.net框架在Math类中提供了一个为double提供动力的方法。但是根据精度要求，我需要将小数提高到十进制幂[Pow（十进制a，十进制b）]。框架是否具有这样的功能？有没有人知道有这种功能的图书馆？
答案 0 :(得分：11)
为了解决我的问题，我发现了一些expansion series，我实现了它们来解决公式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，因为幂总是一个int（在de power系列中）。
答案 1 :(得分：7)
对于正整数Exponent和小数基数，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;
}
答案 2 :(得分：4)
这是一个C＃程序，用于手动实现Math.Pow（），其精度高于.NET基于双重的实现。 剪切并粘贴到linqpad中立即运行，或将.Dump（）更改为Console.WriteLines。
我已经对结果进行了测试。测试如下：
我测试了3种速率的实现：（1）手动计算（2）Excel（3）Math.Pow
手动计算具有最高的准确度。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 - 我曾经用它来确定最高因子可以适应ulong（= 22）。
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;
}
}
*/
答案 3 :(得分：0)
我认为这很大程度上取决于您计划插入的数量。如果'a'和'b'不是'nice'数字，那么您可能会得到一个无法终止的值，无法存储如果C＃BigDecimal像Java BigDecimal一样行为，它可能会在这种情况下引发异常。
答案 4 :(得分：-3)
你确定你真的想这样做吗？ decimal乘法比double慢大约40倍，所以我希望小数Math.Pow()几乎无法使用。
但是，如果您只期望整数幂，我建议您使用已在SO上讨论过的基于整数的幂算法。