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I'm coding some numerical methods and I do not know what the correct analysis would be for choosing the implementation for $arcsin$ and $arctan$ for real numbers. Here's what I know:

Both functions have Taylor series about the origin that converge in $(-1, 1)$. The actual domain of $\arcsin$ is $[1, 1]$, and the domain of $\arctan$ is $(-\infty, \infty)$.

  1. For $\arcsin$, I can cover the whole domain by using $\arcsin(x) = \frac{\pi}2 - \arcsin\sqrt{1 - x^2}$. This guarantees $x^2 < \frac 12$, and the Taylor series can be used.

  2. For $\arctan$, I can reduce $|x|$ by a factor greater than $2$ by using $\arctan(x) = 2\arctan\left(\frac x{1 + \sqrt{1 + x^2}}\right)$. This reduction can be repeated until $|x|$ is smaller than a prespecified value, but one reduction is enough to get inside the convergent region. Note also that the Taylor series of $\arctan$ has alternating signs, but the Taylor series of $arcsin$ has only one sign.

I also know that

  • $\arctan(x) = \arcsin\left(\frac x{\sqrt{x^2 + 1}}\right)$
  • $\arcsin(x) = 2\arctan\left(\frac x{1 + \sqrt{1 - x^2}}\right)$

That means I can use the first equation to reduce $\arctan$ to $\arcsin$ then use method 1, or use the second equation to reduce $\arcsin$ to $\arctan$ then use method 2.

My question is what should I use? I am under the impression that the Taylor series of $\arcsin(x)$ converges a little faster than $\arctan(x)$ termwise, so method 1 may be better. But method 2 allows arbitrary reduction of $|x|$ (at the cost of one square root per reduction), so convergence could be made faster(?). I'm a bit concerned about alternating signs in the Taylor series of $\arctan$ too.

I think I may not know enough numerical analysis to decide. Please help...

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  • $\begingroup$ I don't think this question has a research angle, and research is what this website is about. $\endgroup$ – Gerry Myerson Aug 9 '12 at 5:30
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    $\begingroup$ Any decent numerical mathematical functions library should have implementations of arctan and arcsin. Don't try to reinvent the wheel. $\endgroup$ – Robert Israel Aug 9 '12 at 6:53
  • $\begingroup$ I second Robert's advice 200% :-) $\endgroup$ – Suvrit Aug 9 '12 at 8:44
  • $\begingroup$ mathoverflow.net/questions/19946 $\endgroup$ – Steve Huntsman Aug 9 '12 at 10:50
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    $\begingroup$ Standard libraries normally use Chebyshev polynomials, not Taylor series. Taylor series have undersirable properties. With Chebyshev polynomials, you have the same upper bound on the error throughout its domain. A useful book covering this sort of thing is Numerical Recipes: The Art of Scientific Computing, by Press et al. $\endgroup$ – Ben Crowell Aug 9 '12 at 15:19
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What is your actual goal? To get a fixed accuracy? Arbitrary accuracy? An interesting method is CORDIC. It involves a limited amount of table lookup, addition and division by 2 (so bit shift if computations are in binary) but no multiplication. You seem willing to store $\pi$ so at least some lookup seems allowed. This is most directly used for $\tan$ or $\sin$ but you can as easily run it in reverse to find inverse functions.

Still, with a hardware multiply a power series may be more efficient.

I'd guess that the series $\arcsin(x)=x+{\frac {1}{6}}{x}^{3}+{\frac {3}{40}}{x}^{5}+{\frac {5}{112}}{x}^{7 }+{\frac {35}{1152}}{x}^{9}+{\frac {63}{2816}}{x}^{11}+ \dots$ with all coefficients positive and rapidly decreasing is better than the alternating $\arctan(x)=x-{\frac {1}{3}}{x}^{3}+{\frac {1}{5}}{x}^{5}-{\frac {1}{7}}{x}^{7}+{ \frac {1}{9}}{x}^{9}-{\frac {1}{11}}{x}^{11}+\dots.$

Then there are Pade series such as $$\arcsin(x) \approx \frac{ {\frac {69049}{922320}}{x}^{5}-{\frac {1709}{2196}}{x}^{3} +x }{ {\frac {1075}{6832} }{x}^{4}-{\frac {2075}{2196}}{x}^{2}+1 } $$

which can be more accurate for some values (Perhaps $x \gt 0.2$ compared to the Taylor series above.)

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If you want a competitive implementation you have to know what you're doing. This book looks good, though I haven't read it myself:

http://perso.ens-lyon.fr/jean-michel.muller/SecondEdition.html

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