Roots of a function

Roots of a function
You are encouraged to solve this task according to the task description, using any language you may know.

Task

Create a program that finds and outputs the roots of a given function, range and (if applicable) step width.

The program should identify whether the root is exact or approximate.

For this task, use:     ƒ(x)   =   x³ - 3x² + 2x

11l

Translation of: Python

F f(x)
   R x^3 - 3 * x^2 + 2 * x

-V step = 0.001
-V start = -1.0
-V stop = 3.0

V sgn = f(start) > 0
V x = start

L x <= stop
   V value = f(x)

   I value == 0
      print(‘Root found at ’x)
   E I (value > 0) != sgn
      print(‘Root found near ’x)

   sgn = value > 0
   x += step

Output:

Root found near 8.812395258e-16
Root found near 1
Root found near 2.001

Ada

with Ada.Text_Io; use Ada.Text_Io;
 
procedure Roots_Of_Function is
   package Real_Io is new Ada.Text_Io.Float_Io(Long_Float);
   use Real_Io;
   
   function F(X : Long_Float) return Long_Float is
   begin
      return (X**3 - 3.0*X*X + 2.0*X);
   end F;
   
   Step  : constant Long_Float := 1.0E-6;
   Start : constant Long_Float := -1.0;
   Stop  : constant Long_Float := 3.0;
   Value : Long_Float := F(Start);
   Sign  : Boolean := Value > 0.0;
   X     : Long_Float := Start + Step;
 
begin
   if Value = 0.0 then
      Put("Root found at ");
      Put(Item => Start, Fore => 1, Aft => 6, Exp => 0);
      New_Line;
   end if;
   while X <= Stop loop
      Value := F(X);
      if (Value > 0.0) /= Sign then
         Put("Root found near ");
         Put(Item => X, Fore => 1, Aft => 6, Exp => 0);
         New_Line;
      elsif Value = 0.0 then
         Put("Root found at ");
         Put(Item => X, Fore => 1, Aft => 6, Exp => 0);
         New_Line;
      end if;
      Sign := Value > 0.0;
      X := X + Step;
   end loop;
end Roots_Of_Function;

ALGOL 68

Works with: ALGOL 68 version Revision 1 - no extensions to language used

Works with: ALGOL 68G version Any - tested with release 1.18.0-9h.tiny

Finding 3 roots using the secant method:

MODE DBL = LONG REAL;
FORMAT dbl = $g(-long real width, long real width-6, -2)$;

MODE XY = STRUCT(DBL x, y);
FORMAT xy root = $f(dbl)" ("b("Exactly", "Approximately")")"$;

MODE DBLOPT = UNION(DBL, VOID);
MODE XYRES = UNION(XY, VOID);

PROC find root = (PROC (DBL)DBL f, DBLOPT in x1, in x2, in x error, in y error)XYRES:(
  INT limit = ENTIER (long real width / log(2)); # worst case of a binary search) #
  DBL x1 := (in x1|(DBL x1):x1|-5.0), # if x1 is EMPTY then -5.0 #
      x2 := (in x2|(DBL x2):x2|+5.0), 
      x error := (in x error|(DBL x error):x error|small real), 
      y error := (in y error|(DBL y error):y error|small real);
  DBL y1 := f(x1), y2;
  DBL dx := x1 - x2, dy;

  IF y1 = 0 THEN
    XY(x1, y1) # we already have a solution! #
  ELSE
    FOR i WHILE
      y2 := f(x2);
      IF y2 = 0 THEN stop iteration FI;
      IF i = limit THEN value error FI;
      IF y1 = y2 THEN value error FI;
      dy := y1 - y2;
      dx := dx / dy * y2;
      x1 := x2; y1 := y2; # retain for next iteration #
      x2 -:= dx;
  # WHILE # ABS dx > x error AND ABS dy > y error DO 
      SKIP
    OD;
    stop iteration: 
      XY(x2, y2) EXIT
    value error:
      EMPTY
  FI
);

PROC f = (DBL x)DBL: x UP 3 - LONG 3.1 * x UP 2 + LONG 2.0 * x;

DBL first root, second root, third root;

XYRES first result = find root(f, LENG -1.0, LENG 3.0, EMPTY, EMPTY);
CASE first result IN
  (XY first result): (
    printf(($"1st root found at x = "f(xy root)l$, x OF first result, y OF first result=0));
    first root := x OF first result
  )
  OUT printf($"No first root found"l$); stop
ESAC;

XYRES second result = find root( (DBL x)DBL: f(x) / (x - first root), EMPTY, EMPTY, EMPTY, EMPTY);
CASE second result IN
  (XY second result): (
    printf(($"2nd root found at x = "f(xy root)l$, x OF second result, y OF second result=0));
    second root := x OF second result
  )
  OUT printf($"No second root found"l$); stop
ESAC;

XYRES third result = find root( (DBL x)DBL: f(x) / (x - first root) / ( x - second root ), EMPTY, EMPTY, EMPTY, EMPTY);
CASE third result IN
  (XY third result): (
    printf(($"3rd root found at x = "f(xy root)l$, x OF third result, y OF third result=0));
    third root := x OF third result
  )
  OUT printf($"No third root found"l$); stop
ESAC

Output:

1st root found at x =  9.1557112297752398099031e-1 (Approximately)
2nd root found at x =  2.1844288770224760190097e 0 (Approximately)
3rd root found at x =  0.0000000000000000000000e 0 (Exactly)

Arturo

f: function [n]->
    ((n^3) - 3*n^2) + 2*n


step: 0.01
start: neg 1.0
stop: 3.0
sign: positive? f start
x: start

while [x =< stop][
    value: f x

    if? value = 0 -> 
        print ["root found at" to :string .format:".5f" x]
    else ->
        if sign <> value > 0 -> print ["root found near" to :string .format:".5f" x]
    
    sign: value > 0
    'x + step
]

Output:

root found near 0.00000 
root found near 1.00000 
root found near 2.00000

ATS

#include
"share/atspre_staload.hats"

typedef d = double

fun
findRoots
(
  start: d, stop: d, step: d, f: (d) -> d, nrts: int, A: d
) : void = (
//
if
start < stop
then let
  val A2 = f(start)
  var nrts: int = nrts
  val () =
  if A2 = 0.0
    then (
      nrts := nrts + 1;
      $extfcall(void, "printf", "An exact root is found at %12.9f\n", start)
    ) (* end of [then] *)
  // end of [if]
  val () =
  if A * A2 < 0.0
    then (
      nrts := nrts + 1;
      $extfcall(void, "printf", "An approximate root is found at %12.9f\n", start)
    ) (* end of [then] *)
  // end of [if]
in
  findRoots(start+step, stop, step, f, nrts, A2)
end // end of [then]
else (
  if nrts = 0
    then $extfcall(void, "printf", "There are no roots found!\n")
  // end of [if]
) (* end of [else] *)
//
) (* end of [findRoots] *)

(* ****** ****** *)

implement
main0 () =
findRoots (~1.0, 3.0, 0.001, lam (x) => x*x*x - 3.0*x*x + 2.0*x, 0, 0.0)

AutoHotkey

Poly(x) is a test function of one variable, here we are searching for its roots:

• roots() searches for intervals within given limits, shifted by a given “step”, where our function has different signs at the endpoints.
• Having found such an interval, the root() function searches for a value where our function is 0, within a given tolerance.
• It also sets ErrorLevel to info about the root found.

discussion

MsgBox % roots("poly", -0.99, 2, 0.1, 1.0e-5)
MsgBox % roots("poly", -1, 3, 0.1, 1.0e-5)

roots(f,x1,x2,step,tol) { ; search for roots in intervals of length "step", within tolerance "tol"
   x := x1, y := %f%(x), s := (y>0)-(y<0)
   Loop % ceil((x2-x1)/step) {
      x += step, y := %f%(x), t := (y>0)-(y<0)
      If (s=0 || s!=t)
         res .= root(f, x-step, x, tol) " [" ErrorLevel "]`n"
      s := t
   }
   Sort res, UN ; remove duplicate endpoints
   Return res
}

root(f,x1,x2,d) { ; find x in [x1,x2]: f(x)=0 within tolerance d, by bisection
   If (!y1 := %f%(x1))
      Return x1, ErrorLevel := "Exact"
   If (!y2 := %f%(x2))
      Return x2, ErrorLevel := "Exact"
   If (y1*y2>0)
      Return "", ErrorLevel := "Need different sign ends!"
   Loop {
      x := (x2+x1)/2, y := %f%(x)
      If (y = 0 || x2-x1 < d)
         Return x, ErrorLevel := y ? "Approximate" : "Exact"
      If ((y>0) = (y1>0))
         x1 := x, y1 := y
      Else
         x2 := x, y2 := y
   }
}

poly(x) {
   Return ((x-3)*x+2)*x
}

Axiom

Using a polynomial solver:

expr := x^3-3*x^2+2*x
solve(expr,x)

Output:

  (1)  [x= 2,x= 1,x= 0]
                          Type: List(Equation(Fraction(Polynomial(Integer))))

Using the secant method in the interpreter:

digits(30)
secant(eq: Equation Expression Float, binding: SegmentBinding(Float)):Float ==
  eps := 1.0e-30
  expr := lhs eq - rhs eq
  x := variable binding
  seg := segment binding
  x1 := lo seg
  x2 := hi seg
  fx1 := eval(expr, x=x1)::Float
  abs(fx1)<eps => return x1
  for i in 1..100 repeat
    fx2 := eval(expr, x=x2)::Float
    abs(fx2)<eps => return x2
    (x1, fx1, x2) := (x2, fx2, x2 - fx2 * (x2 - x1) / (fx2 - fx1))
  error "Function not converging."

The example can now be called using:

secant(expr=0,x=-0.5..0.5)

BBC BASIC

      function$ = "x^3-3*x^2+2*x"
      rangemin = -1
      rangemax = 3
      stepsize = 0.001
      accuracy = 1E-8
      PROCroots(function$, rangemin, rangemax, stepsize, accuracy)
      END
      
      DEF PROCroots(func$, min, max, inc, eps)
      LOCAL x, sign%, oldsign%
      oldsign% = 0
      FOR x = min TO max STEP inc
        sign% = SGN(EVAL(func$))
        IF sign% = 0 THEN
          PRINT "Root found at x = "; x
          sign% = -oldsign%
        ELSE IF sign% <> oldsign% AND oldsign% <> 0 THEN
            IF inc < eps THEN
              PRINT "Root found near x = "; x
            ELSE
              PROCroots(func$, x-inc, x+inc/8, inc/8, eps)
            ENDIF
          ENDIF
        ENDIF
        oldsign% = sign%
      NEXT x
      ENDPROC

Output:

Root found near x = 2.29204307E-9
Root found near x = 1
Root found at x = 2

C

Secant Method

#include <math.h>
#include <stdio.h>

double f(double x)
{
    return x*x*x-3.0*x*x +2.0*x;
}

double secant( double xA, double xB, double(*f)(double) )
{
    double e = 1.0e-12;
    double fA, fB;
    double d;
    int i;
    int limit = 50;

    fA=(*f)(xA);
    for (i=0; i<limit; i++) {
        fB=(*f)(xB);
        d = (xB - xA) / (fB - fA) * fB;
        if (fabs(d) < e) 
            break;
        xA = xB;
        fA = fB;
        xB -= d;
    }
    if (i==limit) {
        printf("Function is not converging near (%7.4f,%7.4f).\n", xA,xB);
        return -99.0;
    }
    return xB;
}

int main(int argc, char *argv[])
{
    double step = 1.0e-2;
    double e = 1.0e-12;
    double x = -1.032;		// just so we use secant method
    double xx, value;

    int s = (f(x)> 0.0);

    while (x < 3.0) {
        value = f(x);
        if (fabs(value) < e) {
            printf("Root found at x= %12.9f\n", x);
            s = (f(x+.0001)>0.0);
        }
        else if ((value > 0.0) != s) {
            xx = secant(x-step, x,&f);
            if (xx != -99.0)   // -99 meaning secand method failed
                printf("Root found at x= %12.9f\n", xx);
            else
                printf("Root found near x= %7.4f\n", x);
            s = (f(x+.0001)>0.0);
        }
        x += step;
    }
    return 0;
}

GNU Scientific Library

#include <gsl/gsl_poly.h>
#include <stdio.h>

int main(int argc, char *argv[])
{
    /* 0 + 2x - 3x^2 + 1x^3 */
    double p[] = {0, 2, -3, 1};
    double z[6];
    gsl_poly_complex_workspace *w = gsl_poly_complex_workspace_alloc(4);
    gsl_poly_complex_solve(p, 4, w, z);
    gsl_poly_complex_workspace_free(w);

    for(int i = 0; i < 3; ++i)
        printf("%.12f\n", z[2 * i]);

    return 0;
}

One can also use the GNU Scientific Library to find roots of functions. Compile with

gcc roots.c -lgsl -lcblas -o roots

C#

Translation of: C++

using System;

class Program
{
    public static void Main(string[] args)
    {
        Func<double, double> f = x => { return x * x * x - 3 * x * x + 2 * x; };

        double step = 0.001; // Smaller step values produce more accurate and precise results
        double start = -1;
        double stop = 3;
        double value = f(start);
        int sign = (value > 0) ? 1 : 0;
 
        // Check for root at start
        if (value == 0)
            Console.WriteLine("Root found at {0}", start);

        for (var x = start + step; x <= stop; x += step)
        {
            value = f(x);
 
            if (((value > 0) ? 1 : 0) != sign)
                // We passed a root
                Console.WriteLine("Root found near {0}", x);
            else if (value == 0)
                // We hit a root
                Console.WriteLine("Root found at {0}", x);
 
