Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!

share|improve this question

2 Answers 2

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 $h^{-1}$.

share|improve this answer
1  
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

$f(F)=f(−x\log_{2}x−(1−x)\log_{2}(1−x))$.

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.