Question

我有两种方法:权力和要素:

public static long pow(int x, int n) {
    long p = x;
    for (int i = 1; i < n; i++) {
        p *= x;
    }
    return p;
}

public static long fact(int n) {
    long s = n;
    for (int i = 1; i < n; i++ ) {
        s *= i;
    }
    return s;
}

页: 1 当我想在评价Exponential function i 得出与Math.exp(x)比较的错误结果。我的法典:

public static void exp(int x, double eps) {
    int i = 1;
    double pow = 1.0;
    double fact = 1.0;
    double sum = 0.0;
    double temp;
    do {
        temp = pow/fact;
        sum += temp;
        pow = pow(x, i);
        fact = fact(i);
        i++;
    }
    while (temp > eps);
    System.out.println("Check: " + Math.exp(x));
    System.out.println("My: " + sum);
}

public static void main() {
    int x = 10;
    double eps = 0.0000000000001;

    exp(x, eps);
}

x=10的产出是:

检查:22026.465794806718

我:21798.734894914145

较大的X,“精度损失”越大(并非确切的,因为你可以真正称其准确......)。

当方法/power和:/strong>返回double时,产出是正确的。谁能解释我如何工作?

<方法>pow和fact>必须很长一段时间才能返回,我必须在<>exp上加以使用(联合任务)。

Answer 1

如果你尝试这一方法:

public static long pow(int x, int n) {
    long p = x;
    System.out.println("Pow: "+x+","+n);
    for (int i = 1; i < n; i++) {
        p *= x;
        System.out.println(p);
    }
    return p;
}

You get this output:

...
Pow: 10,20
100
1000
10000
...
...
1000000000000000
10000000000000000
100000000000000000
1000000000000000000
-8446744073709551616
7766279631452241920

长期价值超支:10^20太大,无法长期适应。

Methods pow and fact must return long and I must use them in exp (college assignment).

Then there is not much you can do to fix it. You could throw an exception if eps is too small.

Answer 2

<条码>x通常有多大? 这可能是累进。更改<代码>int中的所有论点:pow和fact,改为long。

Answer 3

长期的数据类型可能无法精确地处理,因此,你对数值的重新估价是错误的。为什么你只是要恢复双重价值观?

<><>Edit>: 我在此提出:

  public static long pow(int x, int n) 
  {
    double p = x;
    for (int i = 1; i < n; i++) {
      p *= x;
    }
    return (long)p;
  }

  public static long fact(int n) 
  {
    double s = n;
    for (int i = 1; i < n; i++ ) {
      s *= i;
    }
    return (long)s;
  }


  public static void exp(int x, double eps) 
  {
    double pow = 1.0;
    double fact = 1.0;
    double sum = 0.0;
    double temp;
    for(int ii=1; ii < 100; ii++)
    {
      pow = pow(x, ii);
      fact = fact(ii);
      temp = (double)pow/(double)fact;
      temp = temp == 1 ? 0 : temp;
      sum += temp;
    }

    System.out.println("Check: " + Math.exp(x));
    System.out.println("My: " + sum);
  }

  public static void main(final String[] args)
  {
    int x = 10;
    double eps = 0.0000000000001;

    exp(x, eps);
  }

这关系到你们最接近的人,不使用mal子。

Check: 22026.465794806718
My: 21946.785573087538

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