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Math Help - Radius of Convergence

  1. #1
    CSM
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    Wink Radius of Convergence

    So I had to calculate the radius of convergence of
    \frac{(z-i)}{z^3-z} around the point z=2i

    This is how I tried it. I'd love to hear what is wrong with my attempt:
    First I rewrote it
    \frac{(z-i)}{z^3-z}=\frac{(z-i)}{z(z-i)(z+i)}=\frac{1}{z(z+i)}
    Then with partial fractioning:
    \frac{a}{z}+\frac{b}{z+1}\Rightarrow a(z+i)+bz=z-i\Rightarrow a=-1,b=2
    So we can write are original function as :
    \frac{-1}{z}+\frac{2}{z+1}=\frac{1}{2i-(z-2i)}-\frac{2}{-i-(z-2i)}
    And use the geometric series formula:
    =2i\sum(\frac{z-2i}{2i})^n+i\sum(\frac{z-2i}{i})^n
    And finally apply the ratio test to the coefficients:
    a_n=(2i)^{n-1}, a_m=i^{n-1}
    |\frac{a_n}{a_{n+1}}|=|\frac{(2i)^{n-1}}{(2i)^n}|=|2i|=2
    |\frac{a_m}{a_{m+1}}|=|\frac{(i)^{n-1}}{(i)^n}|=|i|=1

    So I guess R would be one.
    Please, not only advise me how to find the radius of convergence. Please tell me if I made any mistakes above (particularly in method.)
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  2. #2
    MHF Contributor Also sprach Zarathustra's Avatar
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    Re: Radius of Convergence

    Quote Originally Posted by CSM View Post
    So I had to calculate the radius of convergence of
    \frac{(z-i)}{z^3-z} around the point z=2i



    This is how I tried it. I'd love to hear what is wrong with my attempt:
    First I rewrote it
    \frac{(z-i)}{z^3-z}=\frac{(z-i)}{z(z-i)(z+i)}=\frac{1}{z(z+i)}
    Then with partial fractioning:
    \frac{a}{z}+\frac{b}{z+1}\Rightarrow a(z+i)+bz=z-i\Rightarrow a=-1,b=2
    So we can write are original function as :
    \frac{-1}{z}+\frac{2}{z+1}=\frac{1}{2i-(z-2i)}-\frac{2}{-i-(z-2i)}
    And use the geometric series formula:
    =2i\sum(\frac{z-2i}{2i})^n+i\sum(\frac{z-2i}{i})^n
    And finally apply the ratio test to the coefficients:
    a_n=(2i)^{n-1}, a_m=i^{n-1}
    |\frac{a_n}{a_{n+1}}|=|\frac{(2i)^{n-1}}{(2i)^n}|=|2i|=2
    |\frac{a_m}{a_{m+1}}|=|\frac{(i)^{n-1}}{(i)^n}|=|i|=1

    So I guess R would be one.
    Please, not only advise me how to find the radius of convergence. Please tell me if I made any mistakes above (particularly in method.)
    Mistake:
    \frac{(z-i)}{z^3-z}=\frac{(z-i)}{z(z-i)(z+i)}=\frac{1}{z(z+i)}

    z^3-z=z(z^2-1)=z(z-1)(z+1)
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  3. #3
    CSM
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    Re: Radius of Convergence

    Ouch! Of course. I'll give it a re-run.
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  4. #4
    CSM
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    Re: Radius of Convergence

    Hmmmm but how do I proceed with that?
    \frac{(z-i)}{z^3-z} around the point z=2i
    =\frac{(z-i)}{z^3-z}=\frac{(z-i)}{z(z-1)(z+1)}
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  5. #5
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    Re: Radius of Convergence

    Partial Fractions.
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  6. #6
    MHF Contributor chisigma's Avatar
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    Re: Radius of Convergence

    Quote Originally Posted by CSM View Post
    So I had to calculate the radius of convergence of
    \frac{(z-i)}{z^3-z} around the point z=2i



    This is how I tried it. I'd love to hear what is wrong with my attempt:
    First I rewrote it
    \frac{(z-i)}{z^3-z}=\frac{(z-i)}{z(z-i)(z+i)}=\frac{1}{z(z+i)}
    Then with partial fractioning:
    \frac{a}{z}+\frac{b}{z+1}\Rightarrow a(z+i)+bz=z-i\Rightarrow a=-1,b=2
    So we can write are original function as :
    \frac{-1}{z}+\frac{2}{z+1}=\frac{1}{2i-(z-2i)}-\frac{2}{-i-(z-2i)}
    And use the geometric series formula:
    =2i\sum(\frac{z-2i}{2i})^n+i\sum(\frac{z-2i}{i})^n
    And finally apply the ratio test to the coefficients:
    a_n=(2i)^{n-1}, a_m=i^{n-1}
    |\frac{a_n}{a_{n+1}}|=|\frac{(2i)^{n-1}}{(2i)^n}|=|2i|=2
    |\frac{a_m}{a_{m+1}}|=|\frac{(i)^{n-1}}{(i)^n}|=|i|=1

