Elementary Taylor Series

We recall that for an infinitely often differentiable function $f$ at $x=0$ ,

$\color{#EC7300}{\boxed{f(x)=f(0)+xf'(0)+\frac{x^2}{2!}f''(0)+\frac{x^3}{3!}f^{(3)}(0)+...}}$ .

If $f(x)=\sin x$ , then:

• $f'(x)=\cos x\implies f'(0)=1$ ;
• $f''(x)=-\sin x\implies f''(0)=0$ ;
• $f^{(3)}(x)=-\cos x\implies f^{(3)}(0)=-1$ ;
• $f^{(4)}(x)=\sin x\implies f^{(4)}(0)=0$ ;
• $f^{(5)}(x)=\cos x\implies f^{(5)}(0)=1$ .

Taking the first three nonzero derivatives, we see that $\sin x=\color{#20A900}{\boxed{x-\frac{x^3}{3!}+\frac{x^5}{5!}}}-...$

Taking this further,

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

Using the example above (of cos (x)) will vastly help with this question. Firstly we find the first five derivatives of cos(x) and then we substitute our results into the Taylor Series Expansion, which will give us x+(-1) $\frac{x^3}{3!}$ + $\frac{x^5}{5!}$ . We can simplify this to x- $\frac{x^3}{3!}$ + $\frac{x^5}{5!}$ and get our final answer!