            // Update our sign
            sign = (value > 0) ? 1 : 0;
        }
    }
}

Translation of: Java

using System;

class Program
{
    private static int Sign(double x)
    {
        return x < 0.0 ? -1 : x > 0.0 ? 1 : 0;
    }

    public static void PrintRoots(Func<double, double> f, double lowerBound,
        double upperBound, double step)
    {
        double x = lowerBound, ox = x;
        double y = f(x), oy = y;
        int s = Sign(y), os = s;

        for (; x <= upperBound; x += step)
        {
            s = Sign(y = f(x));
            if (s == 0)
            {
                Console.WriteLine(x);
            }
            else if (s != os)
            {
                var dx = x - ox;
                var dy = y - oy;
                var cx = x - dx * (y / dy);
                Console.WriteLine("~{0}", cx);
            }

            ox = x;
            oy = y;
            os = s;
        }
    }

    public static void Main(string[] args)
    {
        Func<double, double> f = x => { return x * x * x - 3 * x * x + 2 * x; };
        PrintRoots(f, -1.0, 4, 0.002);
    }
}

Brent's Method

Translation of: C++

using System;

class Program
{
    public static void Main(string[] args)
    {
        Func<double, double> f = x => { return x * x * x - 3 * x * x + 2 * x; };
        double root = BrentsFun(f, lower: -1.0, upper: 4, tol: 0.002, maxIter: 100);
    }

    private static void Swap<T>(ref T a, ref T b)
    {
        var tmp = a;
        a = b;
        b = tmp;
    }

    public static double BrentsFun(Func<double, double> f, double lower, double upper, double tol, uint maxIter)
    {
        double a = lower;
        double b = upper;
        double fa = f(a); // calculated now to save function calls
        double fb = f(b); // calculated now to save function calls
        double fs;

        if (!(fa * fb < 0))
            throw new ArgumentException("Signs of f(lower_bound) and f(upper_bound) must be opposites");

        if (Math.Abs(fa) < Math.Abs(b)) // if magnitude of f(lower_bound) is less than magnitude of f(upper_bound)
        {
            Swap(ref a, ref b);
            Swap(ref fa, ref fb);
        }

        double c = a;      // c now equals the largest magnitude of the lower and upper bounds
        double fc = fa;    // precompute function evalutation for point c by assigning it the same value as fa
        bool mflag = true; // boolean flag used to evaluate if statement later on
        double s = 0;      // Our Root that will be returned
        double d = 0;      // Only used if mflag is unset (mflag == false)

        for (uint iter = 1; iter < maxIter; ++iter)
        {
            // stop if converged on root or error is less than tolerance
            if (Math.Abs(b - a) < tol)
            {
                Console.WriteLine("After {0} iterations the root is: {1}", iter, s);
                return s;
            } // end if

            if (fa != fc && fb != fc)
            {
                // use inverse quadratic interopolation
                s =   (a * fb * fc / ((fa - fb) * (fa - fc)))
                    + (b * fa * fc / ((fb - fa) * (fb - fc)))
                    + (c * fa * fb / ((fc - fa) * (fc - fb)));
            }
            else
            {
                // secant method
                s = b - fb * (b - a) / (fb - fa);
            }

            // checks to see whether we can use the faster converging quadratic && secant methods or if we need to use bisection
            if (    ( (s < (3 * a + b) * 0.25) || (s > b)) ||
                    (  mflag && (Math.Abs(s - b) >= (Math.Abs(b - c) * 0.5)) ) ||
                    ( !mflag && (Math.Abs(s - b) >= (Math.Abs(c - d) * 0.5)) ) ||
                    (  mflag && (Math.Abs(b - c) < tol) ) ||
                    ( !mflag && (Math.Abs(c - d) < tol))    )
            {
                // bisection method
                s = (a + b) * 0.5;

                mflag = true;
            }
            else
            {
                mflag = false;
            }

            fs = f(s);// calculate fs
            d = c;    // first time d is being used (wasnt used on first iteration because mflag was set)
            c = b;    // set c equal to upper bound
            fc = fb;  // set f(c) = f(b)

            if (fa * fs < 0) // fa and fs have opposite signs
            {
                b = s;
                fb = fs; // set f(b) = f(s)
            }
            else
            {
                a = s;
                fa = fs; // set f(a) = f(s)
            }

            if (Math.Abs(fa) < Math.Abs(fb)) // if magnitude of fa is less than magnitude of fb
            {
                Swap(ref a, ref b);          // swap a and b
                Swap(ref fa, ref fb); // make sure f(a) and f(b) are correct after swap
            }
        } // end for

        throw new AggregateException("The solution does not converge or iterations are not sufficient");
    }
    // end brents_fun
}

C++

#include <iostream>

double f(double x)
{
	return (x*x*x - 3*x*x + 2*x);
}

int main()
{
	double step = 0.001; // Smaller step values produce more accurate and precise results
	double start = -1;
	double stop = 3;
	double value = f(start);
	double sign = (value > 0);
	
	// Check for root at start
	if ( 0 == value )
		std::cout << "Root found at " << start << std::endl;

	for(	double x = start + step;
			x <= stop;
			x += step )
	{
		value = f(x);
		
		if ( ( value > 0 ) != sign )
			// We passed a root
			std::cout << "Root found near " << x << std::endl;
		else if ( 0 == value )
			// We hit a root
			std::cout << "Root found at " << x << std::endl;
		
		// Update our sign
		sign = ( value > 0 );
	}
}

Brent's Method

Brent's Method uses a combination of the bisection method, inverse quadratic interpolation, and the secant method to find roots. It has a guaranteed run time equal to that of the bisection method (which always converges in a known number of steps (log2[(upper_bound-lower_bound)/tolerance] steps to be precise ) unlike the other methods), but the algorithm uses the much faster inverse quadratic interpolation and secant method whenever possible. The algorithm is robust and commonly used in libraries with a roots() function built in.

The algorithm is coded as a function that returns a double value for the root. The function takes an input that requires the function being evaluated, the lower and upper bounds, the tolerance one is looking for before converging (i recommend 0.0001) and the maximum number of iterations before giving up on finding the root (the root will always be found if the root is bracketed and a sufficient number of iterations is allowed).

The implementation is taken from the pseudo code on the wikipedia page for Brent's Method found here: https://en.wikipedia.org/wiki/Brent%27s_method.

#include <iostream>
#include <cmath>
#include <algorithm>
#include <functional>

double brents_fun(std::function<double (double)> f, double lower, double upper, double tol, unsigned int max_iter)
{
	double a = lower;
	double b = upper;
	double fa = f(a);	// calculated now to save function calls
	double fb = f(b);	// calculated now to save function calls
	double fs = 0;		// initialize 

	if (!(fa * fb < 0))
	{
		std::cout << "Signs of f(lower_bound) and f(upper_bound) must be opposites" << std::endl; // throws exception if root isn't bracketed
		return -11;
	}

	if (std::abs(fa) < std::abs(b))	// if magnitude of f(lower_bound) is less than magnitude of f(upper_bound)
	{
		std::swap(a,b);
		std::swap(fa,fb);
	}

	double c = a;			// c now equals the largest magnitude of the lower and upper bounds
	double fc = fa;			// precompute function evalutation for point c by assigning it the same value as fa
	bool mflag = true;		// boolean flag used to evaluate if statement later on
	double s = 0;			// Our Root that will be returned
	double d = 0;			// Only used if mflag is unset (mflag == false)

	for (unsigned int iter = 1; iter < max_iter; ++iter)
	{
		// stop if converged on root or error is less than tolerance
		if (std::abs(b-a) < tol)
		{
			std::cout << "After " << iter << " iterations the root is: " << s << std::endl;
			return s;
		} // end if
		
		if (fa != fc && fb != fc)
		{
			// use inverse quadratic interopolation
			s =	  ( a * fb * fc / ((fa - fb) * (fa - fc)) )
				+ ( b * fa * fc / ((fb - fa) * (fb - fc)) )
				+ ( c * fa * fb / ((fc - fa) * (fc - fb)) );
		}
		else
		{
			// secant method
			s = b - fb * (b - a) / (fb - fa);
		}

			// checks to see whether we can use the faster converging quadratic && secant methods or if we need to use bisection
		if (	( (s < (3 * a + b) * 0.25) || (s > b) ) ||
				( mflag && (std::abs(s-b) >= (std::abs(b-c) * 0.5)) ) ||
				( !mflag && (std::abs(s-b) >= (std::abs(c-d) * 0.5)) ) ||
				( mflag && (std::abs(b-c) < tol) ) ||
				( !mflag && (std::abs(c-d) < tol))	)
		{
			// bisection method
			s = (a+b)*0.5;

			mflag = true;
		}
		else
		{
			mflag = false;
		}

		fs = f(s);	// calculate fs
		d = c;		// first time d is being used (wasnt used on first iteration because mflag was set)
		c = b;		// set c equal to upper bound
		fc = fb;	// set f(c) = f(b)

		if ( fa * fs < 0)	// fa and fs have opposite signs
		{
			b = s;
			fb = fs;	// set f(b) = f(s)
		}
		else
		{
			a = s;
			fa = fs;	// set f(a) = f(s)
		}

		if (std::abs(fa) < std::abs(fb)) // if magnitude of fa is less than magnitude of fb
		{
			std::swap(a,b);		// swap a and b
			std::swap(fa,fb);	// make sure f(a) and f(b) are correct after swap
		}

	} // end for

	std::cout<< "The solution does not converge or iterations are not sufficient" << std::endl;

} // end brents_fun

Clojure

Translation of: Haskell

(defn findRoots [f start stop step eps]
      (filter #(-> (f %) Math/abs (< eps)) (range start stop step)))

> (findRoots #(+ (* % % %) (* -3 % %) (* 2 %)) -1.0 3.0 0.0001 0.00000001)
(-9.381755897326649E-14 0.9999999999998124 1.9999999999997022)

CoffeeScript

Translation of: Python

print_roots = (f, begin, end, step) ->  
  # Print approximate roots of f between x=begin and x=end,
  # using sign changes as an indicator that a root has been
  # encountered. 
  x = begin
  y = f(x)
  last_y = y
  
  cross_x_axis = ->
    (last_y < 0 and y > 0) or (last_y > 0 and y < 0)
  
  console.log '-----'
  while x <= end
    y = f(x)
    if y == 0
      console.log "Root found at", x
    else if cross_x_axis()
      console.log "Root found near", x
    x += step
    last_y = y

do ->
  # Smaller steps produce more accurate/precise results in general,
  # but for many functions we'll never get exact roots, either due
  # to imperfect binary representation or irrational roots.
  step = 1 / 256

  f1 = (x) -> x*x*x - 3*x*x + 2*x
  print_roots f1, -1, 5, step
  f2 = (x) -> x*x - 4*x + 3
  print_roots f2, -1, 5, step
  f3 = (x) -> x - 1.5
  print_roots f3, 0, 4, step
  f4 = (x) -> x*x - 2
  print_roots f4, -2, 2, step

output

> coffee roots.coffee 
-----
Root found at 0
Root found at 1
Root found at 2
-----
Root found at 1
Root found at 3
-----
Root found at 1.5
-----
Root found near -1.4140625
Root found near 1.41796875

Common Lisp

Translation of: Perl

find-roots prints roots (and values near roots) and returns a list of root designators, each of which is either a number n, in which case (zerop (funcall function n)) is true, or a cons whose car and cdr are such that the sign of function at car and cdr changes.

(defun find-roots (function start end &optional (step 0.0001))
  (let* ((roots '())
         (value (funcall function start))
         (plusp (plusp value)))
    (when (zerop value)
      (format t "~&Root found at ~W." start))
    (do ((x (+ start step) (+ x step)))
        ((> x end) (nreverse roots))
      (setf value (funcall function x))
      (cond
       ((zerop value)
        (format t "~&Root found at ~w." x)
        (push x roots))
       ((not (eql plusp (plusp value)))
        (format t "~&Root found near ~w." x)
        (push (cons (- x step) x) roots)))
      (setf plusp (plusp value)))))

> (find-roots #'(lambda (x) (+ (* x x x) (* -3 x x) (* 2 x))) -1 3)
Root found near 5.3588345E-5.
Root found near 1.0000072.
Root found near 2.000073.
((-4.6411653E-5 . 5.3588345E-5)
 (0.99990714 . 1.0000072)
 (1.9999729 . 2.000073))

D

import std.stdio, std.math, std.algorithm;

bool nearZero(T)(in T a, in T b = T.epsilon * 4) pure nothrow {
    return abs(a) <= b;
}

T[] findRoot(T)(immutable T function(in T) pure nothrow fi,
                in T start, in T end, in T step=T(0.001L),
                T tolerance = T(1e-4L)) {
    if (step.nearZero)
        writefln("WARNING: step size may be too small.");

    /// Search root by simple bisection.
    T searchRoot(T a, T b) pure nothrow {
        T root;
        int limit = 49;
        T gap = b - a;

        while (!nearZero(gap) && limit--) {
            if (fi(a).nearZero)
                return a;
            if (fi(b).nearZero)
                return b;
            root = (b + a) / 2.0L;
            if (fi(root).nearZero)
                return root;
            ((fi(a) * fi(root) < 0) ? b : a) = root;
            gap = b - a;
        }

        return root;
    }

    immutable dir = T(end > start ? 1.0 : -1.0);
    immutable step2 = (end > start) ? abs(step) : -abs(step);
    T[T] result;
    for (T x = start; (x * dir) <= (end * dir); x += step2)
        if (fi(x) * fi(x + step2) <= 0) {
            immutable T r = searchRoot(x, x + step2);
            result[r] = fi(r);
        }

    return result.keys.sort().release;
}

void report(T)(in T[] r, immutable T function(in T) pure f,
               in T tolerance = T(1e-4L)) {
    if (r.length) {
        writefln("Root found (tolerance = %1.4g):", tolerance);

        foreach (const x; r) {
            immutable T y = f(x);

            if (nearZero(y))
                writefln("... EXACTLY at %+1.20f, f(x) = %+1.4g",x,y);
            else if (nearZero(y, tolerance))
                writefln(".... MAY-BE at %+1.20f, f(x) = %+1.4g",x,y);
            else
                writefln("Verify needed, f(%1.4g) = " ~
                         "%1.4g > tolerance in magnitude", x, y);
        }
    } else
        writefln("No root found.");
}

void main() {
    static real f(in real x) pure nothrow {
        return x ^^ 3 - (3 * x ^^ 2) + 2 * x;
    }

    findRoot(&f, -1.0L, 3.0L, 0.001L).report(&f);
}

Output:

Root found (tolerance = 0.0001):
.... MAY-BE at -0.00000000000000000080, f(x) = -1.603e-18
... EXACTLY at +1.00000000000000000020, f(x) = -2.168e-19
.... MAY-BE at +1.99999999999999999950, f(x) = -8.674e-19

NB: smallest increment for real type in D is real.epsilon = 1.0842e-19.

