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let $f(x)$ be continuous on $[a,b]$ and diffrentiable $n+1$ times on $(a,b)$, let $0\leq i \leq n$ be $x_i$ points on $[a,b]$ then there is $c\in(a,b)$ such that:

$$e(x)=\frac{f^{(n+1)(c)}}{(n+1)!}(x-x_0)(x-x_1)\cdots(x-x_n)$$

  1. In the general when $|\frac{f^{(n+1)(c)}}{(n+1)!}|\leq M$ and $b-a=h$ we can bound it by $$e(x)\leq Mh^{n+1}$$

Is it correct?

  1. If the $x_i$ points are equally spaced, meaning that $x_i=a+ih$ for $0\leq i \leq n$ and $h$ a constant then:

$$(x-x_0)(x-x_1)\cdots(x-x_n)\Rightarrow(x-a)(x-a-h)(x-a-2h)\cdots(x-a-nh)$$

How can I bound this function?

  1. If the $x_i$ points are not equally spaced, meaning that $x_i=a+rh$ for $0\leq i \leq n$ and $r_i$ is a number such that $r_0<r_1<...<r_n$ an $h$ is a constant then:

$$(x-x_0)(x-x_1)\cdots(x-x_n)\Rightarrow(x-r_0a)(x-a-r_1h)(x-a-r_2h)\cdots(x-a-r_nh)$$

How can I bound this function?

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  • $\begingroup$ Your first assumption is correct, but normally useless. A More usefull estimate would be $|e(x)| \leq \frac{M_{n+1} h^{n+1}}{(n+1)!}$, where $|f^{(n+1)}(x)|\leq M_{n+1}$. This way there is a chance that your upper bound goes to zero (as $n \to \infty$), even when $b-a\ge 1$. $\endgroup$ – PierreCarre Jan 17 at 10:00
  • $\begingroup$ Regarding your first question, just try with a small number of intervals and get the bounds for $x$ in each sub interval. You'll see that $|(x-x_0) \cdots (x-x_n)| \leq n! h^n$. $\endgroup$ – PierreCarre Jan 17 at 10:46

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