Roots of a function

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 example, use f(x)=x³-3x²+2x.

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: <lang algol68>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</lang> 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)

C++

<lang cpp>#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 ); } }</lang>

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.

<lang lisp>(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)))))</lang>

> (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

<lang d>module findroot ; import std.stdio ; import std.math ;

void report(T)(T[] r, T function(T) f, T tolerance = cast(T) 1e-4L) {

 if (r.length) {
   writefln("Root found (tolerance = %1.4g) :", tolerance) ;
   foreach(x ; r) {
     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.") ;

}

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

T[] findroot(T)(T function(T) f, T start, T end, T step = cast(T) 0.001L,

                                T tolerance = cast(T) 1e-4L) {
 T[T] result ;

 if (nearZero(step))
   writefln("WARNING: step size may be too small.") ;

 T searchRoot(T a, T b) { // search root by simple bisection
   T root ;
   int limit = 49 ;
   T gap = b - a ;
   while (!nearZero(gap) && limit--) {
     if (nearZero(f(a))) return a ;
     if (nearZero(f(b))) return b ;
     root = (b + a)/2.0L ;
     if (nearZero(f(root))) return root ;
     if (f(a) * f(root) < 0)
       b = root ;
     else
       a = root ;
     gap = b - a ;           
   }
   return root ;
 }
 
 T dir = cast(T) (end > start ? 1.0 : -1.0) ;
 step = (end > start) ? abs(step) : - abs(step) ;
 for(T x = start ; x*dir <= end*dir ; x = x + step)
   if (f(x)*f(x + step) <= 0) {
     T r = searchRoot(x, x+ step) ;        
     result[r] = f(r) ;
   } 
 return result.keys.sort ; // reduce duplacated root, if any

}

real f(real x){

 return x*x*x - 3*x*x + 2*x ;

}

void main(){

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

}</lang>

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

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

Fortran

Works with: Fortran version 90 and later

<lang fortran>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</lang> 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. <lang fortran>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</lang>

Haskell

<lang 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]</lang>

Executed in GHCi: <lang haskell>*Main> findRoots (-1.0) 3.0 0.0001 0.000000001 [-9.381755897326649e-14,0.9999999999998124,1.9999999999997022]</lang>

Or using package hmatrix from HackageDB. <lang haskell>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</lang> No complex roots, so: <lang haskell>*Main> mapM_ (print.realPart) $ polySolve [0,2,-3,1] -5.421010862427522e-20 2.000000000000001 0.9999999999999996</lang>

HicEst

HicEst's SOLVE function employs the Levenberg-Marquardt method: <lang HicEst>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</lang> <lang HicEst>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;</lang>

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:

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

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.

<lang J> blackbox=: 0 2 _3 1&p.

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

0 1 2

  0=blackbox 0 1 2

1 1 1</lang>

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

<lang 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);

}</lang> Produces this output:

~2.616794878713638E-18
~1.0000000000000002
~2.000000000000001

Maple

<lang maple>f := x^3-3*x^2+2*x; roots(f,x);</lang>

outputs:

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

There are multiple obvious ways to do this in Mathematica.

Solve

This requires a full equation and will perform symbolic operations only: <lang mathematica>In[1]:= Solve[x^3-3*x^2+2*x==0,x] Out[1]= {{x->0},{x->1},{x->2}}</lang>

NSolve

This requires merely the polynomial and will perform numerical operations if needed: <lang mathematica>In[2]:= NSolve[x^3 - 3*x^2 + 2*x , x] Out[2]= {{x->0.},{x->1.},{x->2.}}</lang> (note that the results here are floats)

FindRoot

This will numerically try to find one(!) local root from a given starting point: <lang mathematica>In[3]:= FindRoot[x^3 - 3*x^2 + 2*x , {x, 1.5}] Out[3]= {x->0.} In[4]:= FindRoot[x^3 - 3*x^2 + 2*x , {x, 1.1}] Out[4]= {x->1.}</lang> (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: <lang mathematica>In[5]:= FindInstance[x^3 - 3*x^2 + 2*x == 0, x] Out[5]= {{x->0}}</lang>

Reduce

This will (symbolically) reduce a given expression to the simplest possible form, solving equations and performing substitutions in the process: <lang mathematica>In[6]:= Reduce[x^3 - 3*x^2 + 2*x == 0, x] Out[6]= x==0||x==1||x==2</lang> (note that this doesn't yield a "solution" but a different expression that expresses the same thing as the original)

OCaml

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

<lang ocaml>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);;</lang>

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:

<lang octave>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</lang>

Otherwise we can program our (simple) method:

Translation of: Python

<lang octave>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</lang>

Perl

<lang 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 );

}</lang>

PL/I

<lang 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; </lang>

Python

Translation of: Perl

<lang python>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</lang>

R

Translation of: Octave

<lang R>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)</lang>

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 is, <lang RLaB> f = function(x) {

 rval = x .^ 3 - 2 * x .^ 2 + 2 * x;
 return rval;

};

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

</lang>

Ruby

Translation of: Python

<lang ruby>def sign(x)

 x <=> 0

end

def find_roots(f, range, step)

 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, 0.001)</lang>

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:

<lang ruby>class Numeric

 def sign
   zero? ? 0 : self / abs
 end

end

def find_roots(range, step = 1e-3)

 range.step( step ).inject( yield(range.begin).sign ) { |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

find_roots(-1..3) { |x| x**3 - 3*x**2 + 2*x }</lang>

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). <lang Tcl>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}}}]</lang> 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. <lang Tcl>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}}}]</lang>

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.