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.

  • 1
    $\begingroup$ Looking at the Lagrange form of the remainder would be easier, I think. $\endgroup$ – David Mitra Jun 9 '14 at 20:22
  • $\begingroup$ It converges pointwise on a compact interval. $\endgroup$ – Arthur Jun 9 '14 at 20:22
  • $\begingroup$ Take the absolute value of the remainder and note that the $N$-th derivative of $\sin$ is bounded. $\endgroup$ – Beni Bogosel Jun 9 '14 at 20:24
  • $\begingroup$ Is the Lagrange form (not been taught to us) equivalent to the remainder term in my original post? $\endgroup$ – 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.

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


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.

  • $\begingroup$ Could you add some explanations? I don't understand where these calculations have come from. $\endgroup$ – Mr Croutini Jun 9 '14 at 20:30
  • $\begingroup$ 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 $\int_0^1 (1-t)^n dt = 1/(n+1)$ $\endgroup$ – mookid Jun 9 '14 at 20:31

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