29

.net 框架在 Math 类中提供了一种为 double 供电的方法。但是根据精度要求,我需要将小数提高到小数幂 [ Pow(decimal a, decimal b) ]。框架有这样的功能吗?有谁知道具有这种功能的库?

4

5 回答 5

14

为了解决我的问题,我找到了一些扩展级数,我让它们实现了求解方程 X^n = e^(n * ln x)。

// Adjust this to modify the precision
public const int ITERATIONS = 27;

// power series
public static decimal DecimalExp(decimal power)
{
    int iteration = ITERATIONS;
    decimal result = 1; 
    while (iteration > 0)
    {
        fatorial = Factorial(iteration);
        result += Pow(power, iteration) / fatorial;
        iteration--;
    }
    return result;
}

// natural logarithm series
public static decimal LogN(decimal number)
{
    decimal aux = (number - 1);
    decimal result = 0;
    int iteration = ITERATIONS;
    while (iteration > 0)
    {
        result += Pow(aux, iteration) / iteration;
        iteration--;
    }
    return result;
}

// example
void main(string[] args)
{
    decimal baseValue = 1.75M;
    decimal expValue = 1/252M;
    decimal result = DecimalExp(expValue * LogN(baseValue));
}

Pow() 和 Factorial() 函数很简单,因为幂始终是 int(在 de 幂级数内)。

于 2009-01-21T18:30:02.217 回答
8

对于正整数 Exponent 和十进制底,这应该是最快的:

// From http://www.daimi.au.dk/~ivan/FastExpproject.pdf
// Left to Right Binary Exponentiation
public static decimal Pow(decimal x, uint y){
    decimal A = 1m;
    BitArray e = new BitArray(BitConverter.GetBytes(y));
    int t = e.Count;

    for (int i = t-1; i >= 0; --i) {
        A *= A;
        if (e[i] == true) {
            A *= x;
        }
    }
    return A;
}
于 2012-09-13T14:45:51.683 回答
5

这是一个 C# 程序,用于手动实现 Math.Pow(),其精度高于 .NET 基于 double 的实现。剪切并粘贴到 linqpad 中以立即运行,或将 .Dump()s 更改为 Console.WriteLines。

我已经对结果进行了测试。测试如下:

  1. 目标 = 0.4% pa,每日复利 10 000
  2. 答案 = 应该是 10 040
  3. 如何 = 十进制 b=10000;for (int i = 0; i<365; i++) { b *= rate; } 其中速率 = (1.004)^(1/365)

我测试了 3 种速率的实现:(1)手动计算(2)Excel(3)Math.Pow

手动计算的准确度最高。结果是:

Manually calculated rate:   1.0000109371043837652682334292
Excel rate:                 1.000010937104383712500000M [see formula =(1.004)^(1/365)]
Math.Pow rate:              1.00001093710438

Manual - .4%pa on R10,000:  10040.000000000000000000000131 
Excel - .4%pa on R10,000:   10039.999999999806627646709094 
Math.Pow - .4%pa on R10,000:10039.999999986201948942509648

我还在那里留下了一些额外的工作 - 我用来确定可以适合 ulong (= 22) 的最高阶乘。

Linqpad代码:

/*
a^b = exp(b * ln(a))
    ln(a) = log(1-x) = - x - x^2/2 - x^3/3 - ...   (where |x| < 1)
        x: a = 1-x    =>   x = 1-a = 1 - 1.004 = -.004
    y = b * ln(a)
    exp(y) = 1 + y + y^2/2 + x^3/3! + y^4/4! + y^5/5! + ...
        n! = 1 * 2 * ... * n        
*/

/*
//
// Example: .4%pa on R10,000 with daily compounding
//

Manually calculated rate:   1.0000109371043837652682334292
Excel rate:                 1.000010937104383712500000M =(1.004)^(1/365)
Math.Pow rate:              1.00001093710438

Manual - .4%pa on R10,000:  10040.000000000000000000000131 
Excel - .4%pa on R10,000:   10039.999999999806627646709094 
Math.Pow - .4%pa on R10,000:10039.999999986201948942509648 

