I wrote the following code to plot the intensity of light exiting an optical components, which is basically a spherical Fourier integral on the incident field, so it has a Bessel function. The argument of which depends on the integrating variable (x) and the plotting variable (r).
from sympy import *
import matplotlib.pyplot as plt
import numpy as np
from scipy.integrate import quad
from scipy.special import jn
#constants
mm = 1
um = 1e-3 * mm
nm = 1e-6 * mm
wavelength = 532*nm
klaser = 2*np.pi / wavelength
waist = 3.2*mm
angle = 2 #degrees
focus = 125 * mm
ng = 1.5 # refractive index of axicon
upperintegration = 5
#integrals
def b(angle):
radians = angle* np.pi/180
return klaser * (ng-1) * np.tan(radians)
def delta(angle):
return np.pi/(b(angle)*waist)
def integrand(x, r):
return klaser/focus * waist**2 * np.exp(-x**2) * np.exp(-np.pi * 1j * x/delta(angle)) * jn(0, waist*klaser*r*x/focus) * x
def intensity1D(r):
return np.sqrt(quad(lambda x: np.real(integrand(x, r)), 0, upperintegration)[0]**2 + quad(lambda x: np.imag(integrand(x, r)), 0, upperintegration)[0]**2)
fig = plt.figure()
ax = fig.add_subplot(111)
t = np.linspace(-3.5, 3.5, 25)
plt.plot(t, np.vectorize(intensity1D)(t))
The issue is that the plot changes drastically as I change the number of points I am using in my linspace, when I plot it.
I suspect this may be because of the oscillatory nature of the integral, so the step-size taken can dramatically change the value of the exponent and hence of the integral.
How does quad deal with this? Are there better methods to integrate numerically for this particular application?
In the call to quad, set the limit argument to a large number. This increases the maximum number subintervals that quad is allowed to use to estimate the integral. When I use
def intensity1D(r):
re = quad(lambda x: np.real(integrand(x, r)), 0, upperintegration, limit=8000)[0]
im = quad(lambda x: np.imag(integrand(x, r)), 0, upperintegration, limit=8000)[0]
return np.sqrt(re**2 + im**2)
and compute the function with the array t defined as
t = np.linspace(1.5, 3, 1000)
I get the following plot:
(I also removed the line from sympy import *. sympy does not appear to be used in
your script.)
You should always check the error estimate that is the second return value of quad.
For example:
In [14]: r = 3.0
In [15]: val, err = quad(lambda x: np.real(integrand(x, r)), 0, upperintegration, limit=8000)
In [16]: val
Out[16]: 2.975500141416676e-11
In [17]: err
Out[17]: 1.4590630152807049e-08
As you can see, the error estimate is much larger than the approximate integral. The estimates returned by quad might be conservative, but a result with such a large error estimate should still be treated with caution. Let's take a look at the corresponding imaginary part:
In [25]: val, err = quad(lambda x: np.imag(integrand(x, r)), 0, upperintegration, limit=8000)
In [26]: val
Out[26]: 0.0026492702707317257
In [27]: err
Out[27]: 1.4808416189183e-08
val is now orders of magnitude larger than the estimated error. So when the magnitude of the complex value is computed in intensity1D(), we end up with estimated relative error on the order of 1e-5. That may be sufficient for your calculation.
At the peak near r=2.1825, the magnitude of the error estimate is still small, and it is much smaller than the computed integral:
In [32]: r = 2.1825
In [33]: quad(lambda x: np.real(integrand(x, r)), 0, upperintegration, limit=8000)
Out[33]: (6.435730031424414, 8.801375195176556e-08)
In [34]: quad(lambda x: np.imag(integrand(x, r)), 0, upperintegration, limit=8000)
Out[34]: (-6.583055286038913, 9.211333259956749e-08)
There are specific methods for integration of oscillatory integrands that actually increase in accuracy as the frequency increases. Filon and Levin methods are described here:
https://www.sciencedirect.com/science/article/pii/S0377042706005929
Mathematica should use one of these if you specify LevinRule as method in
NIntegrate. This is perhaps simple enough that -- if your integrand has the form, apparently common in optics calculations-- that you could even write a short program in your favorite efficient numerical programming language.
I suspect that using usual quadrature for oscillatory integrands is going to be painfully slow if you want to get accurate results.
Related
My problem is the following. I measured a bunch of different physical properties and propagated the methodic and measurement uncertainties all the way to some kind of efficiency ratio. For all my physical properties, a normal distribution seemed to be a good choice, and for the first few calculations and corresponding propagations I had fairly low uncertainties. I forwarded all uncertainties as an extended uncertainty of kp=2 meaning coverage of 95.45% of all possible values.
