Can anyone think of a good approximation to: $$ \arcsin\left(\frac{a}{a+x}\right)\ $$ accurate at $x = 0$? The Taylor series is not available...perhaps some other kind of method?

  • $\begingroup$ $\left|\frac{a}{a+x}\right|\leq 1$, so I can't see way we can't use expansion for $\arcsin$? (which is valid when |argument| of $\arcsin$ is $\leq 1$) $\endgroup$
    – user110822
    Nov 23, 2013 at 18:52
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    $\begingroup$ At $x=0$, $\arcsin$ has an "infinite" left-derivative. $\endgroup$ Nov 23, 2013 at 18:56
  • $\begingroup$ that's right - therein lies the problem. arcsin(x) has an expansion at 0, but doesn't have one at x = 1 in terms of real numbers. Question: if I get a a expansion in terms of complex numbers, can I use it for real x? $\endgroup$
    – apg
    Nov 23, 2013 at 19:05
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    $\begingroup$ I don't think you should look at complex numbers, I think you should abandon the exigence of a Taylor series. Just relax to a power series, which would allow for non-integer powers and I think it's possible to cook up something. $\endgroup$ Nov 23, 2013 at 19:07
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    $\begingroup$ cant you convert $arcsin$ to$arctan$ and further try taylor series. $\endgroup$
    – Suraj M S
    Nov 23, 2013 at 19:13

3 Answers 3


To answer my own question, the most accurate I can get is by using

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


$$ \arcsin\left(\frac{a}{a+x}\right)=\frac{\pi}{2}-\frac{\sqrt{2}}{\sqrt{a}}x^{1/2}+\frac{5}{6a^{3/2}\sqrt{2}}x^{3/2}...\ $$

  • $\begingroup$ For my purpose, I need to raise e to this lot, so this should be a bit easier, using some asymptotics. Suraj's answer works v.well though. $\endgroup$
    – apg
    Nov 23, 2013 at 20:30

$$arcsin(t)=arctan(\frac{t}{\sqrt{1-t^2}})$$further find taylor series.

on evaluating you get $$arcsin(\frac{a}{a+x})=\frac{a}{(2ax+x^2)^\frac{1}{2}}-\frac{a^3}{3(2ax+x^2)\frac{3}{2}}+ \frac{a^5}{6(2ax+x^2)\frac{5}{2}}-.....$$this works for $a$ in$(0,\infty)$.

example : when $x=a ,arcsin(1/2)= 0.5235$ and the formula gives the answer as $0.52389$.


EDIT: As per request, here's how I did it:

I took the derivative of $\arcsin(a/(a+x))$ to get

$$\frac{-a}{(a+x)\sqrt{2ax+x^2}} = \frac{-1}{\sqrt{2ax}\sqrt{1+\frac{x}{2a}}\left(1+\frac{x}{a}\right)}$$

It is then possible to make a Taylor expansion of the last two factors

$$\frac{1}{\sqrt{1+\frac{x}{2a}}}=\sum_{n=0}^{\infty}{-1/2\choose n}\left(\frac{x}{2a}\right)^n$$



which when working out the first couple of terms of the product gives


Integrating and adjusting the integration constant gives


The original formula I had contained some mistakes. Now it agrees with Alex Giles' formula.

  • $\begingroup$ could be better, see my answer $\endgroup$
    – apg
    Nov 23, 2013 at 20:02
  • $\begingroup$ you should show you work $\endgroup$
    – EhBabay
    Nov 23, 2013 at 20:15

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