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comp.lang.c++

how to compute roots of a cubic function with minimal truncation errors?

Peng Yu

9/10/2008 4:24:00 AM

Hi,

I have the following program which computes roots of a cubic function.
The solution is sensitive to the type, which is due to the truncation
error. 'long double T' gives three solutions, and 'typedef double T'
gives one solutions. The correct number of solutions should be two, 1
and 2.

I know there is some trick to reduce the chance of under or overflow.
For example, std::abs(z) shall be implemented as
|x| * sqrt(1 + (y/x)*(y/x)) if |x|>|y|
and
|y| * sqrt(1 + (x/y)*(x/y)) if |y|>|x|,
where z is a complex number, and x and y are its real and complex
parts.

I'm wondering what trick can be played to reduce the truncation error
when solving the cubic polynomial equations.

BTW, I use the algorithm is shown at http://www.hawaii.edu/suremath/jroots...

Thanks,
Peng

#include <vector>
#include <complex>
#include <cmath>
#include <iostream>

template <typename T>
std::vector<T> roots_of_cubic_function(const T &a2, const T &a1, const
T &a0) {

const T one = static_cast<T>(1);
const T three = static_cast<T>(3);

T q = one / 3 * a1 - one / 9 * a2 * a2;
T r = one / 6 * (a1 * a2 - three * a0) - one / 27 * std::pow(a2, 3);

T Delta = std::pow(q, 3) + r * r;
std::cout << "Delta = " << Delta << std::endl;

std::vector<T> v;
if(Delta >= T()) {
T s1 = std::pow(r + sqrt(Delta), one/3);
T s2 = std::pow(r - sqrt(Delta), one/3);
v.push_back(s1 + s2 - a2 / 3);
if(Delta == T()) {
v.push_back(-.5 * (s1 + s2) - a2 / 3);
}
}
else {
std::complex<T> temp = sqrt(std::complex<T>(Delta));
std::complex<T> s1 = std::pow(r + temp, one/3);
std::complex<T> s2 = std::pow(r - temp, one/3);
const T minus_half = - static_cast<T>(1)/2;
v.push_back((s1 + s2 - a2 / 3).real());
v.push_back((minus_half * (s1 + s2) - a2 / 3 + std::complex<T>(0,
sqrt(three)/2) * (s1 - s2)).real());
v.push_back((minus_half * (s1 + s2) - a2 / 3 - std::complex<T>(0,
sqrt(three)/2) * (s1 - s2)).real());
}
return v;
}

int main () {
//typedef long double T;
typedef double T;
const T a2 = -4;
const T a1 = 5;
const T a0 = -2;

std::vector<T> v = roots_of_cubic_function<T>(a2, a1, a0);

std::cout << "Solutions:" << std::endl;
for(std::vector<T>::const_iterator it = v.begin(); it != v.end(); ++
it) {
T x = *it;
T f = ((x + a2) * x + a1) * x + a0;
std::cout << x << " " << f << std::endl;
}
}
2 Answers

Martin Eisenberg

9/10/2008 9:34:00 AM

0

[Hi Peng, followup to topical group sci.math.num-analysis]

From: Peng Yu <PengYu.UT@gmail.com>
Date: Tue, 9 Sep 2008 21:24:10 -0700 (PDT)

Hi,

I have the following program which computes roots of a cubic
function.
The solution is sensitive to the type, which is due to the truncation
error. 'long double T' gives three solutions, and 'typedef double T'
gives one solutions. The correct number of solutions should be two, 1
and 2.

I know there is some trick to reduce the chance of under or overflow.
For example, std::abs(z) shall be implemented as
|x| * sqrt(1 + (y/x)*(y/x)) if |x|>|y|
and
|y| * sqrt(1 + (x/y)*(x/y)) if |y|>|x|,
where z is a complex number, and x and y are its real and complex
parts.

I'm wondering what trick can be played to reduce the truncation error
when solving the cubic polynomial equations.

BTW, I use the algorithm is shown at
http://www.hawaii.edu/suremath/jroots...

Thanks,
Peng

#include <vector>
#include <complex>
#include <cmath>
#include <iostream>

template <typename T>
std::vector<T> roots_of_cubic_function(const T &a2, const T &a1,
const
T &a0) {

const T one = static_cast<T>(1);
const T three = static_cast<T>(3);

T q = one / 3 * a1 - one / 9 * a2 * a2;
T r = one / 6 * (a1 * a2 - three * a0) - one / 27 * std::pow(a2,
3);

T Delta = std::pow(q, 3) + r * r;
std::cout << "Delta = " << Delta << std::endl;

std::vector<T> v;
if(Delta >= T()) {
T s1 = std::pow(r + sqrt(Delta), one/3);
T s2 = std::pow(r - sqrt(Delta), one/3);
v.push_back(s1 + s2 - a2 / 3);
if(Delta == T()) {
v.push_back(-.5 * (s1 + s2) - a2 / 3);
}
}
else {
std::complex<T> temp = sqrt(std::complex<T>(Delta));
std::complex<T> s1 = std::pow(r + temp, one/3);
std::complex<T> s2 = std::pow(r - temp, one/3);
const T minus_half = - static_cast<T>(1)/2;
v.push_back((s1 + s2 - a2 / 3).real());
v.push_back((minus_half * (s1 + s2) - a2 / 3 + std::complex<T>(0,
sqrt(three)/2) * (s1 - s2)).real());
v.push_back((minus_half * (s1 + s2) - a2 / 3 - std::complex<T>(0,
sqrt(three)/2) * (s1 - s2)).real());
}
return v;
}

int main () {
//typedef long double T;
typedef double T;
const T a2 = -4;
const T a1 = 5;
const T a0 = -2;

std::vector<T> v = roots_of_cubic_function<T>(a2, a1, a0);

std::cout << "Solutions:" << std::endl;
for(std::vector<T>::const_iterator it = v.begin(); it != v.end();
++
it) {
T x = *it;
T f = ((x + a2) * x + a1) * x + a0;
std::cout << x << " " << f << std::endl;
}
}


--
Quidquid latine scriptum est, altum videtur.

Raymond Toy

9/12/2008 1:25:00 PM

0

>>>>> "Peng" == Peng Yu <PengYu.UT@gmail.com> writes:

Peng> Hi,
Peng> I have the following program which computes roots of a cubic function.
Peng> The solution is sensitive to the type, which is due to the truncation
Peng> error. 'long double T' gives three solutions, and 'typedef double T'
Peng> gives one solutions. The correct number of solutions should be two, 1
Peng> and 2.

Peng> I know there is some trick to reduce the chance of under or overflow.
Peng> For example, std::abs(z) shall be implemented as
Peng> |x| * sqrt(1 + (y/x)*(y/x)) if |x|>|y|
Peng> and
Peng> |y| * sqrt(1 + (x/y)*(x/y)) if |y|>|x|,
Peng> where z is a complex number, and x and y are its real and complex
Peng> parts.

Peng> I'm wondering what trick can be played to reduce the truncation error
Peng> when solving the cubic polynomial equations.

Peng> BTW, I use the algorithm is shown at http://www.hawaii.edu/suremath/jroots...

Look at the formula for s1 and s2. You might want to factor the r out
from sqrt(q^3+r^2) to abs(r)*sqrt(1+q^3/r^2). This might give you
better accuracy for s1 and s2.

Or choose one of the real roots, use Newton iteration to refine it,
and then solve the resulting quadratic separately. This requires a
bit of care too.

Ray