# How could I improve this approximation?

In a computer application, I need to solve trillions of times an equation which can be reduced to $$f(x)=\sin(x)-a x=0$$ Newton methods (quadratic and higher orders) are used for the solution. Parameter $a$ is random and the solution $x$ looked for is the one between $0$ and $\pi$ if it exists. The retained solution is immediate ($x=0$) if $a \geq 1$ or $a \leq 0$.

For the other cases ($0 \lt a \lt 1$), since I need to save as many iterations as I can, I focused on how to establish a good and unexpensive approximation of the solution. After some empirical experiments, what I found is that writing $$\sin(x) \simeq \frac{4}{\pi^2} x(\pi-x)$$ is a quite good approximation if $0 \leq a \leq 0.7$ leading to $$x \simeq \pi -\frac{\pi ^2 a}{4}$$

For the remaining interval, using Pade approximation $$\sin(x) \simeq \frac{x-\frac{7 x^3}{60}}{1+\frac{x^2}{20}}$$ which leads to $$x \simeq \frac{2 \sqrt{15} \sqrt{1-a}}{\sqrt{3 a+7}}$$ seems interesting.

I wonder if this could be improved, the goal being a simple explicit expression for the estimate of the solution.

Any idea and/or suggestion would really be welcome.

In comments and answers, Bhoot suggested me to look at the approximation $$\sin(x) \simeq \frac{16 (\pi -x) x}{5 \pi ^2-4 (\pi -x) x}$$ proposed by Mahabhaskariya of Bhaskara I, a seventh-century Indian mathematician. This spendid approximation leads to $$x \simeq \frac{2 \sqrt{-\pi ^2 a^2+2 \pi a+4}+\pi a-4}{2 a}$$ which is effectively very good except very close to $a=1$. This already makes a significant improvement.

Temporary improvement

Extremely impressed by the quality of the approximation proposed by Mahabhaskariya of Bhaskara I (more than 1400 years ago), I tried to undertand why it was very good except in the vicinity of $a=1$. I suspected that the derivative could be in error at the end points. Effectively, this formula gives a slope equal to $\frac{16}{5\pi} \simeq 1.01859$ instead of $1$. On the other side, the area under the curve is given by $$A=\pi \left(-4+\pi +\tan ^{-1}\left(\frac{3116}{237}\right)\right) \simeq 1.99955$$ So, modestly, I built an approximation which is $$\sin(x) \simeq \frac{\pi x(\pi-x)}{\pi^2+(\pi-4)x(\pi-x)}$$ which allows to match exactly the function and derivative values at $x=0,\frac{\pi}{2},\pi$; with respect to accuracy, it is not as good as the original showing a maximum error of $0.0052$ instead of $0.0016$. The area under the curve is then given by $$A=\frac{\pi \left(-4 \pi +\pi ^2+4 \sqrt{(4-\pi ) \pi } \tan ^{-1}\left(\sqrt{\frac{4}{\pi }-1}\right)\right)}{(\pi -4)^2} \simeq 1.99161$$ quite significantly worse than the original.

The estimate of the solution is given by $$x \simeq \frac{\pi \left((\pi -4) a-\sqrt{(\pi -4) a (\pi a-2)+1}+1\right)}{2 (\pi -4) a}$$ which makes Newton scheme converging in less than two iterations for the whole range.

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Maybe of interest: en.wikipedia.org/wiki/… –  user105475 Jul 19 at 7:03
@Bhoot. This is a superb approximation (I did not know it). Please, put it as an answer to my post. I shall edit and refer to you. –  Claude Leibovici Jul 19 at 7:20
@ClaudeLeibovici: You are right! It is a nice approximation formula. –  Mhenni Benghorbal Jul 19 at 9:22
@MhenniBenghorbal. When you think that is was done almost $1400$ years ago ! Cheers :) –  Claude Leibovici Jul 19 at 9:24
@ClaudeLeibovici: He is a seventh-century Indian mathematician. –  Mhenni Benghorbal Jul 19 at 9:25

Another option---since you need to do this trillions of times---is to save the solutions for many values of $a$, and then look up/interpolate solutions as you need them.