# Laurent series by long division of trig function

1. Apr 6, 2017

### binbagsss

1. The problem statement, all variables and given/known data

Hi

I am trying to understand this http://math.stackexchange.com/quest...-laurent-series-for-fz-frac-1-cosz4-1-about-0

So the long division yields:

$f(z)=\frac{1}{cos(z^4)-1}=-2z^{-8}-\frac{1}{6}-\frac{1}{120}z^8+\frac{z^{16}}{240(\frac{-z^8}{2}+\frac{z^16}{24}-\frac{z^24}{720})}$

QUESTIONS

1) $cos(z^4)-1$ has a zero where $z^4=0$, so $f(z)$ has a pole at $z^4=0$, from the series above I see that this is a pole of order $8$, however I thought $z^4=0$ has 4 'solutions', not 8, order 4? how is this order 8? (I think I may be confusing it with the case $(cos(z))^4-1)$ which has a zero of order $4$ at $z=0$ right...?)

2) I'm rather confused by this.. looking at $\frac{z^{16}}{240(\frac{-z^8}{2}+\frac{z^16}{24}-\frac{z^24}{720})}$ ; I would have thought it would contribute to the $z^{8}$ and the $z^0$ term, just comparing the powers(from the $z^{16}$ in the numerator 'divided' by the $z^{16}$ and $z^{24}$ in the denominator) however looking at the solution which is $-2z^{-8}-\frac{1}{6}-\frac{1}{120}z^8-\frac{1}{3024}z^{16}$ it doesnt.

And how to compute the contributions from this remaining term, do I need to do long-divison again?? (however http://www.sosmath.com/algebra/factor/fac01/fac01.html here it says the degree of the denominator must be less to do long division, but it isn't so how to I get the contributions from this remaining term?)

3) Isn't the long division done backwards compared to here http://www.sosmath.com/algebra/factor/fac01/fac01.html i.e., lines 3 and 5 on the attached above,divide by the lowest order term, rather than the largest as here? when should you do which, or rather how how are we okay to do $1/g(z)$ , $g(z)$ some trig function by long division since the degree of the denominator is not lower?(I don't know whether the reason is to do with the above comment, that in this simple, polynomial example the degree of the denominator is lower- but in that case how are we okay to do $1/g(z)$ , $g(z)$ some trig function by long division since the degree of the denominator is not lower?

2. Relevant equations

see above

3. The attempt at a solution

see above

2. Apr 6, 2017

### vela

Staff Emeritus
Is there a reason you think the multiplicity of the root of $z^4$ should be equal to the order of the pole of $1/(\cos z^4-1)$?

The SOSmath page is talking about polynomial division. You're not dividing polynomials here; the denominator is an infinite series.

3. Apr 7, 2017

### binbagsss

when $cos(z^4)=1$ at $z^4=0 \implies cos(z^4)-1=0$ at $z^4=0$, pretty sure I reasoned my thoughts above.

Regarding not dividing a polynomial but an infinite series should you always be dividing by the lower degree terms rather than the higher , as done in the example lines 3 and 5?

4. Apr 7, 2017

### binbagsss

And since this is okay, (one) method of computing the contributions from the remaining terms is long-divison? so to get the laurent series you need to perform long-divison twice?

5. Apr 7, 2017

### vela

Staff Emeritus
You didn't explain why multiplicity=order; you simply asserted it. It's that assertion that I'm saying you should question. Suppose you have $f(w) = 1/w^2$. Then $f(z^n)$ has a singularity at $z=0$. Is the pole of order $n$? Or is it of order $2$ because that's the exponent of $w$? Or is it $2n$? And what about it if $f(w) = 1/w^3$? With the original problem, you're basically jumping to "it's of order $n$". Doesn't $f$ have something to do with it as well?

Suppose you want to calculate the series for $1/\sin x$. In order to divide by the highest-order term, you have to identify it first. What's the highest order term in the Taylor series for $\sin x$?

You divide as many times as you need to to obtain the number of terms you want.

6. Apr 8, 2017

### binbagsss

Okay so if we take the domain $[0,2\pi]$ it is order 8, whereas if the domain is assumed to be $(0,2\pi]$ or$[0,2\pi)$ it is order 4?

it's an infinite series, so...?

But whatever you choose to divide by you have to be consistent? so the above has the division by the $z^8$ term, but equally I could have chose to do it by $z^{24}?$

7. Apr 9, 2017

### vela

Staff Emeritus

How is the order of a pole is defined?

Try it and find out. Experimenting with the calculations yourself, you'll understand better why some choices work better than others.

8. Apr 10, 2017

### binbagsss

a) If you write the function as $1/f(z)$ it is defined by the order of the zeros of $f(z)$. so if I wasn't taking $f(z)$ into account as you said, I would have concluded the order is $4$, however since $cos(z)$ takes $1$ twice on $[0,2\pi]$ I conclude it is of order 8... (is this not the reason you were getting at?) whereas if I go back to a domain of either just one of $0,2\pi$, e.g $(0,2\pi]$ $cos(z)$ takes the value $1$ only once on this domain, so I conclude the order is that of $z^4$ i.e. back to order $4$

b) I wasn't asking which is better I was asking whether you need to be consistent with what you choose or not

thanks

Last edited: Apr 10, 2017
9. Apr 11, 2017

### binbagsss

There must be some sort of rule I mean, because if I choose to divide by $z^{16}/24$ first I get the term $24 z^{-16}$ and so would conclude the pole is of order 16 etc...

10. May 1, 2017

### binbagsss

11. May 6, 2017

### vela

Staff Emeritus
The points $z=0$ and $z=2\pi$ are different points in the complex plane, so the fact that $\cos z=1$ at both points isn't really pertinent to the order of the pole at one of the points. You're probably thinking about the fact that $e^{0i} = e^{2\pi i}$, but unless you're dealing with multi-valued functions, like $\log z$, those expressions refer to the same point.

12. May 7, 2017

### Ray Vickson

We have
$\cos(t)-1 = 1 - t^2/2! + t^4/4! - \cdots - 1 = - t^2/2 + \cdots,$
so when $t = z^4$ we have $\cos(z^4) -1 = - z^8/2 + \cdots$. In other words, $\cos(t)-1$ has a zero of order 2, so when $t = z^4$ we get a zero of order 8 in $z$. If we took $t = z^{100}$ we would have a zero of order 200 in $z$.

Thus, $1/(\cos(z^n)-1)$ has a pole of order $2n$ at $z = 0$. This is the case whether we restrict $z$ to $(-0.01, 0.01)$ or $(-\pi/2,\pi/2)$ or $(-2 \pi, 2 \pi)$, or the whole real line or the whole complex plane, etc. We are talking about what happens near $z = 0$, and that is not affected by how far out we let $z$ range when it goes a finite distance away from 0.

Last edited: May 7, 2017
13. May 7, 2017