By expanding the derivative f'(x) in a power series in x
By expanding the derivative $f'(x)$ in a power series in $x$ and integrating term by term, write the power series expansion in $x$ for $f(x)=\arcsin x$.
Explanation of Method
The derivative $(1-x^2)^{-1/2}$ is expanded using the binomial series with exponent $-\frac12$; the resulting central-binomial-coefficient series is then integrated term by term.
- Differentiate to get $f'(x)=(1-x^2)^{-1/2}$.
- Expand using the binomial series with $\alpha=-\frac12$, $t=-x^2$.
- Integrate term by term and fix the constant.
Step 1 · Differentiate
$$ f'(x) = \frac1{\sqrt{1-x^2}} = (1-x^2)^{-1/2}. $$
Step 2 · Binomial expansion
Using $\binom{-1/2}n=\dfrac{(-1)^n(2n)!}{4^n(n!)^2}$: $$ (1-x^2)^{-1/2} = \sum_{n=0}^{\infty}\binom{-1/2}n(-x^2)^n = \sum_{n=0}^{\infty}\frac{(2n)!}{4^n(n!)^2}x^{2n} = 1+\frac{x^2}2+\frac{3x^4}8+\frac{5x^6}{16}+\cdots $$
Step 3 · Integrate term by term
$$ f(x) = \sum_{n=0}^{\infty}\frac{(2n)!}{4^n(n!)^2(2n+1)}x^{2n+1} + C. $$ Since $f(0)=\arcsin0=0$, $C=0$: $$ \arcsin x = x+\frac{x^3}6+\frac{3x^5}{40}+\frac{5x^7}{112}+\cdots $$ Both endpoints $x=\pm1$ converge (slowly), with $\arcsin(1)=\pi/2$.
Summary Table
| Property | Value |
|---|---|
| Series | $\sum\frac{(2n)!}{4^n(n!)^2(2n+1)}x^{2n+1}$ |
| At $x=1$ | $\pi/2$ |
| Region of validity | $|x|\le1$ |
Additional Notes
- All coefficients are positive, unlike $\arctan x$ (Problem 267), reflecting that $\arcsin$ is monotonically increasing with an unbounded derivative growth rate as $x\to1$.
- The coefficients are exactly $\binom{2n}n/(4^n(2n+1))$, central binomial coefficients scaled down — these same coefficients appear throughout combinatorics and in the arcsine distribution of probability theory.
- Problem 269's $\sinh^{-1}x=\ln(x+\sqrt{1+x^2})$ has an almost identical series but with alternating signs, reflecting the relation $\sinh^{-1}(x)=-i\arcsin(ix)$.