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Sum derivation

  1. Nov 17, 2009 #1
    suppose, [tex] s_{n}(f;t) = \sum_{k=-n}^{n}\widehat{f}(k)e^{ikt}[/tex]
    and
    [tex]\sigma_{N}(f;t)= \frac{1}{N+1}\sum_{n=0}^{N}s_{n}(f;t)[/tex].

    how do i get from this [tex]\sigma_{N}(f;t)= \frac{1}{N+1}\sum_{n=0}^{N}s_{n}(f;t)[/tex].

    to this


    [tex]\sigma_{N}(f;t)= \sum_{n=-N}^{N}(1-\frac{|n|}{N+1})\widehat{f}(n)e^{int}[/tex]

    obviously one starts with:

    [tex]\sigma_{N}(f;t)=\frac{1}{N+1}\sum_{n=0}^{N}\sum_{k=-n}^{n}\widehat{f}(k)e^{ikt}[/tex]

    thanks!
     
  2. jcsd
  3. Nov 17, 2009 #2
    And what happens when you reverse the order of summation ... the sum on k outside, the sum on n inside?
     
  4. Nov 17, 2009 #3
    ?














    ?
     
  5. Nov 17, 2009 #4
    wow this must be slowest forum on the face of the planet
     
  6. Nov 17, 2009 #5

    CRGreathouse

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    Counterexample: http://www.myfinanceforum.com/
     
  7. Nov 17, 2009 #6
    Perhaps, but remember we're not all free to check forums 25 hours a day, 8 days a week. Two hours 40 for what looks like a hint seems pretty good to me. Have you tried it?
     
  8. Nov 18, 2009 #7
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