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.
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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.
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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).
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4) Newton-Raphson 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 Newton-Raphson, instead of long division.
Let's first see how you can use Newton-Raphson 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 non-linear equation:
1/x - y = 0
Newton-Raphson 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 Newton-Raphson 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.
