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Define $$f_a(x) = ax - \log \left[ \frac{x}{5(1-x)} + 1\right], \ x < 1,$$ for some constant $a > 0$.

Let $x'$ be the maximum-point of this function, and $f_a(x')$ the max-value. Clearly, both these values will depend on the choice of $a$.

I want to invert the function $a \mapsto f_a(x')$. That is, given some $y$, what should $a$ be such that $f_a(x') = y$? In other words, I want to solve for $a$ in $$ax' - \log\left[ \frac{x'}{5(1-x')} + 1\right] = y.$$

I have tried this myself, but the fact that $x'$ depends on $a$ in a complicated way has made it very hard for me to invert this function. Is there a trick I can use, anything that might make it possible to analytically invert this?

Jaood
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  • Have you found the max point in terms of $a$ ? If so what's the formula for it? – coffeemath Jan 12 '19 at 02:10
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    I did it on pen and paper, threw that away now. Of the top of my head, I think it was $$x' = \frac{-\frac{1}{5} - \sqrt{\frac{1}{25} - 4( \frac{1}{5} - 2)(1 - \frac{1}{5} a^{-1})}}{2(1 - \frac{1}{5})}$$At this point, I stopped, since I don't see how to simplify the square-root part involving $a$, and even if I could, plugging $x'$ back into $f(x)$ is still quite complicated. – Jaood Jan 12 '19 at 03:16
  • https://en.wikipedia.org/wiki/Lambert_W_function – John Wayland Bales Jan 12 '19 at 03:52

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Using pen and paper $$f_a(x) = ax - \log \left[ \frac{x}{5(1-x)} + 1\right]$$ $$f'_a(x) =a-\frac 1 {4 x^2-9 x+5}$$ $$f'_a(x) =0 \implies x_\pm=\frac{9 a\pm\sqrt{a(a+16)} }{8 a}\implies x'=\frac{9 a-\sqrt{a(a+16)} }{8 a}$$ $$f(x')=\frac{9 }{8}a-\frac{1}{8} \sqrt{a(a+16)} -\log \left(a+8+\sqrt{a(a+16)} \right)+\log (10)=y$$ which cannot be inversed even using special functions and numerical methods (such as Newton) would be required. This would converge quite fast since, plotting $y$ as a function of $a$, you could notice that, as soon as $a>1$, $y$ is "almost" linear. Starting with $a_0=y$ would probably be a good idea.

Let us try with $y=12.345$. The iterates would be $$\left( \begin{array}{cc} 0 & 12.345000 \\ 1 & 14.650953 \\ 2 & 14.639921 \end{array} \right)$$

The case where $a$ would be large is interesting too. Expanding $y$ for infinitely large values of $a$ would give $$y=a+\left(\log \left(\frac{5}{a}\right)-1\right)+O\left(\frac{1}{a}\right)$$ and the solution would be $$a \approx -W_{-1}\left(-5 e^{-(y+1)}\right)$$ where appears Lambert function. For the worked example, this would give $a=14.403$ which is not too bad.

Claude Leibovici
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