Trying to better understand the derivation of the Taylor remainder function

Question

For some function $f(x)$ let $T_n(x)$ be its Taylor polynomial of degree $n$ computed in a neighborhood of $x=a$:

$$T_n(x) = \sum_{k=0}^{n}\frac{f^{(k)}(a)}{k!}(x-a)^k$$

Then for the remainder/error function $R_n(x)$ we have

$$f(x) = T_n(x) + R_n(x)$$

At the $(n+1)$th derivative:

$$f^{(n+1)}(x) = T_n^{(n+1)}(x) + R_n^{(n+1)}(x) = 0 + R_n^{(n+1)}(x) = R_n^{(n+1)}(x)$$

Suppose $|R_n^{(n+1)}(x)| < M$ for some constant $M$. For instance say we want to estimate $f(x)$ to $5$ decimal places so we want to know how many terms to compute, $n$. So we need $M = 10^{-5}$. Or, actually, maybe we don't know $M$ yet because what we really want is $|R_n(x)| < 10^{-5}$ so $M$ is unknown currently.

Is it then valid for me to say:

$$\int|R_n^{(n+1)}(x)| dx < \int M dx$$

$$|R_n^{(n)}(x)| < Mx$$

And then:

$$ \int |R_n^{(n)}(x)| dx < \int Mx dx$$

$$|R_n^{(n-1)}(x)| < M\frac{x^2}{2}$$

$$...$$

$$|R_n^{0}(x)| = |R_n(x)| < M\frac{x^{n+1}}{(n+1)!}$$

I mean is this even right so far?

I'm trying to gain some intuition behind this remainder function but when I look online I feel like so many steps are skipped because it's "easy to show" or "obvious" when to me it isn't.

Am I on the right track or missing some core concept or assumption here? I'm not really sure what I'm missing or skipping or where I should go from here.

Answer 1

Here's a couple of hints:

• The Taylor Series for $f^{(n+1)}(x)$ is $\Sigma_{k=0}^{\infty}\frac{f^{(k+n+1)}(a)}{k!}(x-a)^k$
• $(k+n+1)!\geq k!(n+1)!$

If you re-index the remainder formula you should be able to derive the proper inequality.