Hello there,

How would I find the laurent series about 0 of 1/sinh(z) up to and including the z^5 term?

Thank you in advance!

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- Oct 27th 2010, 03:27 PMmathmaniac234Laurent Series and Number of Terms
Hello there,

How would I find the laurent series about 0 of 1/sinh(z) up to and including the z^5 term?

Thank you in advance! - Oct 28th 2010, 10:22 AMOpalg
This is not a particularly pleasant calculation. You know (presumably) that the power series for sinh(z) is . Therefore . Suppose that the inverse of that last expression in parentheses is (assuming, as is fairly obvious from the series in parentheses, that its inverse must also consist only of even powers of z). Then

.

Now compare coefficients of the constant terms on both sides to see that . Then compare successively the coefficients of and to find . After dividing by z, you have the coefficients of and in the Laurent series for . The numbers get fairly unwieldy, so check your results with the answer given here for csch(x). - Oct 28th 2010, 11:56 AMchisigma
The calculation is a little 'more pleasant' if we start considering the function...

(1)

... where...

(2)

Now we consider , that is analitic in so that we can write...

(3)

The can be computed considering that is...

(4)

... so that is [details of the calculation are omitted] ...

(5)

The (5) permit us to write...

(6)

... and from (6) we derive...

(7)

Kind regards