*/

static uint _LOOPS = 10;    // Max = 22, no improvement in accuracy after 10 in this example scenario
//  8: 1.0000109371043837652682333497
//  9: 1.0000109371043837652682334295
// 10: 1.0000109371043837652682334292
// ...
// 21: 1.0000109371043837652682334292
// 22: 1.0000109371043837652682334292

// http://www.daimi.au.dk/~ivan/FastExpproject.pdf
// Left to Right Binary Exponentiation
public static decimal Pow(decimal x, uint y)
{
    if (y == 1)
        return x;

    decimal A = 1m;
    BitArray e = new BitArray(BitConverter.GetBytes(y));
    int t = e.Count;

    for (int i = t-1; i >= 0; --i) {
        A *= A;
        if (e[i] == true) {
            A *= x;
        }
    }
    return A;
}

// http://stackoverflow.com/questions/429165/raising-a-decimal-to-a-power-of-decimal
// natural logarithm series
public static decimal ln(decimal a)
{
    /*
    ln(a) = log(1-x) = - x - x^2/2 - x^3/3 - ...   (where |x| < 1)
        x: a = 1-x    =>   x = 1-a = 1 - 1.004 = -.004
    */
    decimal x = 1 - a;
    if (Math.Abs(x) >= 1)
        throw new Exception("must be 0 < a < 2");

    decimal result = 0;
    uint iteration = _LOOPS;
    while (iteration > 0)
    {
        result -= Pow(x, iteration) / iteration;
        iteration--;
    }
    return result;
}

public static ulong[] Fact = new ulong[] {
    1L,
    1L * 2,
    1L * 2 * 3,
    1L * 2 * 3 * 4,
    1L * 2 * 3 * 4 * 5,
    1L * 2 * 3 * 4 * 5 * 6,
    1L * 2 * 3 * 4 * 5 * 6 * 7,
    1L * 2 * 3 * 4 * 5 * 6 * 7 * 8,
    1L * 2 * 3 * 4 * 5 * 6 * 7 * 8 * 9,
    1L * 2 * 3 * 4 * 5 * 6 * 7 * 8 * 9 * 10,
    1L * 2 * 3 * 4 * 5 * 6 * 7 * 8 * 9 * 10 * 11,
    1L * 2 * 3 * 4 * 5 * 6 * 7 * 8 * 9 * 10 * 11 * 12,
    1L * 2 * 3 * 4 * 5 * 6 * 7 * 8 * 9 * 10 * 11 * 12 * 13,
    1L * 2 * 3 * 4 * 5 * 6 * 7 * 8 * 9 * 10 * 11 * 12 * 13 * 14,
    1L * 2 * 3 * 4 * 5 * 6 * 7 * 8 * 9 * 10 * 11 * 12 * 13 * 14 * 15,
    1L * 2 * 3 * 4 * 5 * 6 * 7 * 8 * 9 * 10 * 11 * 12 * 13 * 14 * 15 * 16,
    1L * 2 * 3 * 4 * 5 * 6 * 7 * 8 * 9 * 10 * 11 * 12 * 13 * 14 * 15 * 16 * 17,
    1L * 2 * 3 * 4 * 5 * 6 * 7 * 8 * 9 * 10 * 11 * 12 * 13 * 14 * 15 * 16 * 17 * 18,
    1L * 2 * 3 * 4 * 5 * 6 * 7 * 8 * 9 * 10 * 11 * 12 * 13 * 14 * 15 * 16 * 17 * 18 * 19,
    1L * 2 * 3 * 4 * 5 * 6 * 7 * 8 * 9 * 10 * 11 * 12 * 13 * 14 * 15 * 16 * 17 * 18 * 19 * 20,
    14197454024290336768L, //1L * 2 * 3 * 4 * 5 * 6 * 7 * 8 * 9 * 10 * 11 * 12 * 13 * 14 * 15 * 16 * 17 * 18 * 19 * 20 * 21,        // NOTE: Overflow during compilation
    17196083355034583040L, //1L * 2 * 3 * 4 * 5 * 6 * 7 * 8 * 9 * 10 * 11 * 12 * 13 * 14 * 15 * 16 * 17 * 18 * 19 * 20 * 21 * 22    // NOTE: Overflow during compilation
};

