Finding Taylor's series of the function: $\frac{e^{a \sin^{-1}x}}{\sqrt{1-x^2}}$ Show that $$\frac{e^{a \sin^{-1}x}}{\sqrt{1-x^2}}=1+\frac{ax}{1!}+\frac{(a^2+1^2)x^2}{2!}+\frac{a(a^2+2^2)x^3}{3!}+\frac{(a^2+1^2)(a^2+3^2)x^4}{4!}+\cdots$$
My attempt:
I integrated the function and got $\frac{e^{a \sin^{-1}x}}{a}$ then I wrote the series of $e^{a \sin^{-1}x}$ but it contained terms like $(\sin^{-1}x)^2$, $(\sin^{-1}x)^3$ and so on so I could not find the series. My idea was to find the series of the anti derivative of the function and then to derivate the obtained series. Any other way to do it?
 A: Let
$$
f(x) = e^{a\arcsin x}  = \sum\limits_{n = 0}^\infty  {f_n x^n } .
$$
As you already observed, $f(x)$ satisfies the non-linear ODE $(1 - x^2 )(f'(x))^2  = a^2f^2 (x)$. Differentiating this equation and dividing through by $2f'(x)$ yields
$$
(1 - x^2 )f''(x) - xf'(x) - a^2 f(x) = 0.
$$
Substituting the power series into this equation gives $f_0  = 1$, $f_1  = a$ (you can see from the definition that $f(0) = 1$, $f'(0) = a$) and
$$
f_{n + 2}  = \frac{{a^2+n^2}}{{(n + 1)(n + 2)}}f_n 
$$
for $n\geq 0$. The power series expansion you are asking for then follows since
$$
\frac{{e^{a\arcsin x} }}{{\sqrt {1 - x^2 } }} = \frac{1}{a}f'(x) = \sum\limits_{n = 0}^\infty  {\frac{{(n + 1)f_{n + 1} }}{a}x^n } .
$$
A: It is possible to follow your idea, by showing that the series for $s(x)=e^{a\sin^{-1}x}$ conjectured from the RHS (via integration) satisfies $s(0)=1$ and $s'(x)=as(x)(1-x^2)^{-1/2}$, considering the last expression as the Cauchy product of $as(x)$ and the binomial series $(1-x^2)^{-1/2}$. This results in a recurrence for the coefficients of $s(x)$, which is possible (but not at all easy) to verify.
An easier (and more straightforward) approach for me is via complex analysis. Let $$f(z)=\frac{e^{a\sin^{-1}z}}{\sqrt{1-z^2}}=\sum_{n=0}^\infty a_n z^n,\qquad a_n=\frac{1}{2\pi i}\oint\frac{f(z)}{z^{n+1}}\,dz;$$ here $f(z)$ is analytic on $|z|<1$, and the integral is taken along any simple (positively oriented) contour encircling $z=0$ closely enough. If we substitute $z=\sin w$ in the integral (and note that simple contours encircling $w=0$ closely enough map to similar contours in the $z$-plane), we get $$a_n=\frac{1}{2\pi i}\oint\frac{e^{aw}\,dw}{\sin^{n+1}w}.$$
Now we use integration by parts, in the form of $\oint f'(w)g(w)\,dw=-\oint f(w)g'(w)\,dw$, where $f(w)$ and $g(w)$ are analytic on a domain containing the path of integration. This gives, for $n>1$, $$\oint\frac{e^{aw}\,dw}{\sin^{n-1}w}=\frac{n-1}{a}\oint e^{aw}\frac{\cos w}{\sin^n w}\,dw=\frac{n-1}{a^2}\oint e^{aw}\left(\frac{\sin w}{\sin^n w}+n\frac{\cos^2 w}{\sin^{n+1}w}\right)dw,$$ i.e. $a^2 a_{n-2}=(n-1)\big(a_{n-2}+n(a_n-a_{n-2})\big)$ or $\color{blue}{n(n-1)a_n=\big(a^2+(n-1)^2\big)a_{n-2}}$.
