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Uniqueness of function

  1. Aug 19, 2008 #1

    nicksauce

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    1. The problem statement, all variables and given/known data
    Find all functions f(z) satisfying a) f(z) is analytic in the disc |z-1| < 1, and b) f(n/(n+1)) = 1 - 1 / (2n^2 + 2n+1).


    2. Relevant equations



    3. The attempt at a solution
    One can deduce by algebraic re-arrangement that one solution is f(z) = 2z / (1+z^2). But how can I show that this function is the unique function satisfying this condition? Any hints?
     
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  3. Aug 19, 2008 #2

    Dick

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    Let g(z)=2z/(1+z^2). Then let h(z)=f(z)-g(z). The zeros of h(z) have a cluster point at z=1. The zeros of an analytic function which is not zero are isolated. So what is h(z)?
     
  4. Aug 19, 2008 #3

    nicksauce

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    Hi Dick, thanks for the reply. A few things: If you let g(z) = 2z/(1+z^2) and you have already f(z) = 2z/(1+z^2), then isn't h(z) identically 0? I don't see what you meant here.

    Second, this was from a fairly unrigorous class intended for physicists, so we never defined 'cluster point'. Perhaps you could re-phrase in more intuitive language?
     
  5. Aug 19, 2008 #4

    Dick

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    Hi nicksauce,

    You don't have f(z)=2z/(1+z^2), that's what you are trying to prove, isn't it? And proving h(z)=0 will do that. Sorry to use words without defining them, it didn't look like a physics question. g(1)=1 and by continuity f(1)=1. So h(1)=0. If h(z) is analytic and nonzero, and it IS analytic since f(z) and g(z) are, then the zero at z=1 must be isolated. Which means there is a disk around z=1 where h(z) is nonzero. Since the limit of n/(n+1) is 1 (i.e. 1 is a cluster point) and h(n/(n+1))=0, that contradicts h(z) having an isolated zero at z=1. Hence, h(z)=0.
     
    Last edited: Aug 19, 2008
  6. Aug 19, 2008 #5

    nicksauce

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    Ok that seems to make sense. Thanks!
     
  7. Aug 19, 2008 #6

    Dick

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    'seems to make sense'?? REALLY!!?? It does make sense! :). I don't know exactly what field of physics you're in, but there is a surprising amount of info you can get out of complex analysis. Look at the S-matrix. If you assume it's analytic with branch cuts in the complex energy plane you get all kinds of stuff.
     
  8. Aug 19, 2008 #7

    nicksauce

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    Allow me to clarify a few things...

    I didn't mean it was physics questions. What I meant was that as a physics student, I took the complex variables course that is intended for physics students, thus with more of a focus on calculations and applications and a bit of a lack of rigorous definitions, rather than a course intended for math students.

    Well I am just an undergrad and I have no idea what an S-matrix is, but I'll try to keep that in mind. :p
     
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