However, for the calculations I am performing right now in order to get the efficiency I end up with results like (16+/-31)%, which is not possible since my efficiency can only spread from 0 to 1. My assumption would be, that I found the "correct" expectation value for the efficiency, but my probability distribution should be a positive skewed gamma-distribution rather than a normal-distribution. With the assumption that the intervals [0; 0.16+0.31], [0; 0.16+0.31*3/2] and [0;1] cover 95.45%, 99.73% and 100% of all possible values I should be able to calculate the gamma-distribution parameters alpha and beta analytically. Unfortunately, the following code using sympy doesn't work because sympy can't handle it.
Does someone have an idea how to solve my problem?
import sympy as sy
# from sympy import init_printing
# from sympy import symbols
# from sympy import Eq
# from sympy import var
# from sympy import integrate
# from sympy import gamma
# from sympy import Pow
# from sympy import exp
R_std3 = float(31 * 3/2 * 1/100)
R = float(16 * 1/100)
R_plus = R + R_std3
alpha = sy.symbols('alpha', positive = True)
beta = sy.symbols('beta', positive = True)
r = sy.symbols('r', positive = True)
z = sy.Pow(beta, alpha) * sy.Pow(r, alpha-1) * sy.exp(-beta*r)
n = sy.gamma(alpha)
pdf = sy.Pow(n, -1) * z
system = [sy.Eq(R, sy.Pow(beta, -1) * alpha ),
sy.Eq(0.9973, sy.integrate(pdf, (r,0,R_plus) ))
]
sy.solve( system )
I wouldn't expect the second (transcendental) equation to have a nice analytic solution unless there's some clever identity that you can use to simplify it. You can solve this numerically though:
In [46]: system
Out[46]:
⎡ α α⋅γ(α, 0.625⋅β)⎤
⎢0.16 = ─, 0.9973 = ───────────────⎥
⎣ β Γ(α + 1) ⎦
In [47]: nsolve(system, [alpha, beta], [1, 10])
Out[47]:
⎡2.16365408317279⎤
⎢ ⎥
⎣ 13.52283801983 ⎦
On another thread, I saw someone manage to integrate the length of a arc using mathematica.They wrote:
In[1]:= ArcTan[3.05*Tan[5Pi/18]/2.23]
Out[1]= 1.02051
In[2]:= x=3.05 Cos[t];
In[3]:= y=2.23 Sin[t];
In[4]:= NIntegrate[Sqrt[D[x,t]^2+D[y,t]^2],{t,0,1.02051}]
Out[4]= 2.53143
How exactly could this be transferred to python using the imports of numpy and scipy? In particular, I am stuck on line 4 in his code with the "NIntegrate" function. Thanks for the help!
Also, if I already have the arc length and the vertical axis length, how would I be able to reverse the program to spit out the original paremeters from the known values? Thanks!
To my knowledge scipy cannot perform symbolic computations (such as symbolic differentiation). You may want to have a look at http://www.sympy.org for a symbolic computation package. Therefore, in the example below, I compute derivatives analytically (the Dx(t) and Dy(t) functions).
>>> from scipy.integrate import quad
>>> import numpy as np
>>> Dx = lambda t: -3.05 * np.sin(t)
>>> Dy = lambda t: 2.23 * np.cos(t)
>>> quad(lambda t: np.sqrt(Dx(t)**2 + Dy(t)**2), 0, 1.02051)
(2.531432761012828, 2.810454936566873e-14)
EDIT: Second part of the question - inverting the problem
From the fact that you know the value of the integral (arc) you can now solve for one of the parameters that determine the arc (semi-axes, angle, etc.) Let's assume you want to solve for the angle. Then you can use one of the non-linear solvers in scipy, to revert the equation quad(theta) - arcval == 0. You can do it like this:
>>> from scipy.integrate import quad
>>> from scipy.optimize import broyden1
>>> import numpy as np
>>> a = 3.05
>>> b = 2.23
>>> Dx = lambda t: -a * np.sin(t)
>>> Dy = lambda t: b * np.cos(t)
>>> arc = lambda theta: quad(lambda t: np.sqrt(Dx(t)**2 + Dy(t)**2), 0, np.arctan((a / b) * np.tan(np.deg2rad(theta))))[0]
>>> invert = lambda arcval: float(broyden1(lambda x: arc(x) - arcval, np.rad2deg(arcval / np.sqrt((a**2 + b**2) / 2.0))))
Then:
>>> arc(50)
2.531419526553662
>>> invert(arc(50))
50.000031008458365
If you prefer a pure numerical approach, you could use the following barebones solution. This worked well for me given that I had two input numpy.ndarrays, x and y with no functional form available.
import numpy as np
def arclength(x, y, a, b):
"""
Computes the arclength of the given curve
defined by (x0, y0), (x1, y1) ... (xn, yn)
over the provided bounds, `a` and `b`.