With $a_0=1$ and $a_1=a$ computed any way you like, the result follows by induction.
A: Hint
You face a problem of composition of series.
Start with
$$\sin ^{-1}(x)=x+\frac{x^3}{6}+\frac{3 x^5}{40}+O\left(x^7\right)$$
$$a\sin ^{-1}(x)=ax+a\frac {x^3}{6}+a\frac{3 x^5}{40}+O\left(x^7\right)$$ Now, use
$$e^{a \sin ^{-1}(x)}=\exp\Big[ax+a\frac {x^3}{6}+a\frac{3 x^5}{40}+O\left(x^7\right) \Big]$$ when done, work the denominator and use long division.
Edit
Your idea of using the antiderivative is very good. You could even continue using the logarithm of it and then go backward.
A: We can use your idea of finding the antiderivative. We have:
\begin{align}
e^{a\sin^{-1}(x)}& =1+a\sin^{-1}x+\dfrac{(a\sin^{-1}x)^2}{2!}+...+\dfrac{(a\sin^{-1}x)^5}{5!}+...\\
&=1+a\left(x+\dfrac{x^3}{6}+\dfrac{3x^5}{40}+...\right)+\dfrac{a^2}{2!}\left(x+\dfrac{x^3}{6}+...\right)^2+\dfrac{a^3}{3!}\left(x+\dfrac{x^3}{6}+...\right)^3+\dfrac{a^4}{4!}x^4+\dfrac{a^5}{5!}x^5+...\\
&=1+a\left(x+\dfrac{x^3}{6}+\dfrac{3x^5}{40}+...\right)+\dfrac{a^2}{2!}\left(x^2+\dfrac{x^4}{3}+...\right)+\dfrac{a^3}{3!}\left(x^3+\dfrac{x^5}{2}+...\right)+\dfrac{a^4}{4!}x^4+\dfrac{a^5}{5!}x^5+...\\
&=1+ax+\dfrac{a^2x^2}{2!}+\left(\dfrac{ax^3}{3!}+\dfrac{a^3x^3}{3!}\right)+\left(\dfrac{a^2x^4}{3!}+\dfrac{a^4x^4}{4!}\right)+\left(\dfrac{3ax^5}{40}+\dfrac{a^3x^5}{2\cdot3!}+\dfrac{a^5x^5}{5!}\right)+...\\
&=1+ax+\dfrac{a^2x^2}{2!}+\dfrac{(a+a^3)x^3}{3!}+\dfrac{(4a^2+a^4)x^4}{4!}+\dfrac{(9a+10a^3+a^5)x^5}{5!}+...\\
\end{align}
Thus, $\dfrac{1}{a}e^{a\sin^{-1}(x)}=\dfrac{1}{a}+x+\dfrac{ax^2}{2!}+\dfrac{(1+a^2)x^3}{3!}+\dfrac{(4a+a^3)x^4}{4!}+\dfrac{(9+10a^2+a^4)x^5}{5!}+...$
Finally, take the derivative on both sides and we are done.
A: For $m\in\mathbb{N}=\{1,2,\dotsc\}$ and $|t|<1$, the function $\bigl(\frac{\arcsin t}{t}\bigr)^{m}$, whose value at $t=0$ is defined to be $1$, has Maclaurin's series expansion
\begin{equation}\label{arcsin-series-expansion-unify}\tag{QS}
\boxed{\biggl(\frac{\arcsin t}{t}\biggr)^{m}
=1+\sum_{k=1}^{\infty} (-1)^k\frac{Q(m,2k;2)}{\binom{m+2k}{m}}\frac{(2t)^{2k}}{(2k)!}},
\end{equation}
where
\begin{equation}\label{Q(m-k)-sum-dfn}
Q(m,k;\alpha)=\sum_{\ell=0}^{k} \binom{m+\ell-1}{m-1} s(m+k-1,m+\ell-1)\biggl(\frac{m+k-\alpha}{2}\biggr)^{\ell}
\end{equation}
for $m,k\in\mathbb{N}$ and $\alpha\in\mathbb{R}$ such that $m+k\ne\alpha$ and $s(m+k-1,m+\ell-1)$ is the Stirling number of the first kind.
The above texts are excerpted from Theorem 1 in the paper [1] below.
Combining the nice general form \eqref{arcsin-series-expansion-unify} for the series expansion $(\arcsin t)^m$ with other incomplete answers above, one can derive alternative answers to the question.
References

