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Thread: Singularities

  1. #1
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    Singularities

    Hey guys
    Im having a panic as i should be revising, but instead im stuck on a question from my coursework, can anyone help

    Find and classify the singularities of
    (pi*cot(pi*x))/x^2

    thanks
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  2. #2
    Moo
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    Hello,
    Quote Originally Posted by pols89 View Post
    Hey guys
    Im having a panic as i should be revising, but instead im stuck on a question from my coursework, can anyone help

    Find and classify the singularities of
    (pi*cot(pi*x))/x^2

    thanks
    cot=cos/sin

    $\displaystyle \frac{\pi \cot(\pi x)}{x^2}=\frac{\pi \cos(\pi x)}{x^2 \sin(\pi x)}$

    For any $\displaystyle x \in \mathbb{Z}$, there is a singularity, because thereafter, $\displaystyle \sin(\pi x)=0$
    Now I believe there is a triple pole for 0, and simple pole (not too sure) for any integer in $\displaystyle \mathbb{Z}^\times$
    you can check it by finding the Laurent series near these points.
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  3. #3
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    Quote Originally Posted by Moo View Post
    Hello,

    cot=cos/sin

    $\displaystyle \frac{\pi \cot(\pi x)}{x^2}=\frac{\pi \cos(\pi x)}{x^2 \sin(\pi x)}$

    For any $\displaystyle x \in \mathbb{Z}$, there is a singularity, because thereafter, $\displaystyle \sin(\pi x)=0$
    Now I believe there is a triple pole for 0, and simple pole (not too sure) for any integer in $\displaystyle \mathbb{Z}^\times$
    you can check it by finding the Laurent series near these points.
    The smallest value of $\displaystyle m$ such that $\displaystyle \lim_{x \rightarrow 0} x^m f(x) \neq \infty$ is $\displaystyle m = 3$ so $\displaystyle x = 0$ is a pole of order 3.

    The smallest value of $\displaystyle m$ such that $\displaystyle \lim_{x \rightarrow n} (x-n)^m f(x) \neq \infty$ is $\displaystyle m = 1$ so $\displaystyle x = n$ is a simple pole (the more PC expression is pole of order 1).
    Last edited by mr fantastic; Jan 4th 2009 at 11:14 PM. Reason: Typos fixed from rushed cut and paste
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  4. #4
    Moo
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    Quote Originally Posted by mr fantastic View Post
    The smallest value of $\displaystyle m$ such that $\displaystyle \lim_{x \rightarrow 0} x^m f(x) \neq \infty$ is $\displaystyle m = 3$ so $\displaystyle x = 0$ is a pole of order 3.

    The smallest value of $\displaystyle m$ such that $\displaystyle \lim_{x \rightarrow n} x^m f(x) \neq \infty$ is $\displaystyle n = 1$ so $\displaystyle x = n$ is a simple pole (the more PC expression is pole of order 1).
    And the limit has to be $\displaystyle \neq 0$, isn't it ?

    Actually, I couldn't be sure for the second poles, because I didn't study the limit
    And I thought it was $\displaystyle \lim_{x \to n} (x-n)^m f(x)$ ?
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  5. #5
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    Quote Originally Posted by Moo View Post
    And the limit has to be $\displaystyle \neq 0$, isn't it ? Mr F says: No. Think about the Laurent series of f(x) around $\displaystyle {\color{red}x = x_0}$ and what you need so that $\displaystyle {\color{red}(x - x_0) f(x)}$ is analytic ....

    Actually, I couldn't be sure for the second poles, because I didn't study the limit
    And I thought it was $\displaystyle \lim_{x \to n} (x-n)^m f(x)$ ? Mr F says: Yes, it is.
    Whoops. Actually there were a couple of typos from my rushed cut and paste. Fixed now. Thanks.
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  6. #6
    Moo
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    Woops sorry. Actually, it's just that $\displaystyle f(n)$ has to be $\displaystyle \neq 0$
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  7. #7
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    Singularities

    thanks guys
    Last edited by mr fantastic; Jan 5th 2009 at 11:26 AM. Reason: Moved new question to new thread
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