Complex Numbers

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RobertPaulson

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Hi, I was reading a previous topic on laurent series and I don't understand the implied math between two steps

It basically was cot(z) = cos(z)/sin(z)

= (1 - (z[sup:21t19b6l]2[/sup:21t19b6l])/2 + Oz[sup:21t19b6l]4[/sup:21t19b6l]) / (z + (z[sup:21t19b6l]3[/sup:21t19b6l])/6 + Oz[sup:21t19b6l]5[/sup:21t19b6l]) I think

and then this went straight to

= (1 - (z[sup:21t19b6l]2[/sup:21t19b6l])/2 + Oz[sup:21t19b6l]4[/sup:21t19b6l])(1/z + z/6 + Oz[sup:21t19b6l]3[/sup:21t19b6l])

Am i missing something stupidly simple?
 
\(\displaystyle That\ term\ needs\ to\ be\ a\ denominator\ factor, it\ should\ be\ multiplied\ by\ the\ 2nd\ denominator.\)
 
\(\displaystyle There\ was\ a\ certain\ expansion\ of\ terms\ there,\ involving\ a\ division\ by\ z^4.\)

\(\displaystyle The\ instructor\ was\ working\ a\ certain\ context\ there.\)

\(\displaystyle You\ should\ quote\ from\ that\ one\ specifying\ which\ line\ seems\ unclear\ to\ you.\)
 
The Laurent expansion for cot(z) is

\(\displaystyle \frac{1}{z}-\frac{1}{3}z-\frac{1}{45}z^{3}-\frac{2}{945}z^{5}-................\)

In general: \(\displaystyle \sum_{n=0}\frac{(-1)^{n}2^{2n}B_{2n}}{(2n)!}z^{2n-1}\)

\(\displaystyle B_{2n}\) are the Bernoulli numbers. The series for cot(z)/z^4 then comes from dividing this by z^4
 
Firstly I just want to say thanks for all the help so far and sorry for the big delay but i'm still having issues with this.

Basically, galactus i don't really fully understand the laurent series yet so i'm not sure what you've done there; I understand most of what was said in the previous topic except for this one step:

\(\displaystyle cot(z)=\frac{cos(z)}{sin(z)}=\left(1-\frac{z^{2}}{2}+O(z^{4})\right)\left(\frac{1}{z}+\frac{z}{6}+O(z^{3})\right)\)

I don't understand how you went from the taylor series expansion for sin(z) and cos(z) to what you have above. If someone could outline the steps that were taken that would be great.

Thanks
 
Think of it as \(\displaystyle \frac{cos(z)}{sin(z)}=cos(z)\cdot \frac{1}{sin(z)}=cos(z)csc(z)\)

The series for cos(z) is multiplied by the series for csc(z)

The series for csc(z) is \(\displaystyle \frac{1}{z}+\frac{z}{6}+\frac{7z^{3}}{360}+................\)

The Taylor series for cot, csc, sec are more complicated than the common ones with sin, cos, tan. They involve Bernouilli numbers and all that.

The series for sec involves the Euler numbers instead of the Bernouilli numbers, though.

Go to Mathworld or Wikipedia and read about Bernouilli and Euler numbers. They are interesting and pop up in unexpected places.
 
\(\displaystyle Galactus\ will\ probably\ show\ you\ more\ of\ this.\)

\(\displaystyle The\ series\ in\ brackets\ on\ the\ left\ is\ the\ series\ for\ cos(z)\)

\(\displaystyle and\ the\ one\ on\ the\ right\ is\ the\ one\ for\ \frac{1}{sin(z)}.\)

\(\displaystyle cot(z)=\frac{cos(z)}{sin(z)}=cos(z)\frac{1}{sin(z)}\)
 
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