有没有办法在Javascript中计算pi？我知道你可以使用Math.PI找到这样的馅饼：
var pie = Math.PI;
alert(pie); // output "3.141592653589793"
但这不准确。我想要的是能够计算它，拥有任意数量的数字，而不是像pie = 3.141592...那样的数字。但是，我想要的不仅仅是拥有更多数字，而是尽可能多（比如拥有一千个数字，但我需要更多）。
答案 0 :(得分：2)
我在this website上找到了这段代码：
<html>
<head>
<title>Pi</title>
<script type="text/javascript">
mess = "";
Base = Math.pow(10, 11);
cellSize = Math.floor(Math.log(Base) / Math.LN10);
a = Number.MAX_VALUE;
MaxDiv = Math.floor(Math.sqrt(a));
function makeArray(n, aX, Integer) {
var i = 0;
for (i = 1; i < n; i++) aX[i] = null;
aX[0] = Integer
}
function isEmpty(aX) {
var empty = true
for (i = 0; i < aX.length; i++) if (aX[i]) {
empty = false;
break
}
return empty
}
function Add(n, aX, aY) {
carry = 0
for (i = n - 1; i >= 0; i--) {
aX[i] += Number(aY[i]) + Number(carry);
if (aX[i] < Base) carry = 0;
else {
carry = 1;
aX[i] = Number(aX[i]) - Number(Base)
}
}
}
function Sub(n, aX, aY) {
for (i = n - 1; i >= 0; i--) {
aX[i] -= aY[i];
if (aX[i] < 0) {
if (i > 0) {
aX[i] += Base;
aX[i - 1]--
}
}
}
}
function Mul(n, aX, iMult) {
carry = 0;
for (i = n - 1; i >= 0; i--) {
prod = (aX[i]) * iMult;
prod += carry;
if (prod >= Base) {
carry = Math.floor(prod / Base);
prod -= (carry * Base)
} else carry = 0;
aX[i] = prod
}
}
function Div(n, aX, iDiv, aY) {
carry = 0;
for (i = 0; i < n; i++) {
currVal = Number(aX[i]) + Number(carry * Base);
theDiv = Math.floor(currVal / iDiv);
carry = currVal - theDiv * iDiv;
aY[i] = theDiv
}
}
function arctan(iAng, n, aX) {
iAng_squared = iAng * iAng;
k = 3;
sign = 0;
makeArray(n, aX, 0);
makeArray(n, aAngle, 1);
Div(n, aAngle, iAng, aAngle);
Add(n, aX, aAngle);
while (!isEmpty(aAngle)) {
Div(n, aAngle, iAng_squared, aAngle);
Div(n, aAngle, k, aDivK);
if (sign) Add(n, aX, aDivK);
else Sub(n, aX, aDivK);
k += 2;
sign = 1 - sign
}
mess += "aArctan=" + aArctan + "<br>"
}
function calcPI(numDec) {
var ans = "";
t1 = new Date();
numDec = Number(numDec) + 5;
iAng = new Array(10);
coeff = new Array(10);
arrayLength = Math.ceil(1 + numDec / cellSize);
aPI = new Array(arrayLength);
aArctan = new Array(arrayLength);
aAngle = new Array(arrayLength);
aDivK = new Array(arrayLength);
coeff[0] = 4;
coeff[1] = -1;
coeff[2] = 0;
iAng[0] = 5;
iAng[1] = 239;
iAng[2] = 0;
makeArray(arrayLength, aPI, 0);
makeArray(arrayLength, aAngle, 0);
makeArray(arrayLength, aDivK, 0);
for (var i = 0; coeff[i] != 0; i++) {
arctan(iAng[i], arrayLength, aArctan);
Mul(arrayLength, aArctan, Math.abs(coeff[i]));
if (coeff[i] > 0) Add(arrayLength, aPI, aArctan);
else Sub(arrayLength, aPI, aArctan)
}
Mul(arrayLength, aPI, 4);
sPI = "";
tempPI = "";
for (i = 0; i < aPI.length; i++) {
aPI[i] = String(aPI[i]);
if (aPI[i].length < cellSize && i != 0) {
while (aPI[i].length < cellSize) aPI[i] = "0" + aPI[i]
}
tempPI += aPI[i]
}
for (i = 0; i <= numDec; i++) {
if (i == 0) sPI += tempPI.charAt(i) + ".<br>";
