Viete formula in cpp - c++

Question

I need to make a code for the Viete's formula for pi for my extended essay. In the code i have the individual terms, but I don't know how to combine them all and use the product function to get the product of all the terms.
This is what i have till now:
#include <stdio.h>
#include <math.h>
int main()
{
int n, i;
n = 10;
i = 1;
double an = sqrt(2);
while(i <= n)
{
printf("a%d = %.10f\n", i, an);
an = sqrt(2 + an);
++i;
}
return 0;
}

Answer 1

Although you've tagged C++, the code you present is strictly old-school C. I'll skip the mini-code review I usually do as you presented a bunch of variable declarations and nothing more.
Here's a program that utilizes both the "product formula" and limit expression. Both expressions are products, so that's where the confusion came from. However, it helps to read up on the stuff you're intending to work with.
// Someone who knows ranges better than me might be able to shorten this up.
#include <cmath> // Preferred over <math.h> when writing C++
#include <iomanip>
#include <iostream>
#include <numbers> // inv_pi and sqrt(2) as of C++20
// Note the lack of a sqrt() function call or use of std::numbers::sqrt2
double viete_product(int n) {
double product = 1.0;
for (int i = 1; i <= n; ++i) {
product *= std::cos(std::numbers::pi / std::pow(2.0, i + 1));
}
return product;
}
// The limit expression does require a sqrt(2) as an initial term.
// Both are product formulas, which likely is the cause of confusion.
double viete_limit_expression(int n) {
/*
* I'm not in the business of doing people's homeowork for them
* so I'm excluding the code I assume you have to implement.
* The other product formula provides enough of a foundation
* that you should be able to figure it out from there.
*/
}
int main() {
const int numRounds = 10;
std::cout << std::setw(25) << std::right
<< "2 / pi : " << (2.0 * std::numbers::inv_pi) << "\n"
<< std::setw(25) << std::right
<< "Viete product: " << viete_product(numRounds) << "\n"
<< std::setw(25) << std::right
<< "Viete limit expression: " << viete_limit_expression(numRounds)
<< '\n';
}
Output:
❯ ./a.out
2 / pi : 0.63662
Viete product: 0.63662
Viete limit expression: 0.63662
At this precision, 10 rounds gets a convergence. I didn't try to find out what a minimum number of rounds was at this precision, 10 or a 100 rounds ran essentially the same on my machine.
As far your code is concerned, you have the setup. You just need to do the math. It's a division and a multiplication in a loop. Write out how you would hand calculate the answer, and the adapt that algorithm to code.

Answer 2

Given the posted code, it seems that the OP is trying to calculate π using the Viète's formula in the form of infinite product of nested radicals.
I'd start rearrenging that formula to make the intent more explicit.
I have the individual terms, but I don't know how to combine them all and use the product function to get the product of all the terms.
I assume that by "product function" the OP means the product of a sequence of factors.
The easiest way to translate that in C++ is to write a simple loop.
double product = starting_value; // Hopefully not 0.
for ( int i = 1; i <= n; ++i ) {
// Evaluate the factor somehow...
product *= factor_i; // product = product * factor_i
}
The OP already figured out the following recurrence relation.
The missing piece is the sequence of products.
Given that this is a converging sequence and that a double has only a limited precision, you may consider to stop the iterations when a certain accuracy has been reached, instead of relying on a fixed number of iterations.
Moreover, at some point (after 26 iterations, in my experiments) the numeric value of the product won't change anymore.