*

*Bai-Ni Guo, Dongkyu Lim, and Feng Qi, Series expansions of powers of arcsine, closed forms for special values of Bell polynomials, and series representations of generalized logsine functions, AIMS Mathematics 6 (2021), no. 7, 7494--7517; available online at https://doi.org/10.3934/math.2021438.

*B.-N. Guo, D. Lim, and F. Qi, Maclaurin's series expansions for positive integer powers of inverse (hyperbolic) sine and tangent functions, closed-form formula of specific partial Bell polynomials, and series representation of generalized logsine function, Appl. Anal. Discrete Math. 16 (2022), in press; available online at https://doi.org/10.2298/AADM210401017G.

*Feng Qi, Explicit formulas for partial Bell polynomials, Maclaurin's series expansions of real powers of inverse (hyperbolic) cosine and sine, and series representations of powers of Pi, Research Square (2021), available online at https://doi.org/10.21203/rs.3.rs-959177/v3.

*Feng Qi, Taylor's series expansions for real powers of functions containing squares of inverse (hyperbolic) cosine functions, explicit formulas for special partial Bell polynomials, and series representations for powers of circular constant, arXiv (2021), available online at https://arxiv.org/abs/2110.02749v2.

*Feng Qi and Mark Daniel Ward, Closed-form formulas and properties of coefficients in Maclaurin's series expansion of Wilf's function, arXiv (2021), available online at https://arxiv.org/abs/2110.08576v1.

