# Thread: Nested loop Problems

1. ## 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. 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;
}```

3. Thank you very much!!

Popular pages Recent additions