Taylor's theorem and the remainder

Level UniversityDifficulty ★★★★★Theorem⌖ Open in the map

What is it?

The exact size of the approximation error: $f(x) - T_n(x) = \frac{f^{(n+1)}(\xi)}{(n+1)!}(x - a)^{n+1}$ for some $\xi$ between $a$ and $x$ . It is how numerical analysts certify accuracy.

Statement

If $f \in C^{n+1}$ on an interval containing $a$ and $x$ , then

f(x) = T_n(x) + \frac{f^{(n+1)}(\xi)}{(n+1)!}(x - a)^{n+1}

for some $\xi$ between $a$ and $x$ (Lagrange form).

Idea of the proof

Apply Rolle's theorem $n + 1$ times to $g(t) = f(t) - T_n(t) - K(t - a)^{n+1}$ , with $K$ chosen so that $g(x) = 0$ .

Formulas

|R_n(x)| \le \frac{M_{n+1}}{(n+1)!}|x - a|^{n+1}, \qquad M_{n+1} = \max|f^{(n+1)}|

Where it shows up in computing

Scientific computing★★★★★fundamentalScientific computing and algorithms
Truncation errors of finite differences and ODE integrators are Taylor remainders.
Floating point (IEEE 754)★★★★★frequentScientific computing and algorithms
Library designers bound the remainder to decide how many terms reach full double precision.