# Numerical methods challenge: Day 14

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.

## Trapezoidal rule

Today we have the trapezoidal rule. This is the simplest of Newton-Cotes formulas for numerical integration. Newton-Cotes is related to Lagrange interpolation with equidistant points.

Following are the codes.

### Python

from __future__ import division, print_function import numpy as np def trapz(y, x=None, dx=1.0): if x is None: inte = 0.5*dx*(2*np.sum(y[1:-1]) + y[0] + y[-1]) else: dx = x[1:] - x[:-1] inte = 0.5*np.dot(y[:-1] + y[1:], dx) return inte dx = 0.001 x = np.arange(0, 10, dx) y = np.sin(x) print(trapz(y, dx=dx)) print(1 - np.cos(10))

with result

1.83961497726 1.83907152908

### Julia

function trapz(y; x=nothing, dx=1.0) if x == nothing inte = 0.5*dx*(2*sum(y[2:end-1]) + y[1] + y[end]) else dx = x[2:end] - x[1:end-1] inte = 0.5* (y[1:end-1] + y[2:end])' * dx end return inte end dx = 0.001 x = 0:dx:10 y = sin.(x) println(trapz(y, dx=dx)) println(1 - cos(10))

with results

1.8390713758204895 1.8390715290764525

### Comparison Python/Julia

Regarding number of lines we have: 18 in Python and 16 in Julia. The comparison
in execution time is done with `%timeit` magic command in IPython and
`@benchmark` in Julia.

For Python:

%timeit trapz(y, dx=dx)

with result

100000 loops, best of 3: 16.9 µs per loop

For Julia:

@benchmark trapz(y, dx=dx)

with result

BenchmarkTools.Trial: memory estimate: 78.31 KiB allocs estimate: 4 -------------- minimum time: 13.080 μs (0.00% GC) median time: 16.333 μs (0.00% GC) mean time: 20.099 μs (12.66% GC) maximum time: 963.732 μs (90.60% GC) -------------- samples: 10000 evals/sample: 1

In this case, we can say that the Python code is roughly as fast as Julia.

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