0
$\begingroup$

would anyone tell me how to solve this?

Use differentiation to find a power series representation for

$$\frac{1}{(3+x)^{2}}$$

What is the radius of convergence, R?

$\endgroup$
  • $\begingroup$ Expanding around x = 0? $\endgroup$ – Ben Lansdell Nov 21 '16 at 6:26
  • $\begingroup$ @ lansdell yes. I think so. $\endgroup$ – Tmm Nov 21 '16 at 6:28
  • $\begingroup$ What is the derivative of $-\dfrac{1}{x+3}$? Can you find the power series representation of $-\dfrac{1}{x+3}$? $\endgroup$ – John Wayland Bales Nov 21 '16 at 6:28
2
$\begingroup$

Possible hint: We have $$1/(3 + x)=(1/3)\frac{1}{(1-(-x/3))}$$ Use this fact that $1/(1-t)=\sum_0^{\infty}t^n~~~(*)$ where $|t|<1$. I mean take $t=-x/3$ and...

Note that $|-t|=|t|<1$ is the radius of convergence of $(*)$

$\endgroup$
1
$\begingroup$

The series of course depends on where you expand around.

Taking $x = 0$ as an example: $$ f(x) = \frac{1}{(x+3)^2} = \sum_{n=0}^\infty f^{(n)}(0)\frac{x^n}{n!}, $$ where $$ f^{(n)}(x) = \frac{(-1)^n(n+1)!}{(x+3)^{n+2}}. $$ So the series is $$ \frac{1}{(x+3)^2} = \sum_{n=0}^\infty \frac{(-1)^n(n+1)!}{(3)^{n+2}}\frac{x^n}{n!} = \sum_{n=0}^\infty \frac{(-1)^n(n+1)}{(3)^{n+2}}x^n, $$ This would have a radius of convergence of 3.

Edit:

As the problem stipulates you should use differentiation, note that another way to get to the same series, as hinted at, is by noting: $$ \frac{1}{(x+3)^2} = -\frac{d}{dx}\frac{1}{(x+3)} $$ where $$ \frac{1}{(x+3)} = \frac{1}{3}\frac{1}{1-(-x/3)} =\frac{1}{3}\sum^\infty_{n=0}[-x/3]^n $$ provided $|-x/3|<1$ ie provided $|x|<3$. Thus differentiating this series gives you the same answer as above.

$\endgroup$

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.