// http://stackoverflow.com/questions/429165/raising-a-decimal-to-a-power-of-decimal
// power series
public static decimal exp(decimal y)
{
    /*
    exp(y) = 1 + y + y^2/2 + x^3/3! + y^4/4! + y^5/5! + ...
    */

    uint iteration = _LOOPS;
    decimal result = 1; 
    while (iteration > 0)
    {
        //uint fatorial = Factorial(iteration);
        ulong fatorial = Fact[iteration-1];
        result += (Pow(y, iteration) / fatorial);
        iteration--;
    }
    return result;
}

void Main()
{   
    decimal a = 1.004M;
    decimal b = 1/365M;

    decimal _ln = ln(a);
    decimal y = b * _ln;
    decimal result = exp(y);
    result.Dump("Manual rate");

    decimal excel = 1.000010937104383712500000M;    // =(1.004)^(1/365)
    excel.Dump("Excel rate");


    decimal m = (decimal)Math.Pow((double)a,(double)b);
    m.Dump("Math.Pow rate");

    //(result - excel).Dump("Diff: Manual - Excel");
    //(m - excel).Dump("Diff: Math.Pow - Excel");

    var f = new DateTime(2013,1,1);
    var t = new DateTime(2014,1,1);
    Test(f, t, 10000, result, "Manual - .4%pa on R10,000");
    Test(f, t, 10000, excel, "Excel - .4%pa on R10,000");
    Test(f, t, 10000, m, "Math.Pow - .4%pa on R10,000");
}

decimal Test(DateTime f, DateTime t, decimal balance, decimal rate, string whichRate)
{
    int numInterveningDays = (t.Date - f.Date).Days;
    var value = balance;
    for (int i = 0; i < numInterveningDays; ++i)
    {
        value *= rate;
    }
    value.Dump(whichRate);
    return value - balance;
}

/*

// Other workings:

//
// Determine maximum Factorial for use in ln(a)
//

ulong max    =  9,223,372,036,854,775,807 * 2   // see http://msdn.microsoft.com/en-us/library/ctetwysk.aspx
Factorial 21 = 14,197,454,024,290,336,768
Factorial 22 = 17,196,083,355,034,583,040
Factorial 23 = 8,128,291,617,894,825,984 (Overflow)

public static uint Factorial_uint(uint i)
{
    // n! = 1 * 2 * ... * n
    uint n = i;
    while (--i > 1)
    {
        n *= i;
    }
    return n;
}

public static ulong Factorial_ulong(uint i)
{
    // n! = 1 * 2 * ... * n
    ulong n = i;
    while (--i > 1)
    {
        n *= i;
    }
    return n;
}

void Main()
{
    // Check max ulong Factorial
    ulong prev = 0;
    for (uint i = 1; i < 24; ++i)
    {
        ulong cur = Factorial_ulong(i);
        cur.Dump(i.ToString());
        if (cur < prev)
        {
            throw new Exception("Overflow");
        }
        prev = cur;
    }
}
*/
于 2014-05-25T13:28:51.263 回答
0

我认为这在很大程度上取决于您计划插入的数字。如果“a”和“b”不是“好”数字,那么您可能会得到一个无法存储的非终止值,如果 C# BigDecimal 的行为与 Java BigDecimal 完全一样,在这种情况下它可能会引发异常。

于 2009-01-09T18:37:07.947 回答
-3

你确定你真的想这样做吗?decimal乘法比 's 慢大约 40 倍,double所以我希望小数Math.Pow()实际上无法使用。

但是,如果您只期望整数幂,我建议您使用已经在 SO 上讨论过的基于整数的幂算法。

于 2009-01-09T18:36:39.227 回答