I am working on a power series representation from Stewart's book (an old edition). I am asked to find the power series representation of

\begin{align*} f(x) = \frac{x^3}{(x-2)^2} \end{align*}

I would like to know if I am on the right track, and/or if my answer is correct. It follows.

My answer

\begin{align*} \frac{1}{1-x} &= 1 + x + x^2 + \cdots = \sum_{n=0}^{\infty} x^n\\ \frac{1}{1-(-x+3)} &= 1 + (-x+3) + (-x+3)^2 + \cdots = \sum_{n=0}^{\infty} (-x+3)^n\\ \frac{1}{x-2} &= 1 + (-x+3) + (-x+3)^2 + \cdots = \sum_{n=0}^{\infty} (-x+3)^n \end{align*}

Differentiating both sides,

\begin{align*} \frac{-1}{(x-2)^2} &= -1 +2 (-x+3)(-1) -3 (-x+3)^2 (-1) + \cdots = \sum_{n=0}^{\infty} -n \cdot (-x+3)^{n-1}\\ \frac{1}{(x-2)^2} &= \sum_{n=0}^{\infty} n \cdot (-x+3)^{n-1} \end{align*}

Multiplying both sides by $x^3$,

\begin{align*} x^3 \cdot \frac{1}{(x-2)^2} &= x^3 \cdot \sum_{n=0}^{\infty} n \cdot (-x+3)^{n-1}\\ \frac{x^3}{(x-2)^2} &= \sum_{n=0}^{\infty} n \cdot x^3 \cdot (-x+3)^{n-1} \end{align*}

Given we have the series, we may proceed to find the radius and interval of convergence:

\begin{align*} a_n &= n \cdot x^3 \cdot (-x+3)^{n-1} \quad a_{n+1} = (n+1) \cdot x^3 \cdot (-x+3)^{n} \quad \frac{a_{n+1}}{a_n} = \frac{(n+1)(-x+3)}{n}\\ \end{align*}

\begin{align*} \lim_{n \to \infty} \left\vert \frac{a_{n+1}}{a_n} \right\vert &= \vert -x+3 \vert \end{align*}

So we find that $R = 1$ and $I = ]2,4[$

(I'll still work on the "edges" of the interval).

Can someone please verify what I've calculated?

  • $\begingroup$ Any range is given on $x$ ? Because the geometric sum works if $(-x+3)\le 1$.. $\endgroup$ – Qwerty Nov 30 '17 at 19:27
  • $\begingroup$ Hi Qwerty. No, no range is given. $\endgroup$ – bru1987 Nov 30 '17 at 19:28
  • 1
    $\begingroup$ If not range or pivot point is given, why did you decide to give a power series about the point $\;x=3\;$ ?? Why not around $\;x=2,\,\frac12\;$ or something else? $\endgroup$ – DonAntonio Nov 30 '17 at 19:29
  • $\begingroup$ Well, let's say that I started with the $x^n$ series and worked my way with it. Would you suggest otherwise @DonAntonio? $\endgroup$ – bru1987 Nov 30 '17 at 19:33
  • $\begingroup$ For example$$\frac1{x-2}=-\frac12\frac1{1-\frac x2}=-\frac12\left(1+\frac x2+\frac{x^2}4+\ldots\right)\;,\;\;etc.$$ $\endgroup$ – DonAntonio Nov 30 '17 at 19:38

Your Answer

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

Browse other questions tagged or ask your own question.