Thread: Proving the Taylor Series of Sin(x)

1. Proving the Taylor Series of Sin(x)

Could someone help me with this problem? Thanks.

"Prove that the Taylor series of sin(x) at a=pi/2 represents sin(x) for all x."

I already know how to find the Taylor series, but I have no idea how to prove anything.

2. Originally Posted by Leaf
Could someone help me with this problem? Thanks.

"Prove that the Taylor series of sin(x) at a=pi/2 represents sin(x) for all x."

I already know how to find the Taylor series, but I have no idea how to prove anything.
Use Taylors' remainder theorem

Taylor's theorem - Wikipedia, the free encyclopedia

Note that for all $\displaystyle x \in \mathbb{R}$

$\displaystyle |\sin(x)|\le 1$ and $\displaystyle |\cos(x)|\le 1$. Also that for each x that the remainder (the error) goes to $\displaystyle 0 \text{ as } n \to \infty$

3. I don't think it works. The Taylor Series only represents sin(x) when x is between (pi/2,pi/2] when the problem asks for all x. I can't prove something that's not true.

4. Originally Posted by Leaf
I don't think it works. The Taylor Series only represents sin(x) when x is between (pi/2,pi/2] when the problem asks for all x. I can't prove something that's not true.
I think you are mistaken. Both $\displaystyle \sin(x)$ and $\displaystyle \cos(x)$ are real analytic entire functions.

Consider the remainder term from the link above it gives

$\displaystyle \displaystyle R_n(x)=\frac{f^{n+1}(c)}{(n+1)!}\left(x-\frac{\pi}{2} \right)^{n+1}$

Since all derivatives of $\displaystyle \sin(x)$ are less than 1

$\displaystyle \displaystyle |R_n(x)| \le \frac{1}{(n+1)!}\left|x-\frac{\pi}{2} \right|^{n+1}$

now for any $\displaystyle y \in \mathbb{R}$ we can choose an $\displaystyle n$ so large that the remainder is as small as we wish

$\displaystyle \displaystyle |R_n(y)| \le \frac{1}{(n+1)!}\left|y-\frac{\pi}{2} \right|^{n+1} < \epsilon$

5. Alright, thanks for the help. I think I got it now.