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Numerical Differentiation and Integration

Numerical Differentiation

The derivative represents the rate of change of a dependent variable $y$ or $f(x)$ with respect to an independent variable $x$. (derivative = $f^{\prime }(x)$)

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There are three basic ways to estimate derivatives: Forward, Backward and Center finite divided difference All based on the Taylor Series

$$f(x+h)=f(x)+f^{\prime }(x)h+\frac{f^{″}(x)}{2!}{h}^2+\dots$$

Another form of this result is the following, where $c$ is some point between $x$ and $x+h$.

$$f(x+h)=f(x)+f^{\prime }(x)h+\frac{f^{″}(c)}{2!}{h}^2$$

Rearranging this last equation gives

$$f^{\prime }(x)=\frac{f(x+h)-f(x)}{h}-\frac{f^{″}(c)}{2}h$$

We will estimate the derivative by dropping the last term. If we can bound the second derivative by some constant, then this last term is $O(h)$ and we get

$$f^{\prime }(x)=\frac{f(x+h)-f(x)}{h}+O(h)$$

Thus,

$$f^{\prime }(x)\approx \frac{f(x+h)-f(x)}{h}$$

This formula is called the first forward divide difference formula and the error in this formula is of order $O(h)$.

The error in this approximating is due to truncation of the last term (the second derivative term).

The truncation error can be made small by making $h$ small. However, as $h$ gets smaller, precision will be lost in this equation due to subtractive cancellation.

The error in calculation for small $h$ is called round-off error. Generally the round-off error will increase as $h$ decreases.

EXAMPLE:

use first forward divide difference formula with $h=0.1$ to approximate the derivative of $f(x)=1/x$ at $x=2$.

SOLUTION:

$$f^{\prime }(x)\approx \frac{f(x+h)-f(x)}{h}=\frac{\frac{1}{2.1}-\frac{1}{2}}{0.1}\approx -0.2381.$$

the error is

$$-0.2381-(-\frac{1}{2^2})=0.0119.$$

Compare this to the error predicted by the formula, which is $f^{″}(c)h/2$ for some $c$ between 2 and 2.1. Since $f^{″}(x)=2x^{-3}$, the error must be between $(0.1)2^{-3}\approx 0.0125$ and $(0.1)(2.1{)}^{-3}\approx 0.0108$, which is consistent with our result. However, this information is usually not available.

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More precise solution

Consider the Taylor Series

$$f(x-h)=f(x)-f^{\prime }(x)h+\frac{f^{″}(c)}{2!}{h}^2$$

Rearranging this last equation gives

$$f^{\prime }(x)=\frac{f(x)-f(x-h)}{h}+\frac{f^{″}(c)}{2}h$$

Or,

$$f^{\prime }(x)=\frac{f(x)-f(x-h)}{h}+O(h)$$

This is called the backward divide difference formula for $f^{\prime }(x)$.

Consider the Taylor series expansions for the forward and backward approximations, extended to the degree 3 terms:

$$f(x+h)=f(x)+f^{\prime }(x)h+\frac{f^{″}(x)}{2!}{h}^2+\frac{f^{‴}(c_1)}{3!}{h}^3$$$$f(x-h)=f(x)-f^{\prime }(x)h+\frac{f^{″}(x)}{2!}{h}^2-\frac{f^{‴}(c_2)}{3!}{h}^3$$

If we subtract these two and solve for $f^{\prime }(x)$ we get:

$$f^{\prime }(x)=\frac{f(x+h)-f(x-h)}{2h}-\left[\frac{f^{‴}(c_1)+f^{‴}(c_2)}{2}\right]\frac{h^2}{6}$$

THEOREM Generalized intermediate value theorem

Let $f$ be a continuous function on the interval $[a,b]$. Let $x_1,\dots ,x_n$ be points in $[a,b]$, and $a_1,\dots ,a_n>0$. Then there exists a number $c$ between $a$ and $b$ such that

$$(a_1+\cdots +a_n)f(c)=a_{1}f(x_1)+\cdots +a_{n}f(x_n).$$

Based on Generalized intermediate value theorem, we have

$$f^{\prime }(x)=\frac{f(x+h)-f(x-h)}{2h}-\frac{h^2}{6}{f}^{‴}(c),$$

where $x-h<c<x+h$.

$$f^{\prime }(x)=\frac{f(x+h)-f(x-h)}{2h}+O(h^2)$$

This is called the centered divide difference formula for $f^{\prime }(x)$.

