C++ Pi Estimating Program Not Working Correctly

You have three problems:

First, you do not implement the Euler formulae correctly.

π²/6 = 1/1² + 1/2² + 1/3² + ...

eulerall = sqrt(6/pow(counter+1,2)) + eulerall; 

The square root of the sum is not the sum of the square roots.

π³/32 = 1/1³ + 1/3³ + 1/5³ + ...

eulerodd = (sqrt(32)*pow(-1, counter)) / (2*counter + 1) + eulerodd; 

This is just... wrong.

Second, you increment counter three times in the loop, instead of once:

while(terms > counter){ ... counter++; ... counter++; ... counter++; ... } 

Third, and most fundamental, you didn't follow the basic rule of software development: start small and simple, add complexity as little at a time, test at every step, and never add to code that doesn't work.

my first problem is that the second and third of my equations are wrong and I can't figure out why

Use counter++ just once. Apart from this Leibniz looks fine. Eulerall is not correct, you should sum all factors and then do sqrt and multiplication at the end:

eulerall = 1/pow(counter+1,2) + eulerall; // do sqrt and multiplication at the end to get Pi 

The similar thing with eulerodd: you should sum all factors and then do sqrt and multiplication at the end.

My second problem is that the program only reads the first input value and stops there.

In fact this is your first problem. This is because you are incrementing counter multiple times:

while(terms > counter){ leibniz = 4*(pow(-1, counter)) / (2*counter+1) + leibniz; counter++; // << increment ^^^^^^^^^^ eulerall = sqrt(6/pow(counter+1,2)) + eulerall; counter++; // << increment ^^^^^^^^^^ eulerodd = (sqrt(32)*pow(-1, counter)) / (2*counter + 1) + eulerodd; counter++; // << increment ^^^^^^^^^^ cout << terms << " " << leibniz << " " << eulerall << " " << eulerodd <<endl; } 

You should increment counter just once.

You're using the same counter and incrementing it after each calculation. So each technique is only accounting for every third term. You should increment counter only once, at the end of the loop.

Also note that it is generally bad form to use a floating-point value as a loop counter. It only takes on integer values in your program, so you can just make it an int. Nothing else needs to change; the math will run the same because the int will promote to a double when you combine the two in math operations.

 #include<iostream> #include<conio.h> #include<cmath> using namespace std; char* main() { while(1) { int Precision; float answer = 0; cout<<"Enter your desired precision to find pi number : "; cin>>Precision; for(int i = 1;i <= Precision;++i) { int sign = (pow((-1),static_cast<float>(i + 1))); answer += sign * 4 * ( 1 / float( 2 * i - 1)); } cout<<"Your answer is equal to : "<<answer<<endl; _getch(); _flushall(); system("cls"); } return "That is f..."; }