# Numerical methods challenge: Day 2

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.

## Regula falsi

The second method to be considered is the
false position method,
or *regula falsi*. This method is used to solve the equation \(f(x) = 0\)
for \(x\) real, and \(f\) continuous. It starts with an interval
\([a,b]\), where \(f(a)\) and \(f(b)\) should have opposite signs.
The method is similar to bisection method but instead of halving the original
interval it takes as a new point of the interval the intercept of the line
that connect the function evaluated at the two extreme points. Then, the new
point is computed from

As in bisection method, we keep the interval where the solution appears, i.e., the sign of the function changes.

We will use the function \(f(x) = \cos(x) - x\) to test the codes, and the initial interval is \([0.5, \pi/4]\).

Following are the codes.

### Python

from __future__ import division, print_function from numpy import abs, cos, pi def regula_falsi(fun, a, b, niter=50, ftol=1e-12, verbose=False): if fun(a) * fun(b) > 0: c = None msg = "The function should have a sign change in the interval." else: for cont in range(niter): qa = fun(a) qb = fun(b) c = (a*qb - b*qa)/(qb - qa) qc = fun(c) if verbose: print("n: {}, c: {}".format(cont, c)) msg = "Maximum number of iterations reached." if abs(qc) < ftol: msg = "Root found with desired accuracy." break elif qa * qc < 0: b = c elif qb * qc < 0: a = c return c, msg def fun(x): return cos(x) - x print(regula_falsi(fun, 0.5, 0.25*pi))

### Julia

function regula_falsi(fun, a, b, niter=50, ftol=1e-12, verbose=false) if fun(a) * fun(b) > 0 c = nothing msg = "The function should have a sign change in the interval." else for cont = 1:niter qa = fun(a) qb = fun(b) c = (a*qb - b*qa)/(qb - qa) qc = fun(c) if verbose println("n: $(cont), c: $(c)") end if abs(fun(c)) < ftol msg = "Root found with desired accuracy." break elseif qa * qc < 0 b = c elseif qb * qc < 0 a = c end msg = "Maximum number of iterations reached." end end return c, msg end function fun(x) return cos(x) - x end println(regula_falsi(fun, 0.5, 0.25*pi))

### Comparison

Regarding number of lines we have: 29 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:

with result

For Julia:

with result

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