Two versions of Maclaurin series for $\arcsin(x)$

Question

In comparing the calculus textbook by Soo T. Tan, and https://mathworld.wolfram.com/MaclaurinSeries.html, there are different entries for the Maclaurin series for inverse sine.

From the book, it seems that it is $$\sum_{n=0}^{\infty} \frac{(2n)!}{(2^n n!)^2 (2n+1)}x^{2n+1}$$ but on the website it is $$\sum_{n=0}^{\infty} \frac{\Gamma(n+\frac{1}{2})}{\sqrt{\pi}(2n+1)n!}x^{2n+1}$$

My hunch is that the website's answer applies to a more general $x$-value, but I'm not sure why the difference (or if they're actually two ways of showing the same thing?)

A follow-up question would be, omitting the hyperbolic functions, does this apply to any of the other series that are listed? Like is there an simpler series for $\tan x$ (which might not actually need Bernoulli numbers?)

Answer 1

The two expressions are equivalent; neither is more general than the other, because $n$ is always a nonnegative integer. In such a case,

$$\Gamma(n+\tfrac{1}{2}) = (n+\tfrac{1}{2})\Gamma(n-1+\tfrac{1}{2}) = \cdots = \Gamma(\tfrac{1}{2})\prod_{k=1}^n \frac{2k-1}{2}. \tag{1}$$ Since $\Gamma(\frac{1}{2}) = \sqrt{\pi}$, it follows that $$\begin{align} \frac{\Gamma(n+\tfrac{1}{2})}{\sqrt{\pi}} &= \prod_{k=1}^n \frac{2k-1}{2} \\ &= \frac{1}{2^n} \prod_{k=1}^n (2k-1) \\ &= \frac{1}{2^n} \prod_{k=1}^n (2k-1)(2k) \prod_{k=1}^n \frac{1}{2k} \\ &= \frac{1}{2^n} (2n)! \frac{1}{2^n n!} \\ &= \frac{(2n)!}{2^{2n} n!}. \tag{2} \end{align}$$ This demonstrates the desired equivalence.

For a proof of $\Gamma(\tfrac{1}{2}) = \sqrt{\pi}$, we let $z = u^2$, $dz = 2u \, du$ in the definition of the gamma function: $$\begin{align} \Gamma(\tfrac{1}{2}) &= \int_{z=0}^\infty z^{1/2-1} e^{-z} \, dz \\ &= \int_{u=0}^\infty u^{-1} e^{-u^2} (2u) \, du \\ &= 2 \int_{u=0}^\infty e^{-u^2} \, du \\ &= \int_{u=-\infty}^\infty e^{-u^2} \, du. \tag{3} \end{align}$$

Then using the familiar rectangular to polar coordinate conversion trick,

$$\begin{align} \Gamma(\tfrac{1}{2})^2 &= \int_{x=-\infty}^\infty e^{-x^2} \, dx \int_{y=-\infty}^\infty e^{-y^2} \, dy \\ &= \int_{x=-\infty}^\infty \int_{y=-\infty}^\infty e^{-(x^2+y^2)} \, dy \, dx \\ &= \int_{\theta=0}^{2\pi} \int_{r=0}^\infty e^{-r^2} r \, dr \, d\theta \\ &= 2\pi \left[-\frac{e^{-r^2}}{2}\right]_{r=0}^\infty \\ &= \pi. \tag{4} \end{align}$$

Answer 2

but on the website it is $$\sum_{n=0}^{\infty} \frac{\Gamma(n+\frac{1}{2})}{\sqrt{\pi}(2n+1)n!}x^{2n+1}$$

Use $\Gamma(z)=(z-1)\Gamma(z-1)$, and $\Gamma(1/2)=\sqrt\pi$, we get

$$\Gamma(n+\frac{1}{2})=(n-\frac{1}2)\cdots\frac{3}2\cdot\frac{1}2\cdot\sqrt\pi=\frac{(2n-1)!!}{2^n}\sqrt\pi=\frac{(2n)!}{(2n)!!\cdot2^n}\sqrt\pi$$

Note that $(2n)!!=2^n n!$, we get

$$\begin{align}\sum_{n=0}^{\infty} \frac{\Gamma(n+\frac{1}{2})}{\sqrt{\pi}(2n+1)n!}x^{2n+1}&=\sum_{n=0}^{\infty} \frac{(2n)!}{2^n n!\cdot2^n}\sqrt\pi\cdot\frac{1}{\sqrt{\pi}(2n+1)n!}x^{2n+1}\\ \\ &=\sum_{n=0}^{\infty} \frac{(2n)!}{(2^n n!)^2 (2n+1)}x^{2n+1}\end{align}$$

So they are equivalent.