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Turn 1/(1-cos(z)) into a laurent series.

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- Thread starter Mappe
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- #1

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Turn 1/(1-cos(z)) into a laurent series.

- #2

Dick

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- #3

vela

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A Laurent series has both positive and negative powers of the variable while the geometric series has just positive powers.I don't quite understand a few details here. First, What is the difference between geometric series and laurent series?

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HallsofIvy

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A "geometric series" is not even a series of functions, but a numerical series:

A geometric series is always of the form

[tex]\sum_{n=0}^\infty a r^n[/tex]

where a and r are given

A "power series" is a series of functions, each of the form \(\displaystyle a_nx^n\) for each "a_n" a number and x a variable. A "Taylor's series", for a specific function, is a power series with specific coefficients, [itex]f^{(n)}(x_0)/n![/itex]. Even if we take the "r" in the geometric series to be a variable, we get a power series specifically of the form [itex]\sum_{n=0}^\infty ax^n[/itex] which is the Taylor's series for a/(1- x) around 0.

A "Laurent" series is like a Taylor's series with negative as well as positive exponents.

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vela

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Vela may have thought you were comparing a Laurent series to a Taylor's series (which you probably intended).

Yes, indeed. I should be more careful with the terminology. Thanks for clarifying.

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vela

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HallsofIvy

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I have no idea what you are doing here. If your series has no negative powers in it, then its value at z= 0 is just its constant term (or 0 if there is no constant term). But 1/(1- cos(z)) is not defined at z= 0.^{z^-1}terms, as it's one taylor expansion of cos, goes from 0 -> inf, and one geometric series of 1/(1-f(x)), goes from 0 -> inf. That is, no -inf. And the cos series only produces even number powered z:s. And is there any easy rules for finding coefficients for sum in sum?

- #10

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Here is how:

[tex]\sum 1/(1-cos(x))[/tex] = [tex]\sum 1/(1-(\sum(-1)^n(x)^{2n}/(2n!)))[/tex] = [tex]\sum(\sum(-1)^n(x)^{2n-2}/((2n-2)!))^m[/tex]

hope it doesnt look like a mess :)

[tex]\sum 1/(1-cos(x))[/tex] = [tex]\sum 1/(1-(\sum(-1)^n(x)^{2n}/(2n!)))[/tex] = [tex]\sum(\sum(-1)^n(x)^{2n-2}/((2n-2)!))^m[/tex]

hope it doesnt look like a mess :)

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- #11

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ohh. it did

- #12

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1/(1-cos(x)) = 1/(1-E(-1)^n*x^(2n)/(2n!)) = 2*E(E((-1)^n*x^(2n-2))/((2n+2)!))^m

where both E's (sums) goes from 0 -> inf, and m is for the outer sum.

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