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

Question

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

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Answer 1 · 2017-04-10T10:00:07

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

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

$f(x) = sum_(k=0)^oo f^((k))(a)/(k!) (x-a)^k$

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

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

where:

$P_n(x) = sum_(k=0)^n f^((k))(a)/(k!) (x-a)^k$

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

$R_n(x) = 1/(n!) int _a^x f^((n+1))(t) (x-t)^n dt$

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

$R_n(x) = 1/((n+1)!) f^((n+1))(xi) (x-a)^(n+1)$

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

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

$abs(f^((n+1))(xi)) < M$

then:

$abs( R_n(x)) <= M/((n+1)!)abs(x-a)^(n+1)$