NA Trapezoidal rule ($n=1$)

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Section 8.4 Trapezoidal rule ( $n=1$ )

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In this case we are using a 2-points interpolation in

$[a,b]\text{.}$ These two points must be the endpoints

$x_0=a$ and

$x_1=b\text{.}$ The interpolating polynomial will have degree 1. The Lagrange polynomials are

$\begin{equation*} L_0(x) = \frac{x-x_1}{x_0-x_1},\;L_1(x) = \frac{x-x_0}{x_1-x_0} \end{equation*}$

and the corresponding weights are

$\begin{equation*} w_0 = \int_a^b\frac{x-a}{b-a}dx = \frac{b-a}{2} = w_1. \end{equation*}$

Hence the Trapezoidal rule formula gives

$\begin{equation*} Q_1(f,[a,b]) = \frac{f(a)+f(b)}{2}\left(b-a\right). \end{equation*}$

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Local error. From the Newton-Cotes error formula, we get that

$\begin{equation*} \left|\int_a^b f(x)dx - Q_1(f,[a,b])\right|\leq \frac{(b-a)^3}{2}\displaystyle\max_{[a,b]}|f''(x)|\int_0^1s(s-1)ds=\frac{(b-a)^3}{12}\displaystyle\max_{[a,b]}|f''(x)|. \end{equation*}$

As usual, this shows that this approximation can be lousy unless

$b-a\ll1\text{.}$

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Trapezoidal rule formula. We bypass as usual the error problem above by subdiving the interval as shown in the previous two sections. In this case, we get that

$\begin{equation*} Q_{1,h}(f,[a,b]) = h\sum_{k=0}^{n-1} \frac{f(x_k)+f(x_{k+1})}{2} \end{equation*}$

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Trapezoidal rule error. From the error formula above, we get that

$\begin{equation*} \left|\int_{x_k}^{x_{k+1}}f(x)dx - Q_1(f,[x_k,x_{k+1}])\right|\leq\frac{h^3}{12}\displaystyle\max_{[a,b]}|f''(x)| \end{equation*}$

and therefore

$\begin{equation*} \left|\int_{a}^{b}f(x)dx - Q_{1,h}(f,[a,b])\right|\leq h^2\frac{b-a}{12}\displaystyle\max_{[a,b]}|f''(x)|. \end{equation*}$

Convergence Speed. The formula above shows that the truncation error of the Trapezoidal rule is

$O(h^2)\text{,}$ like for the Midpoint rule. Below we visualize this scaling law by evaluating the error in the computation of

$\int_0^2\sin(x)dx$ with

$h=10^{-j}, 1\leq j\leq7\text{.}$ As in the Midpoint case, the round-off error effects shows already at

$h=10^{-7}$ and the error achievable on this integral is about

$h=10^{-13}\text{.}$

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from numpy import linspace,array,frompyfunc,sin,zeros                                                                                                                                         
from matplotlib import pyplot as plt 
n=9; a=0.; b=2.; N=1
s = zeros(n); err = zeros(n)
​
for i in range(n):
    h = (b-a)/N
    x = linspace(a,b,N+1)                 # these are the x_i
    t = ( sin(x[:-1]) + sin(x[1:]) ) / 2; # average of sine on endpoints
    s[i] = sum(h*t)                       # this is the integral
    err[i] = abs(1-cos(2)-s[i])           # and this is the actual error
    N *= 10
​
I = range(n)
h = array([(b-a)/10**i for i in I])
plt.clf()
plt.loglog(h,err[I],label=r'$y=error(h)$')
plt.loglog(h,h**2,label=r'$y=h^2$')
plt.xlabel('log h') 
plt.ylabel('log y') 
plt.legend()
plt.show()

Section 8.4 Trapezoidal rule (n=1n=1)

Section 8.4 Trapezoidal rule ( $n=1$ )