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I don't need an answer, as this was a question I got wrong on a problem set, but could someone explain this?

So, we have to represent f(x)= $x^3$/$(2-x)^3$

My professor writes consider g(x) = $x^2/$(2-x)

I don't understand where she got this from?

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1 Answer

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You are allowed to differentiate termwise a power series for all $u$ such that $|u|<R$, where $R$ is the radius of convergence.

Then starting with $$ {1 \over (1-u)} =\sum_{n=0}^{\infty}u^{n}, \qquad|u|<1, \tag1 $$ by differentiating twice, you have successively $$ {1 \over (1-u)^2} =\sum_{n=1}^{\infty}nu^{n-1}, \qquad|u|<1,\tag2 $$ $$ {2 \over (1-u)^3} =\sum_{n=2}^{\infty}n(n-1)u^{n-2}, \qquad|u|<1. \tag3 $$ Applying $(3)$ with $u=\dfrac x2$ gives $$ {2 \over (1-\frac x2)^3} =\sum_{n=2}^{\infty}n(n-1)\frac{x^{n-2}}{2^{n-2}}, \qquad \left|\frac x2\right|<1, \tag4 $$ multiplying out by $x^3$ and simplifying $$ {16\:x^3 \over (2-x)^3} =\sum_{n=2}^{\infty}n(n-1)\frac{x^{n+1}}{2^{n-2}}, \qquad \left|x\right|<2, \tag5 $$ or equivalently with a change of indice you get the power series representation

$$ {x^3 \over (2-x)^3} =\sum_{n=3}^{\infty}(n-1)(n-2)\frac{x^n}{2^{n+1}}, \qquad \left|x\right|<2. \tag6 $$

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