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Thread: Laurent's series for a function

  1. #1
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    Laurent's series for a function

    Hi. Can you help me with deriving the Laurent series for $\displaystyle f(z) = e^{1/z} $ . I can derive it by substituting $\displaystyle 1/z $ for $\displaystyle z $ in the taylor series expression for $\displaystyle f(z) = e^z $ but I was trying to derive the Laurent Series using the integral formulae for the coefficients. When I do that, the integrals of the coefficients of the positive powers of z fail to vanish.

    I would appreciate it a lot if someone took the trouble of pointing out how they vanish. Thank you!
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  2. #2
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    Quote Originally Posted by sashikanth View Post
    Hi. Can you help me with deriving the Laurent series for $\displaystyle f(z) = e^{1/z} $ . I can derive it by substituting $\displaystyle 1/z $ for $\displaystyle z $ in the taylor series expression for $\displaystyle f(z) = e^z $ but I was trying to derive the Laurent Series using the integral formulae for the coefficients. When I do that, the integrals of the coefficients of the positive powers of z fail to vanish.

    I would appreciate it a lot if someone took the trouble of pointing out how they vanish. Thank you!

    I don't understand: this section is abstract algebra and this questions doesn't belong here...is that very hard to understand?!

    Tonio
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  3. #3
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by sashikanth View Post
    Hi. Can you help me with deriving the Laurent series for $\displaystyle f(z) = e^{1/z} $ . I can derive it by substituting $\displaystyle 1/z $ for $\displaystyle z $ in the taylor series expression for $\displaystyle f(z) = e^z $ but I was trying to derive the Laurent Series using the integral formulae for the coefficients. When I do that, the integrals of the coefficients of the positive powers of z fail to vanish.

    I would appreciate it a lot if someone took the trouble of pointing out how they vanish. Thank you!
    Are you aware of using the 'residue at infinity' for computation of integrals?
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  4. #4
    MHF Contributor FernandoRevilla's Avatar
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    There is a kind of vicious circle about the question. Consider:

    $\displaystyle f(z)=e^{1/z}$

    then, $\displaystyle z=0$ is the only singularity of $\displaystyle f$. Besides, $\displaystyle \nexists\;{\lim_{z \to 0 }f(z)$

    this means that $\displaystyle z=0$ is an essential singulariy and its residue is usually found by means of the Laurent expansion. For example if:

    $\displaystyle f(z)=\ldots+\dfrac{A_{-2}}{z^2}+\dfrac{A_{-1}}{z}+A_0+A_1z+A_2z^2+\ldots$

    then,

    $\displaystyle \dfrac{f(z)}{z}=\ldots+\dfrac{A_{-2}}{z^3}+\dfrac{A_{-1}}{z^2}+\dfrac{A_0}{z}+A_1+A_2z+\ldots$

    and

    $\displaystyle A_0=\dfrac{1}{2\pi i}\displaystyle\int_{|z|=R}\dfrac{f(z)dz}{z}=\dfra c{1}{2\pi i}\cdot 2\pi i \;\textrm{Res}(f,z=0)=\textrm{coef}\left(\dfrac{1} {z}\right)=A_0$

    Fernando Revilla
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