指數近似不是小的或大的輸入良好

Question

指數近似代碼基於裁縫系列https://en.wikipedia.org/wiki/Taylor_series圍繞零可以很好地用於輸入，但移動在兩個方向上更遠時是完全無用的。下面是我的小測試代碼的輸出，相比的std :: EXP結果，並在邊境的錯誤是巨大的內-12至12的範圍和打印錯誤計算輸入EXP。對於-12輸入例如誤差約爲高達148255571469％：

in = -12 error = 148255571469.28% 
in = -11.00 error = 18328703925.31% 
in = -10.00 error = 2037562880.10% 
in = -9.00 error = 199120705.27% 
in = -8.00 error = 16588916.06% 
in = -7.00 error = 01128519.76% 
in = -6.00 error = 00058853.00% 
in = -5.00 error = 00002133.29% 
in = -4.00 error = 00000045.61% 
in = -3.00 error = 00000000.42% 
in = -2.00 error = 00000000.00% 
in = -1.00 error = 00000000.00% 
in = 0.00 error = 00000000.00% 
in = 1.00 error = 00000000.00% 
in = 2.00 error = 00000000.00% 
in = 3.00 error = 00000000.00% 
in = 4.00 error = 00000000.03% 
in = 5.00 error = 00000000.20% 
in = 6.00 error = 00000000.88% 
in = 7.00 error = 00000002.70% 
in = 8.00 error = 00000006.38% 
in = 9.00 error = 00000012.42% 
in = 10.00 error = 00000020.84% 
in = 11.00 error = 00000031.13% 
in = 12.00 error = 00000042.40% 

我需要一個誤差近似小於跨越儘可能大的範圍內1％的誤差。任何想法如何實現這一目標？

我的小測試，代碼如下：

#include <cmath> 
    #include <iostream> 
    #include <iomanip> 

    double my_exp(double x) //based on https://en.wikipedia.org/wiki/Exponential_function 
    { 
     double res = 1. + x, t = x; 
     unsigned long factorial = 1; 
     for (unsigned char i = 2; i <= 12; ++i) 
     { 
      t *= x, factorial *= i; 
      res += t/factorial; 
     } 
     return res; 
    } 

    int main(int argc, char* argv[]) 
    { 
     for (double in = -12; in <= 12; in += 1.) 
     { 
      auto error = std::abs(my_exp(in) - std::exp(in)); 
      auto percent = error * 100./std::exp(in); 
      std::cout << "in = " << in << " error = " 
       << std::fixed << std::setw(11) << std::setprecision(2) 
       << std::setfill('0') << percent << "%" << std::endl; 
     } 
     return 0; 
    } 

從看似類似的問題的解決方案從「大約E 1 X」 Approximate e^x不解決這個問題：基於雷米茲和帕德逼近

• 解決方案只有有限的範圍內提供準確度（https://stackoverflow.com/a/6985347/5750612）
• E 1 X = 2×/ LN（2）歸結POW的toapproximation和我無法找到精確的一個
• 裁縫系列不爲小和大的投入工作
• expf_fast溶液產生更均勻的誤差在所有範圍內，但它仍然是過大（〜在範圍的端部20％）

Answer 1

的所述一個發現與泰勒展開近似簡單的方法是檢查收斂速度只是增加一個新的更高的期限和確認結果。

這裏是我的C++代碼，以支持上述觀點。

#include <iostream> 
#include <cmath> 

double my_exp(const double x, const double x0 = 0, const double ncut = 1e-3) 
{ 
    double res = 1.; 
    double addterm = (x - x0); 
    size_t norder = 1; 

    while(true) 
    { 
     double res_update = res + addterm; 
     if(std::abs(res_update - res)/std::abs(res) < ncut){ 
      break; 
     } 
     norder += 1; 
     addterm *= (x - x0)/norder; 
     res = res_update; 
    } 

    return res; 
} 


int main(int argc, char* argv[]) 
{ 
    const double x = std::atof(argv[1]); 
    const double approxi = my_exp(x); 

    const double exactResult = std::exp(x); 

    std::cout<<"approxi : "<< approxi<<std::endl; 
    std::cout<<"exact : "<< exactResult<<std::endl; 

    std::cout<<"err: "<< (1 - std::abs(approxi/exactResult))*100 <<std::endl; 

    return 0; 
}