Thread: did i developed these teylor series correctly..

1. did i developed these teylor series correctly..

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

2. Originally Posted by transgalactic
A.
$\displaystyle sin^2 x=0+x^2-\frac{x^4}{3!}+\frac{(-1)^{2n}}{{(2n+1)!}^2}{x^{4n+2}}+O(4n+2)$
Remember that $\displaystyle 2\sin^2 x = 1 - \cos 2x \implies 2\sin^2 x = 1 - \sum_{n=0}^{\infty} \frac{(2x)^{2n}(-1)^n}{(2n+1)!}$.

$\displaystyle xsin^3 x=0+x^4-\frac{x^6}{3!}+\frac{(-1)^{2n}}{{(2n+1)!}^2}{x^{5n+6}}+O(5n+4)$
This one is similar but you need an identity for $\displaystyle \sin^3x$ like in the first one.

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

regarding b:
$\displaystyle xe^x=x+x^2+\frac{x^3}{2!}+\frac{x^4}{3!}++\frac{x^ {n+1}}{n!}+O(n+1)$
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??

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

5. can you tell me if i solved correctly A,B
ad where is my mistake in C??

6. Originally Posted by transgalactic
regarding C:
$\displaystyle sin3x=3sinx-4sin^3x\\$
$\displaystyle f(x)=sin^3x=\frac{3sinx-sin3x}{4}\\$
$\displaystyle f(0)=0\\$
$\displaystyle f'(x)=\frac{3cosx-\frac{cos3x}{3}}{4}\\$
$\displaystyle f'(0)=0$
$\displaystyle f''(x)=\frac{-3sinx-\frac{sin3x}{9}}{4}\\$
$\displaystyle f''(0)=0$
for every member i get 0
??
You stopped too soon! Since sin(x) starts with x, the lowest power of sin^3(x) is x^3.

7. it doesnt matter
both sinx and sin3x transform into cosx and cos3x and vise versa
in both cases i get zero for every derivative
where did i stop early
i transformed power 3 function into a much easier to differentiate
power 1 function.
whats the problem??