Actually i dont have any problem with coding. Im just stucked at any method which can solve equations of higer degree (upto degree =5)

for example solution of
ax^3+bx^2+cx+d=0


i want to Solve the equations in C Programming.
i want to numerically solve the equations an i only want real values not imaginary.

This post has been edited by umer_alvi: 21 Feb, 2007 - 04:15 PM

Did you want to symbolicly solve these equations, or numericly solve the equation? If numericly did you want to include the imaginary results?

You could of course solve this ingeneral and then use the formulas to calculate the results... one of the roots looks like this:

x=((((-2) * b^3 + 9 * a * c * b - 27 * a^2 * d + sqrt((4 * (3 * a * c - b^2))^3 + ((-2) * b^3 + 9 * a * c * b - 27 * a^2 * d)))^2)^(1/3))/(3 * 2^(1/3) * a) - b/(3 * a) - (2^(1/3) * (3 * a * c - b^2))/(3 * a * ((-2) * b^3 + 9 * a * c * b - 27 * a^2 * d + sqrt((4 * (3 * a * c - b^2))^3 + ((-2) * b^3 + 9 * a * c * b - 27 * a^2 * d)^2))^(1/3))

(as done by mathematica anyway).

however there are lots of numerical methods to solve this much easier than the algebreic sollution.

Newton's method.
x(n+1) <- x(n) - F(x(n))/F'(x(n))

or in the case x(n+1)=(ax(n)^3+bx(n)^2+cx(n)+d)/(3ax(n)^2+2bx(n)+c)

where each x(n) is an approximation of the root and x(0) is an initial guess.

Bisection
Works like a binary search for the root in a interval [a, b].

If f(a)f(b ) < 0 then f(x) crosses the x-axis in this intevel

B1. define c=(a+b )/2
B2. if b-c <= some error tollerance the accept c and stop
B3. else if the sign of f(b ) time the sign of f(c ) <= the a=c else b=c
B4. return to step B1.

This is nice because works with any function and you don't have to calculate a derivitive and you can write a routine to search for a valid interval [a, b].

There are also the secant method, and other Fixed point interation algorithems (both Newton's and Secant methods are versions of this).

A google search on any of those should set you on the path.

This post has been edited by NickDMax: 21 Feb, 2007 - 03:58 PM

i am trying to create a function for bisection so i can plug in different eqautions and intervals to find the root.
Here's what i have so far. i am very confused.


#include <iostream>
#include <cmath>

using namespace std;

void equation(double z, double y)
{

double fa = (z*z*z) - (7*z*z) + (5*z) + 3;
double fb = (y*y*y) - (7*y*y) + (5*y) + 3;

}




void Bisec(float ab, float bc)
{

float fa;
float fb;

double error = .001;
float root;
float c;
if (fa*fb < 0)
{
c = (ab + bc) / 2;

if (ab - bc <= error)
{
c = root;

cout << "The root is " << root << endl;

if (fb*fa > 0)
{
bc = c;
c = root;
cout << "The root is " << root << endl;
}
else
{
ab = c;
c = root;
cout << "The root is " << root << endl;

}}}}


int main()
{
cout << "The root in the equation x^3 - 7x^2 + 5x +3 in the interval [0, 2] is " << endl;
cout << Bisec(0, 2) << endl;



}