EXP到泰勒系列 [英] EXP to Taylor series

问题描述

我试图将exp（x）函数展开为泰勒级数。这里是代码：

I'am trying to expand exp(x) function to Taylor series. Here is code:

double CalcExp(){
double eps = 0.0000000000000000001;
double elem = 1.0;
double sum = 0.0;
int i = 1;
sum = 0.0;
do {
    sum += elem;
    elem *= x / i;
    i++;
} while (elem >= eps);
return sum;

}

输入大X或负X我的程序崩溃。
当我输入X像0.00000000001时，结果是-1。

The problem is when I enter big X or negative X my program crashes. And when I enter X like "0.00000000001" the result is -1.

需要建议。

推荐答案

对于大X值（大约700及以上），您将达到双精度10 ^ 308），并导致无限循环。你不能做太多，你应该限制X输入范围或使用一些大数字库来扩展范围。

For big X values (around 700 and above), you'll hit the range limit for doubles (10^308) and cause an infinite loop. You can't do much about it, you should either limit X input range or use some big number library to have extended range.

另一个解决方法是将它添加到您的循环中：

Another workaround is to add this to your loop:

if (sum > 1E305) {
  // we'll most likely run into an infinite loop
  break;
}

注意，你应该在循环后处理这种情况，不正确的结果。

Note you should handle this case outside the loop afterwards to avoid printing a very large incorrect result.

我无法重现 0.00000000001 的问题，负值也运行良好，尽管结果是错误的，这似乎是算法中的错误/限制。编辑：为了纠正这一点，我们可以使用 e ^ -x 与 1 / e ^ x 。

I can't reproduce the problem for 0.00000000001, this just returns 1 for me. Negative values run fine, too, although the result is wrong which seems to be an error/limitation in the algorithm. To correct this, we can use the fact that e^-x is the same as 1 / e^x.

代码：

#include <stdio.h>

double CalcExp(double x){
  double eps = 0.0000000000000000001;
  double elem = 1.0;
  double sum = 0.0;
  bool negative = false;
  int i = 1;
  sum = 0.0;
  if (x < 0) {
    negative = true;
    x = -x;
  }
  do {
    sum += elem;
    elem *= x / i;
    i++;
    if (sum > 1E305) break;
  } while (elem >= eps);
  if (sum > 1E305) {
    // TODO: Handle large input case here
  }

  if (negative) {
    return 1.0 / sum;
  } else {
    return sum;
  }
}

int main() {
  printf("%e\n", CalcExp(0.00000000001)); // Output: 1.000000e+000
  printf("%e\n", CalcExp(-4));            // Output: 1.831564e-002
  printf("%e\n", CalcExp(-45));           // Output: 2.862519e-020
  printf("%e\n", CalcExp(1));             // Output: 2.718282e+000
  printf("%e\n", CalcExp(750));           // Output: 1.375604e+305
  printf("%e\n", CalcExp(7500000));       // Output: 1.058503e+305
  printf("%e\n", CalcExp(-450000));       // Output: 9.241336e-308
  return 0;
}

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