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Thread: [SOLVED] Poles, order of it and residue

  1. #1
    MHF Contributor arbolis's Avatar
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    [SOLVED] Poles, order of it and residue

    I must prove that the singularities of $\displaystyle f$ are poles. I also must determine their order and the corresponding residue.

    $\displaystyle f(z)=\frac{z}{\cos z}$.

    My attempt :
    I'm stuck on the first part. (So I'd appreciate help only on this part, for the moment).
    So I wrote $\displaystyle \cos (z)= \sum _{n=0}^{\infty} \frac{(-1)^n z^{2n}}{(2n)!}$.
    So if I can find the order of the zeros of $\displaystyle \frac{1}{f(z)}$ I'm done.
    I've found out that $\displaystyle \frac{1}{f(z)}=\frac{1}{z}-\frac{z}{2}+\frac{z^3}{4}-...$
    The first term of the series really bothers me. If it wasn't there then $\displaystyle \frac{1}{f(z)}$ would be equal to $\displaystyle z\cdot h(z)$ with $\displaystyle h(0) \neq$ $\displaystyle 0$ and so $\displaystyle z=0$ would be a $\displaystyle 0$ of order $\displaystyle 1$ of $\displaystyle \frac{1}{f}$, thus a pole of order $\displaystyle 1$ of $\displaystyle f$. But of course I should take in count the $\displaystyle \frac{1}{z}$ term and look for $\displaystyle z=k \frac{\pi}{2}$, $\displaystyle k \in \mathbb{Z}$, but I'm stuck.
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  2. #2
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    Quote Originally Posted by arbolis View Post
    I must prove that the singularities of $\displaystyle f$ are poles. I also must determine their order and the corresponding residue.

    $\displaystyle f(z)=\frac{z}{\cos z}$.

    My attempt :
    I'm stuck on the first part. (So I'd appreciate help only on this part, for the moment).
    So I wrote $\displaystyle \cos (z)= \sum _{n=0}^{\infty} \frac{(-1)^n z^{2n}}{(2n)!}$.
    So if I can find the order of the zeros of $\displaystyle \frac{1}{f(z)}$ I'm done.
    I've found out that $\displaystyle \frac{1}{f(z)}=\frac{1}{z}-\frac{z}{2}+\frac{z^3}{4}-...$
    The first term of the series really bothers me. If it wasn't there then $\displaystyle \frac{1}{f(z)}$ would be equal to $\displaystyle z\cdot h(z)$ with $\displaystyle h(0) \neq$ $\displaystyle 0$ and so $\displaystyle z=0$ would be a $\displaystyle 0$ of order $\displaystyle 1$ of $\displaystyle \frac{1}{f}$, thus a pole of order $\displaystyle 1$ of $\displaystyle f$. But of course I should take in count the $\displaystyle \frac{1}{z}$ term and look for $\displaystyle z=k \frac{\pi}{2}$, $\displaystyle k \in \mathbb{Z}$, but I'm stuck.
    f(z) does not have a singularity at z = 0 so the series you're looking at (around z = 0) is irrelevant.

    The singularities are $\displaystyle z = z_m = \frac{(2m+1)\pi}{2}$ where m is an integer.

    So you need to show that $\displaystyle \lim_{z \rightarrow z_m} (z - z_m)^n f(z)$ is finite for some positive integer value of n (the smallest value of n will be the order of the pole).

    eg. see - Wolfram|Alpha
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  3. #3
    MHF Contributor arbolis's Avatar
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    Quote Originally Posted by mr fantastic View Post
    f(z) does not have a singularity at z = 0 so the series you're looking at (around z = 0) is irrelevant.

    The singularities are $\displaystyle z = z_m = \frac{(2m+1)\pi}{2}$ where m is an integer.