Parameters
----------
x: numpy.ndarray
The array of x values
y: numpy.ndarray
The array of y values corresponding to each value of x
a: int
The lower limit to integrate from
b: int
The upper limit to integrate to
Returns
-------
numpy.float64
The arclength of the curve
"""
bounds = (x >= a) & (y <= b)
return np.trapz(
np.sqrt(
1 + np.gradient(y[bounds], x[bounds])
) ** 2),
x[bounds]
)
Note: I spaced the return variables out that way just to make it more readable and clear to understand the operations taking place.
As an aside, recall that the arc-length of a curve is given by:
If I were to run something like the code below where I pass in the minor and major arc of an ellipse as 3.05 and 2.23 with the 50 degree angle formed by the arc, how would I be able to take the output of 2.531432761012828 as the arc length and pass it back through to solve for t? Thanks!
import math
from scipy.integrate import quad
import numpy as np
t = math.atan(3.05*math.tan(5*math.pi/18)/2.23)
Dx = lambda t: -3.05 * np.sin(t)
Dy = lambda t: 2.23 * np.cos(t)
quad(lambda t: np.sqrt(Dx(t)**2 + Dy(t)**2), 0, t)
The output of the last one was: (2.531432761012828, 2.810454936566873e-14)
To find the upper limit of integration, given the value of the integral, one can apply fsolve to the function which calculates that integral for variable upper limits. Example (not repeating the lines you already have):
from scipy.optimize import fsolve
target = 2.531432761012828
fun = lambda s: quad(lambda t: np.sqrt(Dx(t)**2 + Dy(t)**2), 0, s)[0] - target
s0 = fsolve(fun, 0)[0]
print(s0)
This prints 1.02051.
I dislike having both the variable of integration and the upper limit denoted by the same letter, so in my code the upper limit is called s.
Say I compute the density of Beta(4,8):
from scipy.stats import beta
rv = beta(4, 8)
x = np.linspace(start=0, stop=1, num=200)
my_pdf = rv.pdf(x)
Why does the integral of the pdf not equal one?
> my_pdf.sum()
199.00000139548044
The integral over the pdf is one. You can see this by using numerical integration from scipy
>>> from scipy.integrate import quad
>>> quad(rv.pdf, 0, 1)
(0.9999999999999999, 1.1102230246251564e-14)
or by writing your own ad-hoc integration (with a trapezoidal rule in this example)
>>> x = numpy.linspace(start=0, stop=1, num=201)
>>> (0.5 * rv.pdf(x[0]) + rv.pdf(x[1:-1]).sum() + 0.5 * rv.pdf(x[-1])) / 200.0
1.0000000068732813
rv.pdf returns the value of the pdf at each value of x. It doesn't sum to one because your aren't actually computing an integral. If you want to do that, you need to divide your sum by the number of intervals, which is len(x) - 1, which is 199. That would then give you a result very close to 1.
Short summary: How do I quickly calculate the finite convolution of two arrays?
Problem description
I am trying to obtain the finite convolution of two functions f(x), g(x) defined by
To achieve this, I have taken discrete samples of the functions and turned them into arrays of length steps:
xarray = [x * i / steps for i in range(steps)]
farray = [f(x) for x in xarray]
garray = [g(x) for x in xarray]
I then tried to calculate the convolution using the scipy.signal.convolve function. This function gives the same results as the algorithm conv suggested here. However, the results differ considerably from analytical solutions. Modifying the algorithm conv to use the trapezoidal rule gives the desired results.
To illustrate this, I let
f(x) = exp(-x)
g(x) = 2 * exp(-2 * x)
the results are:
Here Riemann represents a simple Riemann sum, trapezoidal is a modified version of the Riemann algorithm to use the trapezoidal rule, scipy.signal.convolve is the scipy function and analytical is the analytical convolution.
Now let g(x) = x^2 * exp(-x) and the results become:
Here 'ratio' is the ratio of the values obtained from scipy to the analytical values. The above demonstrates that the problem cannot be solved by renormalising the integral.
The question
Is it possible to use the speed of scipy but retain the better results of a trapezoidal rule or do I have to write a C extension to achieve the desired results?
An example
Just copy and paste the code below to see the problem I am encountering. The two results can be brought to closer agreement by increasing the steps variable. I believe that the problem is due to artefacts from right hand Riemann sums because the integral is overestimated when it is increasing and approaches the analytical solution again as it is decreasing.
EDIT: I have now included the original algorithm 2 as a comparison which gives the same results as the scipy.signal.convolve function.
import numpy as np
import scipy.signal as signal
import matplotlib.pyplot as plt
import math
def convolveoriginal(x, y):
'''
The original algorithm from http://www.physics.rutgers.edu/~masud/computing/WPark_recipes_in_python.html.
