Mid-point rule - Numerical Analysis

from pylab import *

9.7.1Single interval¶

The trapezoidal rule approximates the integral as the area below the straight line approximation

\int_a^b f(x) \ud x \approx (b-a) \frac{f(a) + f(b)}{2}

(1)

Another approximation to the area is to use the function value at the mid-point

\int_a^b f(x) \ud x \approx (b-a) f\left( \frac{a+b}{2} \right)

(2)

which is the area of the rectangle.

a, b = 0.0, 1.0
c = 0.5*(a + b)
f = lambda x: 1.0 + x + 0.5* sin(pi*x)
x = linspace(a,b,100)
plot([a,a],[0,f(c)],'k--')
plot([b,b],[0,f(c)],'k--')
plot([c,c],[0,f(c)],'sk--')
plot([a,b],[0,0],'b-')
plot(x, f(x), 'r-', label='f(x)')
plot([a,b],[f(c),f(c)],'k-')
d = 0.05
text(a,-d,"a",ha='center',va='top')
text(b,-d,"b",ha='center',va='top')
text(c,-d,"c",ha='center',va='top')
text(c,f(c)+0.05,"f(c)",ha='center',va='bottom')
axis([-0.1, 1.1, -0.3, 2.3])
legend(loc="upper right"), xlabel('x');

Let $c=(a+b)/2$ . Using Taylor formula

f(x) = f(c) + (x-c) f'(c) + \half (x-c)^2 f''(\xi_x), \qquad \textrm{$\xi_x$ between $x$ and $c$}

(3)

Integrating both sides

\int_a^b f(x) \ud x - (b-a) f\left( \frac{a+b}{2} \right) = \half \int_a^b (x-c)^2 f''(\xi_x) \ud x

(4)

Using integral mean value theorem, the error is

\begin{align*} \int_a^b f(x) \ud x - (b-a) f\left( \frac{a+b}{2} \right) &= \half f''(\eta) \int_a^b (x-c)^2 \ud x \\ &= \frac{(b-a)^3}{24} f''(\eta), \qquad \eta \in [a,b] \end{align*}

(5)

9.7.2Composite rule¶

Let us partition $[a,b]$ into $n$ intervals each of width $h = (b-a)/n$ . The mid-points of these intervals are

x_j = a + (j-\shalf)h, \qquad j=0, 1, \ldots, n-1

(6)

The integral is given by

\begin{aligned} \int_a^b f(x) \ud x &= \sum_{j=1}^{n-1} \int_{x_{j-1}}^{x_j} f(x) \ud x \\ &= \clr{red}{ \sum_{j=1}^{n-1} h f(x_j) } + \clr{blue}{ \frac{h^3}{24} \sum_{j=1}^n f''(\eta_j)}, \qquad \eta_j \in [x_{j-1}, x_j] \\ &=: \clr{red}{I_n(f)} + \clr{blue}{E_n(f)} \end{aligned}

(7)

where

I_n(x) = h[f_0 + f_1 + \ldots + f_{n-1}]

(8)

and the error is

\begin{aligned} E_n(f) &= \frac{h^3}{24} \sum_{j=1}^n f''(\eta_j) = \frac{n h^3}{24} \clr{red}{\left[ \frac{1}{n} \sum_{j=1}^n f''(\eta_j) \right]} \\ &= \frac{(b-a)h^2}{24} \clr{red}{f''(\eta)}, \qquad \eta \in [a,b] \end{aligned}

(9)

9.7.3Integral error estimate¶

Using Taylor formula around $x=c$ with integral remainder term

f(x) = f(c) + (x-c)f'(c) + \int_c^x (x-t) f''(t) \ud t

(11)

The error in mid-point rule is

\int_a^b f(x) \ud x - (b-a) f(c) = \int_a^b \int_c^x (x-t) f''(t) \ud t \ud x

(12)

By change of order of integration

\begin{aligned} \int_a^b \int_c^x (x-t) f''(t) \ud t \ud x &= \int_c^b \int_t^b (x-t) f''(t) \ud x \ud t - \int_a^c \int_a^t (x-t) f''(t) \ud x \ud t \\ &= \int_c^b \frac{(b-t)^2}{2} f''(t) \ud t + \int_a^c \frac{(a-t)^2}{2} f''(t) \ud t \end{aligned}

(13)

Using this estimate, the error of the composite rule is

|E_n(f)| \le \frac{h^2}{4} \int_a^b |f''(t)| \ud t

(14)

9.7 Mid-point rule

9.7.1Single interval¶

9.7.2Composite rule¶

9.7.3Integral error estimate¶