# Numerical methods challenge: Day 3

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

Today's method is Newton's method. This method is used to solve the equation \(f(x) = 0\) for \(x\) real, and \(f\) a differentiable function. It starts with an initial guess \(x_0\) and it succesively refine it by finding the intercept of the tangent line to the function with zero. The new approximation is computed from the old one using

Convergence of this method is generally faster than bisection method. Nevertheless, the convergence is not guaranteed. Another drawback is the need of the derivative of the function.

We will use the function \(f(x) = \cos(x) - x\) to test the codes, and the initial guess is 1.0.

Following are the codes.

### Python

from __future__ import division, print_function from numpy import abs, cos, sin def newton(fun, grad, x, niter=50, ftol=1e-12, verbose=False): msg = "Maximum number of iterations reached." for cont in range(niter): if abs(grad(x)) < ftol: x = None msg = "Derivative near to zero." break if verbose: print("n: {}, x: {}".format(cont, x)) x = x - fun(x)/grad(x) if abs(fun(x)) < ftol: msg = "Root found with desired accuracy." break return x, msg def fun(x): return cos(x) - x def grad(x): return -sin(x) - 1.0 print(newton(fun, grad, 1.0))

### Julia

function newton(fun, grad, x, niter=50, ftol=1e-12, verbose=false) msg = "Maximum number of iterations reached." for cont = 1:niter if abs(grad(x)) < ftol x = nothing msg = "Derivative near to zero." break end if verbose println("n: $(cont), x: $(x)") end x = x - fun(x)/grad(x) if abs(fun(x)) < ftol msg = "Root found with desired accuracy." break end end return x, msg end function fun(x) return cos(x) - x end function grad(x) return -sin(x) - 1.0 end println(newton(fun, grad, 1.0))

### Comparison

Regarding number of lines we have: 28 in Python and 32 in Julia. The comparison
in execution time is done with `%timeit`

magic command in IPython and
`@benchmark`

in Julia.

For Python:

%timeit newton(fun, grad, 1.0)

with result

10000 loops, best of 3: 27.3 µs per loop

For Julia:

@benchmark newton(fun, grad, 1.0)

with result

BenchmarkTools.Trial: memory estimate: 48 bytes allocs estimate: 2 -------------- minimum time: 327.925 ns (0.00% GC) median time: 337.956 ns (0.00% GC) mean time: 351.064 ns (0.80% GC) maximum time: 8.118 μs (92.60% GC) -------------- samples: 10000 evals/sample: 226

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