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

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?

• Looking at the Lagrange form of the remainder would be easier, I think. – David Mitra Jun 9 '14 at 20:22
• It converges pointwise on a compact interval. – Arthur Jun 9 '14 at 20:22
• Take the absolute value of the remainder and note that the $N$-th derivative of $\sin$ is bounded. – Beni Bogosel Jun 9 '14 at 20:24
• Is the Lagrange form (not been taught to us) equivalent to the remainder term in my original post? – Mr Croutini Jun 9 '14 at 20:26

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.
You just have to write: $$\sup_{[x_0,x_1]} |R_n| \le \frac{\sup_{[x_0,x_1]}|x-x_0|^{N+1}}{N!} \int_0^1 (1-t)^N\sup_{[x_0,x_1]}|f^{(N+1)}(x_0+t(x-x_0))|dt \\ = \frac{|x_1-x_0|^{N+1}}{N!} \int_0^1 (1-t)^N dt = \frac{|x_1-x_0|^{N+1}}{(N+1)!}\to 0$$
• you just use the triangle inequality and $\sup A(x)B(x) \le \sup A(x)\sup B(x)$ when $A,B\ge 0$, then the fact that $\sup \sin = \sup\cos = 1$ and $\int_0^1 (1-t)^n dt = 1/(n+1)$ – mookid Jun 9 '14 at 20:31