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Laurent Series expansion for the following function 
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#1
Dec609, 12:06 AM

P: 10

1. The problem statement, all variables and given/known data
Find the Laurent series expansion of f(z) = (e^z  1) / (sinz)^3 at z = 0. 3. The attempt at a solution Ok, so I'm confused on a number of fronts here. For e^z  1, I assume you just use the standard power series expansion of e^z and then tack on a 1 at the end, which would give you the Taylor series of that part. For 1/(sinz)^3 though, I'm really confused. I'm not even sure about (sinz)^3, nevermind the inverse. Do you take the power series of sinz multiplied three times to get (sinz)^3 and then somehow inverse it? I mean, nevermind the Laurent series, I'm not even sure what the Taylor series of (sinz)^3 is. I guess you could try to find a formula for the nth derivative and then use that formula for the Taylor series, but that looks like it would be pretty messy or even impossible. I'm sure I'm missing something obvious here but I don't know what. 


#2
Dec609, 03:17 AM

P: 939

There is something wrong with your question. f(z=0) = infinite, which does not work well with a Laurent series.



#3
Dec609, 10:27 AM

P: 10

Are you sure it's infinite? If you substitute z = 0, then the function is of the form 0/0 where it's not immediately clear what's going on.



#4
Dec709, 01:34 AM

P: 939

Laurent Series expansion for the following function
Yes I am. sin^3 z behaves like z^3 and e^z1 behaves like z near z=0.



#5
Dec709, 08:31 AM

P: 10

I don't understand your point that it doesn't "work well" with a Laurent series. As far as I know Laurent series are supposed to be used in these cases where a function fails to be analytic at a point, and when you find the Laurent series for the function you'll see what type the singularity is from the principal part (i.e. a removable singularity, a simple pole, a pole of order n, or an essential singularity).



#6
Dec709, 09:31 AM

P: 2,158

This is a good exercise to practice Laurent expansions using different methods.
1) Direct substitution of the Taylor expansions in the expression. You can write: f(z) = [exp(z)  1]/[sin(z)]^3 = [z + z^2/2 + z^3/6 + ...]/[z  z^3/6 + z^5/5!  ...]^3 = (take out factor of z in the numerator and a factor of z from the brackets in the denominator) = 1/z^2 [1+ z/2 + z^2/6 + ...]/[1  z^2/6 + z^3/5!  ...]^3 You can now expand this in 3 possible way: 1.1) Use the formula for 1/(1+x)^n and substitute x =  z^2/6 + z^3/5!  1.2) Write [1+ z/2 + z^2/6 + ...]/[1  z^2/6 + z^3/5!  ...]^3 = a + b z + c z^2 + ... mulitply both sides by [1  z^2/6 + z^3/5!  ...]^3 expand out the product on the r.h.s and solve for a, b, c,... 1.3) Expand [1  z^2/6 + z^3/5!  ...]^3 and perform a long division.  2) It is easy to see that f(z) behaves as 1/z^2 near z = 0, so you could multiply f(z) by z^2 and expand that in the Taylor expansion. However, you do then have to consider quite nasty limits that are best computed by substituting Taylor expansions.  3) Directly using the method (1.2): Write: [exp(z)  1]/[sin(z)]^3 = a/z^2 + b/z + c + d z + ..... Multiply both sides by sin^3(z) insert the Taylor expansion of sin(z) and expand ot everything, solve for a, b, c, etc. In all of the above, you can also rewrite sin^3(z) in terms of sin(3z) and sin(z) which saves you from having to take the third power of the Taylor expansion of sin(z).  4) NewtonRaphson method. Useful if you want find many terms of the expansion using just a few steps. You have seen that almost all of the above methods involve performing a division. Long division will yield the terms of the expansions one by one, which means performing a billion computations, if you want to find the billionth term. To speed of things, you can use use division using NewtonRaphson, instead of long division. Let's first see how you can use NewtonRaphson when dividing ordinary numbers (instead of Taylor expansions or Polynomials). If you want to compute: x = 1/y for given y, you can consider this has solving the nonlinear equation: 1/x  y = 0 NewtonRaphson then gives: x_{n+1} = x_{n}  [1/x_{n}  y]/(1/x_{n}^2) = 2 x_{n} y x_{n}^2 There are no divisions to be performed in this algorithm, so it is a proper division algorithm. Also the covergence is quadratic, much faster than long division. In case of poynomials/Taylor expansions, we want to compute: P(x) = 1/Q(x) Then you simply take the above NewtonRaphson algorithm: P_{n+1} = 2 P_{n}  Q P_{n}^2 You take P_{0} to be correct to lowest order and then the above algorithm will double the number of correct expansion coefficients in each iteration. 


#7
Dec709, 10:31 AM

P: 10

Edit: I think I managed to do the long division correctly (possibly?). Can someone confirm whether the coefficient of the z^1 term is 1/2? It's important because the next several parts all depend on the value of the residue at z = 0. 


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