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$f(z)$ is defined like this: $$ f(z) = \frac{z}{(z-1)(z-3)} $$ I need to find a series for $f(z)$ that involves positive and negative powers of $(z-1)$, which converges to $f(z)$ when $0 \leq |z - 1| \leq 2$.

What I understand from question is I must expand $f(z)$ Laurent series.

$$ f(z) = \sum_{m=0}^{\infty}a_{m}(z-1)^{m} + \sum_{m=1}^{\infty}b_{m}(z-1)^{-m}$$

where,

$$ a_{m} = \frac{1}{j2\pi}\oint_{C}\frac{f(z)}{(z-1)^{m+1}}dz $$

$$ b_{m} = \frac{1}{j2\pi}\oint_{C}\frac{f(z)}{(z-1)^{1-m}}dz $$

This is what theory tells me.

But I apply partial fraction method to this function like this:

$$ f(z) = \frac{z}{(z-1)(z-3)} = \frac{z^{-1}}{(1-z^{-1})(1-3z^{-1})} = \frac{-1/2}{(1-z^{-1})} + \frac{1/2}{(1-3z^{-1})} $$

And I know this series expansion from z-transform like this:

$$ f(z) = -\frac{1}{2} \sum_{k=0}^{\infty}z^{-k} + \frac{1}{2} \sum_{k=0}^{\infty}3^{k}z^{-k}$$

I obtain a series expansion but it looks like Mclaurin series not a Laurent series.

Here, my first question an expression may have different type of series expansion?

And second, how to find a Laurent series for $ f(z) $

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No need for contour integrals, just give a name to the quantity you want a Laurent series in, and expand. So with $x=z-1$: $$ \frac z{(z-1)(z-3)}=\frac{x+1}{x(x-2)} =x^{-1}\left(1-\frac3{2-x}\right) \\=x^{-1}\left(1-\frac32\sum_{i\geq0}\bigl(\frac x2\bigr)^i\right) =-\frac12x^{-1}+\sum_{i\geq0}\frac{-3}{4\times2^i}x^i. $$ You can now substitute $x:=z-1$ if you like.

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    $\begingroup$ This is a good one except "-2" part (I think I must be directly "2"). Thanks. $\endgroup$ – mehmet Mar 9 '15 at 17:45
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    $\begingroup$ Soryy, I forgot to say I mean in sum operation. I believe it must be positive sign of 2. $\endgroup$ – mehmet Mar 9 '15 at 18:16
  • $\begingroup$ Also the sign of the sum operation must be negative, I think? $\endgroup$ – mehmet Mar 9 '15 at 18:26
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    $\begingroup$ @mehmet: OK, I see now. Corrected that, thanks. $\endgroup$ – Marc van Leeuwen Mar 10 '15 at 5:28
  • $\begingroup$ Out of interest how would you do Laurent Expansions with Integrals if I may ask $\endgroup$ – Zophikel Sep 11 '17 at 20:39
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The problem is that if you use $\frac 1{1-z} = \sum z^n$ you are essentially writing the Laurent expansion in a neighborhood of $0$. Since you want powers of $z-1$, it means that you want an expansion in a neighborhood of $1$!

Hence the $\frac {1}{z-1} = (z-1)^{-1}$ term is already "good" (just like $\frac 1z$ would be in a Laurent expansion in a neighborhood of $0$).

You have to expand in a neighborhood of $1$ the expression $\frac z{z-3}$. You can set $t = z - 1 \implies z = t + 1$ so your expression becomes $\frac{t+1}{t-2}$.

Now use partial fractions, and find the Laurent series in a neighborhood of $0$ with respect to $t$ of $\frac{t+1}{t-2}$ (This means you can use the geometric series formula), substitute back $t = z-1$ and divide everything by $z-1$ to get the final result.

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  • $\begingroup$ Thanks for your explanation! (: $\endgroup$ – mehmet Mar 9 '15 at 18:19
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    $\begingroup$ @mehmet You're welcome! :) $\endgroup$ – Ant Mar 9 '15 at 18:20
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First, yes. The same function can have different Laurent series, depending on the center of annulus in question.

Consider now the given function. Clearly, the question will be solved with ease once we find the series of $$g(z)=\frac{z}{z-3}.$$Note that$$g(z)=1+\frac{3}{z-3}=1+\frac{3}{(z-1)-2}=1+\frac{3/2}{\frac{z-1}{2}-1},$$and the last expression can be represented as the sum of a geometric series.

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  • $\begingroup$ Thank you for your answer (: $\endgroup$ – mehmet Mar 9 '15 at 17:42

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