Dart

Translation of: Scala

double fn(double x) => x * x * x - 3 * x * x + 2 * x;

findRoots(Function(double) f, double start, double stop, double step, double epsilon) sync* {
  for (double x = start; x < stop; x = x + step) {
    if (fn(x).abs() < epsilon) yield x;
  }
}

main() {
  // Vector(-9.381755897326649E-14, 0.9999999999998124, 1.9999999999997022)
  print(findRoots(fn, -1.0, 3.0, 0.0001, 0.000000001));
}

Delphi

See Pascal.

DWScript

Translation of: C

type TFunc = function (x : Float) : Float;

function f(x : Float) : Float;
begin
   Result := x*x*x-3.0*x*x +2.0*x;
end;

const e = 1.0e-12;

function Secant(xA, xB : Float; f : TFunc) : Float;
const
   limit = 50;
var
   fA, fB : Float;
   d : Float;
   i : Integer;
begin
   fA := f(xA);
   for i := 0 to limit do begin
      fB := f(xB);
      d := (xB-xA)/(fB-fA)*fB;
      if Abs(d) < e then
         Exit(xB);
      xA := xB;
      fA := fB;
      xB -= d;
   end;
   PrintLn(Format('Function is not converging near (%7.4f,%7.4f).', [xA, xB]));
   Result := -99.0;
end;

const fstep = 1.0e-2;

var x := -1.032;		// just so we use secant method
var xx, value : Float;
var s := f(x)>0.0;

while (x < 3.0) do begin
   value := f(x);
   if Abs(value)<e then begin
      PrintLn(Format("Root found at x= %12.9f", [x]));
      s := (f(x+0.0001)>0.0);
   end else if (value>0.0) <> s then begin
      xx := Secant(x-fstep, x, f);
      if xx <> -99.0 then   // -99 meaning secand method failed
         PrintLn(Format('Root found at x = %12.9f', [xx]))
      else PrintLn(Format('Root found near x= %7.4f', [xx]));
      s := (f(x+0.0001)>0.0);
   end;
   x += fstep;
end;

EasyLang

func f x .
   return x * x * x - 3 * x * x + 2 * x
.
numfmt 6 0
proc findroot start stop step . .
   x = start
   while x <= stop
      val = f x
      if val = 0
         print x & " (exact)"
      elif sign val <> sign0 and sign0 <> 0
         print x & " (err = " & step & ")"
      .
      sign0 = sign val
      x += step
   .
.
proc drawfunc start stop . .
   linewidth 0.3
   drawgrid
   x = start
   while x <= stop
      line x * 10 + 50 f x * 10 + 50
      x += 0.1
   .
.
drawfunc -1 3
findroot -1 3 pow 2 -20
print ""
findroot -1 3 1e-6

EchoLisp

We use the 'math' library, and define f(x) as the polynomial : x³ -3x² +2x

(lib 'math.lib)
Lib: math.lib loaded.
(define fp ' ( 0 2 -3 1))
(poly->string 'x fp) → x^3 -3x^2 +2x 
(poly->html 'x fp) → x<sup>3</sup> -3x<sup>2</sup> +2x
(define (f x) (poly x fp))
(math-precision 1.e-6) → 0.000001

(root f -1000 1000) → 2.0000000133245677 ;; 2
(root f -1000 (- 2 epsilon)) → 1.385559938161431e-7 ;; 0
(root f epsilon (- 2 epsilon)) → 1.0000000002190812 ;; 1

Elixir

Translation of: Ruby

defmodule RC do
  def find_roots(f, range, step \\ 0.001) do
    first .. last = range
    max = last + step / 2
    Stream.iterate(first, &(&1 + step))
    |> Stream.take_while(&(&1 < max))
    |> Enum.reduce(sign(first), fn x,sn ->
         value = f.(x)
         cond do
           abs(value) < step / 100 ->
             IO.puts "Root found at #{x}"
             0
           sign(value) == -sn ->
             IO.puts "Root found between #{x-step} and #{x}"
             -sn
           true -> sign(value)
         end
       end)
  end
  
  defp sign(x) when x>0, do: 1
  defp sign(x) when x<0, do: -1
  defp sign(0)         , do: 0
end

f = fn x -> x*x*x - 3*x*x + 2*x end
RC.find_roots(f, -1..3)

Output:

Root found at 8.81239525796218e-16
Root found at 1.0000000000000016
Root found at 1.9999999999998914

Erlang

% Implemented by Arjun Sunel
-module(roots).
-export([main/0]).
main() ->
	F = fun(X)->X*X*X - 3*X*X + 2*X end,
	Step = 0.001,	 % Using smaller steps will provide more accurate results
	Start = -1,
	Stop = 3,
	Sign = F(Start) > 0,
	X = Start,
	while(X, Step, Start, Stop, Sign,F).

while(X, Step, Start, Stop, Sign,F) ->	
	Value = F(X),
	if 
		Value == 0  ->		% We hit a root
        	io:format("Root found at ~p~n",[X]),
        	while(X+Step, Step, Start, Stop,  Value > 0,F);

		(Value < 0) == Sign ->	% We passed a root
		io:format("Root found near ~p~n",[X]),
		while(X+Step , Step, Start, Stop,  Value > 0,F);
		 
		 X > Stop ->
		 	io:format("") ;
		true ->
        	while(X+Step, Step, Start, Stop,  Value > 0,F)
	end.

Output:

Root found near 8.81239525796218e-16
Root found near 1.0000000000000016
Root found near 2.0009999999998915
ok

ERRE

PROGRAM ROOTS_FUNCTION

!VAR E,X,STP,VALUE,S%,I%,LIMIT%,X1,X2,D

FUNCTION F(X)
    F=X*X*X-3*X*X+2*X
END FUNCTION

BEGIN
  X=-1
  STP=1.0E-6
  E=1.0E-9
  S%=(F(X)>0)

  PRINT("VERSION 1: SIMPLY STEPPING X")
  WHILE X<3.0 DO
    VALUE=F(X)
    IF ABS(VALUE)<E THEN
        PRINT("ROOT FOUND AT X =";X)
        S%=NOT S%
      ELSE
        IF ((VALUE>0)<>S%) THEN
          PRINT("ROOT FOUND AT X =";X)
          S%=NOT S%
        END IF
    END IF
    X=X+STP
  END WHILE

  PRINT
  PRINT("VERSION 2: SECANT METHOD")
  X1=-1.0
  X2=3.0
  E=1.0E-15
  I%=1
  LIMIT%=300
  LOOP
    IF I%>LIMIT% THEN
       PRINT("ERROR: FUNCTION NOT CONVERGING")
       EXIT
    END IF
    D=(X2-X1)/(F(X2)-F(X1))*F(X2)
    IF ABS(D)<E THEN
      IF D=0 THEN
         PRINT("EXACT ";)
       ELSE
         PRINT("APPROXIMATE ";)
      END IF
      PRINT("ROOT FOUND AT X =";X2)
      EXIT
    END IF
    X1=X2
    X2=X2-D
    I%=I%+1
  END LOOP
END PROGRAM

Note: Outputs are calculated in single precision.

Output:

VERSION 1: SIMPLY STEPPING X
ROOT FOUND AT X = 8.866517E-07
ROOT FOUND AT X = 1.000001
ROOT FOUND AT X = 2

VERSION 2: SECANT METHOD
EXACT ROOT FOUND AT X = 1

Fortran

Works with: Fortran version 90 and later

PROGRAM ROOTS_OF_A_FUNCTION

  IMPLICIT NONE

  INTEGER, PARAMETER :: dp = SELECTED_REAL_KIND(15)
  REAL(dp) :: f, e, x, step, value
  LOGICAL :: s 
   
  f(x) = x*x*x - 3.0_dp*x*x + 2.0_dp*x
   
  x = -1.0_dp ; step = 1.0e-6_dp ; e = 1.0e-9_dp
 
  s = (f(x) > 0)
  DO WHILE (x < 3.0)
    value = f(x)
    IF(ABS(value) < e) THEN
      WRITE(*,"(A,F12.9)") "Root found at x =", x
      s = .NOT. s
    ELSE IF ((value > 0) .NEQV. s) THEN
      WRITE(*,"(A,F12.9)") "Root found near x = ", x
      s = .NOT. s
    END IF
    x = x + step
  END DO
 
END PROGRAM ROOTS_OF_A_FUNCTION

The following approach uses the Secant Method to numerically find one root. Which root is found will depend on the start values x1 and x2 and if these are far from a root this method may not converge.

INTEGER, PARAMETER :: dp = SELECTED_REAL_KIND(15)
INTEGER :: i=1, limit=100
REAL(dp) :: d, e, f, x, x1, x2
 
f(x) = x*x*x - 3.0_dp*x*x + 2.0_dp*x
 
x1 = -1.0_dp ; x2 = 3.0_dp ; e = 1.0e-15_dp
 
DO 
  IF (i > limit) THEN
    WRITE(*,*) "Function not converging"
    EXIT
  END IF
  d = (x2 - x1) / (f(x2) - f(x1)) * f(x2)
  IF (ABS(d) < e) THEN
    WRITE(*,"(A,F18.15)") "Root found at x = ", x2
    EXIT    
  END IF
  x1 = x2
  x2 = x2 - d
  i = i + 1
END DO

FreeBASIC

Simple bisection method.

#Include "crt.bi"
const iterations=20000000

sub bisect( f1 as function(as double) as double,min as double,max as double,byref O as double,a() as double)
    dim as double last,st=(max-min)/iterations,v
    for n as double=min to max step st
        v=f1(n)
        if sgn(v)<>sgn(last) then
            redim preserve a(1 to ubound(a)+1)
            a(ubound(a))=n
            O=n+st:exit sub
        end if
        last=v
    next
end sub

function roots(f1 as function(as double) as double,min as double,max as double, a() as double) as long
    redim a(0)
    dim as double last,O,st=(max-min)/iterations,v
    for n as double=min to max step st
        v=f1(n)
        if sgn(v)<>sgn(last) and n>min then bisect(f1,n-st,n,O,a()):n=O
        last=v
    next
    return ubound(a)
end function

Function CRound(Byval x As Double,Byval precision As Integer=30) As String
    If precision>30 Then precision=30
    Dim As zstring * 40 z:Var s="%." &str(Abs(precision)) &"f"
    sprintf(z,s,x)
    If Val(z) Then Return Rtrim(Rtrim(z,"0"),".")Else Return "0"
End Function

function defn(x as double) as double
    return x^3-3*x^2+2*x
end function

redim as double r()

print
if roots(@defn,-20,20,r()) then
    print "in range -20 to 20"
    print "All roots approximate"
    print "number","root to 6 dec places","function value at root"
    for n as long=1 to ubound(r)
        print n,CRound(r(n),6),,defn(r(n))
    next n
end if
sleep

Output:

in range -20 to 20
All roots approximate
number        root to 6 dec places        function value at root
 1            0                           -2.929925652002424e-009
 2            1                            1.477781779325033e-009
 3            2                           -2.897852187377925e-009

Go

Secant method. No error checking.

package main

import (
	"fmt"
	"math"
)

func main() {
	example := func(x float64) float64 { return x*x*x - 3*x*x + 2*x }
	findroots(example, -.5, 2.6, 1)
}

func findroots(f func(float64) float64, lower, upper, step float64) {
	for x0, x1 := lower, lower+step; x0 < upper; x0, x1 = x1, x1+step {
		x1 = math.Min(x1, upper)
		r, status := secant(f, x0, x1)
		if status != "" && r >= x0 && r < x1 {
			fmt.Printf("  %6.3f %s\n", r, status)
		}
	}
}

func secant(f func(float64) float64, x0, x1 float64) (float64, string) {
	var f0 float64
	f1 := f(x0)
	for i := 0; i < 100; i++ {
		f0, f1 = f1, f(x1)
		switch {
		case f1 == 0:
			return x1, "exact"
		case math.Abs(x1-x0) < 1e-6:
			return x1, "approximate"
		}
		x0, x1 = x1, x1-f1*(x1-x0)/(f1-f0)
	}
	return 0, ""
}

Output:

   0.000 approximate
   1.000 exact
   2.000 approximate

Haskell

f x = x^3-3*x^2+2*x

findRoots start stop step eps =
  [x | x <- [start, start+step .. stop], abs (f x) < eps]

Executed in GHCi:

*Main> findRoots (-1.0) 3.0 0.0001 0.000000001
[-9.381755897326649e-14,0.9999999999998124,1.9999999999997022]

