Answer to Question #226019 in Math for Fazi

Question

Find McLaurin Series for Sin(x)

Answer 1

$f(x)=sin(x),\;f(0)=0.$

$f'(x)=cos(x),\;f'(0)=1.$

$f''(x)=-sin(x),\;f''(0)=0.$

$f'''(x)=-cos(x),\;f'''(0)=-1.$

$f^{(4)}(x)=-sin(x),\;f^{(4)}(0)=0.$

Since $f^{(4)}=f(x)$ , the cycle of {0, 1, 0, -1} repeats itself, which means that every derivative of even degree gives 0 and that every derivative of odd degree alternates between 1 and -1. So, we have

$f(x)=\Sigma^\infin_{n=0}\frac{f^{(n)}(0)}{n!}x^n=\Sigma^\infin_{n=0}(-1)^n\frac{x^{2n+1}}{(2n+1)!}$