//java program that asks the user to input a number that e^x=1+x+x^2/2! +x^3/3!... e is a mathematical constant equal to 2.718...
import java.util.Scanner;
public class taylor_2 {
public static void main(String args[]) {
Scanner input=new Scanner(System.in);
double x; //input for x
double factorial=1; //initializes factorial
int counter=1; //initializes counter
double result=1; //initializes result
System.out.println("Enter non negative number"); //asks user to enter x
x=input.nextInt();
//output in while loop will continue to be generated if user doesn't entered a negative number
while(x<1){
System.out.println("I said entered a positive number");
x=input.nextInt();
}
while(x>counter){
factorial=factorial*counter;//factorial formula
result=result+(Math.pow(x,counter))/factorial; //equation for e^x=1+x+x^2/2! +x^3/3!
counter++;
}
System.out.println("Taylor series is " +result);//output for taylor equation e^x
}
}
这是我的代码的输出:
输入非负数
2
泰勒系列是4.0
当我输入2时,它应该输出7.3890560983而不是4.0,因为e = 2.718 ...和e^2 = 7.3890560983。我在做什么错?
问题是泰勒系列与e^x不同。它将返回靠近函数e^x的函数。
为了更好地理解它,我建议您查看下一个链接的第二张图片:
https://en.wikipedia.org/wiki/taylor_series
您可以在上一张图片中看到,随着n的越来越大,功能变得越来越准确。
您的代码问题是您的x
值是您的n
值,这不是真的。
x:必须是您现在要的值e^x。
n:是您方程式的准确性。较大意味着更准确。
因此,您必须使用while(n>counter)
更改while(x>counter)
,其中n可以是用户选择精度的变量,或者使用您选择的精度为常数。
我认为在x=100
之前,n=150
应该起作用。
我希望对您有帮助!:)
这里似乎有一个答案:即使算法与您的算法略有不同,to to to t to taylor series for c 。这是其Java版本:
public class TaylorSeries {
public static void main(String args[]) {
Scanner input = new Scanner(System.in);
System.out.println("Enter x:");
double x = input.nextDouble();
double result = calcExp(x);
System.out.println("calcExp(x) = " + result);
System.out.println(" e^x = " + Math.pow(Math.E, x));
}
static double calcExp(double x) {
double eps = 0.0000000000000000001;
double elem = 1.0;
double sum = 0.0;
boolean negative = false;
int i = 1;
sum = 0.0;
if (x < 0) {
negative = true;
x = -x;
}
do {
sum += elem;
elem *= x / i;
i++;
if (sum > Double.MAX_VALUE) {
System.out.println("Too Large");
break;
}
}
while (elem >= eps);
return negative ? 1.0 / sum : sum;
}
}
输出:
Enter x:
2
calcExp(x) = 7.389056098930649
e^x = 7.3890560989306495
所有信用都应在此处转到答案:Exp to Taylor系列。我只将C 代码转换为Java