# Series Expansion of $\arcsin\left(\frac{a}{a+x}\right)$

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?

• $\left|\frac{a}{a+x}\right|\leq 1$, so I can't see way we can't use expansion for $\arcsin$? (which is valid when |argument| of $\arcsin$ is $\leq 1$) Nov 23, 2013 at 18:52
• At $x=0$, $\arcsin$ has an "infinite" left-derivative. Nov 23, 2013 at 18:56
• 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?
– apg
Nov 23, 2013 at 19:05
• 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. Nov 23, 2013 at 19:07
• cant you convert $arcsin$ to$arctan$ and further try taylor series. Nov 23, 2013 at 19:13

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)\$$

so

$$\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}...\$$

• 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.
– 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

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

and

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

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

$$-\frac{1}{\sqrt{2ax}}+\frac{5\sqrt{x}}{4a\sqrt{2a}}+\ldots$$

Integrating and adjusting the integration constant gives

$$\frac{\pi}{2}-\frac{\sqrt{2x}}{\sqrt{a}}+\frac{5(\sqrt{x})^3}{6a\sqrt{2a}}+\ldots$$

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

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