    So I guess R would be one.
    Please, not only advise me how to find the radius of convergence. Please tell me if I made any mistakes above (particularly in method.)
    The function f(z)= \frac{1}{z\ (z+i)} has two singularities in z=0 and z=-i, so that the Taylor series expansion of f(*) around point z=z_{0} has radious of convergence equal to the distance from z_{0} of the 'more near' singularity. If z_{0}= 2 i then clearly is r=2...

    Kind regards

    \chi \sigma
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  7. #7
    CSM
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    Re: Radius of Convergence

    Thanks. Although I made I mistake so the function your looking at isn't correct.
    But I could say \frac{z-i}{z^3-z}=\frac{z-i}{z(z+1)(z-1)} has poles in z=0, z=-1,  z=+1 and then reason like you did...



    I did a partional fractioning and got:
    \frac{i}{z}+\frac{\frac{-1}{2}-\frac{i}{2}}{z+1}+\frac{\frac{1}{2}-\frac{i}{2}}{z-1}
    And then rewrote it to a form with an easy geometric series approach.
    So for example \frac{i}{z}=i\frac{1}{z}=-i\frac{1}{-z}=-i\frac{1}{2i-(z-2i)}=\frac{-i}{2i}\frac{1}{1-(\frac{z-2i}{2i})}
    Gives \frac{-1}{2}\sum(\frac{z-2i}{2i})^n=\frac{-1}{2}\sum(\frac{1}{2i})^n(z-2i)^n
    And that's just for the first part.... that's a lot of work....
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  8. #8
    MHF Contributor FernandoRevilla's Avatar
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    Re: Radius of Convergence

    Quote Originally Posted by CSM View Post
    Thanks. Although I made I mistake so the function your looking at isn't correct. But I could say \frac{z-i}{z^3-z}=\frac{z-i}{z(z+1)(z-1)} has poles in z=0, z=-1,  z=+1 and then reason like you did...
    The nearest singularity of f to 2i is 0 , so by a well known result the radius of convergence for the series expansion is r=d(2i,0)=2 . You needn't use partial fractions .
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  9. #9
    CSM
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    Re: Radius of Convergence

    Another one, -note this is not my homework I am trying to grasp it and got the solutions- which I found hard.

    What is the radius of convergence of f(z)=\frac{z^2-1}{z^3-1} around z=2

    Well I tried and I tried, but couldn't rewrite it nicely.

    Solution:

    Rewrite f(z)=\frac{z+1}{(z-e^{2\pi i/3})(z+e^{2\pi i/3})}
    The singularities are z=e^{2\pi i/3}. z=-e^{2\pi i/3} and the distance from 2 to the nearest is \sqrt{7}


    Well this seems very farfetched to me. How could I have come up with this rewriting myself?
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  10. #10
    MHF Contributor FernandoRevilla's Avatar
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    Re: Radius of Convergence

    Quote Originally Posted by CSM View Post
    Well this seems very farfetched to me. How could I have come up with this rewriting myself?
    f(z)=\frac{z^2-1}{z^3-1}=\frac{(z-1)(z+1)}{(z-1)(z^2+z+1)}=\dfrac{z+1}{(z-z_0)(z-\bar{z_0})} with z_0=-\dfrac{1}{2}+\dfrac{\sqrt{3}}{2}\;i . Then, r=d(2,z_0)=d(2,\bar{z_0})=\sqrt{7} .
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  11. #11
    CSM
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    Re: Radius of Convergence

    Quote Originally Posted by FernandoRevilla View Post
    f(z)=\frac{z^2-1}{z^3-1}=\frac{(z-1)(z+1)}{(z-1)(z^2+z+1)}=\dfrac{z+1}{(z-z_0)(z-\bar{z_0})} with z_0=-\dfrac{1}{2}+\dfrac{\sqrt{3}}{2}\;i . Then, r=d(2,z_0)=d(2,\bar{z_0})=\sqrt{7} .
    Gracias! Makes perfect sense now.
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  12. #12
    MHF Contributor FernandoRevilla's Avatar
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    Re: Radius of Convergence

    Quote Originally Posted by CSM View Post
    Gracias!
    Avec plaisir.
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