Could not understand for loop

[adesh]

Hi ,

The below program finds square root of a given number

double sqrroot(double val)
{
 double i=0;
 double temp1,temp2;
      
       while(i*i<=val)
              i=i+0.1;

       temp1=i;
   
       for(int j=0;j<15;j++)
       {
              temp2=val;
              temp2=temp2/temp1;
              temp2=temp2+temp1;
              temp2=temp2/2;
              temp1=temp2;
       }
return temp2;
}

int main()
{
       cout << sqrroot ( 2.0 ) << endl ;
       cout << sqrroot ( 3.0 ) << endl ;
       cout << sqrroot ( 5.0 ) << endl ;
       cout << sqrroot ( 9.0 ) << endl ;

       _getche();
       return 0;
}

I am unable to understand this part of the code .

for(int j=0;j<15;j++)
       {
              temp2=val;
              temp2=temp2/temp1;
              temp2=temp2+temp1;
              temp2=temp2/2;
              temp1=temp2;
       }

Can anyone give me a simple explanation .
Thankyou for your time .

[Jikax]

don't you understand what a for loop does? or do you want to know what happens inside the for loop?

[adesh]

I know looping constructs .

I want to know how the for loop get you the square root . The loop is made to
run predefined (15 ) no of times . Why is that ?

[roberts]

Looks kind of like the Babylonian method.

http://en.wikipedia.org/wiki/Methods_of_computing_square_roots#Babylonian_method

[adesh]

thanks for the reply .

I think i was able to understand about the babylonian method .

But i dont understand why the loop runs 15 times .

[Gaminic]

I'm guessing it's a precision issue. Taking the squareroot of a non-square usually results in an infinite amount of digits. Depending on the application, the required precision is just a few digits.

An iterated method will narrow down its precision at every iteration by fine-tuning based on the results of the previous step. Here they use 15, but you can probably pick any number depending on how precise you wish to be.

[adesh]

Thanks , i feel now i have understood it after all the trouble with the help of
you all.