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Can the following equation be rewritten as a function, $f(F(x))$, of $F(x)$? I.e. as $y = f(F(x))$?

\begin{equation} F(x) = - x \log_2 x - (1-x) \log_2 (1-x) \end{equation}

where $x = (1+\sqrt{1-y^2})/2$ and $y$ takes values between $0$ and $1$.

I'm thinking the answer is no, but hopefully it's yes!

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up vote 3 down vote accepted

The function $h(x)=-x\,\log_2x-(1-x)\,\log_2(1-x)$ is continuous and strictly decreasing from $[1/2,0]$ onto $[0,1]$. It has a continuous (in fact $C^\infty$) inverse $h^{-1}\colon[0,1]\to[0,1/2]$. Then $$ y=2\,\sqrt{h^{-1}(F)\bigl(1-h^{-1}(F)\bigr)},\quad 0\le y\le1. $$ So the answer is yes, $y$ can be written as a function of $F$. If you want an explicit formula for that function in terms of known functions (elementary or even special functions), then I am afraid that the answer is no. For instance, Mathematica does not find an explicit expression for $h^{-1}$.

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So by this you mean than $h(x)$ has an inverse, but it can't be calculated analytically? Also, how did you get $y = \sqrt{2(1-h^{-1} (F))}$? - I assumed you had rearranged $x=(1+\sqrt{1-y^2})/2$ and then replaced $x$ with $h^{-1}(F)$, but I get $y = 2\sqrt{x(1-x)}$ doing this. – Calvin Feb 14 '12 at 14:38
One more question - since $F$ is the same as $h$, why introduce $h$? ...And in this case what does $h^{-1} (F)$ mean? ..If $F=h$ then $h^{-1} (F) = 1$ (I feel I'm missing something major here!) – Calvin Feb 14 '12 at 14:42
As for the first comment, you are right and I have edited my answer. As for the second, I have used your notation. In your post $F$ is not defined as a function of $x$, but as a variable. Given the value of $F$, you want to know the value of $y$; the first step is to find $x=h^{-1}(F)$ and then $y$ from $x$. – Julián Aguirre Feb 14 '12 at 15:20
Thanks, that was a mistake in the question - I should have written $F(x)$ (I'll change it now). Does that affect your answer? – Calvin Feb 14 '12 at 15:42
No, now you have $y=2\sqrt{F(F^{-1}(x))(1-F(F^{-1}(x)))}$. The problem is hat there is no "explicit" expression for $F^{-1}$. – Julián Aguirre Feb 14 '12 at 21:31


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I have no idea as to why the subscripted "2"s get subscripted in the preview, but not here. – Doug Spoonwood Feb 14 '12 at 13:54

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