Results 1 to 4 of 4

Math Help - [SOLVED] about taylor and laurent series relation

  1. #1
    Bop
    Bop is offline
    Junior Member
    Joined
    Dec 2009
    Posts
    48

    [SOLVED] about taylor and laurent series relation

    Hello, I know a Taylor series for any function must be the same that laurent series of this function with a real variable, but do they have the same form always? I mean:

    We know:

    sin x = \sum \frac{(-1)^n}{(2n+1)!} x^{2n+1}

    and:

    sin z =\sum \frac{(-1)^{n}}{(2n+1)!}z^{2n+1}

    If we know taylor series of f(x) and we want calculate laurent of f(z), can we just replace x for z?


    Thank you.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member Failure's Avatar
    Joined
    Jul 2009
    From
    Zürich
    Posts
    555
    Quote Originally Posted by Bop View Post
    Hello, I know a Taylor series for any function must be the same that laurent series of this function with a real variable, but do they have the same form always? I mean:

    We know:

    sin x = \sum \frac{(-1)^n}{(2n+1)!} x^{2n+1}

    and:

    sin z =\sum \frac{(-1)^{n}}{(2n+1)!}z^{2n+1}

    If we know taylor series of f(x) and we want calculate laurent of f(z), can we just replace x for z?


    Thank you.
    If a Laurent series has no negative powers of z in it, it is the same as the Taylor series.
    If a Laurent series has negative powers of z in it, then a Taylor series does not exist (with radius of convergence > 0).
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Bop
    Bop is offline
    Junior Member
    Joined
    Dec 2009
    Posts
    48
    There is any way to find if all negative powers are zero without resolving all integrals?


    Thank you!
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor Bruno J.'s Avatar
    Joined
    Jun 2009
    From
    Canada
    Posts
    1,266
    Thanks
    1
    Awards
    1
    Quote Originally Posted by Bop View Post
    There is any way to find if all negative powers are zero without resolving all integrals?


    Thank you!
    Well, if the function has a finite value at z_0 and is analytic there, there are certainly no negative powers appearing in the expansion of the function in a neighbourhood of z_0.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Taylor or Laurent series
    Posted in the Advanced Math Topics Forum
    Replies: 5
    Last Post: May 6th 2010, 06:07 PM
  2. Laurent and Taylor Series
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: July 10th 2009, 05:45 AM
  3. Replies: 4
    Last Post: November 17th 2008, 11:53 PM
  4. Laurent + Taylor Series
    Posted in the Calculus Forum
    Replies: 1
    Last Post: January 10th 2008, 05:47 AM
  5. Taylor/Laurent series
    Posted in the Calculus Forum
    Replies: 1
    Last Post: November 8th 2007, 11:50 PM

Search Tags


/mathhelpforum @mathhelpforum