Or using package hmatrix from HackageDB.

import Numeric.GSL.Polynomials
import Data.Complex

*Main> mapM_ print $ polySolve [0,2,-3,1]
(-5.421010862427522e-20) :+ 0.0
2.000000000000001 :+ 0.0
0.9999999999999996 :+ 0.0

No complex roots, so:

*Main> mapM_ (print.realPart) $ polySolve [0,2,-3,1]
-5.421010862427522e-20
2.000000000000001
0.9999999999999996

It is possible to solve the problem directly and elegantly using robust bisection method and Alternative type class.

import Control.Applicative

data Root a = Exact a | Approximate a deriving (Show, Eq)

-- looks for roots on an interval
bisection :: (Alternative f, Floating a, Ord a) =>
             (a -> a) -> a -> a -> f (Root a)
bisection f a b | f a * f b > 0 = empty
                | f a == 0      = pure (Exact a)
                | f b == 0      = pure (Exact b)
                | smallInterval = pure (Approximate c)
                | otherwise     = bisection f a c <|> bisection f c b
  where c = (a + b) / 2
        smallInterval = abs (a-b) < 1e-15 || abs ((a-b)/c) < 1e-15

-- looks for roots on a grid
findRoots :: (Alternative f, Floating a, Ord a) =>
             (a -> a) -> [a] -> а (Root a)
findRoots f []       = empty
findRoots f [x]      = if f x == 0 then pure (Exact x) else empty
findRoots f (a:b:xs) = bisection f a b <|> findRoots f (b:xs)

It is possible to use these functions with different Alternative functors: IO, Maybe or List:

λ> bisection (\x -> x*x-2) 1 2
Approximate 1.414213562373094
λ> bisection (\x -> x-1) 1 2
Exact 1.0
λ> bisection (\x -> x*x-2) 2 3 :: Maybe (Root Double)
Nothing
λ> findRoots (\x ->  x^3 - 3*x^2 + 2*x) [-3..3] :: Maybe (Root Double)
Just (Exact 0.0)
λ> findRoots (\x ->  x^3 - 3*x^2 + 2*x) [-3..3] :: [Root Double]
[Exact 0.0,Exact 0.0,Exact 1.0,Exact 2.0]

To get rid of repeated roots use `Data.List.nub`

λ> Data.List.nub $ findRoots (\x ->  x^3 - 3*x^2 + 2*x) [-3..3]
[Exact 0.0,Exact 1.0,Exact 2.0]
λ> Data.List.nub $ findRoots (\x ->  x^3 - 3*x^2 + x) [-3..3]
[Exact 0.0,Approximate 2.6180339887498967]

HicEst

HicEst's SOLVE function employs the Levenberg-Marquardt method:

OPEN(FIle='test.txt')

1 DLG(NameEdit=x0, DNum=3)

x = x0
chi2 = SOLVE(NUL=x^3 - 3*x^2 + 2*x, Unknown=x, I=iterations, NumDiff=1E-15)
EDIT(Text='approximate exact ', Word=(chi2 == 0), Parse=solution)

WRITE(FIle='test.txt', LENgth=6, Name) x0, x, solution, chi2, iterations
GOTO 1

x0=0.5; x=1; solution=exact; chi2=79E-32 iterations=65;
x0=0.4; x=2E-162 solution=exact; chi2=0; iterations=1E4;
x0=0.45; x=1; solution=exact; chi2=79E-32 iterations=67;
x0=0.42; x=2E-162 solution=exact; chi2=0; iterations=1E4;
x0=1.5; x=1.5; solution=approximate; chi2=0.1406; iterations=14:
x0=1.54; x=1; solution=exact; chi2=44E-32 iterations=63;
x0=1.55; x=2; solution=exact; chi2=79E-32 iterations=55;
x0=1E10; x=2; solution=exact; chi2=18E-31 iterations=511;
x0=-1E10; x=0; solution=exact; chi2=0; iterations=1E4;

Icon and Unicon

Translation of: Java

Works in both languages:

procedure main()
    showRoots(f, -1.0, 4, 0.002)
end

procedure f(x)
    return x^3 - 3*x^2 + 2*x
end

procedure showRoots(f, lb, ub, step)
    ox := x := lb
    oy := f(x)
    os := sign(oy)
    while x <= ub do {
        if (s := sign(y := f(x))) = 0 then write(x)
        else if s ~= os then {
            dx := x-ox
            dy := y-oy
            cx := x-dx*(y/dy)
            write("~",cx)
            }
        (ox := x, oy := y, os := s)
        x +:= step
        }
end

procedure sign(x)
    return (x<0, -1) | (x>0, 1) | 0
end

Output:

->roots
~2.616794878713638e-18
~1.0
~2.0
->

J

J has builtin a root-finding operator, p., whose input is the coeffiecients of the polynomial (where the exponent of the indeterminate variable matches the index of the coefficient: 0 1 2 would be 0 + x + (2 times x squared)). Hence:

   1{::p.  0 2 _3 1    
2 1 0

We can determine whether the roots are exact or approximate by evaluating the polynomial at the candidate roots, and testing for zero:

   (0=]p.1{::p.) 0 2 _3 1 
1 1 1

As you can see, p. is also the operator which evaluates polynomials. This is not a coincidence.

That said, we could also implement the technique used by most others here. Specifically: we can implement the function as a black box and check every 1 millionth of a unit between minus one and three, and we can test that result for exactness.

   blackbox=: 0 2 _3 1&p.
   (#~ (=<./)@:|@blackbox) i.&.(1e6&*)&.(1&+) 3
0 1 2
   0=blackbox 0 1 2
1 1 1

Here, we see that each of the results (0, 1 and 2) are as accurate as we expect our computer arithmetic to be. (The = returns 1 where paired values are equal and 0 where they are not equal).

Java

public class Roots {
    public interface Function {
	public double f(double x);
    }

    private static int sign(double x) {
	return (x < 0.0) ? -1 : (x > 0.0) ? 1 : 0;
    }

    public static void printRoots(Function f, double lowerBound,
				  double upperBound, double step) {
	double x = lowerBound, ox = x;
	double y = f.f(x), oy = y;
	int s = sign(y), os = s;

	for (; x <= upperBound ; x += step) {
	    s = sign(y = f.f(x));
	    if (s == 0) {
		System.out.println(x);
	    } else if (s != os) {
		double dx = x - ox;
		double dy = y - oy;
		double cx = x - dx * (y / dy);
		System.out.println("~" + cx);
	    }
	    ox = x; oy = y; os = s;
	}
    }

    public static void main(String[] args) {
	Function poly = new Function () {
	    public double f(double x) {
		return x*x*x - 3*x*x + 2*x;
	    }
	};
	printRoots(poly, -1.0, 4, 0.002);
    }
}

Produces this output:

~2.616794878713638E-18
~1.0000000000000002
~2.000000000000001

JavaScript

Translation of: Java

Works with: SpiderMonkey version 22

Works with: Firefox version 22

// This function notation is sorta new, but useful here
// Part of the EcmaScript 6 Draft
// developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Functions_and_function_scope
var poly = (x => x*x*x - 3*x*x + 2*x);

function sign(x) {
	return (x < 0.0) ? -1 : (x > 0.0) ? 1 : 0;
}

function printRoots(f, lowerBound, upperBound, step) {
	var  x = lowerBound, ox = x,
		 y = f(x), oy = y,
		 s = sign(y), os = s;

	for (; x <= upperBound ; x += step) {
	    s = sign(y = f(x));
	    if (s == 0) {
			console.log(x);
	    }
	    else if (s != os) {
			var dx = x - ox;
			var dy = y - oy;
			var cx = x - dx * (y / dy);
			console.log("~" + cx);
	    }
	    ox = x; oy = y; os = s;
	}
}

printRoots(poly, -1.0, 4, 0.002);

jq

printRoots(f; lower; upper; step) finds approximations to the roots of an arbitrary continuous real-valued function, f, in the range [lower, upper], assuming step is small enough.

The algorithm is similar to that used for example in the Javascript section on this page, except that a bug has been removed at the point when the previous and current signs are compared.

The function, f, may be an expression (as in the example below) or a defined filter.

printRoots/3 emits an array of results, each of which is either a number (representing an exact root within the limits of machine arithmetic) or a string consisting of "~" followed by an approximation to the root.

def sign:
  if . < 0 then -1 elif . > 0 then 1 else 0 end;

def printRoots(f; lowerBound; upperBound; step):
 lowerBound as $x
 | ($x|f) as $y
 | ($y|sign) as $s
 | reduce range($x; upperBound+step; step) as $x
   # state: [ox, oy, os, roots]
   ( [$x, $y, $s, [] ];
     .[0] as $ox | .[1] as $oy | .[2] as $os
     | ($x|f) as $y
     | ($y | sign) as $s
     | if $s == 0 then  [$x, $y, $s, (.[3] + [$x] )]
       elif $s != $os and $os != 0 then
	  ($x - $ox) as $dx
          | ($y - $oy) as $dy
    	  | ($x - ($dx *  $y / $dy)) as $cx       # by geometry
          | [$x, $y, $s, (.[3] + [ "~\($cx)" ])]  # an approximation
       else [$x, $y, $s, .[3] ] 
       end )
  | .[3] ;

We present two examples, one where step is a power of 1/2, and one where it is not:

Output:

printRoots( .*.*. - 3*.*. + 2*.; -1.0; 4; 1/256)

[
  0,
  1,
  2
]

printRoots( .*.*. - 3*.*. + 2*.; -1.0; 4; .001)
[
  "~1.320318770141425e-18",
  "~1.0000000000000002",
  "~1.9999999999999993"
]

Julia

Assuming that one has the Roots package installed:

using Roots

println(find_zero(x -> x^3 - 3x^2 + 2x, (-100, 100)))

Output:

[0.0,1.0,2.0]

Without the Roots package, Newton's method may be defined in this manner:

function newton(f, fp, x::Float64,tol=1e-14::Float64,maxsteps=100::Int64)
         ##f: the function of x
         ##fp: the derivative of f
 
	 local xnew, xold = x, Inf
	 local fn, fo = f(xnew), Inf
	 local counter = 1
 
	 while (counter < maxsteps) && (abs(xnew - xold) > tol) && ( abs(fn - fo) > tol )
	   x = xnew - f(xnew)/fp(xnew) ## update x
	   xnew, xold = x, xnew
           fn, fo = f(xnew), fn
	   counter += 1
	 end
 
	 if counter >= maxsteps
	    error("Did not converge in ", string(maxsteps), " steps")
         else
	   xnew, counter
         end 
end

Finding the roots of f(x) = x3 - 3x2 + 2x:

f(x) = x^3 - 3*x^2 + 2*x
fp(x) = 3*x^2-6*x+2

x_s, count = newton(f,fp,1.00)

Output:

(1.0,2)

Kotlin

Translation of: C

// version 1.1.2

typealias DoubleToDouble = (Double) -> Double

fun f(x: Double) = x * x * x - 3.0 * x * x + 2.0 * x

fun secant(x1: Double, x2: Double, f: DoubleToDouble): Double {
    val e = 1.0e-12
    val limit = 50
    var xa = x1
    var xb = x2
    var fa = f(xa)
    var  i = 0
    while (i++ < limit) {
        var fb = f(xb)
        val d = (xb - xa) / (fb - fa) * fb
        if (Math.abs(d) < e) break
        xa = xb
        fa = fb
        xb -= d
    }
    if (i == limit) {
        println("Function is not converging near (${"%7.4f".format(xa)}, ${"%7.4f".format(xb)}).")
        return -99.0 
    }
    return xb
}

fun main(args: Array<String>) {
    val step = 1.0e-2
    val e = 1.0e-12
    var x = -1.032
    var s = f(x) > 0.0
    while (x < 3.0) {
        val value = f(x)
        if (Math.abs(value) < e) {
            println("Root found at x = ${"%12.9f".format(x)}")
            s = f(x + 0.0001) > 0.0
        }
        else if ((value > 0.0) != s) {
            val xx = secant(x - step, x, ::f)
            if (xx != -99.0) 
                println("Root found at x = ${"%12.9f".format(xx)}")
            else
                println("Root found near x = ${"%7.4f".format(x)}")
            s = f(x + 0.0001) > 0.0
        }
        x += step
    }
}

Output:

Root found at x =  0.000000000
Root found at x =  1.000000000
Root found at x =  2.000000000

Lambdatalk

1) defining the function:
{def func {lambda {:x} {+ {* 1 :x :x :x} {* -3 :x :x} {* 2 :x}}}}
-> func

2) printing roots:
{S.map {lambda {:x}
               {if {< {abs {func :x}} 0.0001} 
                then {br}- a root found at :x else}} 
       {S.serie -1 3 0.01}}
-> 
- a root found at 7.528699885739343e-16                                                                                                    
- a root found at 1.0000000000000013                                                                                                    
- a root found at 2.000000000000002                                                                                                    

3) printing the roots of the "sin" function between -720° to +720°;

{S.map {lambda {:x}
               {if {< {abs {sin {* {/ {PI} 180} :x}}} 0.01}
                then {br}- a root found at :x° else}} 
       {S.serie -720 +720 10}}
-> 
- a root found at -720°                  
- a root found at -540°                  
- a root found at -360°                  
- a root found at -180°                  
- a root found at 0°                  
- a root found at 180°                  
- a root found at 360°                  
- a root found at 540°                  
- a root found at 720°

Liberty BASIC

'   Finds and output the roots of a given function f(x),
'       within a range of x values.

