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What is the difference between geometric series and laurent series?

  1. Jan 7, 2010 #1
    I don't quite understand a few details here. First, What is the difference between geometric series and laurent series? Than, how do I multiply/divide 2 series with each other? Finally, I have this problem, and I'm really clueless as of what to do.

    Turn 1/(1-cos(z)) into a laurent series.
     
  2. jcsd
  3. Jan 7, 2010 #2

    Dick

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    I really don't want to do this one. It's perfectly straightforward, it's just a pain in the neck to get the coefficients. Power expand cos(x) and cancel the 1. Factor the z^2 out of the numerator. Now you have (2/z^2)*(1/1-f(z)) where f(z) is a power series in z. Use a geometric expansion of (1/(1-f(z))=1+f(z)+f(z)^2+... Now to get a certain number of positive powers of z in the expansion you have to judiciously pick how many terms in the powers of f(z) you need to get to coefficients of the powers of z right.
     
  4. Jan 8, 2010 #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.
     
  5. Jan 8, 2010 #4

    HallsofIvy

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

    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 numbers.

    A "power series" is a series of functions, each of the form [math]a_nx^n[/math] 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.
     
  6. Jan 8, 2010 #5

    vela

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    Yes, indeed. I should be more careful with the terminology. Thanks for clarifying.
     
  7. Jan 8, 2010 #6
    Thanks. I tried this way of solving above, and now I get a sum in a sum. And its from 0 -> inf. This isnt a proper laurentseries, is it? Also, I didn't understand the coefficient for powers of z stuff. And choosing number of terms? Can I not just expand it as it is right now, a geometric series of 1/(1-(series:cos(z)))?
     
  8. Jan 8, 2010 #7

    vela

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    Now you know why Dick didn't want to do it. :) It does sound like you're doing it correctly so far. Now you have to multiply everything out and collect terms. It's not really feasible to do it in general, so in practice, you decide that you want, say, the first 4 terms, which for your problem are the [tex]z^{-2}, z^{-1}, z^0,[/tex] and [tex]z^1[/tex] terms. You have to figure out where all the contributions for those powers of z will come from.
     
  9. Jan 8, 2010 #8
    OK, but the series I come up with doesnt give any 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. Jan 8, 2010 #9

    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.

    How, exactly, did you use the Taylor's series for cos(x) and 1/(1-x)?
     
  11. Jan 8, 2010 #10
    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 :)
     
    Last edited by a moderator: Jan 9, 2010
  12. Jan 8, 2010 #11
    ohh. it did
     
  13. Jan 8, 2010 #12
    with ordinary letters:

    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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