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Jupyter notebook coursenotes/Topic8/Topic8.ipynb

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Kernel: Python 3 (Anaconda)

Topic 8: The Python scientific modules - numpy and scipy

scipy

Scipy is another Python module that offers a large number of pre-written routines for data manipulation including integration, solving differential equations, Fourier analysis and transforms, statistics, image processing, etc.

There is a lot of information about scipy at www.scipy.org

In PHY235 we will limit our use of scipy to differentiation, integration and Fourier Transforms.

Integration

General integration is done with the quad function, part of the scipy.integrate sub-package. To access this function it is necessary to use the following form of the import command


from scipy.integrate import quad

Calls to the quad function are of the general form


result = quad(func, limlo, limho)

where func here is the function to be integrated, it should be a callable Python object (i.e. a function, method or class). For the purposes of this course we will only use a function here. quad returns a tuple containing 2 values, the first is the value of the integration, the second is an upper bound on the error. Here is an example of a short script to perform the integration

∫−π/2+π/2y(x)dxwherey(x)=x3sin(2x)\large{ \large{ \int^{+\pi/2}_{-\pi/2}y(x)dx \hspace{1cm} \mathrm{where} \hspace{1cm} y(x) = x^{3}sin(2x) }}
import math
from scipy.integrate import quad
import matplotlib.pyplot as plt
import numpy as np

def func(x):
    return(x**3*math.sin(2*x))


LimLo=-math.pi/2
LimHi=math.pi/2
intValue,intError=quad(func,LimLo,LimHi)


print('%.4f %g' % (intValue,intError))
1.5196 1.07246e-13 
xAxisValues = np.arange(-np.pi/2,np.pi/2, np.pi/100) 
FuncValues = [ func(x) for x in xAxisValues ] 

plt.plot(xAxisValues, FuncValues)
plt.xlabel('$x$')
plt.ylabel('$x^3 \sin(2X)$')
plt.show()
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?quad

Double, triple and higher level integration can be done with the dblquad, tplquad and nquad fucntions from the scipy.integrate package.

Fourier transforms

scipy offers a way of performing Fourier transforms on data using routines from the fftpack package. The following example performs a fast Fourier transform (FFT) on the function f where

f=5cos⁡(4x)+6cos⁡(3x)+cos⁡(2x)+3cos⁡(x)\large{\large{f=5\cos(4x)+6\cos(3x)+\cos(2x)+3\cos(x)}}

this is a very similar problem to a question in the PHY250 tutorial questions on Fourier Techniques 

%matplotlib inline
import matplotlib.pyplot as plt
import numpy as np
from math import pi,sin, cos
from scipy.fftpack import fft
#
def myfunc(x):
    return(5*cos(4*x)+6*cos(3*x)+cos(2*x)+3*cos(x))

# Create the X/Y data
xAxisValues = np.arange(-10*pi, 10*pi, pi/20) 
FuncValues = [ myfunc(x) for x in xAxisValues ]        

# Visualise the function
plt.plot(xAxisValues, FuncValues)
plt.title('$f=5\cos(4x)+6\cos(3x)+\cos(2x)+3\cos(x)$')
plt.xlabel('[rad]')
plt.show()

# Calculate the fft
FuncFFT=fft(FuncValues)
FuncFFT=np.abs(FuncFFT)
plt.title('FFT(f)')

# Visualise the fft
plt.plot(xAxisValues, FuncFFT)
plt.show()
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?fft

** Notes **

  1. in this example the pyplot plotting package has been used to visualise the data. The entry


%matplotlib inline

causes the plots to be drawn inline in the output cell of the Jupyter notebook. (Otherwise a separate window is created).

