Question

What's an easy way of finding a polynomial expansion of arcsin(x)? I found arctan(x) pretty easily by a simple transformation of the geometric series followed by integration, but I can't seem to use that same method with arcsine because the integral becomes too complicated. I need not necessarily have a Taylor/Maclaurin series, but I would like some sort of a simple summation that could theoretically distributed out into a polynomial. What's an easy way of finding a polynomial expansion of arcsin(x)? I found arctan(x) pretty easily by a simple transformation of the geometric series followed by integration, but I can't seem to use that same method with arcsine because the integral becomes too complicated. I need not necessarily have a Taylor/Maclaurin series, but I would like some sort of a simple summation that could theoretically distributed out into a polynomial. @Mathematics

Answer 1

Additional information for the solvers: How I found arctan(x): We know the geometric series.\[\sum_{n=0}^\infty x^n = \frac{1}{1-x}\]We also know\[\frac{d}{dx}\text{arctan}(x)=\frac{1}{1+x^2}.\]Thus, we can rewrite the right hand side as the following summation.\[\frac{d}{dx}\text{arctan}(x)=\sum_{n=0}^\infty (-x^2)^n=\sum_n^\infty (-1)^n x^{2n}\]Integrating both sides lets us find a polynomial expansion.\[\text{arctan}(x)=\int \sum_{n=0}^\infty (-1)^n x^{2n} dx=\boxed{\displaystyle \sum_{n=0}^\infty \frac{(-1)^n x^{2n+1}}{2n+1}}\ \checkmark\]While this does happen to be a Taylor expansion, I am looking for an expansion for the arcsine function that is polynomial but not necessarily Taylor/Maclaurin. For instance, by the geometric series, we can expand 1/x in the following manner.\[\frac{1}{x}=\sum_{n=0}^{\infty}(1-x)^n\]If an expansion for arcsin(x) is most convenient in this form, I'm totally okay with that.