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Find inverse function of binary entropy

  1. Jul 12, 2009 #1
    1. The problem statement, all variables and given/known data
    Find the inverse function [tex]f^{-1}[/tex] of the binary entropy [tex]f[/tex] (given below) on the domain of definition [0;1/2[ (i.e. where [tex]f[/tex] is continuous strictly increasing).
    The function [tex]f[/tex] is given by:
    [tex]f(x)=-x\log(x)-(1-x)\log(1-x)[/tex]
    (where [tex]\log[/tex] is the logarithm base 2)

    2. Relevant equations
    If I am right with the calculation, this is equivalent to solving:
    [tex]x^{x}(1-x)^{1-x}=2^{-y}[/tex]
    But I have no clue how to solve this either!

    3. The attempt at a solution
    I don't know how to solve this, I also tried with computer programs such as Maple and Mathematica but was not able to compute it either (I don't know much of them so I guess this should be possible (?))
     
  2. jcsd
  3. Jul 12, 2009 #2
    Did it actually say "find" or perhaps some other wording?
    What type of textbook was it?
     
  4. Jul 12, 2009 #3
    Thanks for your answer. I had to translate it from French, it is not in a textbook but part of an assignment I have to do for a physics course.
    Actually I am allowed to use a computer program to get the answer, so it should be enough if Maple, Mathematica or Matlab gives me the symbolic expression of [tex]f^{-1}[/tex] but I am not used to these programs and everything I tried to solve this so far ended up in an error message.
    Any clue?
     
  5. Jul 12, 2009 #4
    Maybe the wording means: Show that this function f defined on [0,1/2[ has an inverse, but does not require you to find a symbolic formula for that inverse.
     
  6. Jul 12, 2009 #5
    Well I need the symbolic expression for the rest of the assignment.
    Do you think this is not solvable?
     
  7. Jul 12, 2009 #6
    I think the inverse is not an elementary function.
     
  8. Jul 12, 2009 #7

    diazona

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    I agree, I don't think it's solvable in the normal sense. But you could find a series expansion for the inverse. Mathematica has a function "InverseSeries" for exactly this purpose.
     
  9. Jul 12, 2009 #8

    epenguin

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    Maybe because it's late at night here but it seems :shy:

    [tex]-x\log(x)-(1-x)\log(1-x)[/tex]

    works out as

    [tex]\log(x)[/tex]
     
  10. Jul 12, 2009 #9

    Dick

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    Must be REALLY late.
     
  11. Jul 13, 2009 #10

    epenguin

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    Yes I will delete that presently.
     
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