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

Let $$f(x) = \sin(m\arcsin x),$$ where $m$ is a non-zero constant, be an example (cf. the cyclometric functions). We form the derivatives $$f'(x) = \frac{m}{\sqrt{1-x^2}}\cos(m\arcsin x),$$ $$f''(x) = -\frac{m^2}{1-x^2}\sin(m\arcsin x) +\frac{mx}{(1-x^2)\sqrt{1-x^2}}\cos(m\arcsin x),$$ which show that $f$ satisfies the differential equation $$(1-x^2)f''-xf'+m^2f = 0.$$ Differentiating this repeatedly gives the equations $$(1-x^2)f'''-3xf''+(m^2-1)f' = 0,$$ $$(1-x^2)f^{(4)}-5xf'''+(m^2-4)f'' = 0,$$ and so on. Using the sum of odd numbers $1+3+5+\cdots+(2n\!-\!1) = n^2$ and induction on $n$ yields the recurrence relation $$(1-x^2)f^{(n+2)}-(2n+1)xf^{(n+1)}+(m^2-n^2)f^{(n)} = 0.$$ Plugging in $x = 0$ yields $$f^{(n+2)}(0) = (n^2-m^2)f^{(n)}(0) \quad (n = 0,\,1,\,2,\,...).$$ Since $f'(0) = m$ , we have that $$f^{(2n+1)}(0) = m(1^2-m^2)(3^2-m^2)\ldots((2n-1)^2-m^2),$$ whereas all even derivatives of $f$ vanish at $x=0$ . (Note that $f$ is an odd function.) Thus, we obtain the Taylor expansion of $f$ : $$\sin(m\arcsin x) = \frac{m}{1!}x+\frac{m(1^2-m^2)}{3!}x^3+ \frac{m(1^2-m^2)(3^2-m^2)}{5!}x^5+\cdots$$ By the ratio test, this series converges for $|x| < 1$ .

Bibliography

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ERNST LINDELÖF: Differentiali- ja integralilasku ja sen sovellutukset I. WSOY. Helsinki (1950).