'   [RC]Roots of an function

    mainwin 80 12

    xMin  =-1
    xMax  = 3
    y     =f( xMin) '   Since Liberty BASIC has an 'eval(' function the fn
    '                       and limits would be better entered via 'input'.
    LastY =y

    eps  =1E-12 '   closeness acceptable

    bigH=0.01

    print
    print " Checking for roots of x^3 -3 *x^2 +2 *x =0 over range -1 to +3"
    print

    x=xMin: dx = bigH
    do
        x=x+dx
        y = f(x)
        'print x, dx, y
        if y*LastY <0 then 'there is a root, should drill deeper
            if dx < eps then    'we are close enough
                print " Just crossed axis, solution f( x) ="; y; " at x ="; using( "#.#####", x)
                LastY = y
                dx = bigH   'after closing on root, continue with big step
            else
                x=x-dx  'step back
                dx = dx/10  'repeat with smaller step
            end if
        end if
    loop while x<xMax

    print
    print " Finished checking in range specified."

    end

    function f( x)
        f =x^3 -3 *x^2 +2 *x
    end function

Lua

-- Function to have roots found
function f (x) return x^3 - 3*x^2 + 2*x end

-- Find roots of f within x=[start, stop] or approximations thereof
function root (f, start, stop, step)
    local roots, x, sign, foundExact, value = {}, start, f(start) > 0
    while x <= stop do
        value = f(x)
        if value == 0 then
            table.insert(roots, {val = x, err = 0})
            foundExact = true
        end
        if value > 0 ~= sign then
            if foundExact then
                foundExact = false
            else
                table.insert(roots, {val = x, err = step})
            end
        end
        sign = value > 0
        x = x + step
    end
    return roots
end

-- Main procedure
print("Root (to 12DP)\tMax. Error\n")
for _, r in pairs(root(f, -1, 3, 10^-6)) do
    print(string.format("%0.12f", r.val), r.err)
end

Output:

Root (to 12DP)  Max. Error

0.000000000008  1e-06
1.000000000016  1e-06
2.000000999934  1e-06

Note that the roots found are all near misses because fractional numbers that seem nice and 'round' in decimal (such as 10^-6) often have some rounding error when represented in binary. To increase the chances of finding exact integer roots, try using an integer start value with a step value that is a power of two.

-- Main procedure
print("Root (to 12DP)\tMax. Error\n")
for _, r in pairs(root(f, -1, 3, 2^-10)) do
    print(string.format("%0.12f", r.val), r.err)
end

Output:

Root (to 12DP)  Max. Error

0.000000000000  0
1.000000000000  0
2.000000000000  0

Maple

f := x^3-3*x^2+2*x;
roots(f,x);

outputs:

[[0, 1], [1, 1], [2, 1]]

which means there are three roots. Each root is named as a pair where the first element is the value (0, 1, and 2), the second one the multiplicity (=1 for each means none of the three are degenerate).

By itself (i.e. unless specifically asked to do so), Maple will only perform exact (symbolic) operations and not attempt to do any kind of numerical approximation.

Mathematica/Wolfram Language

There are multiple obvious ways to do this in Mathematica.

Solve

This requires a full equation and will perform symbolic operations only:

Solve[x^3-3*x^2+2*x==0,x]

Output

 {{x->0},{x->1},{x->2}}

NSolve

This requires merely the polynomial and will perform numerical operations if needed:

 NSolve[x^3 - 3*x^2 + 2*x , x]

Output

 {{x->0.},{x->1.},{x->2.}}

(note that the results here are floats)

FindRoot

This will numerically try to find one(!) local root from a given starting point:

FindRoot[x^3 - 3*x^2 + 2*x , {x, 1.5}]

Output

 {x->0.}

From a different start point:

FindRoot[x^3 - 3*x^2 + 2*x , {x, 1.1}]

Output

{x->1.}

(note that there is no guarantee which one is found).

FindInstance

This finds a value (optionally out of a given domain) for the given variable (or a set of values for a set of given variables) that satisfy a given equality or inequality:

 FindInstance[x^3 - 3*x^2 + 2*x == 0, x]

Output

{{x->0}}

Reduce

This will (symbolically) reduce a given expression to the simplest possible form, solving equations and performing substitutions in the process:

Reduce[x^3 - 3*x^2 + 2*x == 0, x]

 x==0||x==1||x==2

(note that this doesn't yield a "solution" but a different expression that expresses the same thing as the original)

Maxima

e: x^3 - 3*x^2 + 2*x$

/* Number of roots in a real interval, using Sturm sequences */
nroots(e, -10, 10);
3

solve(e, x);
[x=1, x=2, x=0]

/* 'solve sets the system variable 'multiplicities */

solve(x^4 - 2*x^3 + 2*x - 1, x);
[x=-1, x=1]

multiplicities;
[1, 3]

/* Rational approximation of roots using Sturm sequences and bisection */

realroots(e);
[x=1, x=2, x=0]

/* 'realroots also sets the system variable 'multiplicities */

multiplicities;
[1, 1, 1]

/* Numerical root using Brent's method (here with another equation) */

find_root(sin(t) - 1/2, t, 0, %pi/2);
0.5235987755983

fpprec: 60$

bf_find_root(sin(t) - 1/2, t, 0, %pi/2);
5.23598775598298873077107230546583814032861566562517636829158b-1

/* Numerical root using Newton's method */

load(newton1)$
newton(e, x, 1.1, 1e-6);
1.000000017531147

/* For polynomials, Jenkins–Traub algorithm */

allroots(x^3 + x + 1);
[x=1.161541399997252*%i+0.34116390191401,
 x=0.34116390191401-1.161541399997252*%i,
 x=-0.68232780382802]

bfallroots(x^3 + x + 1);
[x=1.16154139999725193608791768724717407484314725802151429063617b0*%i + 3.41163901914009663684741869855524128445594290948999288901864b-1,
 x=3.41163901914009663684741869855524128445594290948999288901864b-1 - 1.16154139999725193608791768724717407484314725802151429063617b0*%i,
 x=-6.82327803828019327369483739711048256891188581897998577803729b-1]

Nim

import math
import strformat

func f(x: float): float = x ^ 3 - 3 * x ^ 2 + 2 * x

var 
  step = 0.01
  start = -1.0
  stop = 3.0
  sign = f(start) > 0
  x = start

while x <= stop:
  var value = f(x)
  
  if value == 0:
    echo fmt"Root found at {x:.5f}"
  elif (value > 0) != sign:
    echo fmt"Root found near {x:.5f}"
  
  sign = value > 0
  x += step

Output:

Root found near 0.00000
Root found near 1.00000
Root found near 2.00000

Objeck

Translation of: C++

bundle Default {
  class Roots {
    function : f(x : Float) ~ Float
    {
      return (x*x*x - 3.0*x*x + 2.0*x);
    }
     
    function : Main(args : String[]) ~ Nil 
    {
      step := 0.001;
      start := -1.0;
      stop := 3.0;
      value := f(start);
      sign := (value > 0);
     
      if(0.0 = value) {
        start->PrintLine();
      };
      
      for(x := start + step; x <= stop;  x += step;)  {
        value := f(x);
        
        if((value > 0) <> sign) {
          IO.Console->Instance()->Print("~")->PrintLine(x);
        }
        else if(0 = value) {
          IO.Console->Instance()->Print("~")->PrintLine(x);
        };
        
        sign := (value > 0);
      };
    }
  }
}

OCaml

A general root finder using the False Position (Regula Falsi) method, which will find all simple roots given a small step size.

let bracket u v =
  ((u > 0.0) && (v < 0.0)) || ((u < 0.0) && (v > 0.0));;

let xtol a b = (a = b);; (* or use |a-b| < epsilon *)

let rec regula_falsi a b fa fb f =
  if xtol a b then (a, fa) else
  let c = (fb*.a -. fa*.b) /. (fb -. fa) in
  let fc = f c in
  if fc = 0.0 then (c, fc) else
  if bracket fa fc then
    regula_falsi a c fa fc f
  else
    regula_falsi c b fc fb f;;

let search lo hi step f =
  let rec next x fx =
    if x > hi then [] else
      let y = x +. step in
      let fy = f y in
      if fx = 0.0 then
        (x,fx) :: next y fy
      else if bracket fx fy then
        (regula_falsi x y fx fy f) :: next y fy
      else
        next y fy in
  next lo (f lo);;

let showroot (x,fx) =
  Printf.printf "f(%.17f) = %.17f [%s]\n" 
    x fx (if fx = 0.0 then "exact" else "approx") in
let f x = ((x -. 3.0)*.x +. 2.0)*.x  in
List.iter showroot (search (-5.0) 5.0 0.1 f);;

Output:

f(0.00000000000000000) = 0.00000000000000000 [exact]
f(1.00000000000000022) = 0.00000000000000000 [exact]
f(1.99999999999999978) = 0.00000000000000000 [exact]

Note these roots are exact solutions with floating-point calculation.

Octave

If the equation is a polynomial, we can put the coefficients in a vector and use roots:

a = [ 1, -3, 2, 0 ];
r = roots(a);
% let's print it
for i = 1:3
  n = polyval(a, r(i));
  printf("x%d = %f (%f", i, r(i), n);
  if (n != 0.0)
    printf(" not");
  endif
  printf(" exact)\n");
endfor

Otherwise we can program our (simple) method:

Translation of: Python

function y = f(x)
  y = x.^3 -3.*x.^2 + 2.*x;
endfunction

step = 0.001;
tol = 10 .* eps;
start = -1;
stop = 3;
se = sign(f(start));

x = start;
while (x <= stop)
  v = f(x);
  if ( (v < tol) && (v > -tol) )
    printf("root at %f\n", x);
  elseif ( sign(v) != se )
    printf("root near %f\n", x);
  endif
  se = sign(v);
  x = x + step;
endwhile

Oforth

: findRoots(f, a, b, st)
| x y lasty |
   a f perform dup ->y ->lasty

   a b st step: x [
      x f perform -> y
      y ==0 ifTrue: [ System.Out "Root found at " << x << cr ]
      else: [ y lasty * sgn -1 == ifTrue: [ System.Out "Root near " << x << cr ] ]
      y ->lasty
      ] ;

: f(x)   x 3 pow x sq 3 * - x 2 * + ;

Output:

findRoots(#f, -1, 3, 0.0001)
Root found at 0
Root found at 1
Root found at 2

findRoots(#f, -1.000001, 3, 0.0001)
Root near 9.90000000000713e-005
Root near 1.000099
Root near 2.000099

ooRexx

/* REXX  program to solve a cubic polynom equation
a*x**3+b*x**2+c*x+d =(x-x1)*(x-x2)*(x-x3)
*/
Numeric Digits 16
pi3=Rxcalcpi()/3
Parse Value '1 -3 2 0' with a b c d
p=3*a*c-b**2
q=2*b**3-9*a*b*c+27*a**2*d
det=q**2+4*p**3
say 'p='p
say 'q='q
Say 'det='det
If det<0 Then Do
  phi=Rxcalcarccos(-q/(2*rxCalcsqrt(-p**3)),16,'R')
  Say 'phi='phi
  phi3=phi/3
  y1=rxCalcsqrt(-p)*2*Rxcalccos(phi3,16,'R')
  y2=rxCalcsqrt(-p)*2*Rxcalccos(phi3+2*pi3,16,'R')
  y3=rxCalcsqrt(-p)*2*Rxcalccos(phi3+4*pi3,16,'R')
  End
Else Do
  t=q**2+4*p**3
  tu=-4*q+4*rxCalcsqrt(t)
  tv=-4*q-4*rxCalcsqrt(t)
  u=qroot(tu)/2
  v=qroot(tv)/2
  y1=u+v
  y2=-(u+v)/2 (u+v)/2*rxCalcsqrt(3)
  y3=-(u+v)/2 (-(u+v)/2*rxCalcsqrt(3))
  End
say 'y1='y1
say 'y2='y2
say 'y3='y3
x1=y2x(y1)
x2=y2x(y2)
x3=y2x(y3)
Say 'x1='x1
Say 'x2='x2
Say 'x3='x3
Exit

qroot: Procedure
Parse Arg a
return sign(a)*rxcalcpower(abs(a),1/3,16)

y2x: Procedure Expose a b
Parse Arg real imag
xr=(real-b)/(3*a)
If imag<>'' Then Do
  xi=(imag-b)/(3*a)
  Return xr xi'i'
  End
Else
  Return xr
::requires 'rxmath' LIBRARY

Output:

p=-3
q=0
det=-108
phi=1.570796326794897
y1=2.999999999999999
y2=-3.000000000000000
y3=0.000000000000002440395154978758
x1=2
x2=0
x3=1.000000000000001

PARI/GP

Gourdon–Schönhage algorithm

polroots(x^3-3*x^2+2*x)

Newton's method

This uses a modified version of the Newton–Raphson method.

polroots(x^3-3*x^2+2*x,1)

Brent's method

solve(x=-.5,.5,x^3-3*x^2+2*x)
solve(x=.5,1.5,x^3-3*x^2+2*x)
solve(x=1.5,2.5,x^3-3*x^2+2*x)

Factorization to linear factors

findRoots(P)={
  my(f=factor(P),t);
  for(i=1,#f[,1],
    if(poldegree(f[i,1]) == 1,
      for(j=1,f[i,2],
        print(-polcoeff(f[i,1], 0), " (exact)")
      )
    );
    if(poldegree(f[i,1]) > 1,
      t=polroots(f[i,1]);
      for(j=1,#t,
        for(k=1,f[i,2],
          print(if(imag(t[j]) == 0.,real(t[j]),t[j]), " (approximate)")
        )
      )
    )
  )
};
findRoots(x^3-3*x^2+2*x)

Factorization to quadratic factors

Of course this process could be continued to degrees 3 and 4 with sufficient additional work.

