# Numerical methods challenge: Day 4

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.

## Newton's method: vector case

Today we have Newton's method one more time. In this case, the function is vector-valued. This implies a slight modification over the original post. The new approximation is computed from the old one using

where we need to use the Jacobian matrix \(J\).

We will test the method with the function \(\mathbf{F}(x, y) = (x + 2y - 2, x^2 + 4x - 4)\) with solution \(\mathbf{x} = (0, 1)\).

Following are the codes.

### Python

from __future__ import division, print_function from numpy import array from numpy.linalg import solve, norm, det def newton(fun, jaco, x, niter=50, ftol=1e-12, verbose=False): msg = "Maximum number of iterations reached." for cont in range(niter): J = jaco(x) f = fun(x) if det(J) < ftol: x = None msg = "Derivative near to zero." break if verbose: print("n: {}, x: {}".format(cont, x)) x = x - solve(J, f) if norm(f) < ftol: msg = "Root found with desired accuracy." break return x, msg def fun(x): return array([x[0] + 2*x[1] - 2, x[0]**2 + 4*x[1]**2 - 4]) def jaco(x): return array([ [1, 2], [2*x[0], 8*x[1]]]) print(newton(fun, jaco, [1.0, 10.0]))

### Julia

function newton(fun, jaco, x, niter=50, ftol=1e-12, verbose=false) msg = "Maximum number of iterations reached." for cont = 1:niter J = jaco(x) f = fun(x) if det(J) < ftol x = nothing msg = "Derivative near to zero." break end if verbose println("n: $(cont), x: $(x)") end x = x - J\f if norm(f) < ftol msg = "Root found with desired accuracy." break end end return x, msg end function fun(x) return [x[1] + 2*x[2] - 2, x[1]^2 + 4*x[2]^2 - 4] end function jaco(x) return [1.0 2.0; 2*x[1] 8*x[2]] end println(newton(fun, jaco, [1.0, 10.0]))

### Comparison

Regarding number of lines we have: 31 in Python and 33 in Julia. The comparison
in execution time is done with `%timeit`

magic command in IPython and
`@benchmark`

in Julia.

For Python:

%timeit newton(fun, jaco, [1.0, 10.0])

with result

1000 loops, best of 3: 284 µs per loop

For Julia:

@benchmark newton(fun, jaco, [1.0, 10.0])

with result

BenchmarkTools.Trial: memory estimate: 10.44 KiB allocs estimate: 192 -------------- minimum time: 6.818 μs (0.00% GC) median time: 7.167 μs (0.00% GC) mean time: 9.607 μs (16.53% GC) maximum time: 2.953 ms (97.40% GC) -------------- samples: 10000 evals/sample: 4

In this case, we can say that the Python code is roughly 40 times slower
than the Julia one. This is an improvement compared to the previous examples,
where the ratio was around 100. The reason for this "improvement" might be
in the inversion of the Jacobian, that calls a `numpy`

routine, doing
the weight-lifting for us.

## Comentarios

Comments powered by Disqus