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Butis really so easy? (Möbius inversion)

  1. May 8, 2007 #1


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    But..is really so easy? (Möbius inversion), let be F(x) and G(x) functions

    [tex] F(x)= G(ax)+G(2ax)+G(3ax)+............... [/tex]

    for n=,1,2,3,4,5,............ a is a fixed real number. then

    [tex] G(ax)= \sum_{n=1}^{\infty}\mu (x) F(nx) [/tex]

    is seems too easy for me, to be true.
  2. jcsd
  3. May 8, 2007 #2

    matt grime

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    Why don't you try it? (You might want to make sure you state it properly.)
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