For class we need to write a program that takes in an input (n) that determines the number of terms in the approximation of the value of pi and outputs the approximation. The program also needs a for loop to calculate this until the user wants to quit using the program.
You should be able to do it pretty easily with a for loop
Basically something like this
1 2 3 4 5 6 7 8 9 10 11 12 13 14
double pi( int n )
{
double pi = 4.0 , decimal = 1.0;
while( n > 2 )
{
decimal -= ( 1.0 / ( 2.0 * n + 1 ) );
--n;
decimal += ( 1.0 / ( 2.0 * n + 1 ) );
--n;
}
if( n > 0 )
decimal -= ( 1.0 / ( 2.0 * n + 1 ) );
return( pi * decimla );
}
I could be wrong though but it looks to me like the formula is
4 * (1 / 1 + 2n ) The inside part would be all the fractions added up ex 1 / 1 - 1 / 3 + 1 / 5 - 1/7
So if n was 3 then it would be ( I gave it the 0 by default so I start at 1 and go to 3 ) When I say I gave it 0 by default I mean I initialized it to 1
4 * ( -1/7 + 1/5 - 1/3 + 1/1 ) = 4 * ( 0.72380952380952380952380952380952 ) = 2.8952380952380952380952380952381
Am I misunderstanding the formula you provided?
Also yes I did do it in reverse order
But basically if n is even say for example 4
That would mean
1 - + - +
And I started with -
And an odd would end in a -
so lets look at 3
1 - + -