    So you need to show that $\displaystyle \lim_{z \rightarrow z_m} (z - z_m)^n f(z)$ is finite for some positive integer value of n (the smallest value of n will be the order of the pole).

    eg. see - Wolfram|Alpha
    I wasn't able to solve the problem using your tip.
    Here's what I've done :
    Considering the singularity at $\displaystyle z=\frac{\pi}{2}$, the series expansion of $\displaystyle \cos z$ around this singularity (singularity for $\displaystyle f$) is $\displaystyle 1+ \left ( z- \frac{\pi}{2} \right )^2 h(z)$ where $\displaystyle h(z)=-\frac{1}{2}+\frac{1}{4!}\left ( z-\frac{\pi}{2} \right )-...$ so $\displaystyle h\left ( \frac{\pi}{2} \right ) \neq 0$.
    --------------------------------------------------------------------
    I've a side-note question : From this can I conclude that the order of the pole of $\displaystyle r(z)=\frac{1}{\cos z} $when $\displaystyle z=\frac{\pi}{2}$ is 2?
    --------------------------------------------------------------------
    Now I wrote that the pole of $\displaystyle f$ occurs when $\displaystyle \frac{1}{f}$ has a zero. That is, when $\displaystyle \frac{1}{z}\left [ 1+ \left ( z- \frac{\pi}{2} \right )^2 h(z) \right ]=0$. I go nowhere.

    No let me try again you way : $\displaystyle \lim_{z \rightarrow \frac{\pi}{2}} (z - z_m)^n f(z)=\lim_{z \rightarrow \frac{\pi}{2}} \left ( z-\frac{\pi}{2} \right ) ^n \cdot \frac{1}{z} \left [ 1+ \left ( z- \frac{\pi}{2} \right )^2 h(z) \right ]$.
    For $\displaystyle n=1$, the limit is worth $\displaystyle 0$. Does that mean the order of the pole of $\displaystyle f$ at $\displaystyle z=\frac{\pi}{2}$ is $\displaystyle 1$, and that I've successfully solved the exercise?
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  4. #4
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    Quote Originally Posted by arbolis View Post
    I wasn't able to solve the problem using your tip.
    Here's what I've done :
    Considering the singularity at $\displaystyle z=\frac{\pi}{2}$, the series expansion of $\displaystyle \cos z$ around this singularity (singularity for $\displaystyle f$) is $\displaystyle 1+ \left ( z- \frac{\pi}{2} \right )^2 h(z)$ where $\displaystyle h(z)=-\frac{1}{2}+\frac{1}{4!}\left ( z-\frac{\pi}{2} \right )-...$ so $\displaystyle h\left ( \frac{\pi}{2} \right ) \neq 0$.
    --------------------------------------------------------------------
    I've a side-note question : From this can I conclude that the order of the pole of $\displaystyle r(z)=\frac{1}{\cos z} $when $\displaystyle z=\frac{\pi}{2}$ is 2?
    --------------------------------------------------------------------
    Now I wrote that the pole of $\displaystyle f$ occurs when $\displaystyle \frac{1}{f}$ has a zero. That is, when $\displaystyle \frac{1}{z}\left [ 1+ \left ( z- \frac{\pi}{2} \right )^2 h(z) \right ]=0$. I go nowhere.

    No let me try again you way : $\displaystyle \lim_{z \rightarrow \frac{\pi}{2}} (z - z_m)^n f(z)=\lim_{z \rightarrow \frac{\pi}{2}} \left ( z-\frac{\pi}{2} \right ) ^n \cdot \frac{1}{z} \left [ 1+ \left ( z- \frac{\pi}{2} \right )^2 h(z) \right ]$.
    For $\displaystyle n=1$, the limit is worth $\displaystyle 0$. Does that mean the order of the pole of $\displaystyle f$ at $\displaystyle z=\frac{\pi}{2}$ is $\displaystyle 1$, and that I've successfully solved the exercise?
    The link I posted shows that the limit exists and is finite. Therefore there is a simple pole. You should consider the general limit $\displaystyle \lim_{z \rightarrow z_m} (z - z_m) f(z)$ to show that f(z) has a simple pole at $\displaystyle z = z_m$.
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  5. #5
    MHF Contributor arbolis's Avatar
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    Ah I see, I should find that the limit is worth $\displaystyle -\frac{\pi}{2}$ instead of $\displaystyle 0$.
    Thanks once again.
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