findRoots(P)={
  my(f=factor(P),t);
  for(i=1,#f[,1],
    if(poldegree(f[i,1]) == 1,
      for(j=1,f[i,2],
        print(-polcoeff(f[i,1], 0), " (exact)")
      )
    );
    if(poldegree(f[i,1]) == 2,
      t=solveQuadratic(polcoeff(f[i,1],2),polcoeff(f[i,1],1),polcoeff(f[i,1],0));
      for(j=1,f[i,2],
        print(t[1]" (exact)\n"t[2]" (exact)")
      )
    );
    if(poldegree(f[i,1]) > 2,
      t=polroots(f[i,1]);
      for(j=1,#t,
        for(k=1,f[i,2],
          print(if(imag(t[j]) == 0.,real(t[j]),t[j]), " (approximate)")
        )
      )
    )
  )
};
solveQuadratic(a,b,c)={
  my(t=-b/2/a,s=b^2/4/a^2-c/a,inner=core(numerator(s))/core(denominator(s)),outer=sqrtint(s/inner));
  if(inner < 0,
    outer *= I;
    inner *= -1
  );
  s=if(inner == 1,
    outer
  ,
    if(outer == 1,
      Str("sqrt(", inner, ")")
    ,
      Str(outer, " * sqrt(", inner, ")")
    )
  );
  if (t,
    [Str(t, " + ", s), Str(t, " - ", s)]
  ,
    [s, Str("-", s)]
  )
};
findRoots(x^3-3*x^2+2*x)

Pascal

Translation of: Fortran

Program RootsFunction;

var
  e, x, step, value: double;
  s: boolean;
  i, limit: integer;
  x1, x2, d: double;

function f(const x: double): double;
  begin
    f := x*x*x - 3*x*x + 2*x;
  end;  

begin
  x    := -1;
  step := 1.0e-6;
  e    := 1.0e-9;
  s    := (f(x) > 0);

  writeln('Version 1: simply stepping x:');
  while x < 3.0 do
  begin
    value := f(x);
    if abs(value) < e then
    begin
      writeln ('root found at x = ', x);
      s := not s;
    end
    else if ((value > 0) <> s) then
    begin
      writeln ('root found at x = ', x);
      s := not s;
    end;
    x := x + step;
  end;
  
  writeln('Version 2: secant method:');
  x1 := -1.0;
  x2 :=  3.0;
  e  :=  1.0e-15;
  i  :=  1;
  limit := 300;
  while true do
  begin
    if i > limit then
    begin
      writeln('Error: function not converging');
      exit;
    end;
    d := (x2 - x1) / (f(x2) - f(x1)) * f(x2);
    if abs(d) < e then
    begin
      if d = 0 then
        write('Exact ')
      else
        write('Approximate ');
      writeln('root found at x = ', x2);
      exit;
    end;
    x1 := x2;
    x2 := x2 - d;
    i  := i + 1;
  end;
end.

Output:

Version 1: simply stepping x:
root found at x =  7.91830063542152E-012
root found at x =  1.00000000001584E+000
root found at x =  1.99999999993357E+000
Version 2: secant method:
Exact root found at x =  1.00000000000000E+000

Perl

sub f
{
        my $x = shift;

        return ($x * $x * $x - 3*$x*$x + 2*$x);
}

my $step = 0.001; # Smaller step values produce more accurate and precise results
my $start = -1;
my $stop = 3;
my $value = &f($start);
my $sign = $value > 0;

# Check for root at start

print "Root found at $start\n" if ( 0 == $value );

for(    my $x = $start + $step;
        $x <= $stop;
        $x += $step )
{
        $value = &f($x);

        if ( 0 == $value )
        {
                # We hit a root
                print "Root found at $x\n";
        }
        elsif ( ( $value > 0 ) != $sign )
        {
                # We passed a root
                print "Root found near $x\n";
        }

        # Update our sign
        $sign = ( $value > 0 );
}

Phix

Translation of: CoffeeScript

procedure print_roots(integer f, atom start, stop, step)
    --
    -- Print approximate roots of f between x=start and x=stop, using 
    --  sign changes as an indicator that a root has been encountered. 
    --
    atom x = start, y = 0
    puts(1,"-----\n")
    while x<=stop do
        atom last_y = y
        y = f(x)
        if y=0
        or (last_y<0 and y>0)
        or (last_y>0 and y<0) then
            printf(1,"Root found %s %.10g\n", {iff(y=0?"at":"near"),x})
        end if
        x += step
    end while
end procedure
 
-- Smaller steps produce more accurate/precise results in general,
-- but for many functions we'll never get exact roots, either due
-- to imperfect binary representation or irrational roots.
constant step = 1/256
 
function f1(atom x) return x*x*x-3*x*x+2*x  end function
function f2(atom x) return x*x-4*x+3        end function
function f3(atom x) return x-1.5            end function
function f4(atom x) return x*x-2            end function
 
print_roots(f1, -1, 5, step)
print_roots(f2, -1, 5, step)
print_roots(f3,  0, 4, step)
print_roots(f4, -2, 2, step)

Output:

-----
Root found at 0
Root found at 1
Root found at 2
-----
Root found at 1
Root found at 3
-----
Root found at 1.5
-----
Root found near -1.4140625
Root found near 1.41796875

PicoLisp

Translation of: Clojure

(de findRoots (F Start Stop Step Eps)
   (filter
      '((N) (> Eps (abs (F N))))
      (range Start Stop Step) ) )

(scl 12)

(mapcar round
   (findRoots
      '((X) (+ (*/ X X X `(* 1.0 1.0)) (*/ -3 X X 1.0) (* 2 X)))
      -1.0 3.0 0.0001 0.00000001 ) )

Output:

-> ("0.000" "1.000" "2.000")

PL/I

f: procedure (x) returns (float (18));
   declare x float (18);
   return (x**3 - 3*x**2 + 2*x );
end f;

declare eps float, (x, y) float (18);
declare dx fixed decimal (15,13);

eps = 1e-12;

do dx = -5.03 to 5 by 0.1;
   x = dx;
   if sign(f(x)) ^= sign(f(dx+0.1)) then
      call locate_root;
end;

locate_root: procedure;
   declare (left, mid, right) float (18);

   put skip list ('Looking for root in [' || x, x+0.1 || ']' );
   left = x; right = dx+0.1;
   PUT SKIP LIST (F(LEFT), F(RIGHT) );
   if abs(f(left) ) < eps then
      do; put skip list ('Found a root at x=', left); return; end;
   else if abs(f(right) ) < eps then
      do; put skip list ('Found a root at x=', right); return; end;
   do forever;
      mid = (left+right)/2;
      if sign(f(mid)) = 0 then
         do; put skip list ('Root found at x=', mid); return; end;
      else if sign(f(left)) ^= sign(f(mid)) then
         right = mid;
      else
         left = mid;
      /* put skip list (left || right); */
      if abs(right-left) < eps then
         do; put skip list ('There is a root near ' ||
            (left+right)/2); return;
         end;
   end;
end locate_root;

PureBasic

Translation of: C++

Procedure.d f(x.d)
  ProcedureReturn x*x*x-3*x*x+2*x
EndProcedure

Procedure main()
  OpenConsole()
  Define.d StepSize= 0.001
  Define.d Start=-1, stop=3
  Define.d value=f(start), x=start
  Define.i oldsign=Sign(value)
  
  If value=0
    PrintN("Root found at "+StrF(start))
  EndIf
  
  While x<=stop
    value=f(x)
    If Sign(value) <> oldsign
      PrintN("Root found near "+StrF(x))
    ElseIf value = 0
      PrintN("Root found at "+StrF(x))
    EndIf
    oldsign=Sign(value)
    x+StepSize
  Wend
EndProcedure

main()

Python

Translation of: Perl

f = lambda x: x * x * x - 3 * x * x + 2 * x

step = 0.001 # Smaller step values produce more accurate and precise results
start = -1
stop = 3

sign = f(start) > 0

x = start
while x <= stop:
    value = f(x)

    if value == 0:
        # We hit a root
        print "Root found at", x
    elif (value > 0) != sign:
        # We passed a root
        print "Root found near", x

    # Update our sign
    sign = value > 0

    x += step

R

Translation of: Octave

f <- function(x) x^3 -3*x^2 + 2*x

findroots <- function(f, begin, end, tol = 1e-20, step = 0.001) {
  se <- ifelse(sign(f(begin))==0, 1, sign(f(begin)))
  x <- begin
  while ( x <= end ) {
    v <- f(x)
    if ( abs(v) < tol ) {
      print(sprintf("root at %f", x))
    } else if ( ifelse(sign(v)==0, 1, sign(v)) != se )  {
      print(sprintf("root near %f", x))
    }
    se <- ifelse( sign(v) == 0 , 1, sign(v))
    x <- x + step
  }
}

findroots(f, -1, 3)

Racket

#lang racket

;; Attempts to find all roots of a real-valued function f
;; in a given interval [a b] by dividing the interval into N parts
;; and using the root-finding method on each subinterval
;; which proves to contain a root.
(define (find-roots f a b 
                    #:divisions [N 10]
                    #:method [method secant])
  (define h (/ (- b a) N))
  (for*/list ([x1 (in-range a b h)] 
              [x2 (in-value (+ x1 h))]
              #:when (or (root? f x1)
                         (includes-root? f x1 x2)))
    (find-root f x1 x2 #:method method)))

;; Finds a root of a real-valued function f
;; in a given interval [a b].
(define (find-root f a b #:method [method secant])
  (cond 
    [(root? f a) a]
    [(root? f b) b]
    [else (and (includes-root? f a b) (method f a b))]))

;; Returns #t if x is a root of a real-valued function f
;; with absolute accuracy (tolerance).
(define (root? f x) (almost-equal? 0 (f x)))

;; Returns #t if interval (a b) contains a root
;; (or the odd number of roots) of a real-valued function f.
(define (includes-root? f a b) (< (* (f a) (f b)) 0))

;; Returns #t if a and b are equal with respect to
;; the relative accuracy (tolerance).
(define (almost-equal? a b)
  (or (< (abs (+ b a)) (tolerance))      
      (< (abs (/ (- b a) (+ b a))) (tolerance))))

(define tolerance (make-parameter 5e-16))

Different root-finding methods

(define (secant f a b)
  (let next ([x1 a] [y1 (f a)] [x2 b] [y2 (f b)] [n 50])
    (define x3 (/ (- (* x1 y2) (* x2 y1)) (- y2 y1)))
    (cond
      ; if the method din't converge within given interval
      ; switch to more robust bisection method
      [(or (not (< a x3 b)) (zero? n)) (bisection f a b)] 
      [(almost-equal? x3 x2) x3]
      [else (next x2 y2 x3 (f x3) (sub1 n))])))

(define (bisection f x1 x2)
  (let divide ([a x1] [b x2])
    (and (<= (* (f a) (f b)) 0)
         (let ([c (* 0.5 (+ a b))])
           (if (almost-equal? a b) 
               c
               (or (divide a c) (divide c b)))))))

Examples:

-> (find-root (λ (x) (- 2. (* x x))) 1 2)
1.414213562373095
-> (sqrt 2)
1.4142135623730951

-> (define (f x) (+ (* x x x) (* -3.0 x x) (* 2.0 x))) 
-> (find-roots f -3 4 #:divisions 50)
'(2.4932181969624796e-33 1.0 2.0)

In order to provide a comprehensive code the given solution does not optimize the number of function calls. The functional nature of Racket allows to perform the optimization without changing the main code using memoization.

Simple memoization operator

(define (memoized f)
  (define tbl (make-hash))
  (λ x
    (cond [(hash-ref tbl x #f) => values]
          [else (define res (apply f x))
                (hash-set! tbl x res)
                res])))

To use memoization just call

-> (find-roots (memoized f) -3 4 #:divisions 50)
'(2.4932181969624796e-33 1.0 2.0)

The profiling shows that memoization reduces the number of function calls in this example from 184 to 67 (50 calls for primary interval division and about 6 calls for each point refinement).

Raku

(formerly Perl 6) Uses exact arithmetic.

sub f(\x) { x³ - 3*x² + 2*x }

my $start = -1;
my $stop = 3;
my $step = 0.001;

for $start, * + $step ... $stop -> $x {
    state $sign = 0;
    given f($x) {
        my $next = .sign;
        when 0.0 {
            say "Root found at $x";
        }
        when $sign and $next != $sign {
            say "Root found near $x";
        }
        NEXT $sign = $next;
    }
}

Output:

Root found at 0
Root found at 1
Root found at 2

REXX

Both of these REXX versions use the   bisection method.

function coded as a REXX function

/*REXX program finds the roots of a specific function:  x^3 - 3*x^2 + 2*x  via bisection*/
parse arg bot top inc .                          /*obtain optional arguments from the CL*/
if bot=='' | bot==","  then bot= -5              /*Not specified?  Then use the default.*/
if top=='' | top==","  then top= +5              /* "       "        "   "   "     "    */
if inc=='' | inc==","  then inc=   .0001         /* "       "        "   "   "     "    */
z= f(bot - inc)                                  /*compute 1st value to start compares. */
!= sign(z)                                       /*obtain the sign of the initial value.*/
           do j=bot  to top  by  inc             /*traipse through the specified range. */
           z= f(j);              $= sign(z)      /*compute new value;  obtain the sign. */
           if z=0  then                               say  'found an exact root at'   j/1
                   else if !\==$  then if !\==0  then say  'passed a root at'         j/1
           != $                                  /*use the new sign for the next compare*/
           end   /*j*/                           /*dividing by unity normalizes J  [↑]  */
exit                                             /*stick a fork in it,  we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
f:  parse arg x;     return x * (x * (x-3) +2)   /*formula used   ──► x^3 - 3x^2  + 2x  */
                                                 /*with factoring ──► x{ x^2 -3x  + 2 } */
                                                 /*more     "     ──► x{ x( x-3 ) + 2 } */

output   when using the defaults for input:

found an exact root at 0
found an exact root at 1
found an exact root at 2

function coded in-line

This version is about   40%   faster than the 1^st REXX version.

