getting Taylor series from differential equation

If a given function $f$ satisfies a differential equation, the Taylor series of $f$ can sometimes be obtained easily.

Let

$$f(x)=\sin (m\arcsin x),$$

where $m$ is a non-zero , be an example (cf. (http://planetmath.org/Cf) the cyclometric functions).  We form the derivatives

$$f^{\prime }(x)=\frac{m}{\sqrt{1-x^2}}\cos (m\arcsin x),$$

$$f^{\prime \prime }(x)=-\frac{m^2}{1-x^2}\sin (m\arcsin x)+\frac{mx}{(1-x^2)\sqrt{1-x^2}}\cos (m\arcsin x),$$

which show that $f$ satisfies the differential equation

$$(1-x^2)f^{\prime \prime }-x{f}^{\prime }+m^{2}f=0.$$

Differentiating this repeatedly gives the equations

$$(1-x^2)f^{\prime \prime \prime }-3x{f}^{\prime \prime }+(m^2-1)f^{\prime }=0,$$

$$(1-x^2)f^{(4)}-5x{f}^{\prime \prime \prime }+(m^2-4)f^{\prime \prime }=0,$$

and so on.  Using the sum of odd numbers  $1+3+5+\mathrm{\cdots }+(2n-1)=n^2$  and induction on $n$ yields the recurrence relation

$$(1-x^2)f^{(n+2)}-(2n+1)x{f}^{(n+1)}+(m^2-n^2)f^{(n)}=0.$$

Plugging in   $x=0$  yields

$$f^{(n+2)}(0)=(n^2-m^2)f^{(n)}(0)\mathit{\hspace{1em}}(n=0,\mathrm{\hspace{0.17em}1},\mathrm{\hspace{0.17em}2},\mathrm{\dots }).$$

Since  $f^{\prime }(0)=m$,  we have that

$$f^{(2n+1)}(0)=m(1^2-m^2)(3^2-m^2)\mathrm{\dots }({(2n-1)}^2-m^2),$$

whereas all even derivatives of $f$ vanish at $x=0$.  (Note that $f$ is an odd function.)  Thus, we obtain the Taylor of $f$:

$$\sin (m\arcsin x)=\frac{m}{1!}x+\frac{m(1^2-m^2)}{3!}{x}^3+\frac{m(1^2-m^2)(3^2-m^2)}{5!}{x}^5+\mathrm{\cdots }$$

By the ratio test, this series converges for  .