Results 1 to 6 of 6

Math Help - Simple Poles

  1. #1
    Junior Member
    Joined
    Nov 2009
    Posts
    27

    Simple Poles

    How can I prove that

    \frac{1}{1-2^{1-s}}

    has a simple pole at s=1.

    Any help is much appreciated.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by signature View Post
    How can I prove that

    \frac{1}{1-2^{1-s}}

    has a simple pole at s=1.

    Any help is much appreciated.
    Here is a clue: Read this limt of (x - 1)/(1 - 2^(1-x)) as x approaches 1 - Wolfram|Alpha (click on Show steps if you don't know how to get the limit).
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Nov 2009
    Posts
    27

    Simple Poles

    I still dont understand it.

    Is there no simple method to approach it,
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by signature View Post
    I still dont understand it.

    Is there no simple method to approach it,
    Go to your classnotes. In them you will find a theorem that says that the singularity z = z_0 is a simple pole of f(z) if \lim_{z \to z_0} ((z - z_0) f(z)) exists and is finite. This is the simple approach I have suggested that you use.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    May 2010
    Posts
    6
    Hi,

    You could show the pole has order 1 directly:

    Pole (complex analysis) - Wikipedia, the free encyclopedia

    Perhaps another possible method is as follows, though I could be wrong:

    I have just posted up a question regarding the domain of the Riemann zeta function and am aware that we can show it has a singularity at s = 1 by rewriting it as the product of an infinite sum by the term you have posted up. The analytic continuation of the Riemann zeta function, valid in Re(z) > 0 gives it as 1 + 1/(s-1) + {some integral well defined integral} and hence we see straight away that is has a simple pole at s = 1.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor chisigma's Avatar
    Joined
    Mar 2009
    From
    near Piacenza (Italy)
    Posts
    2,162
    Thanks
    5
    Lets start considering that is...

    2^{1-s}= e^{(1-s) \ln 2} = \sum_{n=0}^{\infty} \frac{(-1)^{n} \ln^{n} 2}{n!} (s-1)^{n} (1)

    From (1) we derive...

    1-2^{1-s} = \sum_{n=1}^{\infty} \frac{(-1)^{n-1} \ln^{n} 2}{n!} (s-1)^{n}=

    = (s-1) \sum_{n=0}^{\infty} \frac{(-1)^{n} \ln^{n+1} 2}{(n+1)!} (s-1)^{n} = (s-1)\cdot \ln 2 \cdot \varphi(s) (2)

    ... where...

    \varphi(s) = 1 - \frac{\ln 2}{2} \cdot (s-1) + ... (3)

    ... is analytic in s=1 and is \varphi(1)=1. Consequence of that is that \frac{1}{\varphi(s)} is also analytic and is...

    \frac{1}{\varphi(s)} = 1 + \frac{\ln 2}{2} \cdot (s-1) + ... (4)

    ... and finaly that is...

    \frac{1}{1-2^{1-s}}= \frac{1}{\ln 2 \cdot (s-1)\cdot \varphi(s)} = \frac{1}{\ln 2 \cdot (s-1)} \{ 1 + \frac{\ln 2}{2} \cdot (s-1) + ...\} (5)

    If we observe the (5) it is easy tom conclude that \frac{1}{1-2^{1-s}} has a single pole is s=1 and here the residue is \frac{1}{\ln 2}...

    Kind regards

    \chi \sigma
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Proving the simple poles
    Posted in the Differential Geometry Forum
    Replies: 3
    Last Post: May 12th 2011, 10:40 AM
  2. Residues at simple poles
    Posted in the Differential Geometry Forum
    Replies: 3
    Last Post: December 2nd 2009, 03:34 AM
  3. zeros and poles
    Posted in the Calculus Forum
    Replies: 0
    Last Post: October 10th 2009, 06:03 AM
  4. poles
    Posted in the Calculus Forum
    Replies: 0
    Last Post: November 9th 2008, 04:55 PM
  5. Poles
    Posted in the Algebra Forum
    Replies: 1
    Last Post: August 27th 2008, 03:01 AM

Search Tags


/mathhelpforum @mathhelpforum