EXAMPLE

use centered divide difference formula with $h=0.1$ to approximate the derivative of $f(x)=1/x$ at $x=2$.

SOLUTION:

$$f^{\prime }(x)\approx \frac{f(x+h)-f(x-h)}{2h}=\frac{\frac{1}{2.1}-\frac{1}{1.9}}{0.2}\approx -0.2506.$$

the error is

$$-0.2506-(-\frac{1}{2^2})=0.0006.$$

Second derivative

Consider the Taylor series expansions for the forward and backward approximations, extended to the degree 4 terms:

$$f(x+h)=f(x)+f^{\prime }(x)h+\frac{f^{″}(x)}{2!}{h}^2+\frac{f^{‴}(x)}{3!}{h}^3+\frac{f^{(4)}(x)}{4!}{h}^4+\cdots$$$$f(x-h)=f(x)-f^{\prime }(x)h+\frac{f^{″}(x)}{2!}{h}^2-\frac{f^{‴}(x)}{3!}{h}^3+\frac{f^{(4)}(x)}{4!}{h}^4-\cdots$$

If we add these two and solve for $f^{″}(x)$ we get:

$$f^{″}(x)=\frac{f(x+h)-2f(x)+f(x-h)}{h^2}+O(h^2)$$

Try to Improve

The forward difference formula for $f^{\prime }(x)$ had error $O(h)$, while the centered difference formula had error $O(h^2)$. Can we do better?

Consider again the Taylor expansions:

$$f(x+h)=f(x)+f^{\prime }(x)h+\frac{f^{″}(x)}{2!}{h}^2+\frac{f^{‴}(x)}{3!}{h}^3+\frac{f^{(4)}(x)}{4!}{h}^4+\frac{f^{(5)}(x)}{5!}{h}^5+\cdots$$$$f(x-h)=f(x)-f^{\prime }(x)h+\frac{f^{″}(x)}{2!}{h}^2-\frac{f^{‴}(x)}{3!}{h}^3+\frac{f^{(4)}(x)}{4!}{h}^4-\frac{f^{(5)}(x)}{5!}{h}^5+\cdots$$

Let $\varphi (h)=\frac{f(x+h)-f(x-h)}{2h}$ Then,

$$\varphi (h)=f^{\prime }(x)+a_2{h}^2+a_4{h}^4+a_6{h}^6+\cdots$$

where $a_2=\frac{f^{‴}(x)}{3}$, $a_4=\frac{f^{(5)}(x)}{60}$, ... which can just be considered as some constant coefficients.

Also, $\varphi \left(\frac{h}{2}\right)=f^{\prime }(x)+\frac{1}{4}{a}_2{h}^2+\frac{1}{16}{a}_4{h}^4+\frac{1}{64}{a}_6{h}^6+\cdots$

We can use $\varphi (h)$ and $\varphi \left(\frac{h}{2}\right)$ to get:

$$\varphi (h)-4\varphi \left(\frac{h}{2}\right)=-3f^{\prime }(x)+\frac{3}{4}{a}_4{h}^4+\frac{15}{16}{a}_6{h}^6+\frac{63}{64}{a}_8{h}^8+\cdots$$$$\frac{4\varphi \left(\frac{h}{2}\right)-\varphi (h)}{3}=f^{\prime }(x)-\frac{1}{4}{a}_4{h}^4-\frac{5}{16}{a}_6{h}^6-\frac{21}{64}{a}_8{h}^8+\cdots$$

Note: This approximation for $f^{\prime }(x)$ has error of $O(h^4)$!!

$$f^{\prime }(x)\approx \frac{4\varphi \left(\frac{h}{2}\right)-\varphi (h)}{3}$$$$f^{\prime }(x)\approx \frac{4\varphi \left(\frac{h}{2}\right)-\varphi (h)}{3}=\frac{[4\frac{f(x+\frac{h}{2})-f(x-\frac{h}{2})}{h}-\frac{f(x+h)-f(x-h)}{2h}]}{3}=\frac{f(x-h)-8f(x-\frac{h}{2})+8f(x+\frac{h}{2})-f(x+h)}{6h}$$

Richardson Extrapolation

We can use the Richardson method to increase the accuracy of numerical estimates to any series-based quantity. Suppose we want to calculate some quantity $L$ and have found, through theory, an approximation to $L$:

$$\varphi (h)=L+\sum _{k=1}^{\mathrm{\infty }}{a}_{2k}{h}^{2k}$$

Let $D(n,0)=\varphi \left(\frac{h}{2^n}\right)$, define:

$$D(n,m)=\frac{4^{m}D(n,m-1)-D(n-1,m-1)}{4^m-1}$$$$D(n,0)=\varphi \left(\frac{h}{2^n}\right)$$

THEOREM: (Richardson Extrapolation).