/*REXX program finds the roots of a specific function:  x^3 - 3*x^2 + 2*x  via bisection*/
parse arg bot top inc .                          /*obtain optional arguments from the CL*/
if bot=='' | bot==","  then bot= -5              /*Not specified?  Then use the default.*/
if top=='' | top==","  then top= +5              /* "       "        "   "   "     "    */
if inc=='' | inc==","  then inc=   .0001         /* "       "        "   "   "     "    */
x= bot - inc                                     /*compute 1st value to start compares. */
z= x * (x * (x-3) + 2)                           /*formula used   ──► x^3 - 3x^2  + 2x  */
!= sign(z)                                       /*obtain the sign of the initial value.*/
           do x=bot  to top  by  inc             /*traipse through the specified range. */
           z= x * (x * (x-3) + 2);    $= sign(z) /*compute new value;  obtain the sign. */
           if z=0  then                               say  'found an exact root at'   x/1
                   else if !\==$  then if !\==0  then say  'passed a root at'         x/1
           != $                                  /*use the new sign for the next compare*/
           end   /*x*/                           /*dividing by unity normalizes X  [↑]  */

output   is the same as the 1^st REXX version.

Ring

load "stdlib.ring"
function = "return pow(x,3)-3*pow(x,2)+2*x"
rangemin = -1
rangemax = 3
stepsize = 0.001
accuracy = 0.1
roots(function, rangemin, rangemax, stepsize, accuracy)
 
func roots funct, min, max, inc, eps
     oldsign = 0
     for x = min to max step inc
         num = sign(eval(funct))
         if num = 0 
            see "root found at x = " + x + nl
            num = -oldsign
         else if num != oldsign and oldsign != 0
              if inc < eps 
                 see "root found near x = " + x + nl
              else roots(funct, x-inc, x+inc/8, inc/8, eps) ok ok ok
         oldsign = num
     next

Output:

root found near x = 0.00
root found near x = 1.00
root found near x = 2.00

RLaB

RLaB implements a number of solvers from the GSL and the netlib that find the roots of a real or vector function of a real or vector variable. The solvers are grouped with respect whether the variable is a scalar, findroot, or a vector, findroots. Furthermore, for each group there are two types of solvers, one that does not require the derivative of the objective function (which root(s) are being sought), and one that does.

The script that finds a root of a scalar function $f(x)=x^{3}-3\,x^{2}+2\,x$ of a scalar variable x using the bisection method on the interval -5 to 5 is,

f = function(x)
{
  rval = x .^ 3 - 3 * x .^ 2 + 2 * x;
  return rval;
};

>> findroot(f, , [-5,5])
  0

For a detailed description of the solver and its parameters interested reader is directed to the rlabplus manual.

Ruby

Translation of: Python

def sign(x)
  x <=> 0
end

def find_roots(f, range, step=0.001)
  sign = sign(f[range.begin])
  range.step(step) do |x|
    value = f[x]
    if value == 0
      puts "Root found at #{x}"
    elsif sign(value) == -sign
      puts "Root found between #{x-step} and #{x}"
    end
    sign = sign(value)
  end
end

f = lambda { |x| x**3 - 3*x**2 + 2*x }
find_roots(f, -1..3)

Output:

Root found at 0.0
Root found at 1.0
Root found at 2.0

Or we could use Enumerable#inject, monkey patching and block:

class Numeric
  def sign
    self <=> 0
  end
end

def find_roots(range, step = 1e-3)
  range.step( step ).inject( yield(range.begin).sign ) do |sign, x|
    value = yield(x)
    if value == 0
      puts "Root found at #{x}"
    elsif value.sign == -sign
      puts "Root found between #{x-step} and #{x}"
    end
    value.sign
  end
end

find_roots(-1..3) { |x| x**3 - 3*x**2 + 2*x }

Rust

// 202100315 Rust programming solution

use roots::find_roots_cubic;

fn main() {

   let roots = find_roots_cubic(1f32, -3f32, 2f32, 0f32);

   println!("Result : {:?}", roots);
}

Output:

Result : Three([0.000000059604645, 0.99999994, 2.0])

Another without external crates:

use num::Float;

/// Note: We cannot use `range_step` here because Floats don't implement
/// the `CheckedAdd` trait.
fn find_roots<T, F>(f: F, start: T, stop: T, step: T, epsilon: T) -> Vec<T>
where
    T: Copy + PartialOrd + Float,
    F: Fn(T) -> T,
{
    let mut ret = vec![];
    let mut current = start;
    while current < stop {
        if f(current).abs() < epsilon {
            ret.push(current);
        }
        current = current + step;
    }
    ret
}

fn main() {
    let roots = find_roots(
        |x: f64| x * x * x - 3.0 * x * x + 2.0 * x,
        -1.0,
        3.0,
        0.0001,
        0.00000001,
    );

    println!("roots of f(x) = x^3 - 3x^2 + 2x are: {:?}", roots);
}

Output:

roots of f(x) = x^3 - 3x^2 + 2x are: [-0.00000000000009381755897326649, 0.9999999999998124, 1.9999999999997022]

Scala

Imperative version (Ugly, side effects)

Translation of: Java

Output:

Best seen running in your browser either by (ES aka JavaScript, non JVM) or Scastie (remote JVM).

object Roots extends App {
  val poly = (x: Double) => x * x * x - 3 * x * x + 2 * x

  private def printRoots(f: Double => Double,
                         lowerBound: Double,
                         upperBound: Double,
                         step: Double): Unit = {
    val y = f(lowerBound)
    var (ox, oy, os) = (lowerBound, y, math.signum(y))

    for (x <- lowerBound to upperBound by step) {
      val y = f(x)
      val s = math.signum(y)
      if (s == 0) println(x)
      else if (s != os) println(s"~${x - (x - ox) * (y / (y - oy))}")

      ox = x
      oy = y
      os = s
    }
  }

  printRoots(poly, -1.0, 4, 0.002)

}

Functional version (Recommended)

object RootsOfAFunction extends App {
    def findRoots(fn: Double => Double, start: Double, stop: Double, step: Double, epsilon: Double) = {
    for {
      x <- start to stop by step
      if fn(x).abs < epsilon
    } yield x
  }

  def fn(x: Double) = x * x * x - 3 * x * x + 2 * x

  println(findRoots(fn, -1.0, 3.0, 0.0001, 0.000000001))
}

Output:

Vector(-9.381755897326649E-14, 0.9999999999998124, 1.9999999999997022)

Scheme

For R7RS Scheme.

;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;
;;;
;;; Finding (real) roots of a function.
;;;
;;; I follow the model that breaks the task into two distinct ones:
;;; isolating real roots, and then finding the isolated roots. The
;;; former task I will call "isolating roots", and the latter I will
;;; call "rootfinding".
;;;
;;; Isolating real roots of a polynomial can be done exactly, and the
;;; methods can handle infinite domains. Scheme (because it has exact
;;; rationals) is a relatively easy language in which to write such
;;; code.
;;;
;;; I have also isolated the roots of low-degree polynomials on finite
;;; intervals by the following method, in floating-point arithmetic:
;;; rewrite the polynomial in the Bernstein polynomials basis, then
;;; take derivatives to get critical points, working your way back up
;;; in degree from a straight line. This method goes back and forth
;;; between the two subtasks. (You could use the quadratic formula
;;; once you got down to degree two, but I wouldn’t bother. The cubic
;;; and quartic formulas are numerically very poor and should be
;;; avoided.)
;;;
;;; However, these methods require that the function be a
;;; polynomial. Here I will simply use a step size and the
;;; intermediate value theorem.
;;;
;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;

(cond-expand
  (r7rs)
  (chicken (import r7rs)))

;;;
;;; Step 1. Isolation of roots.
;;;
;;; I will simply step over the domain and look for intervals that
;;; contain at least one root. There is a risk of getting the error
;;; message "the root is not bracketed" when you run the
;;; rootfinder. (One could have the rootfinder raise a recoverable
;;; exception or return a special value, instead.)
;;;
(define-library (isolate-roots)

  (export isolate-roots)

  (import (scheme base))

  (begin

    (define (isolate-roots f x-min x-max x-step)
      (define (ith-x i) (+ x-min (* i x-step)))
      (let ((x0 x-min)
            (y0 (f x-min)))
        (let loop ((i1 1)
                   (x0 x0)
                   (y0 y0)
                   (accum (if (zero? y0)
                              `((,x0 . ,x0))
                              '())))
          (let ((x1 (ith-x i1)))
            (if (< x-max x1)
                (reverse accum)
                (let ((y1 (f x1)))
                  (cond
                   ((zero? y1)
                    (loop (+ i1 1) x1 y1 `((,x1 . ,x1) . ,accum)))
                   ((negative? (* y0 y1))
                    (loop (+ i1 1) x1 y1 `((,x0 . ,x1) . ,accum)))
                   (else
                    (loop (+ i1 1) x1 y1 accum)))))))))

    )) ;; end library roots-isolator

;;;
;;; Step 2. Rootfinding.
;;;
;;; I will use the ITP method. I wrote this implementation shortly
;;; after the algorithm was published. See
;;; https://sourceforge.net/p/chemoelectric/itp-root-finder
;;;
;;; Reference:
;;;
;;;   I.F.D. Oliveira and R.H.C. Takahashi. 2020. An Enhancement of
;;;   the Bisection Method Average Performance Preserving Minmax
;;;   Optimality. ACM Trans. Math. Softw. 47, 1, Article 5 (December
;;;   2020), 24 pages. https://doi.org/10.1145/3423597
;;;
(define-library (itp-root-finder)

  (export itp-root-finder-epsilon
          itp-root-finder-extra-steps
          itp-root-finder-kappa1
          itp-root-finder-kappa2

          ;; itp-root-bracket-finder returns two values that form a
          ;; bracket no wider than 2ϵ.
          itp-root-bracket-finder

          ;; itp-root-finder returns the point midway between the ends
          ;; of the final bracket.
          itp-root-finder)

  (import (scheme base))
  (import (scheme inexact))
  (import (scheme case-lambda))
  (import (srfi 143))                   ; Fixnums.
  (import (srfi 144))                   ; Flonums.

  (begin

    (define ϕ
      ;; The Golden Ratio, (1 + √5)/2, rounded down by about
      ;; 0.00003398875.
      1.618)

    (define 1+ϕ (+ 1.0 ϕ))

    (define itp-root-finder-epsilon
      (make-parameter
       (* 1000.0 fl-epsilon)
       (lambda (ϵ)
         (if (positive? ϵ)
             ϵ
             (error 'itp-root-finder-epsilon
                    "a positive value was expected"
                    ϵ)))))

    (define itp-root-finder-extra-steps
      ;; Increase extra-steps above zero, if you wish to try to speed
      ;; up convergence, at the expense of that many more steps in the
      ;; worst case.
      (make-parameter
       0
       (lambda (n₀)
         (if (or (negative? n₀) (not (integer? n₀)))
             (error 'itp-root-finder-extra-steps
                    "a non-negative integer was expected"
                    n₀)
             n₀))))

    (define itp-root-finder-kappa1
      (make-parameter
       0.1
       (lambda (κ₁)
         (if (positive? κ₁)
             κ₁
             (error 'itp-root-finder-kappa1
                    "a positive value was expected"
                    κ₁)))))

    (define itp-root-finder-kappa2
      (make-parameter
       2.0
       (lambda (κ₂)
         (if (or (< κ₂ 1) (< 1+ϕ κ₂))
             ;; We allow <= 1+ϕ (instead of ‘< 1+ϕ’) because we
             ;; already rounded ϕ down.
             (error 'itp-root-finder-kappa2
                    (string-append "a value 1 <= kappa2 <= "
                                   (number->string 1+ϕ)
                                   " was expected")
                    κ₂)
             κ₂))))

    (define (sign x)
      (cond ((negative? x) -1)
            ((positive? x) 1)
            (else 0)))

    (define (apply-sign σ x)
      (cond ((fxnegative? σ) (- x))
            ((fxpositive? σ) x)
            (else 0)))

    (define (itp-root-bracket-finder%% f a b ϵ n₀ κ₁ κ₂)
      (let* ((2ϵ (inexact (* 2 ϵ)))
             (n½ (exact (ceiling (log (/ (inexact (- b a)) 2ϵ) 2))))
             (n_max (+ n½ n₀))
             (ya (f a))
             (yb (f b))
             (σ_ya (sign ya))
             (σ_yb (sign yb)))
        (cond
         ((fxzero? σ_ya) (values a a))
         ((fxzero? σ_yb) (values b b))
         (else
          (when (fxpositive? (* σ_ya σ_yb))
            (error 'itp-root-bracket-finder
                   "the root is not bracketed"
                   a b))
          (let loop ((pow2 (expt 2 n_max))
                     (a (inexact a))
                     (b (inexact b))
                     (ya (inexact ya))
                     (yb (inexact yb)))
            (if (or (= pow2 1) (fl<=? (fl- b a) 2ϵ))
                (values a b)
                (let* ( ;; x½ – the bisection.
                       (x½ (fl* 0.5 (fl+ a b)))

                       ;; xf – interpolation by regula falsi.
                       (xf (fl/ (fl- (fl* yb a) (fl* ya b))
                                (fl- yb ya)))

