1
$\begingroup$

I am currently working on the power series for a homework assignment. I have to find the radius of convergence for the function $$\frac{10}{1+64x^2}$$

By setting up the

$$\frac{1}{1+64x^2}$$

part of the expression as a derivative of

$$\frac{1}{8}arctan(8x)$$

I was able to expand the series term by term and get the correct co-efficients.

The series is:

$$\frac{10}{1+64x^2} = 10 - 640x^2 + 40960x^4 - 2621440x^6 ...$$

This series is correct, since my homework portal accepts it. In order to find the radius of convergence, I need to be able to put this series in the form of a summation notation so that I can apply the ratio tests. How do I extract this series' summation notation? How would I account for the 0 co-efficients for the terms with degree $1, 3$, and so forth (odd-terms). I know it involves an alternating series somewhere, but I am not sure how to account for the terms whose co-efficients are zero, when it comes to getting the summation form notation.

$\endgroup$
3
  • $\begingroup$ Hint: Is there something that you can multiply each term by to get the next? $\endgroup$ – Archaick Apr 12 '15 at 6:21
  • $\begingroup$ Please use latex for maths text.. See here. $\endgroup$ – mattos Apr 12 '15 at 6:23
  • $\begingroup$ 8x? I will learn how to use LaTeX, I'm new to this website and haven't learned it yet. Thanks for the resource. $\endgroup$ – Ferreroire Apr 12 '15 at 6:26
2
$\begingroup$

$$\begin{align} \frac{10}{1+64x^2} &= 10 - 640x^2 + 40960x^4 - 2621440x^6 ... \\ &= 10 [1 - 64x^{2} + 4096x^{4} - 262144x^{6} ... ]\\ &= 10 [(-1)^{0}(8x)^{0} + (-1)^{1}(8x)^{2} + (-1)^{2}(8x)^{4} + ...] \\ &= 10 \sum_{k = 0}^{\infty} (-1)^{k} (8x)^{2k} \\ \end{align}$$

$\endgroup$
1
  • $\begingroup$ Awesome, I was blind! The power ^2k explains so much, thanks! I thought along the lines of how to get it to cycle through the odd terms as well. $\endgroup$ – Ferreroire Apr 12 '15 at 6:33

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.