I was wondering if there is a nice formula (or approximation) for $\arcsin(x)$ which is defined $[-1,1]$?
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How can a formula get nicer than $\arcsin(x)$? (If you mean in terms of the trigonometric functions, polynomials, exponentials and logarithms etc then the answer is no---why would we invent a new notation $\arcsin$ if there were already a nice elementary way to express the inverse of $\sin$ in terms of well-known functions?) – YiFan Tey Nov 08 '19 at 00:48
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@YiFan What about approximations? – WhoAmI Nov 08 '19 at 00:53
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As noted there is no way to explicitly express $\arcsin(x)$ in terms of e.g. the $\sin,\cos,\tan,\exp,\log$ functions. After all, if there were, why would we need to invent the new notation $\arcsin$? There are however nice approximations to the function. The most obvious one which comes to mind is the Taylor approximation. We have the Taylor series expansion $$\arcsin(x)=\sum_{k=0}^\infty\frac{(2k-1)!!}{(2k)!!}\frac{x^{2k+1}}{2k+1}.$$ If we truncate the sum by discarding the higher order terms, we end up with decent polynomial approximations to $\arcsin(x)$ on the interval. Many other approximations exist, for example the Padé approximant, which gives for example $$\arcsin(x)\approx\frac{x-(1709/2196)x^3+(69049/922320)x^5}{1-(2075/2196)x^2+(1075/6832)x^4}.$$ These are not unique to $\arcsin$, of course, and can be applied to any sufficiently nice function.
YiFan Tey
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How many terms of the expansion would you need for it to be accurate between $[-1,1]$? – WhoAmI Nov 08 '19 at 01:07
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Probably if you require that level of accuracy then neither of the two approximations make good sense (to even attain $10^{-4}$ error at $x=1$ requires more than a thousand terms of the Taylor expansion). They are only good for theoretical considerations, not numerical ones---but I am not sure what would be a better way to do this. – YiFan Tey Nov 08 '19 at 01:27