Solving ordinary differential equations (OEDs)

The scipy.integrate package provides a function called odeint which can be used to solve differential equations.

odeint has many options. In its simplest form it can be executed via a function call such as 


y = odeint(func, y0, t)

where

  • y is the output (solution) from odeint

  • func is a Python version of the mathematical function to be solved

  • y0 is the value of y(0)

  • t is a numpy array of values for t

The following is a simple example of solving a first order differential equation

dydt=−2y,y(0)=5,(0≤t≤4)\large{\large{ \frac{dy}{dt} = -2y, \hspace{1cm} y(0)=5, \hspace{1cm} (0 \leq t \leq 4)}}
%matplotlib inline
import matplotlib.pyplot as plt
import numpy as np
from scipy.integrate import odeint
def dydt(y,t):        # see Note 1
    return (-2 * y)  
y0 = 5
tvec = np.linspace(0, 4, 41)
print(tvec)
yvec = odeint(dydt, y0, tvec)
plt.plot(tvec,yvec)
plt.xlabel("t")
plt.ylabel("y")
plt.show()
[ 0. 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1. 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2. 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 3. 3.1 3.2 3.3 3.4 3.5 3.6 3.7 3.8 3.9 4. ]
<matplotlib.figure.Figure at 0x7fcd201e1400>

** Notes **

  1. This function must have 2 input variables even if the output is only dependent on one of those

Solving second order differential equations

It is possible to use scipy to solved second order differential equations. In this case the first step is always to transform any nth-order ODE into a system of n first order ODEs. In addition we also need n initial conditions, one for each variable represented in the first order ODEs. This is best illustrated with an example.

For a damped harmonic oscillator of mass m then

md2xdt2+bdxdt+kx=0\large{\large{ m\frac{d^{2}x}{dt^{2}} + b\frac{dx}{dt} + kx = 0 }}

where b is the damping coefficient and k is the spring constant

This can be rewritten as

d2xdt2=−(bm)dxdt−(km)x\large{\large{\frac{d^{2}x}{dt^{2}} = -\left(\frac{b}{m}\right)\frac{dx}{dt} - \left(\frac{k}{m}\right)x }}

To solve using odeint this needs to be written as two first order ODEs as follows

dxdt=v\large{\large{\frac{dx}{dt} = v}}dvdt=−(bm)v−(km)x\large{\large{\frac{dv}{dt} = -\left(\frac{b}{m}\right)v -\left(\frac{k}{m}\right)x}}

Taking m=l, k=1.2, b=0.3 and the initial conditions as x=1 and dx/dt=v=0 respectively it is then possible to set up a Python script to solve these two equations simultaneously, as follows

%matplotlib inline
from scipy.integrate import odeint
from matplotlib.pyplot import plot, xlabel, ylabel, show, title
from numpy import linspace

def DampedOsc(Z,t,b,k,m):
    x=Z[0]; v=Z[1]
    return(v,-(k*x/m)-(b*v/m))

t = linspace(0,30,300)
U0 = [1,0] # Initial values
b = 0.3    # b is the damping coefficient
k = 1.2    # k is the spring constant
m = 1      # m is the mass

U=odeint(DampedOsc,U0,t,args=(b,k,m)) #constants needed in function can be passed as arguments

plot(t,U) # Note - this plots both columns of the vector U
xlabel('Time'); title('Damped Oscillator')
show()
plot(U[:,0],U[:,1])
xlabel('Position'); ylabel('Velocity')
show()
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The same technique can be used to solve 2 simultaneous first order ODEs.

** Note ** by definition the first argument to this function called by odeint (argument Z in the function DampedOsc in the example above) is a vector which contains values of the 2 variables which odeint are being solved for. So, in the example above Z=[x,v]. You do not have to provide values for Z, odeint will do this automatically. The function must however return the derivatives of the variables, in this case [x',v'].

Interpolation

The following example illustrates the interp1d function (note that this is interp-1-d not interp-l-d).

A sparsely sampled series of data points (just 13 points over the range -2.5$\pito+2.5 to +2.5\pi$) of the cosine function is interpolated using the interp1d function using both Linear and Cubic interpolation. The resulting functions, InterpLin and InterpCub can be used to predict a value of the function for any value of x. For the cubic interpolation the agreement between the function and the actual cosine function is excellent.

