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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.

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  • $\begingroup$ Your argument can be made rigorous. You should either look at proof in Wikipedia which uses an argument similar to yours (search "integral form of the remainder" in wiki page) or refer my blog post which avoids integration and uses Rolle's theorem. $\endgroup$ – Paramanand Singh Feb 28 '18 at 12:07
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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.

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