When does the remainder term in the taylor series go to zero?
Theorem: Let $f\in C^{N+1}([\alpha,\beta])$ and $x,x_0\in(\alpha,\beta)$. Then
$$f(x)=f(x_0)+f'(x_0)(x-x_0)+\frac{1}{2}f''(x_0)(x-x_0)^2+…+\frac{f^{(N)}(x_0)}{N!}(x-x_0)^N+\frac{(x-x_0)^{N+1}}{N!}\int_0^1 (1-t)^Nf^{(N+1)}(x_0+t(x-x_0))dt$$
Where the remainder term, $R_N$, is
$$R_N=\frac{(x-x_0)^{N+1}}{N!}\int_0^1 (1-t)^Nf^{(N+1)}(x_0+t(x-x_0))dt$$
I'm really not understanding this concept, especially because I'm struggling to comprehend what $R_N$ actually means.
Best Answer
There is no easy answer to the question of how to prove that the remainder term goes to zero. It is an art. The art of bounds, the mathematical art known as "Analysis".
If you try some examples you may begin to develop a mastery at this art. For instance, try $f(x) = \cos(x)$. The absolute value of the integrand is $$\bigl| \, (1-t)^Nf^{(N+1)}(x_0+t(x-x_0)) \, \bigr| $$ This is easy to bound. Since all derivatives of $\cos(x)$ are just $\pm\sin(x)$ or $\pm\cos(x)$, and since $0 \le t \le 1$, it follows that the integrand is $\le 1$ in absolute value, for all $t$. So, integrating between $0$ and $1$, and using that $\bigl| \int (blah) \bigr| \le \int |blah|$, we get $$|R_N| \le \frac{|x-x_0|^{N+1}}{N!} $$ Since $|x-x_0|$ is independent of $N$, this comes down to a standard limit problem that you should know from calculus, namely $$\lim_{N \to \infty} a^N/N!=0 $$