1. ## Construct the series

$f(z) = \sum_{n=0}^{\infty}\frac{a_n}{z^n}$ for the function $\frac{z-1}{z+1}$

so i have that $\frac{z-1}{z+1} = 1 - \frac{2}{z+1}$

does that mean i expand the function at $z=-1$? or $z=\infty$ because the sum is to $\infty$

2. ## Re: Construct the series

Try constructing an ordinary power series at w=0 for g(w) = f(1/w), and then revert back to z via w = 1/z.

What you're doing here is creating a power series for f "about the point infinity" (or, equivalently, a Laurent series for f about z=0.)

3. ## Re: Construct the series

I guess it means you replace $z\to \frac{z-1}{z+1}$

4. ## Re: Construct the series

Originally Posted by johnsomeone
Try constructing an ordinary power series at w=0 for g(w) = f(1/w), and then revert back to z via w = 1/z.

What you're doing here is creating a power series for f "about the point infinity" (or, equivalently, a Laurent series for f about z=0.)
Can you elaborate i don't get what you mean.

5. ## Re: Construct the series

Also, we can express:

$f(z)=1-2\;\dfrac{1}{1-(-z)}=1-2\;\displaystyle\sum_{n=0}^{\infty}(-1)^nz^n\quad (0<|z|<1)$

$f(z)=1-\dfrac{2}{z}\;\dfrac{1}{1-(-1/z)}=1-\dfrac{2}{z}\displaystyle\sum_{n=0}^{\infty}\dfrac {(-1)^n}{z^n}\quad (1<|z|<\infty)$

The second expansion provides the solution:

$f(z)=1+\displaystyle\sum_{n=0}^{\infty}\dfrac{2(-1)^{n+1}}{z^{n+1}}=1+\sum_{n=1}^{\infty}\dfrac{2(-1)^{n}}{z^{n}}\quad (1<|z|<\infty)$

That is, $a_n=\begin{Bmatrix} 1 & \mbox{ if }& n=0\\2\;(-1)^n & \mbox{if}& n>0\end{matrix}$

6. ## Re: Construct the series

Instead of trying to directly express it in a (1/z) power series, which is kind of unusual/unnatural compared to our usual efforts of trying to express something in an ordinary power series, first substitute z = 1/w, and then try to find the "ordinary" power series in w. Once you've done that, switch back, using w = 1/z, and you'll have your answer.

You can think of that process in two ways. One way is just an algebraic book keeping exercise. Instead of "confusing" 1/z terms everywhere, we track them as "nice" w terms, and then at the end reveal that those nice w's were actually ugly 1/z's all along. You don't need to do this - FernadoRevilla's post shows how the calculation is doable directly. But it's a suggestion for helping you do that calculation.

The other perspective on this algebraic process I'll put in a separate post, so as not to scare the crap out of you. You can safely ignore it and don't need to know any of it to do this problem.

$f(z) = \frac{z-1}{z+1}$. Let $z = 1/w$.

Then $f(1/w) = \frac{\frac{1}{w}-1}{\frac{1}{w}+1} = \frac{\frac{1}{w}-1}{\frac{1}{w}+1} \ \frac{w}{w} = \frac{1-w}{1+w}$

$= \frac{1-(1+w)+1}{1+w} = \frac{2-(1+w)}{1+w} = \frac{2}{1+w} + \frac{-(1+w)}{1+w} = \frac{2}{1+w} -1$.

Thus $f(1/w) = -1 + \frac{2}{1+w}$.

Now, as is *very* common, try to make it look like the geometric series $\frac{1}{1-a} = 1 + a + a^2 + a^3 +...$.

(You could always start taking derivatives and using Taylor Series. But the geometric series trick is much easier when it's possible. It also gives you the radius of convergence of the series - for the above, it's $\lVert a \rVert < 1$.)

$f(1/w) = -1 + \frac{2}{1+w} = -1 + 2\frac{1}{1-(-w)}$

$= -1 + 2\{ 1 + (-w) + (-w)^2 + (-w)^3 + ... \}$ converging on $\lVert -w \rVert < 1$.

Thus $f(1/w) = -1 + 2\{ 1 - w + w^2 - w^3 + ... \} = (-1 + 2) - 2w + 2w^2 - 2w^3 + ...$

Thus $f(1/w) = 1 - 2w + 2w^2 - 2w^3 + ...$ converging on $\lVert -w \rVert < 1$.

Now w = 1/z, z = 1/w, so get

Thus $f(1/(1/z)) = 1 - 2(1/z) + 2(1/z)^2 - 2(1/z)^3 + ...$ converging on $\lVert -(1/z) \rVert < 1$.

Thus $f(z) = 1 - 2z^{-1} + 2z^{-2} - 2z^{-3} + ...$ converging on $1 < \lVert z \rVert$ (Same as FernandoRevilla's result).