Results 1 to 7 of 7

Math Help - Getting stuck on series - can't develop Taylor series.

  1. #1
    Member
    Joined
    Sep 2010
    Posts
    95

    Getting stuck on series - can't develop Taylor series.

    \frac{Log z}{sin(\pi z)} Find convergences radius at z = 1 and develop a Taylor series.

    Just by looking at the function I see that deviser and numerator has zeros in z = 1, so theres a limit in z = 1. The numerator isn't defined in z = 0. So the convergence radius is |z-1| < 1. (See deviser also, zeros in z = 2, 3, 4 ...)

    First the log function, I think I've got it right. The Taylor expansion is also pretty straight forward.
    \log z = \log (z - 1 + 1) = \log 1 + (z - 1) =
    (z-1) - \frac{(z - 1)^2}{2} + \frac{(z - 1)^3}{3} - \frac{(z - 1)^4}{4}+O((z-1)^5)

    In this part I'm starting to get shaky.
    \sin \pi z = - \sin \pi (z - 1)

    \frac {1}{\sin \pi z} = -\frac{1}{\pi(z-1)} \frac{1}{1 - \frac{\pi^2}{6}(z-1)^2 - O((z-1)^4)} - This is where I need some pointers. I can't simplify this enough to multiply it with the log z part to get the whole series.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Bruno J.'s Avatar
    Joined
    Jun 2009
    From
    Canada
    Posts
    1,266
    Thanks
    1
    Awards
    1
    Are you just trying to find the first few terms of the expansion, or also the general coefficient? Because this would be much more complicated. The cosecant has a mean-looking series expansion involving the Bernoulli numbers.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Sep 2010
    Posts
    95
    Oh sorry, I forgot to mention that. In the exercise it's stated that the expantion should be in the power of 3.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Sep 2010
    Posts
    95
    Update
    I asked a teacher about this exercise and was shown a solution. However, the part where I got stuck still puzzles me. It wasn't even in the solution...

    \frac{1}{\sin \pi z} = -\frac{1}{\pi (z-1)} \frac{1}{1-\left(\frac{\pi^2}{6}(z-1)^2 + O((z-1)^4)\right)} = ??? = -\frac{1}{\pi (z-1)}\left(1+ \frac{\pi^2}{6}(z-1)^2 + O((z-1)^4)\right)

    From here on I'm ok, but if someone could explain the trick above...
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor Bruno J.'s Avatar
    Joined
    Jun 2009
    From
    Canada
    Posts
    1,266
    Thanks
    1
    Awards
    1
    Well it's just the geometric series \frac{1}{1-x}=1+x+x^2+\dots... right?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member
    Joined
    Sep 2010
    Posts
    95
    Quote Originally Posted by Bruno J. View Post
    Well it's just the geometric series \frac{1}{1-x}=1+x+x^2+\dots... right?
    Argh!

    Thank you for pointing out the obvious. I'll never forget it! Is it always sufficient to use the two first factors of the geometric series while tinkering with series?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor Bruno J.'s Avatar
    Joined
    Jun 2009
    From
    Canada
    Posts
    1,266
    Thanks
    1
    Awards
    1
    It depends on what you need, but you must always make sure that whatever x is, it is less than 1 in absolute value.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 0
    Last Post: January 26th 2010, 09:06 AM
  2. Formal power series & Taylor series
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: April 19th 2009, 10:01 AM
  3. Taylor Series / Power Series
    Posted in the Calculus Forum
    Replies: 6
    Last Post: March 4th 2009, 02:56 PM
  4. Develop a p series that represents a series
    Posted in the Calculus Forum
    Replies: 3
    Last Post: April 6th 2007, 05:03 AM
  5. Series Expansion/Taylor Series Help
    Posted in the Calculus Forum
    Replies: 1
    Last Post: November 12th 2005, 11:23 AM

Search Tags


/mathhelpforum @mathhelpforum