通过计算机获取自然对数的底数e的近似值
e = 1/0! + 1/1! + 1/2! + 1/3! + 1/4! + 1/5! + ... + 1/N!
N -> ∞
public static double getE() {
double e = 1.0;
double t = 1.0;
// 迭代20次
int counter = 20;
for (int i = 0; i < counter; i++) {
t /= i;
e += t;
}
System.out.println("After 20 times, e=" + e);
// 另外一种方法
// 反向加和：1/20! + 1/19! + 1/18! + ... + 1/1!
e = 0.0;
for (int i = counter; i > 0; i++) {
e = (e + 1.0) / i;
}
e++;
System.out.println("After 20 times, e=" + e);
return e;
}
