如果你想计算π值的近似值(出于某种原因)，你应该尝试二进制提取算法。Bellard对BBP的改进给出了O(N²)中的PI。

如果你想获得π值的近似值来进行计算，那么:

PI = 3.141592654

当然，这只是一个近似值，并不完全准确。误差略大于0.00000000004102。(4个十万亿分之一，大约4/10,000,000,000)。

如果你想用π做数学运算，那就准备好铅笔和纸，或者电脑代数包，然后使用π的精确值π。

如果你真的想要一个公式，这个很有趣:

π = -i ln(-1)

这是一个“经典”方法，非常容易实现。 这个在python(不是最快的语言)中的实现:

from math import pi
from time import time


precision = 10**6 # higher value -> higher precision
                  # lower  value -> higher speed

t = time()

calc = 0
for k in xrange(0, precision):
    calc += ((-1)**k) / (2*k+1.)
calc *= 4. # this is just a little optimization

t = time()-t

print "Calculated: %.40f" % calc
print "Constant pi: %.40f" % pi
print "Difference: %.40f" % abs(calc-pi)
print "Time elapsed: %s" % repr(t)

你可以在这里找到更多信息。

无论如何，在python中获得精确的圆周率值的最快方法是:

from gmpy import pi
print pi(3000) # the rule is the same as 
               # the precision on the previous code

下面是gpy pi方法的源代码，我认为在这种情况下，代码没有注释那么有用:

static char doc_pi[]="\
pi(n): returns pi with n bits of precision in an mpf object\n\
";

/* This function was originally from netlib, package bmp, by
 * Richard P. Brent. Paulo Cesar Pereira de Andrade converted
 * it to C and used it in his LISP interpreter.
 *
 * Original comments:
 * 
 *   sets mp pi = 3.14159... to the available precision.
 *   uses the gauss-legendre algorithm.
 *   this method requires time o(ln(t)m(t)), so it is slower
 *   than mppi if m(t) = o(t**2), but would be faster for
 *   large t if a faster multiplication algorithm were used
 *   (see comments in mpmul).
 *   for a description of the method, see - multiple-precision
 *   zero-finding and the complexity of elementary function
 *   evaluation (by r. p. brent), in analytic computational
 *   complexity (edited by j. f. traub), academic press, 1976, 151-176.
 *   rounding options not implemented, no guard digits used.
*/
static PyObject *
Pygmpy_pi(PyObject *self, PyObject *args)
{
    PympfObject *pi;
    int precision;
    mpf_t r_i2, r_i3, r_i4;
    mpf_t ix;

    ONE_ARG("pi", "i", &precision);
    if(!(pi = Pympf_new(precision))) {
        return NULL;
    }

    mpf_set_si(pi->f, 1);

    mpf_init(ix);
    mpf_set_ui(ix, 1);

    mpf_init2(r_i2, precision);

    mpf_init2(r_i3, precision);
    mpf_set_d(r_i3, 0.25);

    mpf_init2(r_i4, precision);
    mpf_set_d(r_i4, 0.5);
    mpf_sqrt(r_i4, r_i4);

    for (;;) {
        mpf_set(r_i2, pi->f);
        mpf_add(pi->f, pi->f, r_i4);
        mpf_div_ui(pi->f, pi->f, 2);
        mpf_mul(r_i4, r_i2, r_i4);
        mpf_sub(r_i2, pi->f, r_i2);
        mpf_mul(r_i2, r_i2, r_i2);
        mpf_mul(r_i2, r_i2, ix);
        mpf_sub(r_i3, r_i3, r_i2);
        mpf_sqrt(r_i4, r_i4);
        mpf_mul_ui(ix, ix, 2);
        /* Check for convergence */
        if (!(mpf_cmp_si(r_i2, 0) && 
              mpf_get_prec(r_i2) >= (unsigned)precision)) {
            mpf_mul(pi->f, pi->f, r_i4);
            mpf_div(pi->f, pi->f, r_i3);
            break;
        }
    }

    mpf_clear(ix);
    mpf_clear(r_i2);
    mpf_clear(r_i3);
    mpf_clear(r_i4);

    return (PyObject*)pi;
}

编辑:我在剪切和粘贴和缩进方面有一些问题，你可以在这里找到源代码。

下面是我在高中时学过的计算圆周率的技巧。

我之所以分享它，是因为我认为它足够简单，任何人都可以无限期地记住它，而且它教会了你“蒙特卡罗”方法的概念——这是一种统计方法，可以得到答案，这些答案不会立即通过随机过程演绎出来。

画一个正方形，在这个正方形内画一个象限(半圆的四分之一)(一个半径等于正方形边的象限，这样它就能尽可能多地填充正方形)

现在向正方形投掷飞镖，并记录飞镖落在何处——也就是说，在正方形内任意选择一个点。当然，它落在了正方形内部，但它落在半圆内部吗?记录这个事实。

重复此过程多次，你会发现半圆内的点数量与抛出的总数量之比为x。

由于正方形的面积是r乘以r，可以推导出半圆的面积是x乘以r乘以r(即x乘以r的平方)。因此x乘以4会得到。

这不是一个快速使用的方法。但这是蒙特卡罗方法的一个很好的例子。如果你环顾四周，你可能会发现许多超出你计算能力的问题都可以用这种方法来解决。

使用machine -like公式

176 * arctan (1/57) + 28 * arctan (1/239) - 48 * arctan (1/682) + 96 * arctan(1/12943) 

[; \left( 176 \arctan \frac{1}{57} + 28 \arctan \frac{1}{239} - 48 \arctan \frac{1}{682} + 96 \arctan \frac{1}{12943}\right) ;], for you TeX the World people.

