英文:
determining e to the x in Java using limited features
问题
我正试图确定e的x次方的值，基本上使用if语句和循环。
（公式为ex = 1 + x/1! + x2/2! + x3/3! + ...）
我能够通过以下代码确定e的值：
{Scanner input = new Scanner(System.in);System.out.println("请告诉我们您想计算的项数：");int i = input.nextInt();int n = 1;double e = 1;int counter = 1;long m = 1;while (n <= i){while (counter <= n){{m = m * counter;counter++;}}e += 1 / (double) m;n++;}System.out.println(e);}
英文:
I am trying to determine the value of e to the x, using basically if's and loop's.
(The formula goes e<sup>x</sup> = 1 + x/1! + x<sup>2</sup>/2! + x<sup>3</sup>/3! + ...)
I was able to detemine e, with the following code:
{Scanner input = new Scanner(System.in);System.out.println("Please inform us a number of terms you want to calc: ");int i = input.nextInt();int n = 1;double e = 1;int counter = 1;long m = 1;while (n <= i){while (counter <= n){{m = m * counter;counter++;}}e += 1 / (double) m;n++;}System.out.println(e);}
I was hoping someone would be able to help, with this task.
答案1
得分: 1
如果这个有效（我不确定是否有效），我认为你需要更多地思考你的问题？
如果你想要找到 e^x，找到 e 本身是没有用的，因为你已经有一个公式可以通过已知的 e 和 x 值来计算 e^x...
e^x = 1 + x/1! + x^2/2! + x^3/3! + ...
只需设置一个循环，并计算这个求和值
double total = 0;for (double i = 0; i < something; i++) {total += Math.pow(x, i) / factorial(i);}
但这像是作业题，所以你可能需要自己编写阶乘和幂次计算的方法，我假设是这样的...
英文:
If this works (which idk if it does), I think you need to think more about your question?
if you want to find e^x, finding e itself is useless because you already have a formula to find e^x given e and x...
e^x = 1 + x/1! + x^2/2! + x^3/3! + ...
just set up a for loop and calculate the summation
double total = 0;for (double i = 0; i < something; i++) {total += Math.pow(x, i) / factorial(i);}
but this smells like hw so you're probably going to have to make the factorial and power methods yourself, I'm assuming...
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