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Math Help - Complex Analysis-- Integral

  1. #1
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    Complex Analysis-- Integral

    Evaluate:

    ∫C e^z (z-1)(z-2)^2 / sin^2 pi z

    where C is the circle │z│= 2.5
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by taypez View Post
    Evaluate:

    ∫C e^z (z-1)(z-2)^2 / sin^2 pi z

    where C is the circle │z│= 2.5
    The poles of this inside the circle are at z=-2, -1, 0, 1, and the residues at
    these poles are 0, 0, 0, e/pi^2, so the integral should be:

    ∫C e^z (z-1)(z-2)^2 / sin^2 pi z = e/pi^2

    (but be warned I'm not entirely satisfied with this)

    RonL
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  3. #3
    TD!
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    Are we talking about:

    <br />
\int\limits_C {\frac{{e^z \left( {z - 1} \right)\left( {z - 2} \right)^2 }}{{\sin ^2 \left( {\pi z} \right)}}} dz<br />

    Because then I find non-zero residues for all the poles (-2,-1,0,1).

    Also, from the residue theorem, aren't we forgetting a factor 2pi.i ?
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  4. #4
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    Quote Originally Posted by TD! View Post
    Are we talking about:

    <br />
\int\limits_C {\frac{{e^z \left( {z - 1} \right)\left( {z - 2} \right)^2 }}{{\sin ^2 \left( {\pi z} \right)}}} dz<br />

    Because then I find non-zero residues for all the poles (-2,-1,0,1).

    Also, from the residue theorem, aren't we forgetting a factor 2pi.i ?
    The factor of 2 pi i helps, and I would be happry to be convinced about the
    residues at the other points as I still don't have agreement with my numerical
    integral (he said giving away the secret of why he was not happy with his previous
    post).

    But this is the problem I have. At 0 for instance:

    z^2{\frac{{e^z \left( {z - 1} \right)\left( {z - 2} \right)^2 }}{{\sin ^2 \left( {\pi z} \right)}}}

    is analytic so the residue is (2 pi i times) the coefficient of x in the taylor series
    of this about z=0, which of course now that I compute it I find it non zero
    (where before it seemed to be 0 )

    So now I am with you.

    RonL
    Last edited by CaptainBlack; December 20th 2006 at 01:01 PM.
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  5. #5
    TD!
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    I found the residues to be:

    - at z = -2: -8/(eČ.piČ)
    - at z = -1: 3/(e.piČ)
    - at z = 0: 4/piČ
    - at z = 1: e/piČ
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  6. #6
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    Quote Originally Posted by TD! View Post
    I found the residues to be:

    - at z = -2: -8/(e&#178;.pi&#178
    - at z = -1: 3/(e.pi&#178
    - at z = 0: 4/pi&#178;
    - at z = 1: e/pi&#178;
    Now that agrees with my numerical integration


    Code:
    This is EULER, Version 2.3 RL-06.
    Type help(Return) for help.
    Enter command: (20971520 Bytes free.)
    Processing configuration file.
    Done.
    >
    >
    >dthe=0.001*pi;
    >
    >the=dthe/2:dthe:2*pi;
    >rc=2.5;
    >z=rc*exp(I*the);
    >
    >fz=exp(z)*(z-1)*(z-2)^2/(sin(pi*z))^2;
    >
    >
    >..numerical integral:
    >
    >II=sum(fz*rc*I*exp(I*the))*dthe
          7.26249e-014+4.29033i 
    >
    >
    >e=exp(1)
          2.71828 
    >
    >..TD!'s residues
    >
    >(-8/e^2/pi^2+3/e/pi^2+4/pi^2+e/pi^2)*2*pi*I
                     0+4.29033i 
    >
    RonL
    Last edited by CaptainBlack; December 21st 2006 at 03:51 AM. Reason: remove some blank lines
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  7. #7
    TD!
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    Isn't it wonderful when numerical and analytical results appear to match
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  8. #8
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    Quote Originally Posted by TD! View Post
    Isn't it wonderful when numerical and analytical results appear to match
    Experience suggests that the opeative word is appear

    RonL
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  9. #9
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    Quote Originally Posted by TD! View Post
    Isn't it wonderful when numerical and analytical results appear to match
    I think it is more wonderful when analytical results do not match, and it turns out the numerical results were wrong.
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  10. #10
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    Quote Originally Posted by ThePerfectHacker View Post
    I think it is more wonderful when analytical results do not match, and it turns out the numerical results were wrong.
    Why. They are complementary, a mistake can occur either way and
    their agreement is a sanity check.

    It is always important to be able to have two or more independent
    techniques available. The ability of the mind to see what it wants is
    always with us. An idependent check helps prevent us being carried away
    (resolution disagreements also helps develop technique and understanding
    of the limitations of our methods).

    Finally, it is also instructive to set about computing the integral numericaly
    a few things becom clear that otherwise would remain hidden.

    RonL
    Last edited by CaptainBlack; December 22nd 2006 at 12:18 AM.
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  11. #11
    TD!
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    Quote Originally Posted by CaptainBlack View Post
    Experience suggests that the opeative word is appear

    RonL
    My words were carefully chosen
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