Find inverse function of binary entropy

[emma83]

Homework Statement

Find the inverse function f^{-1} of the binary entropy f (given below) on the domain of definition [0;1/2[ (i.e. where f is continuous strictly increasing).
The function f is given by:
f(x)=-x\log(x)-(1-x)\log(1-x)
(where \log is the logarithm base 2)

Homework Equations

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

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 (?))

 

[g_edgar]

Did it actually say "find" or perhaps some other wording?
What type of textbook was it?

 

[emma83]

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 f^{-1} 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?

 

[g_edgar]

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.

 

[emma83]

Well I need the symbolic expression for the rest of the assignment.
Do you think this is not solvable?

 

[g_edgar]

I think the inverse is not an elementary function.

 

[diazona]

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.

 

[epenguin]

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

-x\log(x)-(1-x)\log(1-x)

works out as

\log(x)

 

[Dick]

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

-x\log(x)-(1-x)\log(1-x)

works out as

\log(x)

Must be REALLY late.

 

[epenguin]

Yes I will delete that presently.