Laurent series by long division of trig function

[binbagsss]

Homework Statement

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?

Homework Equations

see above

The Attempt at a Solution

[/B]
see above

 

[vela]

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...?)

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

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?

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

 

[binbagsss]

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.

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?

 

[binbagsss]

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

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?

 

[vela]

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.

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?

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?

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 $\sin x$?

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?

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

 

[binbagsss]

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?.

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?

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.

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 $z^{24}?$

 

[vela]

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?

How is the order of a pole is defined?

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 $z^{24}?$

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

 

[binbagsss]

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.

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

 

[binbagsss]

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

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...

 

[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 $4$

Was this completely off track since it had no reply? I did try to take on board your comment. ..

 

[vela]

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.

 

[Ray Vickson]

Was this completely off track since it had no reply? I did try to take on board your comment. ..

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.

 

[binbagsss]

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.

what breed is your dog?

 

[Ray Vickson]

what breed is your dog?

His father is a schnauzer and his mother is a bichon-poodle cross. I prefer the term "designer dog" to "mutt".

 

[binbagsss]

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 $z$.

Should this be obvious / is there another way to see this without expanding out cos ? Thanks

 

[fresh_42]

Should this be obvious / is there another way to see this without expanding out cos ? Thanks

Yes. For $f(w)=-1+\cos w$ we have $f(0)=0$ and $f\,'(0)=0$, hence a double zero. With $w=z^4$ it becomes an eight fold zero.

I think the question has been answered, so we can close here.