Roots of a function
You are encouraged to solve this task according to the task description, using any language you may know.
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.
Ada
<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;</ada>
ALGOL 68
Finding 3 roots using the secant method:
MODE DBL = LONG REAL;
FORMAT dbl = $g$;
MODE DBLRES = UNION(DBL, VOID);
MODE DBLOPT = UNION(DBL, VOID);
PROC find root = (PROC (DBL)DBL f, DBLOPT in x1, in x2, in ex)DBLRES:(
INT limit=100;
DBL x1 := (in x1|(DBL x1):x1|-5.0), # if x1 is EMPTY then use -5.0 #
x2 := (in x2|(DBL x2):x2|+5.0), # if x2 is EMPTY then use +5.0 #
ex := (in ex|(DBL ex):ex|small real);
DBL f o x1 := f(x1), f o x2;
DBL dx := x1 - x2;
IF f o x1 = 0 THEN
x1 # we already have a solution! #
ELSE
FOR i WHILE
f o x2 := f(x2);
IF f o x2 = 0 THEN stop iteration FI;
IF i = limit THEN value error FI;
IF f o x1 = f o x2 THEN value error FI;
dx := dx / ( f o x1 - f o x2 ) * f o x2;
x1 := x2; f o x1 := f o x2; # retain for next iteration #
x2 -:= dx;
# WHILE # ABS dx > ex DO
SKIP
OD;
stop iteration:
x2 EXIT
value error:
EMPTY
FI
);
PROC f = (DBL x)DBL: x*x*x - LONG 3.0*x*x + LONG 2.0*x;
DBL first root, second root, third root;
DBLRES first result = find root(f, LENG -1.0, LENG 3.0, EMPTY);
CASE first result IN
(DBL first result): (
printf(($"1st root found at x = "f(dbl)l$, first result));
first root := first result
)
OUT printf($"No first root found"l$)
ESAC;
DBLRES second result = find root( (DBL x)DBL: f(x) / (x - first root), EMPTY, EMPTY, EMPTY);
CASE second result IN
(DBL second result): (
printf(($"2nd root found at x = "f(dbl)l$, second result));
second root := second result
)
OUT printf($"No second root found"l$)
ESAC;
DBLRES third result = find root( (DBL x)DBL: f(x) / (x - first root) / ( x - second root ), EMPTY, EMPTY, EMPTY);
CASE third result IN
(DBL third result): (
printf(($"3rd root found at x = "f(dbl)l$, third result));
third root := third result
)
OUT printf($"No third root found"l$)
ESAC
Output:
1st root found at x = +1.000000000000000000000000000e +0 2nd root found at x = +2.000000000000000000000000000e +0 3rd root found at x = +0.000000000000000000000000000e +0
C++
<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 ); } }</cpp>
D
<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) ;
}</d>
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
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[1] 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
J
J has builtin a root-finding operator, p., whose input is the (reversed) coeffiecients of the polynomial. Hence:
>{:p. 0 2 _3 1
2 1 0
Java
<java>public class Roots{ private static final double epsilon= 1E-10; //error bound, change for more or less accuracy
public double f(double x){ return x * x * x - 3 * x * x + 2 * x; //any formula you want here }
public void roots(double lowerBound, double upperBound, double step){ for(double x= lowerBound;x <= upperBound;x+= step){ double val; if(Math.abs(val= f(x)) < epsilon){ System.out.println(val); } } } }</java>
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
There are multiple obvious ways to do this in Mathematica.
Solve
This requires a full equation and will perform symbolic operations only:
In[1]:= Solve[x^3-3*x^2+2*x==0,x]
Out[1]= {{x->0},{x->1},{x->2}}
NSolve
This requires merely the polynomial and will perform numerical operations if needed:
In[2]:= NSolve[x^3 - 3*x^2 + 2*x , x]
Out[2]= {{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:
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.}
(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:
In[5]:= FindInstance[x^3 - 3*x^2 + 2*x == 0, x]
Out[5]= {{x->0}}
Reduce
This will (symbolically) reduce a given expression to the simplest possible form, solving equations and performing substitutions in the process:
In[6]:= Reduce[x^3 - 3*x^2 + 2*x == 0, x] Out[6]= 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)
Perl
<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 );
}</perl>
Python
From Perl: <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</python>