Did i developed these teylor series correctly

1. Feb 14, 2009

transgalactic

A.
$$sin^2 x=0+x^2-\frac{x^4}{3!}+\frac{(-1)^{2n}}{{(2n+1)!}^2}{x^{4n+2}}+O(4n+2)$$
B.
$$xe^x=x+x^2+\frac{x^3}{2!}+\frac{x^4}{3!}++\frac{x^{n+1}}{n!}+O(n+1)$$
C.
$$xsin^3 x=0+x^4-\frac{x^6}{3!}+\frac{(-1)^{2n}}{{(2n+1)!}^2}{x^{5n+6}}+O(5n+4)$$

Last edited: Feb 14, 2009
2. Feb 15, 2009

transgalactic

A:
$$cos 2x=cos^2x-sin^2x=1-2sin^2x\\$$
$$f(x)=sin^2x=\frac{1-cos 2x}{2}\\$$
$$f(0)=0\\$$
$$f'(x)=\frac{1-cos 2x}{2}=cos 2xsin2x=\frac{sin4x}{2}\\$$
$$f'(0)=0\\$$
$$f''(x)=2cos4x\\$$
$$f^{(3)}=-8sin4x\\$$
$$f^{(4)}=-32cos4x\\$$
i know to to use each derivative to build a member out of it.
how to defne the n'th member??

regarding b:
$$xe^x=x+x^2+\frac{x^3}{2!}+\frac{x^4}{3!}++\frac{x^{n+1}}{n!}+O(n+1)$$
i am sure that its correct because i took the series for e^x and multiplied eah member by x.
so the only error that could be is with the remainder .
i thought that if we multiply the remainder by x we add 1 to it
where is my mistake??

regarding C:
$$sin3x=3sinx-4sin^3x\\$$
$$f(x)=sin^3x=\frac{3sinx-sin3x}{4}\\$$
$$f(0)=0\\$$
$$f'(x)=\frac{3cosx-\frac{cos3x}{3}}{4}\\$$
$$f'(0)=0$$
$$f''(x)=\frac{-3sinx-\frac{sin3x}{9}}{4}\\$$
$$f''(0)=0$$
for every member i get 0
??

did i solved A,B correctly
where is my mistake for C

??