There are constants $a_{k,m}$ such that

$$D(n,m)=L+\sum _{k=m+1}^{\mathrm{\infty }}{a}_{k,m}{\left(\frac{h}{2^n}\right)}^{2k}(0\le m\le n).$$

COROLLARY:

$$D(n,n)=L+O(h^{2(n+1)})$$

$D(n,n)$ Can be found through a pyramid table:

$$\begin{array}{ccccc}D(0,0)& & & & \\ D(1,0)& D(1,1)& & & \\ D(2,0)& D(2,1)& D(2,2)& & \\ ⋮& ⋮& ⋮& \ddots & \\ D(n,0)& D(n,1)& D(n,2)& \cdots & D(n,n)\end{array}$$

EXAMPLE:

Consider $f(x)=\arctan (x)$. Suppose we want to find $f^{\prime }(\sqrt{2})$. Let $\varphi (h)=\frac{1}{2h}[f(\sqrt{2}+h)-f(\sqrt{2}-h)]$, and start with $h=0.01$.

Then, we compute $D(n,m)$ in a pyramid fashion (as we did for Newton’s divided differences). The first column is just $D(n,0)=\varphi \left(\frac{h}{2^n}\right)$

$D(n,m)$	0	1	2
0	0.333339506181068
1	0.333334876543723	0.333333333331274
2	0.333333719135820	0.333333333333186	0.333333333333313

Numerical Integration

Trapezoidal rule

The need often arises for evaluating the definite integral of a function that has no explicit antiderivative or whose antiderivative is not easy to obtain. In this section, we will discuss some numerical method to approximating ${\int }_a^{b}f(x)dx$.

Let $f(x)$ be a function with a continuous second derivative, defined one the interval $[x_0,x_1]$, as shown in the figure. Denote the corresponding function values by $y_0=f(x_0)$ and $y_1=f(x_1)$. Consider the degree 1 interpolating polynomial $P_1(x)$ through $(x_0,y_0)$ and $(x_1,y_1)$. Using the Lagrange formulation, we find that the interpolating polynomial with error term is

$$f(x)=y_0\frac{x-x_1}{x_0-x_1}+y_1\frac{x-x_0}{x_1-x_0}+\frac{(x-x_0)(x-x_1)}{2!}{f}^{″}(c_x)=P(x)+E(x).$$

$c_x$ depends continuously on $x$.

Integrating both sides on the interval of interest $[x_0,x_1]$ yields

$${\int }_{x_0}^{x_1}f(x)dx={\int }_{x_0}^{x_1}P(x)dx+{\int }_{x_0}^{x_1}E(x)dx.$$

Computing the first integral gives

$$\begin{array}{rl}{\int }_{x_0}^{x_1}P(x)dx& =y_0{\int }_{x_0}^{x_1}\frac{x-x_1}{x_0-x_1}dx+y_1{\int }_{x_0}^{x_1}\frac{x-x_0}{x_1-x_0}dx\\ & =y_0\frac{h}{2}+y_1\frac{h}{2}=h\frac{y_0+y_1}{2},\end{array}$$

where we have defined $h=x_1-x_0$ to be the interval length.

The error term is

$$\begin{array}{rl}{\int }_{x_0}^{x_1}E(x)dx& =\frac{1}{2!}{\int }_{x_0}^{x_1}(x-x_0)(x-x_1)f^{″}(c(x))dx\\ & =\frac{f^{″}(c)}{2}{\int }_{x_0}^{x_1}(x-x_0)(x-x_1)dx\\ & =\frac{f^{″}(c)}{2}{\int }_0^{h}u(u-h)du\\ & =-\frac{h^3}{12}{f}^{″}(c).\end{array}$$$${\int }_{x_0}^{x_1}f(x)dx=\frac{h}{2}(y_0+y_1)-\frac{h^3}{12}{f}^{″}(c)$$

where $h=x_1-x_0$ and $c$ is between $x_0$ and $x_1$.

