1
$\begingroup$

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)?

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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 $1+\log\left(\frac{\pi}{2}-\theta\right)=b$).

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

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

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

and then

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

and

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

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.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.