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

Question

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?

Answer 1

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

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

Answer 2

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.

Answer 3

$$arcsin(t)=arctan(\frac{t}{\sqrt{1-t^2}})$$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 $0.52389$.