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Understanding Borel transform (sum) in QM

  1. Oct 25, 2006 #1
    Hello, i have heard tat you can use Borel transform in QM so the divergent perturbation series become convergent..:confused: :confused: i am a bit confused since if you must compute the Borel transform:

    [tex] B(x)=\sum_{n=0}^{\infty}\frac{a(n)x^{n}}{n!} [/tex]

    But this can only be made for a few cases [tex] a(n)=(-1)^{n} [/tex] and
    [tex] a(n)=(-1)^{n} n! [/tex]

    But if the a(n) don't follow a known pattern Borel Transform is useless..:grumpy: :grumpy:
  2. jcsd
  3. Oct 25, 2006 #2
    sorry the first "image" (i really hate this f... Latex :mad: :mad:) should be:

    [tex] \sum_{n=0}^{\infty}(a(n) x^{n})/n! [/tex]

    -the second: [tex]a(n)= (-1)^{n} n! [/tex]

    - And the third [tex]a(n)= (-1)^{n} [/tex]
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