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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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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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    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 $F^{-1}$.2012-02-14
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$f(F)=f(−x\log_{2}x−(1−x)\log_{2}(1−x))$.

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    I have no $i$dea as to why the subscripted "2"s get subscripted in the preview, but not here.2012-02-14