Compute and approximat value for eulers constant
the problem is for me to compute and approxiamte value for euler's constant by summing the series (adding up the terms 1, 1/2, ..., 1/n in that order and then subtracting the value fo In(n)).
I have to do this with both double and float arithmetic.
Here is what I have so far.
import java.io.*;
public class PP2 {
public static void main(String[] args) throws
IOException {
System.out.println("Enter the limit for sum:");
BufferedReader kbd = new BufferedReader(new
InputStreamReader(System.in));
String S = kbd.readLine();
int loops = Integer.parseInt(S.trim());
int n;
for (n = 1; n <= loops; n++)
{
double sum = sum + (double)(1/n);
float sum2 = sum2 + (float)(1/n);
}
double condoub = sum - (double)Math.log(loops);
float conflout = sum2 - (float)Math.log(loops);
System.out.println("Euler's const, double: =" + condoub);
System.out.println("Euler's const, flout: =" + conflout);
}
}
thanks for the help :D
[1070 byte] By [
Mcody2] at [2007-11-11 7:00:31]

# 1 Re: Compute and approximat value for eulers constant
You should declare the sum and sum2 prior to the loop, the way you have
coded it will not accumulate, it doesn't even compile :)
import java.io.*;
public class PP2 {
public static void main(String[] args) throws IOException {
System.out.println("Enter the limit for sum:");
BufferedReader kbd = new BufferedReader(new InputStreamReader(System.in));
String S = kbd.readLine();
int loops = Integer.parseInt(S.trim());
int n;
double sum=0;
float sum2=0;
for (n = 1; n <= loops; n++) {
sum = sum + (double) (1 / n);
sum2 = sum2 + (float) (1 / n);
}
double condoub = sum - (double) Math.log(loops);
float conflout = sum2 - (float) Math.log(loops);
System.out.println("Euler's const, double: =" + condoub);
System.out.println("Euler's const, flout: =" + conflout);
}
}
sjalle at 2007-11-11 22:39:48 >
