1
$\begingroup$

I am trying to find the Taylor series of $$f(x) = \frac{x}{(x-1)^2}$$ around $x=2$. Since the standard technique of taking and evaluating derivatives leads to crazy expression by the second derivative, I was hoping it would be possible to manipulate the geometric series $\frac{1}{1-x} = \sum_{n=0}^{\infty}x^n$ for $|x|<1$. Substituting $(x-2)$ for $x$ gives me $$\frac{1}{1-(x-2)} = \frac{1}{3-x} =\sum_{n=0}^{\infty}(x-2)^n$$ This is where I am stuck. How can I further manipulate this to get my desired function?

$\endgroup$
1
$\begingroup$

Does this help?

$$\begin{align} \frac{x}{(x-1)^2}=-x\frac{-1}{(x-1)^2} &= -x\frac{d}{dx}\frac{1}{x-1}\\ &= -x \frac{d}{dx}\frac{1}{1+(x-2)}\\ &= (-(x-2)-2) \frac{d}{dx}\frac{1}{1+(x-2)} \end{align}$$

$\endgroup$
0
1
$\begingroup$

You have made a slight mistake while replacing $x$ by "$(x-2)$", it should be "$(x+2)$".

Indeed it shows clearly when using an intermediate variable with a different name:

$\phantom{----}$substitute $(x\to 2)$ by $(x=2+u)$ with $(u\to 0)$


$\dfrac 1{1-x}$ becomes $\dfrac 1{1-(u+2)}=\dfrac {1}{-u-1}=\dfrac {-1}{1+u}=\dfrac{-1}{1+(x-2)}$

Now the $x-2$ appears but in the correct expression.

Then you can exploit the derivation trick presented by G.Tony Jacobs

And use $\sum\limits_{n=0}^{\infty}(2-x)^n\quad$ with a changed sign since we deal with $\dfrac 1{1+u}$ and not $\dfrac 1{1-u}$.


Edit: 11/11

$\begin{array}{ll}\displaystyle f(x) &=\dfrac x{(x-1)^2}=\dfrac{u+2}{(1+u)^2}=-(u+2)\mathrm{\dfrac {d}{du}}\left(\dfrac 1{1+u}\right)=-(u+2)\mathrm{\dfrac {d}{du}}\left(\sum\limits_{n=0}^{\infty}(-1)^nu^n\right)\\ \\ &\displaystyle\overset{(0)}{=}-(u+2)\sum\limits_{n=1}^{\infty}(-1)^nnu^{n-1}\\ \\ &\displaystyle\overset{(1)}{=}\sum\limits_{n=1}^{\infty}\bigg(-(-1)^nnu^{n}-2(-1)^nnu^{n-1}\bigg)\\ \\ &\displaystyle\overset{(2)}{=}\sum\limits_{n=1}^{\infty}-(-1)^nnu^{n}+\sum\limits_{n=0}^{\infty}2(-1)^n(n+1)u^n\\ \\ &\displaystyle\overset{(3)}{=}2+\sum\limits_{n=1}^{\infty}(-1)^n(n+2)u^{n}\\ \\ &\displaystyle\overset{(4)}{=}\sum\limits_{n=0}^{\infty}(-1)^n(n+2)u^{n} \end{array}$

$(0)$: derivate the innards of the sum term by term

$(1)$: distribute $(u+2)$ inside the sum

$(2)$: split sum into two parts and re-index the second one so it has $u^n$ as a general term.

$(3)$: isolate orphan terms (here this is the term in $u^0$) and regroup the other ones under the sum.

$(4)$: eventually since $2=(-1)^0(0+2)u^0$ we can re-embed it into the sum and start indexing at zero instead of $1$.

Note: you could have done $(3)$ and $(4)$ in one go, but as a general method, it is better to have separate steps as it happens that some series are not so compliant with orphaned terms.

Finally if you express the Taylor series with the proper $x$ variable you get:

$\displaystyle\dfrac x{(x-1)^2}=\sum\limits_{n=0}^{\infty}(n+2)(2-x)^n$

$\endgroup$
3
  • $\begingroup$ Thanks for pointing this out! Unfortunately I still don’t know how the derivation trick would help me here, since I am looking for the expansion around $x=2$ $\endgroup$ Nov 11 '17 at 0:34
  • $\begingroup$ I have written a full solution while explaining each step. As I have said, it is easier to work with the intermediate variable $u$ and replace by $(x-2)$ only at the very end. $\endgroup$
    – zwim
    Nov 11 '17 at 4:03
  • $\begingroup$ Thank you so much! Now it's clear :) $\endgroup$ Nov 11 '17 at 11:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.