Thread: Nested loop Problems

  1. #1
    Registered User
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    Oct 2001
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    101

    Nested loop Problems

    How can I use nested loop to solve the problem below?

    Calculate the valuse of x from the infinite series
    x=4- 4/3 + 4/5 - 4/7 + 4/9 - 4/11 + .......
    Print a table that shows the value of x approximated by 1,2,3,etc, terms.

    What I wrote is:

    #include <iostream.h>
    #include <iomanip.h>
    #include <math.h>
    void print(double x,int n)
    {
    cout << (long int) (x) << '.';
    for( ; n>0; n--)
    {
    x = x - (long int) (x);
    x = 10. * x;
    cout << int(x);
    }
    }
    double dec(double x, int n)
    {
    double p;
    p = pow(10.,n);
    return ((long int)(x*p + ((x>0.) ? 0.499999 : -0.499999)))/p;
    }
    int main()
    {
    int n;
    double f, x = 3.141515151515;

    for (cout<< '\n', n=0; n<15; n++)
    {cout << '\n' ;print(x, n); }
    cout << '\n';
    return 0;
    }

    However, my teacher said that it's completely wrong! So, can anyone tells me how to do the problem by using nested loop?
    thank you very much!!

  2. #2
    and the hat of int overfl Salem's Avatar
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    Why do you need two loops?

    Code:
    #include <stdio.h>
    
    // x=4- 4/3 + 4/5 - 4/7 + 4/9 - 4/11 + .......
    // re-written as
    // x= 4/1 - 4/3 + 4/5 - 4/7 + 4/9 - 4/11 + .......
    int main ( ) {
        double  denom = 1;
        double  plusminus = 1.0;
        double  x = 0;
        int     i;
        for ( i = 0 ; i < 10 ; i++ ) {
            x = x + 4.0 / ( denom * plusminus );
            printf( "Term=%d, result=%f, denom=%.0f\n", i, x, denom*plusminus );
            denom = denom + 2.0;
            plusminus = -plusminus;
        }
        return 0;
    }
    If you dance barefoot on the broken glass of undefined behaviour, you've got to expect the occasional cut.
    If at first you don't succeed, try writing your phone number on the exam paper.

  3. #3
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    Thank you very much!!
    Last edited by DramaKing; 10-21-2001 at 12:07 PM.

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