Suppose that $f(x) =\frac{x^2+x}{(1-x)^3}$. What would the power series representation for this function? I have found the power series representation for $\frac{1}{(1-x)^3}$ which is $\sum_{n=2}^{\infty} \frac{n(n-1)x^{n-2}}{2}$. Would it be allowed to multiply the series by $x^2+x$ to get $\left(x^2+x\right) \sum_{n=2}^{\infty} \frac{n(n-1)x^{n-2}}{2}$?

  • $\begingroup$ You can do that, however you should take a look where your original function is defined vs where the series expansion is defined (i.e. for what $x$ the series converges). $\endgroup$ – ty. Mar 19 '18 at 20:56
  • $\begingroup$ but doesn't have more to do with the radius of convergence and the endpoints rather then the multiplication $\endgroup$ – user524644 Mar 19 '18 at 21:01
  • $\begingroup$ Maple code: convert((x^2+x)/(1-x)^3, Sum, dummy = n, include = powers) says:$\sum _{n=0}^{\infty } n^2 x^n=\frac{x^2+x}{(1-x)^3}$ $\endgroup$ – Mariusz Iwaniuk Mar 19 '18 at 21:13

This is indeed correct and the result is then given by $\sum_{n=0}^{\infty}n^2x^n$, as stated in M. Iwaniuk's comment.

$$\left(x^2+x\right)\sum_{n=2}^{\infty}\frac{n(n-1)x^{n-2}}{2}=x^2\sum_{n=2}^{\infty}\frac{n(n-1)x^{n-2}}{2}+x\sum_{n=2}^{\infty}\frac{n(n-1)x^{n-2}}{2}=\sum_{n=2}^{\infty}\frac{n(n-1)x^{n}}{2}+\sum_{n=2}^{\infty}\frac{n(n-1)x^{n-1}}{2}=\sum_{n=2}^{\infty}\frac{n(n-1)x^{n}}{2}+\sum_{n=1}^{\infty}\frac{(n+1)nx^{n}}{2}=x+\sum_{n=2}^{\infty}\frac{n(n-1)+n(n+1)}{2}x^{n}=\sum_{n=0}^{\infty}n^2x^n$$, as $0x^0=0$ and $x=1^2x^1$

  • 1
    $\begingroup$ I dont understand how he was able to arrive at this simplification $\endgroup$ – user524644 Mar 19 '18 at 21:41

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