It is of interest to re-run the Python script with varying values of the number of samples (13 in this example) to observe how the accuracy of the interpolation varies. This example also illustrates use of the subplot command.

%matplotlib inline
from scipy.interpolate import interp1d
import numpy as np
from math import cos,pi
import matplotlib.pyplot as plt

# The following plots cos(x) between -2.5pi and +2.5pi with 100 sample points
xAxisHi = np.linspace(-2.5*pi, 2.5*pi, 100)
yCosHi = np.cos(xAxisHi)

# 13 samples of cos(x) in the same interval range
xAxisLo = np.linspace(-2.5*pi, 2.5*pi, 13)
yCosLo = np.cos(xAxisLo)

# interpolation (linear and cubic)
InterpLin = interp1d(xAxisLo, yCosLo, kind='linear')
InterpCub = interp1d(xAxisLo, yCosLo, kind='cubic')

# plot cosine function from -2.5pi to +2.5pi
plt.subplot(311)
plt.plot(xAxisLo, yCosLo, 'bo')
plt.legend(['sample'],loc='center left', bbox_to_anchor=(1, 0.5))
#... and add the interpolation
plt.subplot(312)
plt.plot(xAxisHi, InterpLin(xAxisHi),'ro',xAxisLo, yCosLo, 'bo')
plt.legend(['linear','sample'],loc='center left', bbox_to_anchor=(1, 0.5))
plt.show()
plt.subplot(312)
plt.plot(xAxisHi, InterpCub(xAxisHi),'go',xAxisLo, yCosLo, 'bo')
plt.legend(['cubic','sample'],loc='center left', bbox_to_anchor=(1, 0.5))
plt.show()
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Numerical integration

In the case where we have an array of y values without knowing the explicit functional dependence of y and x, it is still possible to perform a numerical integration using scipy's simps function (within the integrate package). scipy.integrate.simps uses Simpson's rule to perform the numerical integration. In the following the data from the Interpolation example are compared in 3 ways:

  1. Integrate the cosine function

  2. Integrate the function from the cubic interpolation

  3. Numerically integrate the array of points

A high level of agreement between the three methods is observed.

from scipy.interpolate import interp1d
from scipy.integrate import quad,simps
import numpy as np
from math import cos,pi,sin
import matplotlib.pyplot as plt

# The following plots cos(x) between -2.5pi and +2.5pi with 100 sample points
xAxisHi = np.linspace(-2.5*pi, 2.5*pi, 100)
yCosHi = np.cos(xAxisHi)
plt.plot(xAxisHi,yCosHi,label='cos func')
# 13 samples of cos(x) in the same interval range
xAxisLo = np.linspace(-2.5*pi, 2.5*pi, 13)
yCosLo = np.cos(xAxisLo)

# interpolation (linear and cubic)
InterpLin = interp1d(xAxisLo, yCosLo, kind='linear')
InterpCub = interp1d(xAxisLo, yCosLo, kind='cubic')

# First integrate cos(theta) from -2.5*pi to +2.5*pi
LimLo=-2.5*pi
LimHi=2.5*pi
intFunc,intFuncError=quad(cos,LimLo,LimHi)
print('Integrating cosine function: value=%.4f error=%g' % (intFunc,intFuncError))
intCubic,intCubicError=quad(InterpCub,LimLo,LimHi)
print('Integrating cubic interpolation: value=%.4f error=%g' % (intCubic,intCubicError))
intSimpson=simps(yCosLo,xAxisLo)
print('Integrating 13 sample points: value=%.4f' % (intSimpson))

# Plot Data
plt.plot(xAxisHi,InterpCub(xAxisHi),label='cubic interp')
plt.plot(xAxisLo,yCosLo,'o',label='sample')
plt.legend(loc='center left', bbox_to_anchor=(1, 0.5))
plt.show()
Integrating cosine function: value=2.0000 error=1.11662e-13 Integrating cubic interpolation: value=2.0962 error=2.32535e-08 Integrating 13 sample points: value=2.0407
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