如何自己编写幂函数?

2021-12-17 00:00:00 floating-point math c++

我一直想知道如何自己制作一个计算幂的函数(例如 23).在大多数语言中，这些都包含在标准库中，主要是作为 pow(double x, double y)，但我如何自己编写?

I was always wondering how I can make a function which calculates the power (e.g. 23) myself. In most languages these are included in the standard library, mostly as pow(double x, double y), but how can I write it myself?

我在考虑 for 循环，但它认为我的大脑陷入了循环(当我想用非整数指数计算幂时，例如 54.5 或底片 2-21) 然后我疯了 ;)

I was thinking about for loops, but it think my brain got in a loop (when I wanted to do a power with a non-integer exponent, like 54.5 or negatives 2-21) and I went crazy ;)

那么，如何编写一个计算实数幂的函数?谢谢

So, how can I write a function which calculates the power of a real number? Thanks

哦，也许需要注意的是:我不能使用使用幂的函数(例如 exp)，这会使它最终变得毫无用处.

Oh, maybe important to note: I cannot use functions which use powers (e.g. exp), which would make this ultimately useless.

推荐答案

负幂不是问题，它们只是正幂的倒数 (1/x).

Negative powers are not a problem, they're just the inverse (1/x) of the positive power.

浮点运算稍微复杂一点；如您所知，分数幂等效于根(例如 x^(1/2) == sqrt(x))，并且您还知道以相同的基数乘幂等效于将它们的相加指数.

Floating point powers are just a little bit more complicated; as you know a fractional power is equivalent to a root (e.g. x^(1/2) == sqrt(x)) and you also know that multiplying powers with the same base is equivalent to add their exponents.

有了以上所有内容，您可以:

With all the above, you can:

  • 将指数分解为整数部分和有理部分.
  • 使用循环计算整数幂(您可以优化它分解因子并重用部分计算).
  • 使用您喜欢的任何算法计算根(任何迭代近似，如二分法或牛顿法都可以).
  • 将结果相乘.
  • 如果指数为负，则应用倒数.

示例:

2^(-3.5) = (2^3 * 2^(1/2)))^-1 = 1 / (2*2*2 * sqrt(2))

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