Solutions for FCUP's Numerical Analysis Assignments

1 - Numerical Series Approximation/source/java for speed comparison/Question1.java

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+public class Question1 {
+
+	public static void main(String args[]) {
+		System.out.println(machine_eps());
+	}
+
+	public static float machine_eps() {
+		float epsilon = 1.0f;
+
+		while ((1.0f + 0.5f * epsilon) != 1.0f){
+			epsilon = 0.5f * epsilon;
+		}
+
+		return epsilon;
+	}
+}

1 - Numerical Series Approximation/source/java for speed comparison/Question2.java

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+public class Question2 {
+
+        public static void main(String[] args) {
+	        double power_of_ten = Math.pow(10,-1 * Integer.parseInt(args[0]));
+	        compute_serie_2(power_of_ten);
+        }
+
+        /**
+         * Compute serie 2 with absole error below a given epsilon
+         */
+        public static void
+        compute_serie_2(double epsilon)
+        {
+                // Constant factor
+                double factor = 9.0f/(double)(2.0f*Math.sqrt(3));
+                // Taken from D'Alembert criterion for given that L = 0.5 < 1
+                double super_L = 1.0f/(double)(1-0.25);
+
+                // Summation index
+                int k = 0;
+                // Acumulation for the Summation
+                double acc = 0;
+
+                // Previous and current value on the series
+                double a = 1.0f; // a with k = 0
+                // While our absolute error is higher than the given epsilon
+                while(epsilon < factor*a*super_L)
+                {
+                        // Accumulate with previous term
+                        acc += a;
+                        // Compute current term
+                        a = _compute_serie_2_term(k++) * a;
+                }
+                // System.out.println(epsilon);
+                // System.out.println("Value: " + factor*acc);
+                // System.out.println("Error: " + (3.1415926535897932 - factor*acc));
+                // System.out.println("K: " +(k-1) + "\n--------------");
+                System.out.println(factor*acc + " " + (k - 1));
+        }
+
+        /**
+         * Compute k term of Serie 2 given the previous one
+         */
+        public static double
+        _compute_serie_2_term(int k)
+        {
+                return (double)(k+1.0f)/(double)(4.0f*k+6.0f);
+        }
+}

1 - Numerical Series Approximation/source/java for speed comparison/Question3.java

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+public class Question3 {
+
+    public static void main(String[] args) {
+	    double err = Math.pow(10, -Integer.parseInt(args[0].trim()));
+	    compute_serie_2(err);
+    }
+
+    public static double compute_serie_3_term(long n)
+    {
+        return -((double)(2*n+1))/(double)(2*n+3);
+    }
+
+    public static void compute_serie_2(double err)
+    {
+        long k = 0;
+        double ak = 1;
+        double acc = 1;
+
+        while (err < 4 * Math.abs(ak)) {
+	        ak = compute_serie_3_term(k++) * ak;
+	        acc += ak;
+        }
+
+        System.out.println(k + " " + 4 * acc);
+    }
+
+}
+