# Find inverse function of binary entropy

1. Jul 12, 2009

### emma83

1. The problem statement, all variables and given/known data
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)

2. Relevant 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!

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. Jul 12, 2009

### g_edgar

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

3. Jul 12, 2009

### 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?

4. Jul 12, 2009

### 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.

5. Jul 12, 2009

### emma83

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

6. Jul 12, 2009

### g_edgar

I think the inverse is not an elementary function.

7. Jul 12, 2009

### 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.

8. Jul 12, 2009

### epenguin

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

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

works out as

$$\log(x)$$

9. Jul 12, 2009

### Dick

Must be REALLY late.

10. Jul 13, 2009

### epenguin

Yes I will delete that presently.