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Numerical methods challenge: Day 15

During October (2017) I will write a program per day for some well-known numerical methods in both Python and Julia. It is intended to be an exercise then don't expect the code to be good enough for real use. Also, I should mention that I have almost no experience with Julia, so it probably won't be idiomatic Julia but more Python-like Julia.

Simpson's rule

Today we have the Simpson's rule. This is another Newton-Cotes formulas, used for numerical integration. Newton-Cotes is related to Lagrange interpolation with equidistant points.

Following are the codes.


from __future__ import division, print_function
import numpy as np

def simps(y, x=None, dx=1.0):
    n = len(y)
    if x is None:
        inte = np.sum(y[0:n-2:2] + 4*y[1:n-1:2] + y[2:n:2])*dx
        dx = x[1::2] - x[:-1:2]
        inte =[0:n-2:2] + 4*y[1:n-1:2] + y[2:n:2], dx)
    return inte/3

n = 21
x = np.linspace(0, 10, n)
y = np.sin(x)
print(simps(y, x=x))
print(1 - np.cos(10))

with result



function simps(y; x=nothing, dx=1.0)
    n = length(y)
    if x == nothing
        inte = sum(y[1:2:n-2] + 4*y[2:2:n-1] + y[3:2:n])*dx
        dx = x[2:2:end] - x[1:2:end-1]
        inte = (y[1:2:n-2] + 4*y[2:2:n-1] + y[3:2:n])' * dx
    return inte/3

n = 21
x = linspace(0, 10, n)
y = sin.(x)
println(simps(y, x=x))
println(1 - cos(10))

Comparison Python/Julia

Regarding number of lines we have: 19 in Python and 17 in Julia. The comparison in execution time is done with %timeit magic command in IPython and @benchmark in Julia.

For Python:

%timeit simps(y, x=x)

with result

100000 loops, best of 3: 13.8 µs per loop

For Julia:

@benchmark simps(y, x=x)

with result

  memory estimate:  1.23 KiB
  allocs estimate:  14
  minimum time:     1.117 μs (0.00% GC)
  median time:      1.200 μs (0.00% GC)
  mean time:        1.404 μs (7.04% GC)
  maximum time:     222.286 μs (96.45% GC)
  samples:          10000
  evals/sample:     10

In this case, we can say that the Python code is 10 times slower than Julia.


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