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.

For $\alpha \geq 0$ the transformation $x \mapsto \log(x) - \alpha \log(1-x)$ maps the unit interval to the real line (in fact for $\alpha = 0$ the transformation is not surjective). For $\alpha=1$ this is the logit transformation, which has a well-known inverse. It is also not difficult to find an inverse for $\alpha = 0.5$ and I would guess along these lines for some more $\alpha = 1/N$, $N$ positive natural number. But I do not see a closed form for an arbitrary $\alpha > 0$.

Finding an inverse numerically is not too difficult because the function is strictly monotone, but this is at least aesthetically unpleasing ;)

Is there a way to show that for arbitrary $\alpha$ no such solution can exist, or is there maybe one?

share|improve this question
    
You need a minus sign rather than a plus in your formula. The logit transformation is $x \mapsto \log x - \log (1-x)$. –  Rahul Jan 27 '11 at 23:00

2 Answers 2

up vote 0 down vote accepted

According to Wolfram Alpha, there is no analytic form for the inverse.

share|improve this answer
    
How reliable are the answers from Wolfram Alpha? It probably just means Wolfram's algorithm can't find one... –  Philipp Jan 27 '11 at 22:29

If $y = \log x - \alpha \log (1-x)$, then $e^y = x/(1-x)^\alpha$. If you pick $\alpha = 1/5$ or $\alpha = 5$, you will have to solve a quintic equation for $x$. Some of the coefficients will involve $e^y$ which is almost always transcendental, and (I think) this implies that the quintic will not in general be solvable.

share|improve this answer

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.