else {
if (document.getElementById("cbCount").checked) addcount = " (" + (i) + ")";
else addcount = "";
if (document.getElementById("cbSpace").checked) thespace = " ";
else thespace = "";
if ((i) % 50 == 0 && i != 0) sPI += tempPI.charAt(i) + addcount + "<br>";
else if (i % 5 == 0) sPI += tempPI.charAt(i) + thespace;
else sPI += tempPI.charAt(i)
}
}
ans += ("PI (" + numDec + ")=" + sPI + "<br>");
ans += ("Win PI=<br>3.1415926535897932384626433832795<br>");
t2 = new Date();
timeTaken = (t2.getTime() - t1.getTime()) / 1000;
ans += "It took: " + timeTaken + " seconds";
var myDiv = document.getElementById("d1");
myDiv.innerHTML = ans
}
</script>
</head>
<body>
<h1>
Pi Machin
</h1>
<form name="" id="" method="post" action="" enctype="text/plain" onsubmit="calcPI(this.t1.value);return false;">
Number of Digits:<br>
<input type="text" name="t1" id="t1" value="100" size="25" maxlength="25">
<br>Add Count:
<input type="checkbox" name="cbCount" id="cbCount" value="" checked="checked">
<br>Add Spaces:
<input type="checkbox" name="cbSpace" id="cbSpace" value="" checked="checked">
<br>
<input type="button" value="Calculate Pi" onclick="calcPI(this.form.t1.value)">
</form>
<div id="d1">0</div>
</body>
</html>
答案 1 :(得分：2)
您可以通过使用蒙特卡罗模拟来近似π的值。在[-1,1]范围内生成随机X和Y然后，似然（X，Y）在以原点为中心的单位圆中是π/ 4。更多样本可以更好地估计其价值。然后，您可以通过比较单位圆中的样本与样本总数的比率并乘以4来估算π。
this.pi = function(count) {
var inside = 0;
for (var i = 0; i < count; i++) {
var x = random()*2-1;
var y = random()*2-1;
if ((x*x + y*y) < 1) {
inside++
}
}
return 4.0 * inside / count;
}
答案 2 :(得分：1)
您可以将此用于您的目的
Math.PI.toFixed（n）的
其中n是您希望显示的小数位数。
显示pi的舍入值。它可以被认为是相当正确的，最多15位小数。
答案 3 :(得分：1)
以下是使用 无限系列
的实现方式function calculatePI(iterations = 10000){
let pi = 0;
let iterator = sequence();
for(let i = 0; i < iterations; i++){
pi += 4 / iterator.next().value;
pi -= 4 / iterator.next().value;
}
function* sequence() {
let i = 1;
while(true){
yield i;
i += 2;
}
}
return pi;
}
答案 4 :(得分：1)
这是Jeremy Gibbons在Unbounded Spigot Algorithms for the Digits of Pi（2004年）第6章中描述的流算法的实现：
function * generateDigitsOfPi() {
let q = 1n;
let r = 180n;
let t = 60n;
let i = 2n;
while (true) {
let digit = ((i * 27n - 12n) * q + r * 5n) / (t * 5n);
yield Number(digit);
let u = i * 3n;
u = (u + 1n) * 3n * (u + 2n);
r = u * 10n * (q * (i * 5n - 2n) + r - t * digit);
q *= 10n * i * (i++ * 2n - 1n);
t *= u;
}
}
// Demo
let iter = generateDigitsOfPi();
let output = document.querySelector("div");
(function displayTenNextDigits() {
let digits = "";
for (let i = 0; i < 10; i++) digits += iter.next().value;
output.insertAdjacentHTML("beforeend", digits);
scrollTo(0, document.body.scrollHeight);
requestAnimationFrame(displayTenNextDigits);
})();div { word-wrap:break-word; font-family: monospace }<div></div>
答案 5 :(得分：-3)
像这样使用toFixed（）方法。在toFixed里面你可以修复小数点后显示的位数。
var n=22/7;
console.log(n.toFixed(20))