pi

What is the motivation for defining PI as PI=4.D0*DATAN(1.D0) within Fortran 77 code? I understand how it works, but, what is the reasoning? This style ensures that the maximum precision available on ANY architecture is used when assigning a value to PI. Because Fortran does not have a built-in constant for PI . But rather than typing in the number manually and potentially making a mistake or not getting the maximum possible precision on the given implementation, letting the library calculate the result for you guarantees that neither of those downsides happen. These are equivalent and you'll

In a project using SciPy and NumPy, should I use scipy.pi , numpy.pi , or math.pi ? >>> import math >>> import numpy as np >>> import scipy >>> math.pi == np.pi == scipy.pi True So it doesn't matter, they are all the same value. The only reason all three modules provide a pi value is so if you are using just one of the three modules, you can conveniently have access to pi without having to import another module. They're not providing different values for pi. One thing to note is that not all libraries will use the same meaning for pi, of course, so it never hurts to know what you're using. For

I'm using some macros, and observing some strange behaviour. I've defined PI as a constant, and then used it in macros to convert degrees to radians and radians to degrees. Degrees to radians works fine, but radians to degrees does not: piTest.cpp: #include <cmath> #include <iostream> using namespace std; #define PI atan(1) * 4 #define radians(deg) deg * PI / 180 #define degrees(rad) rad * 180 / PI int main() { cout << "pi: " << PI << endl; cout << "PI, in degrees: " << degrees(PI) << endl; cout << "45 degrees, in rad: " << radians(45) << endl; cout << "PI * 180 / PI: " << (PI * 180 / PI) <<

I'm doing a program that aproximate PI and i'm trying to use long long, but it isn't working. Here is the code #include<stdio.h> #include<math.h> typedef long long num; main(){ num pi; pi=0; num e, n; scanf("%d", &n); for(e=0; 1;e++){ pi += ((pow((-1.0),e))/(2.0*e+1.0)); if(e%n==0) printf("%15lld -> %1.16lld\n",e, 4*pi); //printf("%lld\n",4*pi); } } karadoc %lld is the standard C99 way, but that doesn't work on the compiler that I'm using (mingw32-gcc v4.6.0). The way to do it on this compiler is: %I64d So try this: if(e%n==0)printf("%15I64d -> %1.16I64d\n",e, 4*pi); and scanf("%I64d", &n);

I was trying various methods to implement a program that gives the digits of pi sequentially. I tried the Taylor series method, but it proved to converge extremely slowly (when I compared my result with the online values after some time). Anyway, I am trying better algorithms. So, while writing the program I got stuck on a problem, as with all algorithms: How do I know that the n digits that I've calculated are accurate? Mysticial Since I'm the current world record holder for the most digits of pi, I'll add my two cents : Unless you're actually setting a new world record, the common practice

I have been thinking about this issue and I can't figure it out. Perhaps you can assist me. The problem is my code isn't working to output 1000 digits of pi in the python coding language. Here's my code: def make_pi(): q, r, t, k, m, x = 1, 0, 1, 1, 3, 3 while True: if 4 * q + r - t < m * t: yield m q, r, t, k, m, x = (10*q, 10*(r-m*t), t, k, (10*(3*q+r))//t - 10*m, x) else: q, r, t, k, m, x = (q*k, (2*q+r)*x, t*x, k+1, (q*(7*k+2)+r*x)//(t*x), x+2) digits = make_pi() pi_list = [] my_array = [] for i in range(1000): my_array.append(str("hello, I'm an element in an array \n" )) big_string = ""

I need some help calculating Pi. I am trying to write a python program that will calculate Pi to X digits. I have tried several from the python mailing list, and it is to slow for my use. I have read about the Gauss-Legendre Algorithm , and I have tried porting it to Python with no success. I am reading from Here , and I would appreciate any input as to where I am going wrong! It outputs: 0.163991276262 from __future__ import division import math def square(x):return x*x a = 1 b = 1/math.sqrt(2) t = 1/4 x = 1 for i in range(1000): y = a a = (a+b)/2 b = math.sqrt(b*y) t = t - x * square((y-a))