Simpson Rule

We replace degree 1 interpolant by parabola.

$$\begin{array}{r}f(x)\\ & =y_0\frac{(x-x_1)(x-x_2)}{(x_0-x_1)(x_0-x_2)}+y_1\frac{(x-x_0)(x-x_2)}{(x_1-x_0)(x_1-x_2)}+y_2\frac{(x-x_0)(x-x_1)}{(x_2-x_0)(x_2-x_1)}\\ & +\frac{(x-x_0)(x-x_1)(x-x_2)}{3!}{f}^{‴}(c_x)\\ & =P(x)+E(x).\end{array}$$

Integrating gives

$${\int }_{x_0}^{x_2}f(x)dx={\int }_{x_0}^{x_2}P(x)dx+{\int }_{x_0}^{x_2}E(x)dx,$$

where

$$\begin{array}{r}{\int }_{x_0}^{x_2}P(x)dx\\ & =y_0{\int }_{x_0}^{x_2}\frac{(x-x_1)(x-x_2)}{(x_0-x_1)(x_0-x_2)}dx+y_1{\int }_{x_0}^{x_2}\frac{(x-x_0)(x-x_2)}{(x_1-x_0)(x_1-x_2)}dx+y_2{\int }_{x_0}^{x_2}\frac{(x-x_0)(x-x_1)}{(x_2-x_0)(x_2-x_1)}dx\\ & =y_0\frac{h}{3}+y_1\frac{4h}{3}+y_2\frac{h}{3}.\end{array}$$

where $h=x_2-x_1=x_1-x_0$.

The error term can be computed (proof omitted) as

$${\int }_{x_0}^{x_2}E(x)dx=-\frac{h^5}{90}{f}^{(4)}(c)$$

for some $c$ in the interval $[x_0,x_2]$, provided that $f^{(4)}(c)$ exists and is continuous. Concluding the derivation yields Simpson’s rule

Simpson’s Rule

$${\int }_{x_0}^{x_2}f(x)dx=\frac{h}{3}(y_0+4y_1+y_2)-\frac{h^5}{90}{f}^{(4)}(c),$$

where $h=x_1-x_0=x_2-x_1$ and $c$ is between $x_0$ and $x_2$.

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EXAMPLE

Apply the Trapezoid rule and Simpson’s rule to approximate

$${\int }_1^2\ln xdx,$$

and find an upper bound for the error in your approximations.

SOLUTION

$${\int }_1^2\ln xdx\approx \frac{h}{2}(y_0+y_1)=\frac{1}{2}(\ln 1+\ln 2)=\frac{\ln 2}{2}\approx 0.3466.$$

The error for the Trapezoid rule is $-\frac{h^3}{12}{f}^{″}(c)$, where $1<c<2$. Since $f^{″}(x)=-1/x^2$, the magnitude of the error is at most

$$\frac{1^3}{12c^2}\le \frac{1}{12}\approx 0.0834$$

In other words,

$${\int }_1^2\ln xdx=0.3466\pm 0.0834$$$${\int }_1^2\ln xdx\approx \frac{h}{3}(y_0+4y_1+y_2)=\frac{0.5}{3}(\ln 1+4\ln \frac{3}{2}+\ln 2)\approx 0.3858.$$

The error for the Simpson’s rule is $-\frac{h^5}{90}{f}^{(4)}(c)$, where $1<c<2$. Since $f^{(4)}(x)=-6/x^4$, the magnitude of the error is at most

$$\frac{6(0.5{)}^5}{90c^4}\le \frac{6(0.5{)}^5}{90}=\frac{1}{480}\approx 0.0021.$$

In other words,

$${\int }_1^2\ln xdx=0.3858\pm 0.0021.$$

The integral can be computed exactly by using integration by parts:

$${\int }_1^2\ln xdx=x\ln x{|}_1^2-{\int }_1^{2}1dx=2\ln 2-1\ln 1-1\approx 0.386294.$$

The Trapezoid rule and Simpson’s rule error bounds are both consistent with this result.

