计算e ^ x而不使用任何函数 [英] Calculating e^x without using any functions

问题描述

我们应该使用这种公式计算e ^ x：

e ^ x = 1 +（x ^ 1/1！）+ 2/2！）......

我有这样的代码：

  while（result> = 1.0E-20）
 {
 power = power * input; 
 factorial = factorial * counter; 
 result = power / factorial; 
 eValue + = result; 
 counter ++; 
 iterations ++;我现在的问题是，因为factorial是一个类型long long，我可以''''' t真正存储一个大于20的数字！所以发生的是当程序输出有趣的数字，当它到达那个点。
 
 
 
正确的解决方案可以有一个X值至多709，所以e ^ 709应输出：8.21840746155e + 307 
 
 
 
程序是用C ++编写的。
解决方案
 p>两者x ^ n和n！快速增长与n（指数和超分别）和将很快溢出您使用的任何数据类型。另一方面，x ^ n / n！ （最终），你可以停止当它的小。也就是说，使用x ^（n + 1）/（n + 1）！的事实。 =（x ^ n / n！）*（x /（n + 1））。像这样：
  term = 1.0; 
 for（n = 1; term> = 1.0E-10; n ++）
 {
 eValue + = term; 
 term = term * x / n; 
} 
  
（直接在此框中键入代码，但我希望它能工作。 
 
 
 
编辑：请注意，术语x ^ n / n！是，对于大x，增加一段时间，然后减少。对于x = 709，它在上升到0之前上升到〜1e + 306，这正好在 double 可以处理的限度上（ double 的范围是〜1e308和 term * x 将其推回），但 long double 工作正常。当然，您的最终结果 e  x 大于任何条件，因此假设您使用的数据类型足够大，可以容纳结果，您将罚款。
 
 
 
（对于x = 709，如果使用<$ c，你可以使用 double  $ c> term = term / n * x ，但它不适用于710.）
 
We are supposed to calculate e^x using this kind of formula:

e^x = 1 + (x ^ 1 / 1!) + (x ^ 2 / 2!) ......

I have this code so far:
while (result >= 1.0E-20 )
{
    power = power * input;
    factorial = factorial * counter;
    result = power / factorial;
    eValue += result;
    counter++;
    iterations++;
}
My problem now is that since factorial is of type long long, I can't really store a number greater than 20! so what happens is that the program outputs funny numbers when it reaches that point .. 

The correct solution can have an X value of at most 709 so e^709 should output: 8.21840746155e+307

The program is written in C++.
解决方案
Both x^n and n! quickly grow large with n (exponentially and superexponentially respectively) and will soon overflow any data type you use. On the other hand, x^n/n! goes down (eventually) and you can stop when it's small. That is, use the fact that x^(n+1)/(n+1)! = (x^n/n!) * (x/(n+1)). Like this, say:
term = 1.0;
for(n=1; term >= 1.0E-10; n++)
{
    eValue += term;
    term = term * x / n;
}
(Code typed directly into this box, but I expect it should work.)

Edit: Note that the term x^n/n! is, for large x, increasing for a while and then decreasing. For x=709, it goes up to ~1e+306 before decreasing to 0, which is just at the limits of what double can handle (double's range is ~1e308 and term*x pushes it over), but long double works fine. Of course, your final result ex is larger than any of the terms, so assuming you're using a data type big enough to accommodate the result, you'll be fine.

(For x=709, you can get away with using just double if you use term = term / n * x, but it doesn't work for 710.)

                        
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