- #1
binbagsss
- 1,254
- 11
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