Composite Trapezoidal Rule

Consider an evenly spaced grid

$$a=x_0<x_1<x_2<\cdots <x_{m-2}<x_{m-1}<x_m=b$$

Along the horizontal axis, where $h=x_{i+1}-x_i$ for each $i$ as shown in figure. On each subinterval, we make the approximation with error term

$${\int }_{x_i}^{x_{i+1}}f(x)dx=\frac{h}{2}(f(x_i)+f(x_{i+1}))-\frac{h^3}{12}{f}^{″}(c_i),$$

Assuming that $f^{″}$ is continuous, adding up over all subintervals yields

$${\int }_a^{b}f(x)dx=\frac{h}{2}(f(a)+f(b)+2\sum _{i=1}^{m-1}f(x_i))-\sum _{i=0}^{m-1}\frac{h^3}{12}{f}^{″}(c_i),$$

The error term can be written

$$\frac{h^3}{12}\sum _{i=0}^{m-1}{f}^{″}(c_i)=\frac{h^3}{12}m{f}^{″}(c),$$

Since $mh=(b-a)$, the error term is $\frac{(b-a)h^2}{12}{f}^{″}(c)=O(h^2)$.

Composite Trapezoid Rule

$${\int }_a^{b}f(x)dx=\frac{h}{2}(y_0+y_m+2\sum _{i=1}^{m-1}{y}_i)-\frac{(b-a)h^2}{12}{f}^{″}(c),$$

where $h=(b-a)/m$ and $c$ is between $a$ and $b$.

Consider an evenly spaced grid

$$a=x_0<x_1<x_2<\cdots <x_{2m-2}<x_{2m-1}<x_{2m}=b$$

Along the horizontal axis, where $h=x_{i+1}-x_i$ for each $i$ as shown in figure. On each subinterval by the interpolating parabola fit at $x_{2i},x_{2i+1}$, and $x_{2i+2}$, we make the approximation with error term

$${\int }_{x_{2i}}^{x_{2i+2}}f(x)dx=\frac{h}{3}(f(x_{2i})+4f(x_{2i+1})+f(x_{2i+2}))-\frac{h^5}{90}{f}^{(4)}(c_i).$$

Composite Simpson’s Rule

Consider an evenly spaced grid

$$a=x_0<x_1<x_2<\cdots <x_{2m-2}<x_{2m-1}<x_{2m}=b$$

Along the horizontal axis, where $h=x_{i+1}-x_i$ for each $i$ as shown in figure. On each subinterval by the interpolating parabola fit at $x_{2i},x_{2i+1}$, and $x_{2i+2}$, we make the approximation with error term

$${\int }_{x_{2i}}^{x_{2i+2}}f(x)dx=\frac{h}{3}(f(x_{2i})+4f(x_{2i+1})+f(x_{2i+2}))-\frac{h^5}{90}{f}^{(4)}(c_i).$$

This time, the overlapping is over even-numbered $x_j$ only. Adding up over all subintervals yields

$${\int }_a^{b}f(x)dx=\frac{h}{3}(f(a)+f(b)+4\sum _{i=1}^{m}f(x_{2i-1})+2\sum _{i=1}^{m-1}f(x_{2i}))-\sum _{i=0}^{m-1}\frac{h^5}{90}{f}^{(4)}(c_i).$$

The error term can be written

$$\frac{h^5}{90}\sum _{i=0}^{m-1}{f}^{(4)}(c_i)=\frac{h^5}{90}m{f}^{(4)}(c),$$

Since $m\cdot 2h=(b-a)$, the error term is $\frac{(b-a)h^4}{180}{f}^{(4)}(c)=O(h^4)$.

Composite Simpson’s Rule

$${\int }_a^{b}f(x)dx=\frac{h}{3}(y_0+y_{2m}+4\sum _{i=1}^m{y}_{2i-1}+2\sum _{i=1}^{m-1}{y}_{2i})-\frac{(b-a)h^4}{180}{f}^{(4)}(c),$$

where $c$ is between $a$ and $b$.

EXAMPLE

Carry out four-panel approximations of

$${\int }_1^2\ln xdx,$$

Using the composite Trapezoid rule and composite Simpson’s rule.

