# Order of the error for the Trapezoidal and Simpson's method of numerical integration

What are the order of the error for the Trapezoidal and Simpson's method of numerical integration? What is the definition of order of the error of a quadrature formula? Is it true that order of error is one higher than that of degree of precision (I know DOP)

Attempt:

I know that the error term of basic Trapezoidal rule for the interval $[x_0, x_1]$ is $$E_T=-\frac{h^3}{12}f''(\xi), ~x_0<\xi<x_1$$ and that of the composite Trapezoidal rule for the interval $[x_0, x_n]$ is $$E_{TC}=-\frac{h^3n}{12}f''(\xi)=-\frac{h^2(b-a)}{12}f''(\xi), ~x_0<\xi<x_n$$

Then what can we say about the order of the error for the Trapezoidal rule? It is $O(h^3)$ or $O(h^2)$? Or can we say that order of error for basic Trapezoidal rule and composite Trapezoidal rule are $O(h^3)$ and $O(h^2)$, respectively? Which one is true? and Please explore the idea of it.

The order of error only makes sense for the composite rule, since for the simple rule the step size $h=x_f-x_0=b-a$ is constant. Only for the composite rule do you get a variable $h=(b-a)/n$ that allows to consider the asymptotic error behavior. Thus $O(h^2)$ for the composite rule, and no asymptotic error for the simple rule.
Since the (composite) Simpson rule can be seen as Richardson extrapolation (first step of the Romberg method) of the symmetric trapezoidal rule, its error order is automatically $O(h^4)$.
• You have some quantity $Q$ and some approximation method depending on some parameter $h$ giving a value $A(h)$, then the error order is the largest $p$ such that $A(h)-Q=O(h^p)$. Dec 27 '15 at 17:55
• The simple Simpson rule can be considered to integration of an approximation of f by a quadratic or cubic, for if p is a polynomial of degree 3 or less, with p(a)=f(a) & p(b)=f(b) & p((a+b)/2)=f((a+b)/2) then $\int_a^b p(x)dx=(b-a)(f(a)+4f((a+b)/2)+f(b))/6$. A lesser-known one is $(b-a)(5f(a)+8f((a+b)/2)+5f(b))/18$, which equals $\int_a^b f(x)dx$ when f is a polynomial of degree 5 or less. Dec 28 '15 at 7:00
• 1) No. The local error contributions are $O(h^5)$, but you get a sum of $n$ of those, so the composite error is $O((b-a)h^4)$. 2) Taylor expansion around the midpoint of the interval should give rather symmetric formulas. Jul 14 '18 at 5:47
We know that trapezoidal method should not be give any error up to linear polynomial and constant. So error will be starting from second order polynomial thus order of error is $$o(h^2)$$ for trapezoidal.