(n+2)(1+(-1)^n)}{16}.$$

Alternatively, note that $$\frac{1}{(1-z)^3} = \sum_{n \ge 0} \binom{n+2}{2}z^n,$$ so $$\frac{z}{(1-z)^3} = \sum_{n \ge 0} \binom{n+2}{2}z^{n+1} = \sum_{n \ge 0} \binom{n+1}{2}z^n.$$ Now substitute $z=x^2$ to obtain \begin{align} \frac{x^2}{(1-x^2)^3} &= \sum_{n \ge 0} \binom{n+1}{2}(x^2)^n \\ &= \sum_{n \ge 0} \frac{n(n+1)}{2}x^{2n} \\ &= \sum_{n \ge 0} \frac{(n/2)(n/2+1)}{2}\cdot\frac{1+(-1)^n}{2}x^n \\ &= \sum_{n\ge 0}\frac{n(n+2)(1+(-1)^n)}{16} x^n. \end{align}

RobPratt
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  • there might be a typo $$\sum_{n \ge 0} \binom{n+2}{2}z^{n+1} = \sum_{n \ge 1} \binom{n+1}{2}z^n.$$ instead of $$\sum_{n \ge 0} \binom{n+2}{2}z^{n+1} = \sum_{n \ge 0} \binom{n+1}{2}z^n.$$ – Not a Salmon Fish May 13 '22 at 11:35
  • @user955791 Both are correct because the $n=0$ term is $0$. I deliberately kept the $0$ indexing to match the desired final form. – RobPratt May 13 '22 at 12:58
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Start with the generalized binomial theorem $(1+x)^a =\sum_{n=0}^{\infty} \binom{a}{n}x^n $ with $\binom{a}{n} =\dfrac{\prod_{k=0}^{n-1}(a-k)}{k!} $.

From this you can show $(1-x)^{-s} =\sum_{n=0}^{\infty} \binom{s+n-1}{n}x^n $ so $(1-x^2)^{-s} =\sum_{n=0}^{\infty} \binom{s+n-1}{s-1}x^{2n} $ and then $(1-x^2)^{-3} =\sum_{n=0}^{\infty} \binom{n+2}{2}x^{2n} =\sum_{n=0}^{\infty} \frac{(n+2)(n+1)}{2}x^{2n} $.

marty cohen
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  • Thanks, Then how should I deal with the $x^2$ outside of the summation notation – xniwniw May 12 '22 at 22:31
  • This is what I showed in the second part (Alternatively...) of my answer. – RobPratt May 12 '22 at 23:17
  • Right, in alternatively part is a good solution, but I'm wondering if there are different degrees of $x$ in the numerator and denominator, such as $\dfrac{x^3}{(1-x^2)^3}$ or $\dfrac{x^5}{(1-x^2)^3}$, how to solve those – xniwniw May 12 '22 at 23:36
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    Multiplication by $x^k$ shifts the index: $x^k\sum_{n\ge 0} b_n x^n = \sum_{n\ge 0} b_n x^{n+k} = \sum_{n\ge k} b_{n-k} x^n$. – RobPratt May 13 '22 at 01:56