                       (b-a (fl- b a))
                       (δ (fl* κ₁ (flabs (expt b-a κ₂))))
                       (x½-xf (fl- x½ xf))
                       (σ (sign x½-xf))

                       ;; xt – the ‘truncation’ of xf.
                       (xt (if (fl<=? δ (flabs x½-xf))
                               (fl+ xf (apply-sign σ δ))
                               x½))

                       (r (- (* pow2 ϵ) (fl* 0.5 b-a)))

                       ;; xp – the projection of xt onto [x½-r,x½+r].
                       (xp (if (fl<=? (flabs (fl- xt x½)) r)
                               xt
                               (fl- x½ (apply-sign σ r))))

                       (yp (inexact (f xp))))

                  (let ((pow2/2 (truncate-quotient pow2 2))
                        (σ_yp (sign yp)))

                    (cond ((fx=? σ_ya σ_yp)
                           ;; yp has the same sign as ya. Make it the
                           ;; new ya.
                           (loop pow2/2 xp b yp yb))

                          ((fx=? σ_yb σ_yp)
                           ;; yp has the same sign as yb. Make it the
                           ;; new yb.
                           (loop pow2/2 a xp ya yp))

                          (else
                           ;; yp is zero.
                           (values xp xp)))))))))))

    (define (itp-root-bracket-finder% f a b ϵ n₀ κ₁ κ₂)
      (cond
       ((< b a) (itp-root-bracket-finder% b a f ϵ n₀ κ₁ κ₂))
       (else
        (let* ((ϵ (or ϵ (itp-root-finder-epsilon)))
               (n₀ (or n₀ (itp-root-finder-extra-steps)))
               (κ₁ (or κ₁ (itp-root-finder-kappa1)))
               (κ₂ (or κ₂ (itp-root-finder-kappa2))))
          (when (negative? ϵ)
            (error 'itp-root-bracket-finder
                   "a positive value was expected" ϵ))
          (when (negative? κ₁)
            (error 'itp-root-bracket-finder
                   "a positive value was expected" κ₁))
          (when (or (< κ₂ 1) (< 1+ϕ κ₂))
            ;; We allow <= 1+ϕ (instead of ‘< 1+ϕ’) because we already
            ;; rounded ϕ down.
            (error 'itp-root-bracket-finder
                   (string-append "a value 1 <= kappa2 <= "
                                  (number->string 1+ϕ)
                                  " was expected")
                   κ₂))
          (when (or (negative? n₀) (not (integer? n₀)))
            (error 'itp-root-bracket-finder
                   "a non-negative integer was expected" n₀))
          (itp-root-bracket-finder%% f a b ϵ n₀ κ₁ κ₂)))))

    (define itp-root-bracket-finder
      (case-lambda
        ((f a b)
         (itp-root-bracket-finder% f a b #f #f #f #f))
        ((f a b ϵ)
         (itp-root-bracket-finder% f a b ϵ #f #f #f))
        ((f a b ϵ n₀)
         (itp-root-bracket-finder% f a b ϵ n₀ #f #f))
        ((f a b ϵ n₀ κ₁)
         (itp-root-bracket-finder% f a b ϵ n₀ κ₁ #f))
        ((f a b ϵ n₀ κ₁ κ₂)
         (itp-root-bracket-finder% f a b ϵ n₀ κ₁ κ₂))))

    (define (itp-root-finder% f a b ϵ n₀ κ₁ κ₂)
      (call-with-values
          (lambda ()
            (itp-root-bracket-finder f a b ϵ n₀ κ₁ κ₂))
        (lambda (a b)
          (if (= a b)
              a
              (* 0.5 (+ a b))))))

    (define itp-root-finder
      (case-lambda
        ((f a b)
         (itp-root-finder% f a b #f #f #f #f))
        ((f a b ϵ)
         (itp-root-finder% f a b ϵ #f #f #f))
        ((f a b ϵ n₀)
         (itp-root-finder% f a b ϵ n₀ #f #f))
        ((f a b ϵ n₀ κ₁)
         (itp-root-finder% f a b ϵ n₀ κ₁ #f))
        ((f a b ϵ n₀ κ₁ κ₂)
         (itp-root-finder% f a b ϵ n₀ κ₁ κ₂))))

    )) ;; end library itp-root-finder

(import (scheme base))
(import (scheme write))
(import (isolate-roots))
(import (itp-root-finder))

(define (f x)
  ;; x³ - 3x² + 2x, written in Horner form.
  (* x (+ 2 (* x (+ -3 x)))))

(define (find-root f interval)
  (define (display-exactness root)
    (display (if (and (exact? root)
                      (exact? (f root))
                      (zero? (f root)))
                 "  (exact)    "
                 "  (inexact)  ")))
  (let ((x0 (car interval))
        (x1 (cdr interval)))
    (if (= x0 x1)
        (begin
          (let ((root (if (exact? x0) x0 x1)))
            (display-exactness root)
            (display "(rootfinder not used)  ")
            (display root)
            (newline)))
        (begin
          ;;
          ;; I am not careful here to avoid accidentally excluding the
          ;; root from the bracketing interval [x0,x1]. Floating point
          ;; is very tricky to work with.
          ;;
          (let ((root (itp-root-finder f x0 x1)))
            (display-exactness root)
            (display "(rootfinder used)      ")
            (display root)
            (newline))))))

;;; The following two demonstrations find all three roots exactly, as
;;; exact rationals, without the need for a rootfinding step.
(newline)
(display "Stepping by 1/1000 from 0 to 2:")
(newline)
(do ((p (isolate-roots f 0 2 1/1000) (cdr p)))
    ((not (pair? p)))
  (find-root f (car p)))
(newline)
(display "Stepping by 1/1000 from -10 to 10:")
(newline)
(do ((p (isolate-roots f -10 10 1/1000) (cdr p)))
    ((not (pair? p)))
  (find-root f (car p)))

;;; The following demonstration gives inexact results, because the
;;; step size is an inexact number.
(newline)
(display "Stepping by 0.001 from -10.0 to 10.0:")
(newline)
(do ((p (isolate-roots f -10.0 10.0 0.001) (cdr p)))
    ((not (pair? p)))
  (find-root f (car p)))

;;; The following demonstration gives inexact results, because the
;;; rootfinder is needed.
(newline)
(display "Stepping by 13/3333 from -2111/1011 to 33/13:")
(newline)
(do ((p (isolate-roots f -2111/1011 33/13 13/3333) (cdr p)))
    ((not (pair? p)))
  (find-root f (car p)))

(newline)

Output:

$ gosh roots-of-a-function.scm
 
Stepping by 1/1000 from 0 to 2:
  (exact)    (rootfinder not used)  0
  (exact)    (rootfinder not used)  1
  (exact)    (rootfinder not used)  2

Stepping by 1/1000 from -10 to 10:
  (exact)    (rootfinder not used)  0
  (exact)    (rootfinder not used)  1
  (exact)    (rootfinder not used)  2

Stepping by 0.001 from -10.0 to 10.0:
  (inexact)  (rootfinder not used)  0.0
  (inexact)  (rootfinder not used)  1.0
  (inexact)  (rootfinder not used)  2.0

Stepping by 13/3333 from -2111/1011 to 33/13:
  (inexact)  (rootfinder used)      -1.3380129580295458e-15
  (inexact)  (rootfinder used)      0.9999999999998657
  (inexact)  (rootfinder used)      1.9999999999999998

Sidef

func f(x) {
    x*x*x - 3*x*x + 2*x
}
 
var step = 0.001
var start = -1
var stop = 3
 
for x in range(start+step, stop, step) {
    static sign = false
    given (var value = f(x)) {
        when (0) {
            say "Root found at #{x}"
        }
        case (sign && ((value > 0) != sign)) {
            say "Root found near #{x}"
        }
    }
    sign = value>0
}

Output:

Root found at 0
Root found at 1
Root found at 2

Tcl

This simple brute force iteration marks all results, with a leading "~", as approximate. This version always reports its results as approximate because of the general limits of computation using fixed-width floating-point numbers (i.e., IEEE double-precision floats).

proc froots {lambda {start -3} {end 3} {step 0.0001}} {
    set res {}
    set lastsign [sgn [apply $lambda $start]]
    for {set x $start} {$x <= $end} {set x [expr {$x + $step}]} {
        set sign [sgn [apply $lambda $x]]
        if {$sign != $lastsign} {
            lappend res [format ~%.11f $x]
        }
        set lastsign $sign
    }
    return $res
}
proc sgn x {expr {($x>0) - ($x<0)}}

puts [froots {x {expr {$x**3 - 3*$x**2 + 2*$x}}}]

Result and timing:

/Tcl $ time ./froots.tcl
~0.00000000000 ~1.00000000000 ~2.00000000000

real    0m0.368s
user    0m0.062s
sys     0m0.030s

A more elegant solution (and faster, because you can usually make the initial search coarser) is to use brute-force iteration and then refine with Newton-Raphson, but that requires the differential of the function with respect to the search variable.

proc frootsNR {f df {start -3} {end 3} {step 0.001}} {
    set res {}
    set lastsign [sgn [apply $f $start]]
    for {set x $start} {$x <= $end} {set x [expr {$x + $step}]} {
        set sign [sgn [apply $f $x]]
        if {$sign != $lastsign} {
            lappend res [format ~%.15f [nr $x $f $df]]
        }
        set lastsign $sign
    }
    return $res
}
proc sgn x {expr {($x>0) - ($x<0)}}
proc nr {x1 f df} {
    # Newton's method converges very rapidly indeed
    for {set iters 0} {$iters < 10} {incr iters} {
        set x1 [expr {
            [set x0 $x1] - [apply $f $x0]/[apply $df $x0]
        }]
        if {$x0 == $x1} {
            break
        }
    }
    return $x1
}

puts [frootsNR \
    {x {expr {$x**3 - 3*$x**2 + 2*$x}}} \
    {x {expr {3*$x**2 - 6*$x + 2}}}]

TI-89 BASIC

Finding roots is a built-in function: zeros(x^3-3x^2+2x, x) returns {0,1,2}.

In this case, the roots are exact; inexact results are marked by decimal points.

Wren

Translation of: Go

Library: Wren-fmt

import "./fmt" for Fmt

var secant = Fn.new { |f, x0, x1|
    var f0 = 0
    var f1 = f.call(x0)
    for (i in 0...100) {
        f0 = f1
        f1 = f.call(x1)
        if (f1 == 0) return [x1, "exact"]
        if ((x1-x0).abs < 1e-6) return [x1, "approximate"]
        var t = x0
        x0 = x1
        x1 = x1-f1*(x1-t)/(f1-f0)
    }
    return [0, ""]
}

var findRoots = Fn.new { |f, lower, upper, step|
    var x0 = lower
    var x1 = lower + step
    while (x0 < upper) {
        x1 = (x1 < upper) ? x1 : upper
        var res = secant.call(f, x0, x1)
        var r = res[0]
        var status = res[1]
        if (status != "" && r >= x0 && r < x1) {
            Fmt.print(" $6.3f $s", r, status)
        }
        x0 = x1
        x1 = x1 + step
    }
}

var example = Fn.new { |x|  x*x*x - 3*x*x + 2*x }
findRoots.call(example, -0.5, 2.6, 1)

Output:

  0.000 approximate
  1.000 exact
  2.000 approximate

XPL0

Translation of: Wren

include xpllib; \for Print

func real F(X);
real X;
return X*X*X - 3.*X*X + 2.*X;

char Status;

func real Secant(X0, X1);
real X0, X1, F0, F1, T;
int  I;
[F1:= F(X0);
for I:= 0 to 100-1 do
    [F0:= F1;
    F1:= F(X1);
    if F1 = 0. then [Status:= "exact";  return X1];
    if abs(X1-X0) < 1e-6 then [Status:= "approximate";  return X1];
    T:= X0;
    X0:= X1;
    X1:= X1 - F1*(X1-T)/(F1-F0);
    ];
Status:= 0;  return 0.;
];

func FindRoots(Lower, Upper, Step);
real Lower, Upper, Step;
real X0, X1, R;
[X0:= Lower;
X1:= Lower + Step;
while X0 < Upper do
    [X1:= if X1 < Upper then X1 else Upper;
    R:= Secant(X0, X1);
    if Status # 0 and R >= X0 and R < X1 then
        Print(" %2.3f %s\n", R, Status);
    X0:= X1;
    X1:= X1 + Step;
    ];
];

FindRoots(-0.5, 2.6, 1.)

Output:

  0.000 approximate
  1.000 exact
  2.000 approximate

zkl

Translation of: Haskell

fcn findRoots(f,start,stop,step,eps){
   [start..stop,step].filter('wrap(x){ f(x).closeTo(0.0,eps) })
}

fcn f(x){ x*x*x - 3.0*x*x + 2.0*x }
findRoots(f, -1.0, 3.0, 0.0001, 0.00000001).println();

Output:

L(-9.38176e-14,1,2)

Translation of: C

fcn secant(f,xA,xB){
   reg e=1.0e-12;

   fA:=f(xA); if(fA.closeTo(0.0,e)) return(xA);

   do(50){
      fB:=f(xB);
      d:=(xB - xA) / (fB - fA) * fB;
      if(d.closeTo(0,e)) break;
      xA = xB; fA = fB; xB -= d;
   }
   if(f(xB).closeTo(0.0,e)) xB
   else "Function is not converging near (%7.4f,%7.4f).".fmt(xA,xB);
}

step:=0.1;
xs:=findRoots(f, -1.032, 3.0, step, 0.1);
xs.println(" --> ",xs.apply('wrap(x){ secant(f,x-step,x+step) }));

Output:

L(-0.032,0.968,1.068,1.968) --> L(1.87115e-19,1,1,2)