Reputation: 943
I am writing a Java program for Simpson's method. The basic program works as expected, although I cannot get the (absolute) error part to work.
I think I need to reference my while (absError < 0.000001)
loop differently. What am I doing wrong?
First try
public static double function(double x, double s) {
double sech = 1 / Math.cosh(x); // Hyperbolic cosecant
double squared = Math.pow(sech, 2);
return ((Math.pow(x, s)) * squared);
}
// Simpson's rule - Approximates the definite integral of f from a to b.
public static double SimpsonsRule(double a, double b, double s, int n) {
double dx, x, sum4x, sum2x;
double absError = 1.0;
double simpson = 0.0;
double simpson2 = 0.0;
dx = (b-a) / n;
sum4x = 0.0;
sum2x = 0.0;
// 4/3 terms
for (int i = 1; i < n; i += 2) {
x = a + i * dx;
sum4x += function(x,s);
}
// 2/3 terms
for (int i = 2; i < n-1; i += 2) {
x = a + i * dx;
sum2x += function(x,s);
}
// Compute the integral approximation.
simpson = function(a,s) + function(a,b);
simpson = (dx / 3)*(simpson + 4 * sum4x + 2 * sum2x);
while ( absError < 0.000001)
{
simpson2 = SimpsonsRule(a, b, s, n);
absError = Math.abs(simpson2 - simpson) / 15;
simpson = simpson2;
n++;
}
System.out.println("Number of intervals is " + n + ".");
return simpson2;
}
This doesn't work since I did not write
simpson2 = SimpsonsRule(a, b, s, n);
correctly.
I tried doing this a second way, but the solution ultimately also fails.
public static double function(double x, double s) {
double sech = 1 / Math.cosh(x); // Hyperbolic cosecant
double squared = Math.pow(sech, 2);
return ((Math.pow(x, s)) * squared);
}
// Simpson's rule - Approximates the definite integral of f from a to b.
public static double SimpsonsRule(double a, double b, double s, int n) {
double dx, x, sum4x, sum2x;
double absError = 1.0;
double simpson = 0.0;
double simpson2 = 0.0;
dx = (b-a) / n;
sum4x = 0.0;
sum2x = 0.0;
// 4/3 terms
for (int i = 1; i < n; i += 2) {
x = a + i * dx;
sum4x += function(x,s);
}
// 2/3 terms
for (int i = 2; i < n-1; i += 2) {
x = a + i * dx;
sum2x += function(x,s);
}
// Compute the integral approximation.
simpson = function(a,s) + function(a,b);
simpson = (dx / 3)*(simpson + 4 * sum4x + 2 * sum2x);
while ( absError < 0.000001)
{
n++;
dx = (b-a) / n;
// 4/3 terms
for (int i = 1; i < n; i += 2) {
x = a + i * dx;
sum4x += function(x,s);
}
// 2/3 terms
for (int i = 2; i < n-1; i += 2) {
x = a + i * dx;
sum2x += function(x,s);
}
simpson = function(a,s) + function(a,b);
simpson2 = (dx / 3)*(simpson + 4 * sum4x + 2 * sum2x);
absError = Math.abs(simpson2 - simpson) / 15;
simpson = simpson2;
}
System.out.println("Number of intervals is " + n + ".");
return simpson2;
}
I need to write the while loop differently. What is wrong with the way the error is referenced inside the while loop?
The java code up until
while ( absError < 0.000001)
{
simpson2 = SimpsonsRule(a, b, s, n);
absError = Math.abs(simpson2 - simpson) / 15;
simpson = simpson2;
n++;
}
System.out.println("Number of intervals is " + n + ".");
return simpson2;
Works fine and correctly calculates Simpson's method.
Upvotes: 4
Views: 2920
Reputation: 766
Looks like your Simpson's method implementation is not converging. The easiest thing you can do to avoid infinite cycle in while - you have to add another condition - maximum number of iterations.
Something like that:
int n = 0;
while (error < ACCURACY && n++ < MAX_ITERATIONS) {
// while body
}
where ACCURACY
is 0.000001
in your case (or 1e-6
) and MAX_ITERATIONS
is an integer constant, for example 100000
or 1e+6
.
Why your algorithm is not converging - this is another question - look carefully on your formulas - use debugging tools. Good luck!
Upvotes: 3
Reputation: 943
I fixed it. Thanks for your help.
// Simpson's rule - Approximates the definite integral of f from a to b.
public static double SimpsonsRule(double a, double b, double s, int n) {
double simpson, dx, x, sum4x, sum2x;
dx = (b-a) / n;
sum4x = 0.0;
sum2x = 0.0;
// 4/3 terms
for (int i = 1; i < n; i += 2) {
x = a + i * dx;
sum4x += function(x,s);
}
// 2/3 terms
for (int i = 2; i < n-1; i += 2) {
x = a + i * dx;
sum2x += function(x,s);
}
// Compute the integral approximation.
simpson = function(a,s) + function(a,b);
simpson = (dx / 3)*(simpson + 4 * sum4x + 2 * sum2x);
return simpson;
}
// Handles the error for for f(x) = t^s * sech(t)^2. The integration is
// done from 0 to 100.
// Stop Simspson's Method when the relative error is less than 1 * 10^-6
public static double SimpsonError(double a, double b, double s, int n)
{
double futureVal;
double absError = 1.0;
double finalValueOfN;
double numberOfIterations = 0.0;
double currentVal = SimpsonsRule(a,b,s,n);
while (absError / currentVal > 0.000001) {
n = 2*n;
futureVal = SimpsonsRule(a,b,s,n);
absError = Math.abs(futureVal - currentVal) / 15;
currentVal = futureVal;
}
// Find the number of iterations. N starts at 8 and doubles every iteration.
finalValueOfN = n / 8;
while (finalValueOfN % 2 == 0) {
finalValueOfN = finalValueOfN / 2;
numberOfIterations++;
}
System.out.println("The number of iterations is " + numberOfIterations + ".");
return currentVal;
}
Upvotes: 1