在Scheme中实现，例如:

(+ (- (+ (* 176 (atan(第57 / 1)))(* (atan (28 / 1 239))) (* 48 (atan (1 / 682))) (* 96 (12943 atan (/ 1))))

从圆面积计算π:-)

<input id="range" type="range" min="10" max="960" value="10" step="50" oninput="calcPi()"> <br> <div id="cont"></div> <script> function generateCircle(width) { var c = width/2; var delta = 1.0; var str = ""; var xCount = 0; for (var x=0; x <= width; x++) { for (var y = 0; y <= width; y++) { var d = Math.sqrt((x-c)*(x-c) + (y-c)*(y-c)); if (d > (width-1)/2) { str += '.'; } else { xCount++; str += 'o'; } str += "&nbsp;" } str += "\n"; } var pi = (xCount * 4) / (width * width); return [str, pi]; } function calcPi() { var e = document.getElementById("cont"); var width = document.getElementById("range").value; e.innerHTML = "<h4>Generating circle...</h4>"; setTimeout(function() { var circ = generateCircle(width); e.innerHTML = "<pre>" + "π = " + circ[1].toFixed(2) + "\n" + circ[0] +"</pre>"; }, 200); } calcPi(); </script>

正好是3![弗林克教授(辛普森一家)]

开玩笑，但这里有一个在c#(。微软网络框架。

using System;
using System.Text;

class Program {
    static void Main(string[] args) {
        int Digits = 100;

        BigNumber x = new BigNumber(Digits);
        BigNumber y = new BigNumber(Digits);
        x.ArcTan(16, 5);
        y.ArcTan(4, 239);
        x.Subtract(y);
        string pi = x.ToString();
        Console.WriteLine(pi);
    }
}

public class BigNumber {
    private UInt32[] number;
    private int size;
    private int maxDigits;

    public BigNumber(int maxDigits) {
        this.maxDigits = maxDigits;
        this.size = (int)Math.Ceiling((float)maxDigits * 0.104) + 2;
        number = new UInt32[size];
    }
    public BigNumber(int maxDigits, UInt32 intPart)
        : this(maxDigits) {
        number[0] = intPart;
        for (int i = 1; i < size; i++) {
            number[i] = 0;
        }
    }
    private void VerifySameSize(BigNumber value) {
        if (Object.ReferenceEquals(this, value))
            throw new Exception("BigNumbers cannot operate on themselves");
        if (value.size != this.size)
            throw new Exception("BigNumbers must have the same size");
    }

    public void Add(BigNumber value) {
        VerifySameSize(value);

        int index = size - 1;
        while (index >= 0 && value.number[index] == 0)
            index--;

        UInt32 carry = 0;
        while (index >= 0) {
            UInt64 result = (UInt64)number[index] +
                            value.number[index] + carry;
            number[index] = (UInt32)result;
            if (result >= 0x100000000U)
                carry = 1;
            else
                carry = 0;
            index--;
        }
    }
    public void Subtract(BigNumber value) {
        VerifySameSize(value);

        int index = size - 1;
        while (index >= 0 && value.number[index] == 0)
            index--;

        UInt32 borrow = 0;
        while (index >= 0) {
            UInt64 result = 0x100000000U + (UInt64)number[index] -
                            value.number[index] - borrow;
            number[index] = (UInt32)result;
            if (result >= 0x100000000U)
                borrow = 0;
            else
                borrow = 1;
            index--;
        }
    }
    public void Multiply(UInt32 value) {
        int index = size - 1;
        while (index >= 0 && number[index] == 0)
            index--;

        UInt32 carry = 0;
        while (index >= 0) {
            UInt64 result = (UInt64)number[index] * value + carry;
            number[index] = (UInt32)result;
            carry = (UInt32)(result >> 32);
            index--;
        }
    }
    public void Divide(UInt32 value) {
        int index = 0;
        while (index < size && number[index] == 0)
            index++;

        UInt32 carry = 0;
        while (index < size) {
            UInt64 result = number[index] + ((UInt64)carry << 32);
            number[index] = (UInt32)(result / (UInt64)value);
            carry = (UInt32)(result % (UInt64)value);
            index++;
        }
    }
    public void Assign(BigNumber value) {
        VerifySameSize(value);
        for (int i = 0; i < size; i++) {
            number[i] = value.number[i];
        }
    }

    public override string ToString() {
        BigNumber temp = new BigNumber(maxDigits);
        temp.Assign(this);

        StringBuilder sb = new StringBuilder();
        sb.Append(temp.number[0]);
        sb.Append(System.Globalization.CultureInfo.CurrentCulture.NumberFormat.CurrencyDecimalSeparator);

        int digitCount = 0;
        while (digitCount < maxDigits) {
            temp.number[0] = 0;
            temp.Multiply(100000);
            sb.AppendFormat("{0:D5}", temp.number[0]);
            digitCount += 5;
        }

        return sb.ToString();
    }
    public bool IsZero() {
        foreach (UInt32 item in number) {
            if (item != 0)
                return false;
        }
        return true;
    }

    public void ArcTan(UInt32 multiplicand, UInt32 reciprocal) {
        BigNumber X = new BigNumber(maxDigits, multiplicand);
        X.Divide(reciprocal);
        reciprocal *= reciprocal;

        this.Assign(X);

        BigNumber term = new BigNumber(maxDigits);
        UInt32 divisor = 1;
        bool subtractTerm = true;
        while (true) {
            X.Divide(reciprocal);
            term.Assign(X);
            divisor += 2;
            term.Divide(divisor);
            if (term.IsZero())
                break;

            if (subtractTerm)
                this.Subtract(term);
            else
                this.Add(term);
            subtractTerm = !subtractTerm;
        }
    }
}