[SOLVED] Laurent series

Bop

Dec 2009
48
0
Hello, I want to calculate laurent series and its convergence radius of:

\(\displaystyle f(z)=\frac{1}{sin({\pi z})}\)

Here it is what i've done, I'm not sure if we can do it in this way:


\(\displaystyle sin z =\sum \frac{(-1)^{n}}{(2n+1)!}z^{2n+1}=z-\frac{z^3}{3!}+\frac{z^5}{5!}-O(z^7)\) for \(\displaystyle -1<z<1\)

so:

\(\displaystyle sin (\pi z) =\sum \frac{(-1)^{n}}{(2n+1)!}(\pi z)^{2n+1}=\pi z-\frac{(\pi z)^3}{3!}+\frac{(\pi z)^5}{5!}-O((\pi z)^7)\) for \(\displaystyle -1<z<1\)

so:


\(\displaystyle \frac{1}{sin (\pi z)} =\sum \frac{(2n+1)!}{(-1)^{n}}\frac{1}{(\pi z)^{2n+1}}=\frac{1}{\pi z}-\frac{3!}{(\pi z)^3}+\frac{5!}{(\pi z)^5}-\frac{1}{O((\pi z)^7)}\) for \(\displaystyle -1<z<1\)


Is it right?


Thank you.
 
Apr 2010
12
0
warning
\(\displaystyle \sum \frac{A}{B} \neq (\sum \frac{B}{A})^(-1) \)
 

chisigma

MHF Hall of Honor
Mar 2009
2,162
994
near Piacenza (Italy)
We can start from the series expansion...

\(\displaystyle \frac{\sin (\pi z)}{\pi z} = 1 - \frac{(\pi z)^{2}}{3!} + \frac{(\pi z)^{4}}{5!} - \dots\) (1)

... and then, setting...

\(\displaystyle f(z) = \frac{\pi z} {\sin (\pi z)} = a_{0} + a_{1} z + a_{2} z^{2} + \dots\) (2)

... find the \(\displaystyle a_{n}\) imposing...

\(\displaystyle f(z) \frac{\sin (\pi z)}{\pi z} = (a_{0} + a_{1} z + a_{2} z^{2} + \dots) \{ 1 - \frac{(\pi z)^{2}}{3!} + \frac{(\pi z)^{4}}{5!} - \dots\} = 1\) (3)

From (3) now we derive directly...

\(\displaystyle a_{0} = 1\)

\(\displaystyle a_{1} =0\)

\(\displaystyle a_{2} - a_{0} \frac{\pi ^{2}}{3!} = 0 \rightarrow a_{2} = \frac{\pi^{2}}{6}\)

\(\displaystyle a_{3} =0\)

\(\displaystyle a_{4} - a_{2} \frac{\pi^{2}}{3!} + a_{0} \frac{\pi^{4}}{5!}=0 \rightarrow a_{4} = \frac{7 \pi^{4}}{360}\) (4)

... and so one, so that is...

\(\displaystyle \frac{\pi z}{\sin (\pi z)} = 1 + \frac {\pi^{2} z^{2}} {6} + \frac{7 \pi ^{4} z^{4}}{360} + \dots\) (5)

... and now, deviding (5) by \(\displaystyle \pi z\), we obtain...

\(\displaystyle \frac{1}{\sin (\pi z)} = \frac{1}{\pi z} + \frac {\pi z} {6} + \frac{7 \pi ^{3} z^{3}}{360} + \dots\) (6)

The Laurent series (6) converges for \(\displaystyle 0 < |z| < 1\) ...

Kind regards

\(\displaystyle \chi\) \(\displaystyle \sigma\)
 
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