What is Lagrange Error and how do you find the value for $M$ $M$ ?

Question

I know the formula
$R_{n} \leq \frac{M}{(n + 1)!} {| x - a |}^{n + 1}$ $R_n le frac{M}{(n+1)!}|x-a|^(n+1)$ , but I'm confused as to how to find $M$ $M$ .

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Answer 1 · 2017-04-10T10:00:07

$M$ $M$ is the maximum value of $∣ ∣ f^{(n + 1)} (ξ) ∣ ∣$ $abs(f^((n+1))(xi))$ for $ξ$ $xi$ in the interval delimited by $x$ $x$ and $a$ $a$ .

Consider the Taylor series of a function $f (x)$ $f(x)$ around $x = a$ $x=a$ :

$f (x) = \infty \sum k = 0 \frac{f^{(k)} (a)}{k!} {(x - a)}^{k}$ $f(x) = sum_(k=0)^oo f^((k))(a)/(k!) (x-a)^k$

If we stop the Taylor series at $k = n$ $k = n$ we have:

$f (x) = P_{n} (x) + R_{n} (x)$ $f(x) = P_n(x) +R_n(x)$

where:

$P_{n} (x) = n \sum k = 0 \frac{f^{(k)} (a)}{k!} {(x - a)}^{k}$ $P_n(x) = sum_(k=0)^n f^((k))(a)/(k!) (x-a)^k$

and it can be demonstrated that rest can be expressed as:

$R_{n} (x) = \frac{1}{n!} \int_{a}^{x} f^{(n + 1)} (t) {(x - t)}^{n} d t$ $R_n(x) = 1/(n!) int _a^x f^((n+1))(t) (x-t)^n dt$

Applying the second theorem of the mean to this integral we have:

$R_{n} (x) = \frac{1}{(n + 1)!} f^{(n + 1)} (ξ) {(x - a)}^{n + 1}$ $R_n(x) = 1/((n+1)!) f^((n+1))(xi) (x-a)^(n+1)$

where $ξ$ $xi$ is a point between $x$ $x$ and $a$ $a$

Clearly if in the interval delimited by $x$ $x$ and $a$ $a$ we have:

$∣ ∣ f^{(n + 1)} (ξ) ∣ ∣ < M$ $abs(f^((n+1))(xi)) < M$

then:

$| R_{n} (x) | \leq \frac{M}{(n + 1)!} {| x - a |}^{n + 1}$ $abs( R_n(x)) <= M/((n+1)!)abs(x-a)^(n+1)$