'''
P, Q, N = len(x), len(y), len(x) + len(y) - 1
z = []
for k in range(N):
t, lower, upper = 0, max(0, k - (Q - 1)), min(P - 1, k)
for i in range(lower, upper + 1):
t = t + x[i] * y[k - i]
z.append(t)
return np.array(z) #Modified to include conversion to numpy array
def convolve(y1, y2, dx = None):
'''
Compute the finite convolution of two signals of equal length.
#param y1: First signal.
#param y2: Second signal.
#param dx: [optional] Integration step width.
#note: Based on the algorithm at http://www.physics.rutgers.edu/~masud/computing/WPark_recipes_in_python.html.
'''
P = len(y1) #Determine the length of the signal
z = [] #Create a list of convolution values
for k in range(P):
t = 0
lower = max(0, k - (P - 1))
upper = min(P - 1, k)
for i in range(lower, upper):
t += (y1[i] * y2[k - i] + y1[i + 1] * y2[k - (i + 1)]) / 2
z.append(t)
z = np.array(z) #Convert to a numpy array
if dx != None: #Is a step width specified?
z *= dx
return z
steps = 50 #Number of integration steps
maxtime = 5 #Maximum time
dt = float(maxtime) / steps #Obtain the width of a time step
time = [dt * i for i in range (steps)] #Create an array of times
exp1 = [math.exp(-t) for t in time] #Create an array of function values
exp2 = [2 * math.exp(-2 * t) for t in time]
#Calculate the analytical expression
analytical = [2 * math.exp(-2 * t) * (-1 + math.exp(t)) for t in time]
#Calculate the trapezoidal convolution
trapezoidal = convolve(exp1, exp2, dt)
#Calculate the scipy convolution
sci = signal.convolve(exp1, exp2, mode = 'full')
#Slice the first half to obtain the causal convolution and multiply by dt
#to account for the step width
sci = sci[0:steps] * dt
#Calculate the convolution using the original Riemann sum algorithm
riemann = convolveoriginal(exp1, exp2)
riemann = riemann[0:steps] * dt
#Plot
plt.plot(time, analytical, label = 'analytical')
plt.plot(time, trapezoidal, 'o', label = 'trapezoidal')
plt.plot(time, riemann, 'o', label = 'Riemann')
plt.plot(time, sci, '.', label = 'scipy.signal.convolve')
plt.legend()
plt.show()
Thank you for your time!
or, for those who prefer numpy to C. It will be slower than the C implementation, but it's just a few lines.
>>> t = np.linspace(0, maxtime-dt, 50)
>>> fx = np.exp(-np.array(t))
>>> gx = 2*np.exp(-2*np.array(t))
>>> analytical = 2 * np.exp(-2 * t) * (-1 + np.exp(t))
this looks like trapezoidal in this case (but I didn't check the math)
>>> s2a = signal.convolve(fx[1:], gx, 'full')*dt
>>> s2b = signal.convolve(fx, gx[1:], 'full')*dt
>>> s = (s2a+s2b)/2
>>> s[:10]
array([ 0.17235682, 0.29706872, 0.38433313, 0.44235042, 0.47770012,
0.49564748, 0.50039326, 0.49527721, 0.48294359, 0.46547582])
>>> analytical[:10]
array([ 0. , 0.17221333, 0.29682141, 0.38401317, 0.44198216,
0.47730244, 0.49523485, 0.49997668, 0.49486489, 0.48254154])
largest absolute error:
>>> np.max(np.abs(s[:len(analytical)-1] - analytical[1:]))
0.00041657780840698155
>>> np.argmax(np.abs(s[:len(analytical)-1] - analytical[1:]))
6
Short answer: Write it in C!
Long answer
Using the cookbook about numpy arrays I rewrote the trapezoidal convolution method in C. In order to use the C code one requires three files (https://gist.github.com/1626919)
The C code (performancemodule.c).
The setup file to build the code and make it callable from python (performancemodulesetup.py).
The python file that makes use of the C extension (performancetest.py)
The code should run upon downloading by doing the following
Adjust the include path in performancemodule.c.
Run the following
python performancemodulesetup.py build
python performancetest.py
You may have to copy the library file performancemodule.so or performancemodule.dll into the same directory as performancetest.py.
Results and performance
The results agree neatly with one another as shown below:
The performance of the C method is even better than scipy's convolve method. Running 10k convolutions with array length 50 requires
convolve (seconds, microseconds) 81 349969
scipy.signal.convolve (seconds, microseconds) 1 962599
convolve in C (seconds, microseconds) 0 87024
Thus, the C implementation is about 1000 times faster than the python implementation and a bit more than 20 times as fast as the scipy implementation (admittedly, the scipy implementation is more versatile).
EDIT: This does not solve the original question exactly but is sufficient for my purposes.