Taylor series of $\sin(x)$ converges uniformly on $[-\pi,\pi]$?

Question

According to my notes, the Taylor series of $\sin(x)$ converges uniformly on $[-\pi,\pi]$.

I know that the remainder term needs to converge uniformly to $0$ for this to be the case.

But I really don't know how to begin showing that this series converges uniformly. I think it's the domain that really stumps me. I think I should start showing that the remainder term converges to $0$. So we have:

$$R_n= \frac{(x-x_0)^{N+1}}{N!}\int_0^1 (1-t)^Nf^{(N+1)}(x_0+t(x-x_0))dt$$

Where $R_n$ denotes the remainder term.

What should I do?

Thanks in advance.

Answer 1

Since all of the derivatives of $\sin(x)$ satisfy $$|f^{(N+1)}(x)| \le 1$$ for all $x$, we see that $$|R_n| \le \frac{|x-x_0|^{N+1}}{N!} \le \frac{(2\pi)^{N+1}}{N!}$$ and the term on the right converges to zero independently of $x$. Thus we can conclude that the Taylor series converges uniformly.

Here we used that the integrand is bounded in absolute value by 1.

Answer 2

You just have to write: $$\sup_{[x_0,x_1]} |R_n| \le \frac{\sup_{[x_0,x_1]}|x-x_0|^{N+1}}{N!} \int_0^1 (1-t)^N\sup_{[x_0,x_1]}|f^{(N+1)}(x_0+t(x-x_0))|dt \\ = \frac{|x_1-x_0|^{N+1}}{N!} \int_0^1 (1-t)^N dt = \frac{|x_1-x_0|^{N+1}}{(N+1)!}\to 0 $$

This shows that the series is uniformly convergent on every compact interval.