I was tasked with the following:

Letting $h=(b-a)/(m+1)$ and $x_j=a+jh$ where $j=0,1,\dots,m,m+1$, the composite trapezoid rule says \begin{equation} \int^{b}_a\approx h\sum^{m}_{j=0}\frac{g(x_j)+g(x_{j+1})}{2}\,. \end{equation} Assuming $g$ is smooth enough, show that the error in the approximation is $\mathcal{O}(h^2)$.

I proceeded as follows:

Noting that $x_j=a+hj$ and $x_{j+1}=a+h(j+1)$, we can write the Taylor expansions for any subinterval as follows, expanding about $x_j$ and $x_{j+1}$ respectively: \begin{equation*} g(x)=g(a+h j)+(-a-h j+x) g'(a+h j)+\mathcal{O}\left((-a-h j+x)^2\right) \end{equation*} and \begin{equation*} g(x)=g(a+h(j+1))+(-a-h (j+1)+x) g'(a+h j+h)+\mathcal{O}\left((-a-h (j+1)+x)^2\right)\,. \end{equation*} For the $x_i$ case, we have \begin{equation*} \lim_{h\to\infty}\Bigg|\frac{(-a-h j+x)^2}{h^2}\Bigg|=\lim_{h\to\infty}\Bigg|\frac{a^2}{h^2}-\frac{2 a x}{h^2}+\frac{2 a j}{h}+\frac{x^2}{h^2}-\frac{2 j x}{h}+j^2\Bigg|=j^2\,, \end{equation*} and by inspection it is easy to see that any higher powers of $h$ in the denominator send the limit to zero and lower powers send it to infinity. The same is true for the $x_{i+1}$ case: \begin{align*} \lim_{h\to\infty}\Bigg|\frac{-a-h (j+1)+x)^2}{h^2}\Bigg|&=\lim_{h\to\infty}\Bigg|\frac{a^2}{h^2}-\frac{2 a x}{h^2}+\frac{2 a j}{h}+\frac{2 a}{h}+\frac{x^2}{h^2}-\frac{2 j x}{h}-\frac{2 x}{h}+j^2+2 j+1\Bigg|\\ &=j^2+2j+1=(j+1)^2\,. \end{align*} Because $j$ is a constant for each subinterval, we thus know that the error term in either expansion is $\mathcal{O}(h^2)$ and therefore the error in their addition and division by 2 is $\mathcal{O}(h^2)$. However, each term in the summation is multiplied by $h$, leaving us with $\mathcal{O}(h^3)$. But when we take the sum over all the subintervals, we multiply by $\frac{b-a}{h}$, bringing us back to $\mathcal{O}(h^2)$.

Is this a valid solution? I have not seen the limit definition for big-O used much and wonder if this is a good application for it. The TA says that taking $h$ to infinity is improper because it represents the grid spacing, but can't we just adjust $b$ and $a$ to make that happen?

Note: I understand that a proof like Ian's is generally the way this is done. I'm specifically seeking comments about my approach here.


1 Answer 1


This looks a bit long to me for a proof like this. First of all, it is enough to just look at one subinterval, so you are comparing $\int_a^b f(x) dx$ and $\frac{f(a)+f(b)}{2}$. Then just sum up the errors, being careful to remember that the number of summands depends on $h$. For simplicity then let me denote the interval by $(0,h)$.

As for the error on one subinterval, recall that the trapezoidal rule exactly integrates linear functions. Thus you are really dealing with

$$\int_0^h f(x)-\left ( f(0) + \frac{f(h)-f(0)}{h} x \right ) dx.$$

Now write $f(x)=f(0)+f'(0)x+O(h^2)$ to get

$$\int_0^h f(0)-f(0)+f'(0)x+\frac{f(h)-f(0)}{h} x + O(h^2) dx = \int_0^h f'(0) x + \frac{f(h)-f(0)}{h} x dx + O(h^3).$$

Now finish by showing that $\frac{f(h)-f(0)}{h}-f'(0)=O(h)$ and using the simple fact that $\int_0^h x dx = h^2/2$.

This technique of writing the quadrature rule as the exact integral of a "nearby" function is very widely applicable, and similar things show up in other areas of numerical analysis too.

  • $\begingroup$ I've come to understand that this is more or less the "standard" proof. I'm just wondering if my approach is valid. $\endgroup$
    – user460890
    Feb 6, 2018 at 22:06
  • $\begingroup$ @PeterSchilling No, it's not right, but that's only because you're misunderstanding the type of big Oh limit that you have. This is as the argument of the big Oh goes to zero, not infinity. $\endgroup$
    – Ian
    Feb 7, 2018 at 18:52
  • $\begingroup$ I'm just trying to show that the error in my Taylor expansion is $\mathcal{O}(h^2)$. To do so I used the definitions here. $\endgroup$
    – user460890
    Feb 7, 2018 at 18:55
  • $\begingroup$ @PeterSchilling Yes, now I understood after clicking the link. That's not the definition being used. Big Oh notation always has an implicit limit involved which is not always infinity. In real and numerical analysis it is usually zero instead, since it is as some quantity gets very close to another (in this context it is the grid spacing). In computer science the limit is usually infinity (since computational complexity is of primary interest in large problems). $\endgroup$
    – Ian
    Feb 7, 2018 at 18:56
  • $\begingroup$ Okay, taking $h\to0$ was my first intuition, but then the limit is infinity. Is it as simple as inverting the fraction I have so that we get a constant? It just seems a bit hand-wavy to me to assert that the error in the expansion is $\mathcal{O}(h^2)$ without some kind of limit. $\endgroup$
    – user460890
    Feb 7, 2018 at 19:04

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