Power Series Expansion for Real Arcsine Function

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Theorem

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

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

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 $-1 \le x \le 1$.

By Stirling's Formula:

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


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 $-1 \le x \le 1$.

$\blacksquare$


Also see


Sources

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