[Regular Falsi Method] Algorithm to calculate the root of the given equation using swift #swift #regularfalsimethod

regularFalsiMethod.swift

import Foundation
import Darwin
let e = Darwin.M_E
let pi = Double.pi
// attempt to make newton rhapson method derived algo
// note that the trignometric values are in radians
// How to represent certain functions :
/// log base 10 : log10(float)
/// log base e : log(float)
/// power to function : pow(x, y) -> gives x ^ y
// note that you have to provide the values of xn and xnn
var xn = 1.0
// make x a float value
var xnn = 2.0  // higher value
let precision = 3.0 // keep precision as double

var Fn = 0.0 // stores the value of F(xn)
var fn = 0.0 // stores the value of f(xn)
var Fnn = 0.0 // stores the value of F(x(n + 1))
var fnn = 0.0 // stores the value of f(x(n + 1))
 // you need to provide the functions explicitly.. its too complex to do differentiation in swift
func firstExpression(_ x :Double) -> Double{
    //Enter the values of F(x) here
    let res = (x * x * x * x ) - 12
    return res
}

print("x0 : \(xn)")
print("F(x0) = \(firstExpression(xn))")
var flip = 0
var prev = 0.0
var i = 0
var temp = xnn + 1 // done as a security measure
while true
{
 Fnn = firstExpression(xnn)
 Fn = firstExpression(xn)
    print("x\(i + 1) : \(xnn)")
    print("F(x\(i + 1)) = \(Fnn)")



     var digi = xnn - (Fnn * (xnn - xn) / (Fnn - Fn))

   if  firstExpression(digi) > 0 && digi < xnn {
    temp = xnn
    xnn = digi

    }
    else
    if firstExpression(digi) < 0 && xn < digi
    {temp = xn
        xn = digi

    }

 print("\nx\(i + 2) = \(digi)")
    print("F(x\(i + 2)) = \(firstExpression(digi))")



    i += 1
    if flip == 1 {break}
    if round(xnn * pow(10, precision )  ) == round(temp * pow(10, precision )  ) || round(xn * pow(10, precision )  ) == round(temp * pow(10, precision)  ) { flip += 1 }
}