1
$\begingroup$

I've been working on the following problem:

" Compute the Taylor series of the function $f(x)=\frac{1}{1-x}$ at $x=2$ and determine the radius of convergence $r$ "

I know that the given Taylor series is $\sum_{n=0}^{\infty} \frac{f^{(n)}(2)}{n!}(x-2)^n$, but I don't know how to determine the radius of convergence.

I have tried to find an expression for the Taylor series of $-\ln|1-x|$ at $x=2$ instead, because I was hoping to find a nice looking geometric series from which $r$ could easily be found. I have also tried to use the ratio test on $\sum_{n=0}^{\infty} \frac{f^{(n)}(2)}{n!}(x-2)^n$, but I don't know what to do with the derivatives.

$\endgroup$
  • $\begingroup$ Well, have you tried computing the derivatives? $\endgroup$ – Eric Wofsey Jun 3 '18 at 20:50
1
$\begingroup$

Hint: (General.) Radius of convergence of power series $\sum_n a_n(x-x_0)^n$ is given by formula $$r = \frac 1{\limsup_n |a_n|^{1/n}}.$$

Hint:(Specific.) $$\frac 1{1-x} = \frac 1{-1-(x-2)} = -\frac 1{1+(x-2)}$$ and thus the Taylor series around $x_0 = 2$ can be obtained from geometric series. Radius of convergence is the same as well.

Edit: $$\frac 1{1-x} = \frac 1{-1-(x-2)} = -\frac 1{1+(x-2)} =-\sum_{n=0}^\infty (-(x-2))^n = \sum_{n=0}^\infty (-1)^{n+1}(x-2)^n$$

$\endgroup$
  • $\begingroup$ I don't think I understand your last hint. The power series that conv. to $\frac{1}{1-x}=-\frac{1}{1+(x-2)}$ is $\sum x^n$ with $r=1$, but I don't see how the Taylor series around $x_0=2$ can be obtained from this. $\endgroup$ – Tom Jun 3 '18 at 22:01

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.