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!

What about having $\frac 43 x^2$ out ? You get your $\frac 1{1-x/3}=\sum (x/3)^n$... — zwim, May 04 '19 at 03:33
@zwim so then would it look like: 4/3x^2 d/dx (-1) (1/1-x) ? — johnners17, May 04 '19 at 03:34

wjmccann · Accepted Answer · 2019-05-04T03:53:14.293

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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

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

edited May 04 '19 at 03:53

answered May 04 '19 at 03:37

wjmccann

3,085

1

wait, i don't think this works. The first term is definitely 4x^2/9, but after that, the subsequent terms don't work. – johnners17 May 04 '19 at 03:43
You're right after plotting this it is slightly off, I'm not sure where my error is – wjmccann May 04 '19 at 03:45
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@wjmccann you derived $x^n$ to $x^{n-1}$ instead of $nx^{n-1}$. – zwim May 04 '19 at 03:49
@zwim so how does that change the solution? what is n in this case? – johnners17 May 04 '19 at 03:52
You're right @zwim. Turns out that polynomials are the hardest function to differentiate haha – wjmccann May 04 '19 at 03:54
@wjmccann what would the radius of convergence be then? I tried the ratio test but i think i messed up. – johnners17 May 04 '19 at 04:03
@johnners17 what did you get? I am fairly certain the ratio test works in this case – wjmccann May 04 '19 at 04:10
@wjmccann I got 3, which was right! Thanks so much! – johnners17 May 04 '19 at 17:42

score 0 · Answer 2 · answered May 04 '19 at 03:39

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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 $\frac x3$.

answered May 04 '19 at 03:39

Claude Leibovici

260,315

i'm still confused. I did what you said, and ended up with a term with x but no n, so how would it be a power series? – johnners17 May 04 '19 at 03:49
@johnners17. You should end with $(4t^2)\sum_{n=0}^\infty (1+n)t^n$ which makes $\cdots\qquad $??? – Claude Leibovici May 04 '19 at 04:01

score 0 · Answer 3 · answered May 04 '19 at 04:06

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

answered May 04 '19 at 04:06

NoChance

6,427

Writing a function as a power series

3 Answers3