SOLUTION For composite Trapezoid rule on $[1,2]$, four panels means that $h=1/4$.

$${\int }_1^2\ln xdx\approx \frac{1}{8}[y_0+y_4+2\sum _{i=1}^3{y}_i]=\frac{1}{8}[\ln 1+\ln 2+2(\ln \frac{5}{4}+\ln \frac{6}{4}+\ln \frac{7}{4})]\approx 0.3837$$

The error is at most

$$\frac{(b-a)h^2}{12}|f^{″}(c)|=\frac{1}{16}\frac{1}{12}\frac{1}{c^2}\le \frac{1}{(16)(12)(1^2)}=\frac{1}{192}\approx 0.0052$$

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For four-panel Simpson’s rule set $h=1/8$.

$${\int }_1^2\ln xdx\approx \frac{1}{8}\frac{1}{3}[y_0+y_8+4\sum _{i=1}^4{y}_{2i-1}+2\sum _{i=1}^3{y}_{2i}]=\frac{1}{24}[\ln 1+\ln 2+4(\ln \frac{9}{8}+\ln \frac{11}{8}+\ln \frac{13}{8}+\ln \frac{15}{8})+2(\ln \frac{5}{4}+\ln \frac{6}{4}+\ln \frac{7}{4})]\approx 0.386292$$

The error is at most

$$\frac{(b-a)h^4}{180}|f^{(4)}(c)|={\left(\frac{1}{8}\right)}^4\frac{1}{180}\frac{6}{c^4}\le \frac{1}{(8^4)(180)(1^4)}\approx 0.000008$$

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EXAMPLE

Find the number of panels $m$ necessary for the composite Simpson’s rule to approximate

$${\int }_0^{\pi }{\sin }^{2}xdx,$$

within six correct decimal places.

SOLUTION

We require the error to satisfy

$$\frac{(\pi -0)h^4}{180}|f^{(4)}(c)|<0.5\times {10}^{-6}.$$

Since the fourth derivative of ${\sin }^{2}x$ is $-8\cos 2x$, we need

$$\frac{\pi h^4}{180}8<0.5\times {10}^{-6},$$

or $h<0.0435$. therefore, $m=ceil\left(\frac{\pi }{2h}\right)=37$ panels will be sufficient.

Romberg integration

In this section, let’s use the Richardson Extrapolation technique to improve integral approximations.

Consider the Trapezoidal Rule using $2^n$ subintervals of $[a,b]$. Then, $h=\frac{b-a}{2^n}$ and the approximation will be

$$\varphi (n)=\frac{1}{2}\frac{b-a}{2^n}\sum _{i=0}^{2^n-1}[f(x_i)+f(x_{i+1})]$$

We know that this approximation has error $O(h^2)$. Thus,

$$\varphi (n)={\int }_a^{b}f(x)dx+a_2{h}^2+a_4{h}^4+a_6{h}^6+\cdots$$

This is exactly the kind of expression we had for Richardson Extrapolation. Let

$$\begin{array}{rl}R(n,0)& =\varphi (n)\\ R(n,m)& =\frac{4^{m}R(n,m-1)-R(n-1,m-1)}{4^m-1}\end{array}$$

Compute the table of values:

$$\begin{array}{ccccc}R(0,0)& & & & \\ R(1,0)& R(1,1)& & & \\ R(2,0)& R(2,1)& R(2,2)& & \\ ⋮& ⋮& ⋮& \ddots & \\ R(n,0)& R(n,1)& R(n,2)& \cdots & R(n,n)\end{array}$$

This is called Romberg Integration

EXAMPLE

Approximate ${\int }_1^2\ln xdx$ using Romberg integration up to $n=3$.

SOLUEION

$$\begin{array}{rl}R(0,0)& =\varphi (0)=\frac{1}{2}\times (2-1)\times (\ln 1+\ln 2)=0.3466\\ R(1,0)& =\varphi (1)=\frac{1}{2}\times \frac{2-1}{2}\times (\ln 1+2\ln \frac{3}{2}+\ln 2)=0.3760\\ R(2,0)& =\varphi (2)=\frac{1}{2}\times \frac{2-1}{2^2}\times (\ln 1+2\ln \frac{5}{4}+2\ln \frac{3}{2}+2\ln \frac{7}{4}+\ln 2)=0.3837\\ R(1,1)& =\frac{4R(1,0)-R(0,0)}{4-1}=0.3858\\ R(2,1)& =\frac{4R(2,0)-R(1,0)}{4-1}=0.3863\\ R(2,2)& =\frac{4^{2}R(2,1)-R(1,1)}{4^2-1}=0.3863\end{array}$$

	Trapezoid	Simpson's
$h=1$	0.3466
$h=0.50$	0.3760	0.3858
$h=0.25$	0.3837	0.3863	0.3863
error	$3.97\times {10}^{-3}$	$4.94\times {10}^{-4}$	$6\times {10}^{-6}$
	$O(h^2)$	$O(h^4)$	$O(h^6)$