Writing a function as a power series

Question

I've been having some issues with this function.

$$ f(x) =\dfrac{4x^2}{(x-3)^2} $$ I've been able to take $4x^2$ out, and differentiate it to give me $$ 4x^2 \dfrac{d}{dx}(-1)(\dfrac{1}{-3-(-x)}) $$ However, I'm not sure what to do with the $-3$, as to write it as a power series, it has to be $1$. Thanks!

Answer 1

Found the error, turns out I cannot take derivatives of polynomials correctly. Thank you @zwim

Take $$ g(x)=\dfrac{1}{(x-3)^2} $$

and integrate to get $$ \int g(x)dx = \dfrac{-1}{x-3} = \dfrac{1}{3-x} = \dfrac{1}{3}\dfrac{1}{1-\frac{x}{3}} $$

if $u=\frac{x}{3}$ then $$ \dfrac{1}{3}\dfrac{1}{1-u} = \dfrac{1}{3}(1+u+u^2+...) = \dfrac{1}{3}\left(1+\dfrac{x}{3}+\dfrac{x^2}{9}+...\right) $$

Now this is the series for $\int g(x)dx$, so we need to differentiate and get

$$ \dfrac{1}{9}+\dfrac{2x}{27}+\dfrac{3x^2}{81}+... $$

finally we multiply by $4x^2$ to get that

$$ f(x) = \dfrac{1\cdot 4x^2}{9}+\dfrac{2\cdot 4x^3}{27}+... = \sum_{n=0}^\infty \dfrac{(n+1)\cdot 4x^{n+2}}{3^{n+2}} $$

Answer 2

Hint

Define $x=3t$ $$f(x) =\dfrac{4x^2}{(x-3)^2} \implies f(t)=\frac{4 t^2}{(1-t)^2}$$ Now, use the binomial theorem for $$(1-t)^{-2}$$ and multiply the result by $4t^2$.

When finished, replace $t$ by $\frac x3$.

Answer 3

You could use Wiki-Taylor Series to obtain the expansion for example: