Maclaurin series

Posted March 12th, 2008 by damascus

Calculus questions

if $f(x)= \sin(x^3)$ , find the fifteenth derivative of f(x), for Maclaurin series, that means a=0

Comments

Derivative of Maclaurin series

On April 18th, 2008 Structure says:

Let consider the Taylor series of an analytic function

$f(x)=\sum_{n=0}^{+\infty}\frac{f^{(n)}}{n!}(x_0)(x-x_0)^n$

For $x_0=0$ we have Maclaurin series

$f(x)=\sum_{n=0}^{+\infty}\frac{f^{(n)}}{n!}(0)x^n$

We have for g(x)=sin(x) we have

$g'(x)=\cos x=\cos(-x)=\sin(\frac{\pi}{2}+x)=\sin(x+\frac{\pi}{2})$

We can go on and write

$g"(x)=\cos(x+\frac{\pi}{2})=\sin(x+2\frac{\pi}{2})$

and by induction

$g^{(n)}(x)=\cos(x+(n-1)\frac{\pi}{2})=\sin(x+n\frac{\pi}{2})$

For x=0 we have

$g^{(n)}(0)=\sin(n\frac{\pi}{2})$

and also

$g^{(2n)}(0)=\sin(2n\frac{\pi}{2})=0$

and

$g^{(2n+1)}(0)=\sin(2n+1)\frac{\pi}{2}=\sin(n\pi+\frac{\pi}{2})=\cos(n\pi)=(-1)^n$

We can now write

$\sin(x)=\sum_{n=0}^{+\infty}\frac{\sin(n\frac{\pi}{2})}{n!}(0)x^n=$

$=\sum_{n=0}^{+\infty}\frac{\sin(2n\frac{\pi}{2})}{(2n)!}(0)x^{2n}+\sum_{n=0}^{+\infty}\frac{\sin(2n+1)\frac{\pi}{2}}{(2n+1)!}(0)x^{2n+1}=\sum_{n=0}^{+\infty}(-1)^n\frac{1}{(2n+1)!}x^{2n+1}$

So we have

$g(x)=\sin x=\sum_{n=0}^{+\infty}(-1)^n\frac{1}{(2n+1)!}x^{2n+1}$

For $f(x)=\sin(x^3)$ simply replace x by $x^3$ in previous series

$f(x)=\sin x^3=\sum_{n=0}^{+\infty}(-1)^n\frac{1}{(2n+1)!}x^{6n+3}$

Such a power series is derivable and you can take term by term derivative of each monomial .
If you write explicitly the terms of the series you see third ninth fifteenth twenty-first powers of x and so on.
As you need the fifteenth derivative each power decrease by 15 (if it is big enough) the first term is a constant That is why the sum begins from n=2 coresponding to the fifteenth power of x

$f^{(15)}(x)=(\sin x^3)^{(15)}=\sum_{n=2}^{+\infty}(-1)^n\frac{(6n+3)(6n+2)...(6n-11)}{(2n+1)!}x^{6n-12}$

$=\sum_{m=0}^{+\infty}(-1)^m\frac{(6m+15)(6m+14)....(6m+1)}{(2m+5)!}x^{6m}$