A: In [1, p. 3, (2.7)] and [2, pp. 210--211, (10.49.33) and (10.49.34)], the formulas
\begin{equation}\label{arcsin-pochhammer}\tag{Q1}
\sum_{k=0}^{\infty}\frac{(\textrm{i} a)_{k/2}}{(\textrm{i} a+1)_{-k/2}}\frac{(-\textrm{i} x)^k}{k!}
=\exp\biggl(2a\arcsin\frac{x}{2}\biggr)
\end{equation}
and
\begin{equation}\label{JO(833)}\tag{Q2}
\sum_{k=0}^{\infty}\frac{\bigl(\textrm{i} a+\frac{1}{2}\bigr)_{k/2}}{\bigl(\textrm{i} a+\frac{1}{2}\bigr)_{-k/2}} \frac{(-\textrm{i} x)^k}{k!}
=\frac{2}{\sqrt{4-x^2}\,}\exp\biggl(2a\arcsin\frac{x}{2}\biggr)
\end{equation}
were collected, where $\textrm{i}=\sqrt{-1}$ is the imaginary unit, the extended Pochhammer symbol $(z)_\alpha$ for $z,\alpha\in\mathbb{C}$ such that $z+\alpha\ne0,-1,-2,\dotsc$ is defined by
\begin{equation}\label{extended-Pochhammer-dfn}\tag{Q3}
(z)_{\alpha}=\frac{\Gamma(z+\alpha)}{\Gamma(z)},
\end{equation}
and the Euler gamma function $\Gamma(z)$ is defined [3, Chapter 3] by
\begin{equation*}
\Gamma(z)=\lim_{n\to\infty}\frac{n!n^z}{\prod_{k=0}^n(z+k)}, \quad z\in\mathbb{C}\setminus\{0,-1,-2,\dotsc\}.
\end{equation*}
In \eqref{arcsin-pochhammer} and \eqref{JO(833)}, replacing $x$ by $2x$ and employing the extended Pochhammer symbol in \eqref{extended-Pochhammer-dfn} gives
\begin{align}\label{arcsin-pochhammer-binomial}
\textrm{e}^{2a\arcsin x}
&=\sum_{k=0}^{\infty}(-2\textrm{i})^k \frac{\Gamma\bigl(\textrm{i} a+\frac{k}{2}\bigr)}{\Gamma(\textrm{i} a)} \frac{\Gamma(\textrm{i} a+1)}{\Gamma\bigl(\textrm{i} a-\frac{k}{2}+1\bigr)}\frac{x^k}{k!}\\
&=1+\textrm{i} a\sum_{k=1}^{\infty}(-2\textrm{i})^k \binom{\textrm{i} a+\frac{k}{2}-1}{k-1}\frac{x^k}{k}\tag{Q4}
\end{align}
and
\begin{equation}\label{JO(833)-x2(2x)}\tag{Q5}
\boxed{\frac{\textrm{e}^{2a\arcsin x}}{\sqrt{1-x^2}\,}
=\sum_{k=0}^{\infty} (-2\textrm{i})^k\frac{\Gamma\bigl(\textrm{i} a+\frac{1+k}{2}\bigr)}{\Gamma\bigl(\textrm{i} a+\frac{1-k}{2}\bigr)}\frac{x^k}{k!}
=\sum_{k=0}^{\infty} (-2\textrm{i})^k\binom{\textrm{i} a+\frac{k-1}{2}}{k}x^k},
\end{equation}
where the extended binomial coefficient $\binom{z}{w}$ is defined in [4] by
\begin{equation}\label{Gen-Coeff-Binom}\tag{Q6}
\binom{z}{w}=
\begin{cases}
\dfrac{\Gamma(z+1)}{\Gamma(w+1)\Gamma(z-w+1)}, & z\not\in\mathbb{N}_-,\quad w,z-w\not\in\mathbb{N}_-\\
0, & z\not\in\mathbb{N}_-,\quad w\in\mathbb{N}_- \text{ or } z-w\in\mathbb{N}_-\\
\dfrac{\langle z\rangle_w}{w!},& z\in\mathbb{N}_-, \quad w\in\mathbb{N}_0\\
\dfrac{\langle z\rangle_{z-w}}{(z-w)!}, & z,w\in\mathbb{N}_-, \quad z-w\in\mathbb{N}_0\\
0, & z,w\in\mathbb{N}_-, \quad z-w\in\mathbb{N}_-\\
\infty, & z\in\mathbb{N}_-, \quad w\not\in\mathbb{Z}
\end{cases}
\end{equation}
in terms of the gamma function $\Gamma(z)$ and the falling factorial
\begin{equation}\label{Fall-Factorial-Dfn-Eq}
\langle z\rangle_k=
\prod_{\ell=0}^{k-1}(z-\ell)=
\begin{cases}
z(z-1)\dotsm(z-k+1), & k\in\mathbb{N};\\
1,& k=0.
\end{cases}
\end{equation}
The above texts are excerpted from the first proof of Theorem 1 in the paper [5] below.
References

*

*J. M. Borwein and M. Chamberland, Integer powers of arcsin, Int. J. Math. Math. Sci. 2007, Art. ID 19381, 10 pages; available online at https://doi.org/10.1155/2007/19381.

*E. R. Hansen, A Table of Series and Products, Prentice-Hall, Englewood Cliffs, NJ, USA, 1975.

*N. M. Temme, Special Functions: An Introduction to Classical Functions of Mathematical Physics, A Wiley-Interscience Publication, John Wiley & Sons, Inc., New York, 1996; available online at http://dx.doi.org/10.1002/9781118032572.

*C.-F. Wei, Integral representations and inequalities of extended central binomial coefficients, Math. Methods Appl. Sci. (2022), in press; available online at https://doi.org/10.1002/mma.8115.

*B.-N. Guo, D. Lim, and F. Qi, Maclaurin's series expansions for positive integer powers of inverse (hyperbolic) sine and tangent functions, closed-form formula of specific partial Bell polynomials, and series representation of generalized logsine function, Appl. Anal. Discrete Math. 16 (2022), in press; available online at https://doi.org/10.2298/AADM210401017G.

