# Power Series Expansion for Real Arcsine Function

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## Theorem

The (real) arcsine function has a Taylor series expansion:

 $\ds \arcsin x$ $=$ $\ds \sum_{n \mathop = 0}^\infty \frac {\paren {2 n}!} {2^{2 n} \paren {n!}^2} \frac {x^{2 n + 1} } {2 n + 1}$ $\ds$ $=$ $\ds x + \frac 1 2 \frac {x^3} 3 + \frac {1 \times 3} {2 \times 4} \frac {x^5} 5 + \frac {1 \times 3 \times 5} {2 \times 4 \times 6} \frac {x^7} 7 + \cdots$

which converges for $-1 \le x \le 1$.

## Proof

From the General Binomial Theorem:

 $\ds \paren {1 - x^2}^{-1/2}$ $=$ $\ds 1 + \frac 1 2 x^2 + \frac {1 \times 3} {2 \times 4} x^4 + \frac {1 \times 3 \times 5} {2 \times 4 \times 6} x^6 + \cdots$ $\text {(1)}: \quad$ $\ds$ $=$ $\ds \sum_{n \mathop = 0}^\infty \frac {\paren {2 n}!} {2^{2 n} \paren {n!}^2} x^{2 n}$

for $-1 < x < 1$.

From Power Series is Termwise Integrable within Radius of Convergence, $(1)$ can be integrated term by term:

 $\ds \int_0^x \frac 1 {\sqrt{1 - t^2} } \rd t$ $=$ $\ds \sum_{n \mathop = 0}^\infty \int_0^x \frac {\paren {2 n}!} {2^{2 n} \paren {n!}^2} t^{2 n} \rd t$ $\ds \leadsto \ \$ $\ds \arcsin x$ $=$ $\ds \sum_{n \mathop = 0}^\infty \frac {\paren {2 n}!} {2^{2 n} \paren {n!}^2} \frac {x^{2 n + 1} } {2 n + 1}$ Derivative of Arcsine Function

We will now prove that the series converges for $-1 \le x \le 1$.

 $\ds \frac {\paren {2 n}!} {2^{2 n} \paren {n!}^2} \frac {x^{2 n + 1} } {2 n + 1}$ $\sim$ $\ds \frac {\paren {2 n}^{2 n} e^{-2 n} \sqrt {4 \pi n} } {2^{2 n} n^{2 n} e^{-2 n} 2 \pi n} \frac {x^{2 n + 1} } {2 n + 1}$ $\ds$ $=$ $\ds \frac 1 {\sqrt {\pi n} } \frac {x^{2 n + 1} } {2 n + 1}$

Then:

 $\ds \size {\frac 1 {\sqrt {\pi n} } \frac {x^{2 n + 1} } {2 n + 1} }$ $<$ $\ds \size {\frac {x^{2 n + 1} } {n^{3/2} } }$ $\ds$ $\le$ $\ds \frac 1 {n^{3/2} }$

Hence by Convergence of P-Series:

$\ds \sum_{n \mathop = 1}^\infty \frac 1 {n^{3/2} }$

is convergent.

So by the Comparison Test, the Taylor series is convergent for $-1 \le x \le 1$.

$\blacksquare$