EXP to Taylor series | 易学教程

问题

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;

}

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.

Need advice. Thank's for help.

回答1:

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.

Another workaround is to add this to your loop:

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

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

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. EDIT: To correct this, we can use the fact that e^-x is the same as 1 / e^x.

Code:

#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;
}

回答2:

Need advice.

Try stepping through your program in a debugger to see where it's going wrong. If you don't have a debugger, insert print statements within the loop to monitor the values of variables that change.

来源：https://stackoverflow.com/questions/5362573/exp-to-taylor-series

标签

c++

taylor-series

exp