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Let $f:[0, 1] \rightarrow [0, \infty]$ be a function of $x$, with a parameter $\theta > 0$, such that

  • $f$ is continuous
  • $f$ is strictly decreasing
  • $f(0) = \infty$
  • $f(1) = 0$

For example, $f(x) = (- \log(x))^{\theta - 1}$.

For a given $\theta$, is there a way to approximate $f^{-1}(2 f(0.5))$ for any such $f$?

user7064
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  • With just the given information, I'd say no. For any given $x \in (0,0.5)$, there exists a function that satisfies your four bullet points and such that $f^{-1}(2f(0.5)) = x$. Furthermore, I don't really see how $\theta$ is intended to come in to the problem: is there some further dependence on $\theta$ that you forgot to include? – Willie Wong Apr 10 '13 at 14:55

1 Answers1

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That value could be almost anything. Take for $\theta > 0$: $$ f_\theta(x) = x^{- \theta} - 1 $$ Then $f_\theta(1/2) = 2^\theta - 1$, and $$f^{-1}_\theta(2 f_\theta(1/2)) = \left( 2(2^\theta - 1) + 1 \right)^{1/\theta} = (2^{\theta + 1} - 1)^{1/\theta} \approx 2^{(\theta + 1)/\theta} \left(1 - \frac{2^{- \theta - 1}}{\theta} + \ldots\right)$$ by using the first two terms of the binomial expansion.

As $f_\theta(0) \to \infty$ and $f_\theta(1) = 0$, one could also scale this function at will, and mangle it in a myriard other ways.

vonbrand
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  • Thanks! can you explain the approximation please? – user7064 Apr 10 '13 at 16:56
  • @user7064, $2(2^r - 1) \approx 2^{r + 1}$ if $r >> 0$, unless $\theta$ is very small, $2^{r + 1} \theta$ is much larger than 1. Very crude, but gives an idea how it behaves. Looking it over, should have just used $\theta$ instead of $r$, and chucked the factor. I.e., just have done $f_\theta(x) = x^{- \theta} - 1$. Oh, well. – vonbrand Apr 10 '13 at 18:57