Axion004 - 4 months ago 7x

Java Question

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)`

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.

Answer

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!

Source (Stackoverflow)

Comments