Finding Taylor's series of the function: $\frac{e^{a \sin^{-1}x}}{\sqrt{1-x^2}}$

Solution 1:

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 $$ \frac{{e^{a\arcsin x} }}{{\sqrt {1 - x^2 } }} = \frac{1}{a}f'(x) = \sum\limits_{n = 0}^\infty {\frac{{(n + 1)f_{n + 1} }}{a}x^n } . $$

Solution 2:

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.

Solution 3:

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.