# 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

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

Differentiating this repeatedly gives the equations

 $(1-x^{2})f^{\prime\prime\prime}-3xf^{\prime\prime}+(m^{2}-1)f^{\prime}=0,$
 $(1-x^{2})f^{(4)}-5xf^{\prime\prime\prime}+(m^{2}-4)f^{\prime\prime}=0,$

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  $f^{\prime}(0)=m$,  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 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  $|x|<1$.

## References

• 1 Ernst Lindelöf: Differentiali- ja integralilasku ja sen sovellutukset I.  WSOY. Helsinki (1950).
Title getting Taylor series from differential equation GettingTaylorSeriesFromDifferentialEquation 2013-03-22 15:06:07 2013-03-22 15:06:07 Wkbj79 (1863) Wkbj79 (1863) 17 Wkbj79 (1863) Example msc 41A58 ExamplesOnHowToFindTaylorSeriesFromOtherKnownSeries SawBladeFunction SpecialCasesOfHypergeometricFunction