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Numerical Integration#

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Our second topic today will be about numerical integration, which is useful in determining of course the integrals of functions at certain positions. Here we will only refer to 3 different methods with increasing accuracy.

[2]:

import numpy as np
import matplotlib.pyplot as plt

%matplotlib inline
%config InlineBackend.figure_forsmat = 'retina'

plt.rcParams.update({'font.size': 10,
                     'lines.linewidth': 1,
                     'lines.markersize': 5,
                     'axes.labelsize': 10,
                     'xtick.labelsize' : 9,
                     'ytick.labelsize' : 9,
                     'legend.fontsize' : 8,
                     'contour.linewidth' : 1,
                     'xtick.top' : True,
                     'xtick.direction' : 'in',
                     'ytick.right' : True,
                     'ytick.direction' : 'in',
                     'figure.figsize': (4, 3),
                     'figure.dpi': 150 })

Box method#

box

The simplest method for the numerical integration of a function $f (x)$ is the box method. There you approximate the function in a certain intervall $Δ x$ by a horizontal line at the function value of the left edge of the intervall for example.

$\int_{a}^{b} f (x) \approx \sum_{i} f (x_{i}) Δ x$

So lets write a function for that:

[3]:

def f(x):
    return(x)

[4]:

def int_box(f,a,b,N):
    x=np.linspace(a,b,N)
    y=f(x)
    return(np.sum((x[1]-x[0])*y))

[7]:

int_box(f,0,1,100)

[7]:

0.5050505050505051

[11]:

acc=[]
for N in range(10,10000,100):
    acc.append(int_box(f,0,1,N))
plt.loglog(range(10,10000,100),np.array(acc)-0.5)
plt.show()

../../_images/notebooks_L5_2_integration_9_0.png

Trapezoid method#

The trapezoid method is taking the next step of function approximation in the interval $Δ x$ . It is approximating it with a linear function.

$\int_{a}^{b} f (x) d x = \sum_{i = 1}^{N} \frac{f (x_{i}) + f (x_{i - 1})}{2} Δ x$

which is actually the same as

$\int_{a}^{b} f (x) d x = [\frac{f (x_{0} + f (x_{N}))}{2} + \sum_{i = 1}^{N - 1} f (x_{i})] Δ x$

We will use the first formula for coding it, and you may try the second yourself.

[12]:

def int_trap(f,a,b,N):
    x=np.linspace(a,b,N)
    y=f(x)
    return(np.sum((y[1:]+y[:-1])*(x[1]-x[0])/2))

[13]:

## value from the box method
int_box(f,0,1,100)

[13]:

0.5050505050505051

[14]:

## value from the tapez method
int_trap(f,0,1,100)

[14]:

0.5000000000000001

The trapez method therefore seems to give a better accuracy than the box method for the same number of steps.

Simpson method#

The Simpson method now continues with approximating the function now with a collection of parabolas.

$\int_{a}^{b} f (x) d x \approx \sum_{i = 1}^{\frac{N - 1}{2}} \int_{x_{2 i - 1}}^{x_{2 i + 1}} g_{i} (x) d x$

where the function $g_{i} (x)$ is a parabola

$g_{i} (x) = [A] x^{2} + [B] x + [C]$

where the $[A], [B], [C]$ depends only on the function values at the edges of the slice.

Simpson

After some extensive algebra, which we do not want to do in detail, we arrive at

$\int_{a}^{b} f (x) d x \approx \frac{Δ x}{3} \sum_{i = o d d}^{N - 1} (f (x_{i - 1}) + f (x_{i}) + f (x_{i + 1}))$

as a simple formula on how to calculate the integral of a function using the Simpson method. Note that this method requires N being an odd number, which generates an even number of slices. There is a correction for odd number of slices, which we do not consider here.

[15]:

def int_simp(f,a,b,N):
    x=np.linspace(a,b,N)
    y=f(x)
    return(np.sum((y[0:-2:2]+4*y[1:-1:2]+y[2::2])*(x[1]-x[0])/3))

[16]:

## value from the tapez method
int_trap(f,0,1,100)

[16]:

0.5000000000000001

[17]:

## value from the box method
int_box(f,0,1,100)

[17]:

0.5050505050505051

[18]:

## value from the simpson method
## take care, N needs to be odd
int_simp(f,0,1,99)

[18]:

0.5

It turns out, that the Simpson rule is indeed the best among the three methods we have considered. The error is the box method goes as $Δ x$ , the one of the trapezoid method as $Δ x^{2}$ , while the simpson method provides and accuracy going with $Δ x^{4}$ . Thus doubling the number of integration points decreases the error by a factor of 16.