PI迭代算法存在的问题 [英] Problems with iterative algorithm for PI

问题描述

我正在尝试一些近似Pi的公式，但这个 [ ^ ]一个给了我一些麻烦：

I was trying out some of the formulas for approximating Pi, but this[^] one gives me some trouble:

Dim alpha As Decimal = 6D - 4D * Math.Sqrt(CDec(2))
Dim piY As Decimal = Math.Sqrt(CDec(2)) - 1D

''' <summary>
'''
''' </summary>
''' <param name="N">Greater than or equal to 1</param>
''' <returns></returns>
''' <remarks></remarks>
Private Function GetPi(ByVal N As Integer) As Decimal
    Dim result As Decimal

    Dim TempAlpha, TempYPi As Decimal
    TempAlpha = alpha
    TempYPi = piY

    If N = 0 Then
        Return 1D / TempAlpha
    End If

    For i As Integer = 1 To N
        TempYPi = (1D - Math.Pow((1D - Math.Pow(TempYPi, 4D)), 1D / 4D)) / (1D + Math.Pow((1D - Math.Pow(TempYPi, 4D)), 1D / 4D))
        TempAlpha = Math.Pow(1D + TempYPi, 4D) * TempAlpha - Math.Pow(2D, 2D * (CDec(i) - 1D) + 3D) * TempYPi * (1D + TempYPi + Math.Pow(TempYPi, 2D))
    Next
    result = 1D / TempAlpha
    Return result
End Function

仅仅3次迭代后它就会停止变好（理论上它可以改善2 * 4 ^ n的精度），即使我将预制形式改为双精度到十进制也是如此。有人知道为什么或如何解决它？



PS。当然我可以使用Numerics.BigInteger做最后一个这里 [ ^ ]，但这有些不同。

It stops getting better after just 3 iterations (theoretically it shoud improve the precistion with 2*4^n), even when I change the precition form double to decimal. Does anybody know why, or how to fix it?

PS. Sure I could use the Numerics.BigInteger and do the last one here[^], but thats something slightly different.

推荐答案

来自维基百科页面：

From the wikipedia page:

四分之一收敛到PI，分三步给出大约100位数字，20步后超过一万亿位数

converges quartically to PI, giving about 100 digits in three steps and over a trillion digits after 20 steps

你无法获得这样的精度（100位）， double 或十进制数字。

那里实际上你的代码没有错，只是小数不适用于 Math.Pow 和 Math.Sqrt 。经过两次迭代后，所有数字都已正确。至少他们在我的电脑上。 :)



这里我使用的例子（C＃WindowsForms）：

There is actually nothing wrong with your code except that decimals won't work for Math.Pow and Math.Sqrt. After just two iterations all the digits are already correct. At least they were on my PC. :)

Here the example I worked with (C# WindowsForms):

namespace PI
{
    public partial class Form1 : Form
    {
        static double alpha = 6d - 4d * Math.Sqrt(2d);
        static double piY = Math.Sqrt(2d) - 1d;

        public Form1()
        {
            InitializeComponent();
        }

        private void button1_Click(object sender, EventArgs e)
        {
            int iterations = int.Parse(textBox2.Text);
            GetPi(iterations);
        }

        private double GetPi(int n)
        {
            double result;
            double TempAlpha, TempYPi;

            TempAlpha = alpha;
            TempYPi = piY;

            result =  1d / TempAlpha;

            textBox1.Clear();

            for(int i = 1; i<=n; i++)
            {
                TempYPi = (1d - Math.Pow((1d - Math.Pow(TempYPi, 4d)), 1d / 4d)) / (1d + Math.Pow((1d - Math.Pow(TempYPi, 4d)), 1d / 4d));
                TempAlpha = Math.Pow(1d + TempYPi, 4d) * TempAlpha - Math.Pow(2d, 2d * ((double)i - 1d) + 3d) * TempYPi * (1d + TempYPi + Math.Pow(TempYPi, 2d));
                result = 1d / TempAlpha;
                textBox1.Text += String.Format("{0} : {1}\r\n", i, result);
            }

            return result;
        }

    }
}

2次迭代的输出是：

The output for 2 iterations is:

1 : 3,14159264621355
2 : 3,14159265358979
PI: 3,1415926535897932384626433832795028841971693993751058209...

正如你在2次迭代后很容易看到的那样，double的精度已经用完了，所有计算的数字都已经正确了。



问候，



Manfred

As you can easily see after just 2 iteration the precision for double is already exhausted an all the digits calculated are already correct.

Regards,

Manfred

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