How to find a solution to the following equation \begin{align*} \frac{1}{1+x}-\frac{c}{x}-2\log \left( \frac{1+x}{x}\right)+A=0 \end{align*}

where $c$ and $A$ are some constants such that $c\ge 1$ and $A>0$.

I tried a few softwares to solve it but I don't think there is an analytic solution.

Could some one please suggest a way to find solution in some approximate sense? Thank you

  • 2
    $\begingroup$ It is a concave function, so an effective numerical method is Newton's method. $\endgroup$ – Jack D'Aurizio Jul 28 '15 at 17:49
  • $\begingroup$ @JackD'Aurizio Would it always converge? $\endgroup$ – Boby Jul 28 '15 at 17:53
  • $\begingroup$ given a starting point sufficiently close to the solution, yes. You may take a good starting point by considering a polynomial approximation of $\log(x+1)-\log(x)=\int_{x}^{x+1}\frac{dt}{t}$. $\endgroup$ – Jack D'Aurizio Jul 28 '15 at 17:58
  • 1
    $\begingroup$ If it is a concave function on an interval, then it may be shown that newton's method will always converge to the extrema. The issue here is that c and A are parameters, so it's better to specify them before performing the numerical technique. $\endgroup$ – DaveNine Jul 28 '15 at 18:27

Rearranging, probably useless but I'll leave it here:

$$2\log \left(\frac{1+x}{x}\right) = \frac{x}{x^2+x} - \frac{cx + c}{x^2+x} + \frac{Ax^2 + Ax}{x^2 + x} $$

$$\log \left(\frac{1+x}{x}\right) = \frac{Ax^2 + (A+1-c)x - c}{2x^2 + 2x}$$

$$\frac{1+x}{x} = \exp\left(\frac{Ax^2 + (A+1-c)x - c}{2x^2 + 2x}\right)$$

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