Roots of a function

From Rosetta Code
Task
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:     f(x) = x3 - 3x2 + 2x

Ada[edit]

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[edit]

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)

ATS[edit]

 
#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[edit]

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[edit]

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[edit]

      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[edit]

Secant Method[edit]

#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[edit]

#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++[edit]

#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[edit]

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[edit]

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[edit]

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[edit]

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[edit]

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.

DWScript[edit]

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;

EchoLisp[edit]

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[edit]

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[edit]

% 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[edit]

 
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[edit]

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

Go[edit]

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[edit]

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

HicEst[edit]

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[edit]

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[edit]

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.
(#~ (=<./)@:[email protected]) 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[edit]

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[edit]

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[edit]

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[edit]

Assuming that one has the Roots package installed:

using Roots
 
println(fzeros(x -> x^3 - 3x^2 + 2x))
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,tol=1e-14,maxsteps=100)
#f: the function of x
#fp: the derivative of f
 
xnew, xold = x, Inf
fn, fo = f(xnew), Inf
 
 
counter = 1
 
while (counter < maxsteps) && (abs(xnew - xold) > tol) && ( abs(fn - fo) > tol )
x = xnew - f(xnew)/fp(xnew) # update step
xnew, xold = x, xnew
fn, fo = f(xnew), fn
counter = 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)

Liberty BASIC[edit]

'   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[edit]

-- 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[edit]

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[edit]

There are multiple obvious ways to do this in Mathematica.

Solve[edit]

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[edit]

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[edit]

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[edit]

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[edit]

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[edit]

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]

Objeck[edit]

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[edit]

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.

Oforth[edit]

: 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

Octave[edit]

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

PARI/GP[edit]

Gourdon–Schönhage algorithm[edit]

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

Newton's method[edit]

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

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

Brent's method[edit]

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[edit]

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[edit]

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[edit]

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[edit]

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 );
}

Perl 6[edit]

Uses exact arithmetic.

sub f($x) { $x*$x*$x - 3*$x*$x + 2*$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

PicoLisp[edit]

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[edit]

 
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[edit]

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[edit]

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[edit]

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[edit]

 
#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).

REXX[edit]

The   bisection method   is used.

/*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);  !=sign(z) /*use these values for initial compare.*/
 
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

Ring[edit]

 
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[edit]

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 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[edit]

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 }


Scala[edit]

This example is in need of improvement.
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)

Sidef[edit]

func f(x) {
x*x*x - 3*x*x + 2*x;
}
 
var step = 0.001;
var start = -1;
var stop = 3;
 
range(start+step, stop, step).each { |x|
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[edit]

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[edit]

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.

zkl[edit]

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)