1. ## Taylor expansion and radius of convergence

Find the Taylor expansion around z = 0 and the radius of convergence of $\displaystyle \frac {z}{z^2+z-2}$.

I started out by simplify the expression. $\displaystyle \frac {z}{(z-1)(z+2)}$ Furthermore in separated fashion, I don't know the mathematical term. $\displaystyle \frac {1}{3} \frac{1}{(z-1)} + \frac {2}{3} \frac{1}{(z+2)}$

$\displaystyle - \frac {1}{3} \frac{1}{(1-z)} + \frac {2}{6} \frac{1}{(1+\frac{z}{2})}$ = $\displaystyle \frac {1}{3} \left( \frac{1}{(1+\frac{z}{2})} - \frac{1}{(1-z)} \right)$ = Expansion = $\displaystyle \frac {1}{3} \left( {(1-\frac{z}{2} + \frac{z^2}{4} - \frac{z^3}{8} + O (z^4))} - {(1+z+z^2+z^3 + O (z^4))} \right)$ = $\displaystyle \frac {1}{3} \left( 1-\frac{z}{2} + \frac{z^2}{4} - \frac{z^3}{8} - 1-z-z^2-z^3 + O (z^4)\right)$ = $\displaystyle - \frac {z}{2} -\frac{z^2}{4} - \frac{3z^3}{8} + O (z^4)$

Questions
I couldn't find any formula giving the expansion like for instance $\frac{z^n}{n!}$. How would I go about finding it? Is there a golden rule telling me how far I should develop if it isn't stated? I just happened to chose the power of 4 because my table of formulas does. Is the convergence radius = 1, (z-1) i the closest singularity to z = 0, right?

2. I agree that you should use the Partial Fractions method to get $\displaystyle \frac{1}{3}\left(\frac{1}{z - 1}\right) + \frac{2}{3}\left(\frac{1}{z + 2}\right)$.

Then I would remember that the infinite sum of a geometric series $\displaystyle \sum_{n = 0}^{\infty}ar^n = \frac{a}{1 - r}$ for $|r| < 1$.

Note that $\displaystyle \frac{1}{3}\left(\frac{1}{z - 1}\right) = -\frac{1}{3}\left(\frac{1}{1 - z}\right) = -\frac{1}{3}\sum_{n = 0}^{\infty}z^n$ for $|z| < 1$

and $\displaystyle \frac{2}{3}\left(\frac{1}{z + 2}\right) = \frac{1}{3}\left(\frac{1}{\frac{z}{2} + 1}\right) = \frac{1}{3}\left[\frac{1}{1 - \left(-\frac{z}{2}\right)}\right] = \frac{1}{3}\sum_{n = 0}^{\infty}\left(-\frac{z}{2}\right)^n$ for $\displaystyle \left|-\frac{z}{2}\right|<1$, i.e. $|z| < 2$.

3. Argh! Trying to add the two series together only messed things up. Thank you!

Is there some theorem saying that the series with the smallest convergence radius is the convergence radius for several added series? Or is it cool to just take that for granted?

Edit, I meant trying to calculate the terms of the to series. Adding/subtracting is what's supposed to be done.

4. I would say that you need an interval of convergence that makes both series converge, so yes, you can take that for granted.

5. Yes, there is a theorem saying that "the series with the smallest convergence radius is the convergence radius for several added series?" Here, since $\frac{1}{z- 1}$ is defined for all z except z= 1 and $\frac{1}{z+ 2}$ for all z except z= -2, and 1 is closer to 0 than -2 is, the interval of convergence is from -1 to 1 and the radius of convergence is 1.

One problem that I have seen on a couple of different tests is this: Find the radius of convergence of The Taylor's series expansion of $\frac{1}{1+ x^2}$ about x= 1.

Of course, one method would be to actually find the Taylor's series expansion by finding derivatives at x= 1. That's going to take more time than you want to use on a test!

The simplest is to note that the denominator is 0 at x= i and -i and so the fraction is defined for all x except i and -i. Since a series "converges as long as it can", the series for $\frac{1}{1+ x^2}$ will converge for all z closer to 1 than i or -i. The distance from 1 to either i or -i is $\sqrt{2}$ so the radius of convergence is $\sqrt{2}$.

Notice that the original statement of the problem did NOT imply that we were talking about complex numbers- but extending the problem into the complex numbers shows that the "radius of convergence" really is a radius!

6. HallsofIvy - True, sometimes you need to know the context, this particular question it's an exercise in complex analysis.

To clarify an other thing I mentioned. I had a look through some older tests given by my uni. It seems the expansion questions are supposed to be answered somewhat like this, given that the calculations are hunky dory. At least that's what's stated in the solutions.

$\displaystyle \frac {z}{z^2+z-2}$

Taylorexpansion: $\displaystyle - \frac{z}{2} - \frac{z^2}{4} - \frac{3z^3}{8} + O(z^4)$

Sum of the series: $\displaystyle \frac{1}{3} \sum_{n=0}^\infty \left( \left(- \frac{z}{2}\right)^n - z^n\right)$

Convergence: $|z| < 1$