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Hello every one i need to help me in understanding this small program
...you can approximate by using the following series
∏=4(1-1/3+1/5-1/7+1/9-1/11+1/13-.....-1/(2i-1)+1/(2i+1))
write a program that displays the value for i=10000,20000,....,and 100000
I've developed this code to solve the problem

public static void main(String[] args) {
       int count=1;
       double sum=0;
       for(int i=1;i<2*i+1;i+=2)
       {
           if(count % 2==0)
               sum-=1.0/i;
           else
               sum+=1.0/i;
           count++;
       }
       double p=4*sum;
       System.out.println("The value of p is "+p);
    }

my question
is this code I've written solve the problem?
and if yes, what about "compute the value when i is equal to 10000,20000,.....,and 100000"
that is supposed to be implemented in my code

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Last Post by masijade
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Put that code into a different method (not main) that accepts an integer as a paramter (and change i<2*i+1 to i<2*parameter+1) and then call that method from main with 10000, 20000, etc (which can be figured in a for loop from 1 to 10 multiplying this number by 10000).

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Thanks very much but i take another way by using nested loops
and this gave me true result....this is the code

public static void main(String[] args) {
       int count=1;
       double sum=0;
       double p;
       for(int i=10000;i<=100000;i+=10000)
       {
       for(int x=1;x<=2*i+1;x+=2)
       {
           if(count % 2==0)
               sum-=1.0/x;
           else
               sum+=1.0/x;
           count++;
       }
       p=4*sum;
       System.out.println("The value of p is "+p);
       sum=0;
       count=1;
       }
    }
}

I'd be grateful if you give me any comments about my code to improve my writing skills
thanks......

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To each his own. You need to start using methods and the such though, and this was a good opportunity to try it out.

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