# Interpolation Error For Equal And Non Equal Interval

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 $$r_0 an $$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?

• 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 $b-a\ge 1$. – PierreCarre Jan 17 at 10:00
• 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 $|(x-x_0) \cdots (x-x_n)| \leq n! h^n$. – PierreCarre Jan 17 at 10:46