# Maclaurin Series

• April 18th 2010, 11:21 AM
ihr02
Maclaurin Series
How can I find the Maclaurin series for (z/z^3 + 8) and find the radiance of convergence?
• April 18th 2010, 12:28 PM
Failure
Quote:

Originally Posted by ihr02
How can I find the Maclaurin series for (z/z^3 + 8) and find the radiance of convergence?

One way to do it is this:

$\frac{z}{z^3+8}=\frac{z}{8}\cdot\frac{1}{1+\left(\ frac{z}{2}\right)^3}$
Now you can substitute $\left(\frac{z}{2}\right)^3$ for $u$ in $\frac{1}{1+u}=\sum_{k=0}^\infty(-1)^k u^k$ to get

$=\frac{z}{8}\cdot\sum_{k=0}^\infty (-1)^k\left(\frac{z}{2}\right)^{3k} = \sum_{k=0}^\infty \frac{(-1)^k}{2^{3(k+1)}}z^{3(k+1)}$.