I have to solve this

$$a\sqrt{1-x^2} + \log(x) = b \tag{1}$$ for $x \approx 0$ where $a$ and $b$ are two constants and $x>0$.

One (maybe naive) approach is to approximate $\log(x)$, which I asked here. (However, this is not XY problem because I do use the approximation of $\log(x)$ for other problems, so please do not mark it as duplicated)

Another way I tried is to allow $\sqrt{1-x^2} \approx 1$ then $$x \approx e^{b-a} \tag{2}$$

Even though the result of (2) is quite good, it is not satisfactory. Could anyone propose other solutions better than (2)?

  • $\begingroup$ Write $A=e^a, B=e^b$ and exponentiates both sides of the equation. You get $xe^{a\sqrt{1-x^2}} \approx xA(1-\tfrac{1}{2}x^2)=B$ which is a cubic equation in $x$. $\endgroup$ – gammatester Oct 27 '18 at 10:13
  • $\begingroup$ @gammatester Thanks, but I think it should be $xe^{a\sqrt{1-x^2}} \approx xA^{\sqrt{1-x^2}}$, shouldn't it? $\endgroup$ – AlexTP Oct 28 '18 at 10:04

Your equation is equivalent to $a\sin\theta+\log\cos\theta = b $, which can be solved through Newton's method both in the case $\theta\approx 0^+$ (in such a case we consider as a starting point a solution of $a\theta-\frac{\theta^2}{2}=b$) and in the case $\theta\approx\frac{\pi}{2}^-$ (in such a case we consider as a starting point the solution of $1+\log\left(\frac{\pi}{2}-\theta\right)=b$).

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  • $\begingroup$ Nice way, I didn’t see that! $\endgroup$ – user Oct 27 '18 at 18:33
  • $\begingroup$ @Jack D'Aurizio, thanks very nice. However, besides making the equation nicer, could you please tell me what is the advantages of setting $x=\cos(\theta)$? Does it make the rate of convergence faster in comparing to keeping $x$ as unknown? $\endgroup$ – AlexTP Nov 12 '18 at 13:59
  • $\begingroup$ @Jack, your answer is truly elegant, but it converges slower than the direct Newton method on $x$. The conclusion comes from simulations with the same accepted threshold. $\endgroup$ – AlexTP Nov 12 '18 at 17:33

If you are looking for an approximation for $x$ small then we have

$$ \sqrt{1-x^2}\approx 1-\frac12x^2$$

and then

$$a\sqrt{1-x^2} + \log(x) \approx a-\frac12 ax^2+\log x\approx a+\log x $$


$$a+\log x =b \implies x=e^{b-a}$$

is a nice first order approximation.

To obtain a better approximation we need to solve

$$a-\frac12 ax^2+\log x=b$$

which can be solved numerically starting for the solution found by the first order approximation.

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  • $\begingroup$ could you please elaborate how we can conlude that 'we do not have solution' for that problem? Thanks. $\endgroup$ – AlexTP Oct 27 '18 at 10:04
  • $\begingroup$ @AlexTP I mean that when $x$ approach to $0$ (with x>0) $\log x$ diverges to $-\infty$ and therefore we can't find any $a$ and $b$ such that $a\sqrt{1-x^2} + \log(x) \to b$as $x \to b$. Are you looking for that or something different? $\endgroup$ – user Oct 27 '18 at 10:08

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