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Finding the residues of poles

  1. Aug 17, 2015 #1
    Consider the equation ##\mathcal{F}(\lambda)=0\ \ \ \forall\ \lambda = \lambda_{n},\ n \in \mathbb{N}##.

    I understand that the expression ##\frac{d}{d\lambda}\ \ln\mathcal{F}(\lambda)=\frac{\mathcal{F'(\lambda)}}{\mathcal{F}(\lambda)}## has poles of order 1 exactly at ##\lambda_{n}## because ##\frac{\mathcal{F'(\lambda)}}{\mathcal{F}(\lambda)}=\frac{\mathcal{F'(\lambda)}}{(\lambda- \lambda_{1})...(\lambda-\lambda_{n})}##.

    I wonder how I might expand the expression ##\frac{d}{d\lambda}\ \ln\mathcal{F}(\lambda)=\frac{\mathcal{F'(\lambda)}}{\mathcal{F}(\lambda)}## about ##\lambda_{n}## to find out that the residue of the poles is 1.

    Any ideas?
     
  2. jcsd
  3. Aug 18, 2015 #2
    I've found out that, expanding about ##\lambda = \lambda_{n}##, with ##\mathcal{F'}(\lambda_{n})\neq0##, we obtain

    ##\frac{\mathcal{F'}(\lambda)}{\mathcal{F}(\lambda)}=\frac{\mathcal{F'}(\lambda-\lambda_{n}+\lambda_{n})}{\mathcal{F}(\lambda-\lambda_{n}+\lambda_{n})} = \frac{\mathcal{F'}(\lambda_{n})+(\lambda-\lambda_{n})\mathcal{F''}(\lambda_{n})+...}{(\lambda-\lambda_{n})\mathcal{F'}(\lambda_{n})+(\lambda-\lambda_{n})^{2}\mathcal{F''}(\lambda_{n})+...} = \frac{1}{\lambda - \lambda_{n}}+...##,

    so that the residue at all eigenvalues is 1.

    I took this evaluation out from a paper, so I am not really sure about a couple of things I have written - have we used the Taylor series to expand each of ##\mathcal{F'}(\lambda-\lambda_{n}+\lambda_{n})## and ##\mathcal{F}(\lambda-\lambda_{n}+\lambda_{n})## about ##\lambda=\lambda_{n}##?
     
  4. Aug 24, 2015 #3

    HallsofIvy

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    Frankly, I don't understand what you mean by "expanding about [itex]\lambda= \lambda_n[/itex] because you haven't said what you mean by [itex]\lambda_n[/itex]! Apparently you mean "eigenvalues" but then what do you mean by "eigenvalues" for a function of the complex numbers?

    The simplest way to define residues of pole is this: A function, f(z), of a complex variable, has a "pole of order n" at [itex]z= z_0[/itex] if and only if expanding it in a Laurent series in [itex](z- z_0)[/itex], the lowest power if [itex](z- z_0)^{-n}[/itex]. In that case, the residue is the coefficient of [itex](z- z_0)^{-1}[/itex].
     
  5. Aug 25, 2015 #4

    mathwonk

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    a holomorphic function f(z) with a zero of order n at z=p, equals (z-p)^n times a holomorphic function which is non zero at p, say with value a≠0. Its derivative then has a zero of order n-1 at p, and equals (z-p)^(n-1) times a function with value na at p. hence the quotient f'/f looks like 1/(z-p) times a function with value n at p. + a holomorphic function. Hence the residue at p of f'/f is n, the iorder of the zero at p.

    In your case you seem to be assuming the zeroes at the positive integers have order one, so the residue is one at each of them. i.e. you are dividing f' by f = a(z-p) + b(z-p)^2+...., where a≠0. youn can compute this.
     
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