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For all numbers n, N* = 32-n. (n*)*

  1. Sep 7, 2006 #1
    How do you calculate:

    For all numbers n, N* = 32-n.

  2. jcsd
  3. Sep 7, 2006 #2


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    Staff Emeritus
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    Is "N" the same as "n"?

    If you mean that n* is defined as 32- n, then "*" just means "subtract n from 32". Doing it twice, (n*)*= (32-n)*= 32- (32-n)= n.
    If that is not what you mean then I think you need to clarify.

    Hmm, but that doesn't have any thing to do with exponents. Do you mean that n* is defined as 32-n? That is, of course, the same as [itex]/frac{1}{32^n}[/itex]. Doing that twice,
    [tex](n*)*= 32^{-\frac{1}{32^n}}[/tex]
    which is 1 over the 32n root of 32.

    I have a feeling that is also not what you meant. Please clarify!
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