在Java中不使用内置函数计算e^x

Question

我是一名Java初学者，目前正在阅读《如何像计算机科学家一样思考》入门书。我在迭代一章中遇到了一个问题。谁能给我指一下正确的方向吗

当我使用数学时。exp，我得到的答案与代码得到的答案完全不同。 注意，这不是家庭作业

问题是：

One way to calculate e^x is to use the infinite series expansion e^x = 1 + x + x² /2! + x³/3! + x⁴/4! +... If the loop variable is named i, then the ith term is xⁱ/i!.

1. Write a method called myexp that adds up the first n terms of this series.

下面是代码：

public class InfiniteExpansion {
    public static void main(String[] args){

        Scanner infinite = new Scanner(System.in);
        System.out.println("what is the  value of X?");
        double x = infinite.nextDouble();
        System.out.println("what is the power?");
        int power = infinite.nextInt();



        System.out.println(Math.exp(power));//for comparison

        System.out.println("the final value of series is: "+myExp(x, power));
    }

    public static double myExp(double myX, double myPower){

        double firstResult = myX;
        double denom = 1;
        double sum =myX;

        for(int count =1;count<myPower;count++){

            firstResult = firstResult*myX;//handles the numerator

            denom = denom*(denom+1);//handles the denominator

            firstResult = firstResult/denom;//handles the segment

            sum =sum+firstResult;// adds up the different segments
        }

        return (sum+1);//gets the final result
    }

}

Answer 1

这里有几个错误：

1. firstResult应该从1开始，这样它就变成了1+x+x^2，而不是1+x^2+x^3
2. 正如timctran所说，你没有以正确的方式计算阶乘

总而言之，您可以将操作简化为：

firstResult = firstResult * myX / (count+1);
sum += firstResult;

编辑：
-我运行了代码，看到了数学。exp（power）是打印出来的，而不是数学。exp（x）
-我的第一项是错误的，因为sum被初始化为myX

Answer 2

赋值denom = denom*(denom+1)将给出如下序列：1, 1*2=2, 2*3=6, 6*7=42, 42*43=... 但是你想要denom = denom*count

一般来说，我们只想打印第一个n阶乘，从1!：1!, 2!, 3!, ..., n!开始。在第k项，我们取第k-1项并乘以k。这将是在前一项上递归计算k!。具体例子：4!是3!次4，6!是5!次6

在代码中，我们有

var n = 7;
var a = 1;
for (int i = 1; i <= n; i++ ) {
    a = a*i; // Here's the recursion mentioned above.
    System.out.println(i+'! is '+a);
}

试着运行上面的程序，比较一下运行下面的程序会得到什么结果：

var n = 7;
var a = 1;
for (int i = 1; i <= n; i++ ) {
    a = a*(a+1);
    System.out.println('Is '+i+'! equal to '+a+'?');
}

Answer 3

为什么要把它复杂化？我尝试了一个解决方案，结果如下：

//One way to calculate ex is to use the infinite series expansion 
//ex = 1 + x + x2 /2! + x3/3! + x4/4! +... 
//If the loop variable is named i, then the ith term is xi/i!.
//
//Write a method called myexp that adds up the first n terms of this series.
import java.util.Scanner;

public class InfiniteExpansion2 {

    public static void main(String[] args) {

        Scanner infinite = new Scanner(System.in);
        System.out.println("what is the  value of X?");
        double x = infinite.nextDouble();
        System.out.println("what is the value of I?");  // !
        int power = infinite.nextInt();

        System.out.println(Math.exp(power));//for comparison

        System.out.println("the final value of series is: " + myCalc(x, power));
    }

    public static double fac(double myI) {
        if (myI > 1) {
            return myI * fac(myI - 1);
        } else {
            return 1;
        }
    }

    public static double exp(double myX, double myE) {
        double result;
        if (myE == 0) {
            result = 1;
        } else {
            result = myX;
        }
        for (int i = 1; i < myE; i++) {
            result *= myX;
        }
        return result;
    }

    public static double myCalc(double myX, double myI) {
        double sum = 0;
        for (int i = 0; i <= myI; i++) { // x^0 is 1
            sum += (exp(myX, i) / fac(i));
        }
        return sum;
    }

}

如果你想像工程师一样思考，我会这样做：

• 保持简单
• 把它打碎
• 紧紧抓住这项任务（比如我把var命名为myI，而不是myPower——首先，对我来说似乎更清楚——这样你就不会感到困惑）

希望你喜欢

Answer 4

我尝试了一个解决方案，结果如下：

public class Fact {

    public int facto(int n){
            if(n==0)
                return 1;
            else
                return n*facto(n-1);
        }
    }

}

import java.util.Scanner;

public class Ex {

public static void main(String[] args){

        Fact myexp=new Fact();

        Scanner input=new Scanner(System.in);
        int n=1;

        double e=1,i=0,x;
        int j=1;
        System.out.println("Enter n: ");
        n=input.nextInt();
        System.out.println("Enter x: ");
        x=input.nextDouble();
        while(j<=n)
        {
            int a=myexp.facto(j);
            double y=Math.pow(x,j)/(double)a;

            i=i+y;
            ++j;
        }       

        e=e+i;
        System.out.println("e^x= "+ e);

    }
}