Solving polynomial equation in Java

Question

I have the following code which gives the price given a certain rate:

public static void main(String[] args) {
    double term = 5;
    double coupon = 5;
    double rate = 0.0432;

    //should get the price == 103;
    double price;

    double sum = 0;
    for (int i = 1; i <= term; i++) {
        sum = sum + (coupon / Math.pow((1 + rate), i));
    }

    double lastAmount = (100 / Math.pow((1 + rate), (term)));

    price = sum + lastAmount;

    System.out.println("price is:" + price);
}

My question is, how can I do this other way around? where the price is given and the rate is unknown. So I figured out, it's going to be an exponential equation. where it should look like below:

103x^5 - 5x^4 - 5x^3 - 5x^2 - 5x - 105=0

where x = 1+rate

solving the exponential equation, you get x = 1.0432 which means the rate is 0.0432 as expected.

How can I implement this in java to solve the equation.

Your equation `103x^5 - 5x^4 - 5x^3 - 5x^2 - 5x - 105=0` is a polynomial equation, not an exponential equation. — rgettman, Nov 27 '17 at 18:50
It's been proven that a degree five polynomial has no general closed-form solution. You're going to have to choose a numeric method. There's plenty of information about this out there - you just may have to search a little. — Dawood ibn Kareem, Nov 27 '17 at 18:52
Apache Commons has an implementation of Brent's Method for root finding. See: http://commons.apache.org/proper/commons-math/javadocs/api-3.3/org/apache/commons/math3/analysis/solvers/BrentSolver.html — Matthew McPeak, Nov 27 '17 at 18:58

score 0 · Answer 1 · answered Nov 27 '17 at 18:53

You would need to determine if your equation can be solved using radicals, and if so, you could try implementing that algorithm. Otherwise, you could try taking the TI-82 graphing calculator approach of "visualizing" the equation in code (iterating at a "sufficiently" small grain), watching for an inflection from positive to negative (or vice versa), or a point where the equation hits zero exactly, and approximating down to find a close enough value. However, in reading up on Quintic Functions, it seems there may not be a great formulaic solution to this problem (at least not one that my math capabilities can fathom).

The qunitinc equation can indeed not be solved by radicals, and there are specific numerical methods much more efficient than graphing and zooming. — , Nov 27 '17 at 19:02

score 0 · Answer 2 · 2017-11-27T19:01:10.647

The computation of the interest rate from the price is known to be a problem without a closed-form solution (for N>4).

Using the summation formula for geometric progressions, the equation spells

((1+r)^N - 1) / r = R

which amounts to finding the roots of a polynomial of degree N. You could use a polynomial solver, but here Netwon's method is more appropriate.

The function to be cancelled is

(1+r)^N - 1 - R r

and its derivative on r,

N(1+r)^(N-1) - R.

The rate that would correspond to simple interest, R/N, can be a good starting value.

score 0 · Accepted Answer · answered Nov 27 '17 at 21:00

Here is code that does it using the implementation of Brent's Method for root finding (from Apache Commons). It will print any roots between -10000 and +10000. It may not work for functions having multiple roots less than intervalSize (0.01) apart (yours is not such a function).

package sandbox;

import org.apache.commons.math3.analysis.UnivariateFunction;
import org.apache.commons.math3.analysis.solvers.BrentSolver;

public class BrentsMethodRootFinderExample {

    private void solveit () {
        BrentSolver solver = new BrentSolver();
        UnivariateFunction f = new UnivariateFunction() {

            @Override
            public double value(double x) {
                return 103*Math.pow(x, 5.0)-5*Math.pow(x, 4.0)-5*Math.pow(x, 3.0)-5*Math.pow(x, 2.0)-5*x - 105;
            }
        };

        double intervalStart = -10000;
        double intervalSize = 0.01;
        while (intervalStart < 10000) {
            intervalStart+= intervalSize;
            if(Math.signum(f.value(intervalStart)) != Math.signum(f.value(intervalStart+intervalSize))) {
                System.out.println("x = " + solver.solve(1000, f, intervalStart, intervalStart+intervalSize));
            }
        }
    }

    public static void main(String[] args) {
        new BrentsMethodRootFinderExample().solveit();
    }


}

Output:

x = 1.0432004413216969

score 0 · Answer 4 · answered Aug 19 '23 at 14:17

Another option for finding polynomial real roots may be using a library just programed by me.

The output using the command line version is the following:

$ java -jar ./polynomial-roots-v1.0-SNAPSHOT-all.jar -precision 15 -polynomial_coefs "-105 -5 -5 -5 -5 103"
{ 103 * x ^ 5 - 5 * x ^ 4 - 5 * x ^ 3 - 5 * x ^ 2 - 5 * x - 105 = 0 }

Solution[1]: x = 1.04320048314876

The library can be easily included in a Java application (it is offered under lgpl license)

You can download it at:

real polynomial root calculation library download

Solving polynomial equation in Java

4 Answers4