We Keep Coding

Find zeroes of function with simpy.optimize.bisect, with a complex function

I'm a student of mechanical engineering, and this is the first year I've met with the Python environment, or the distribution of it Anaconda.
I was given a task to find the zeroes of this function:
𝐷⋅sin(𝛼)cos(𝛼)+𝑙⋅cos(𝛼)sin(𝛼)2−𝑙⋅cos(𝛼)−ℎ⋅sin(𝛼)=0
With the parameters:
D = 220mm,
h = 1040mm,
l = 1420mm,where
n = 81
is the number of equally distanced points on the function
and the function is limited to :
𝛼∈[0,2𝜋] where 𝛼 is a np.array.
plotted function
The issue is, when I try to insert the function in bisect(fun, a, b), the error says
'numpy.ndarray' object is not callable
Can someone aid a noob programer ? Thanks.

The question is not clear, you should share your code and the title should say scipy, not simpy, if I am correct.
Apart from this, I do not get the same plot of the function, can you check if it is correct?
If you want to use the bisection method you should do something like this:
import numpy as np
from scipy.optimize import bisect
def fun(x, D, h, l):
return D * np.sin(x) * np.cos(x) + l * np.cos(x) * np.sin(x) * 2 - l * np.cos(x) - h * np.sin(x)
D = 220
h = 1040
l = 1420
print(bisect(lambda x: fun(x, D, h, l), 0, 2*np.pi))
Note that the bisection method only finds one zero, and this does not work at all because the two extremes of the function have the same sign. For this particular function, you could run the bisect in the intervals (0, pi) and (pi, 2pi) to find both zeros.

Oscillatory integral in python

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.

How to put an integral in a function in python/matplotlib

So pretty much, I am aiming to achieve a function f(x)
My problem is that my function has an integral in it, and I only know how to construct definite integrals, so my question is how does one create an indefinite integral in a function (or there may be some other method I am currently unaware of)
My function is defined as :
(G is gravitational constant, although you can leave G out of your answer for simplicity, I'll add it in my code)
Here is the starting point, but I don't know how to do the integral portion
import numpy as np
def f(x):
rho = 5*(1/(1+((x**2)/(3**2))))
function_result = rho * 4 * np.pi * x**2
return function_result
Please let me know if I need to elaborate on something.
EDIT-----------------------------------------------------
I made some major progress, but I still have one little error.
Pretty much, I did this:
from sympy import *
x = Symbol('x')
rho = p0()*(1/(1+((x**2)/(rc()**2))))* 4 * np.pi * x**2
fooply = integrate(rho,x)
def f(rx):
function_result = fooply.subs({x:rx})
return function_result
Which works fine when I plug in one number for f; however, when I plug in an array (as I need to later), I get the error:
raise SympifyError(a)
sympy.core.sympify.SympifyError: SympifyError: [3, 3, 3, 3, 3]
(Here, I did print(f([3,3,3,3,3]))). Usually, the function returns an array of values. So if I did f([3,2]) it should return [f(3),f(2)]. Yet, for some reason, it doesn't for my function....
Thanks in advance

how about:
from sympy import *
x, p0, rc = symbols('x p0 rc', real=True, positive=True)
rho = p0*(1/(1+((x**2)/(rc))))* 4 * pi * x**2
fooply = integrate(rho,x)/x
rho, fooply
(4*pi*p0*x**2/(1 + x**2/rc),
4*pi*p0*rc*(-sqrt(rc)*atan(x/sqrt(rc)) + x)/x)
fooply = fooply.subs({p0: 2.0, rc: 3.0})
np_fooply = lambdify(x, fooply, 'numpy')
print(np_fooply(np.array([3,3,3,3,3])))
[ 29.81247362 29.81247362 29.81247362 29.81247362 29.81247362]

To plug in an array to a SymPy expression, you need to use lambdify to convert it to a NumPy function (f = lambdify(x, fooply)). Just using def and subs as you have done will not work.
Also, in general, when using symbolic computations, it's better to use sympy.pi instead of np.pi, as the former is symbolic and can simplify. It will automatically be converted to the numeric pi by lambdify.

How do you invert scipy quad function to pass in an arc length and receive a value for (t)?

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.

Artefacts from Riemann sum in scipy.signal.convolve

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.