如何计算任意一个电源/根？ [英] How to calculate an arbitrary power/root?

问题描述

我有需要提高一个数量的分数功率应用。目标平台是FPGA，我可以得到的估计上FPU的大小，但我需要一个算法提出一个数字来一个分数幂只是进行可行性研究。我假设浮点为最坏的情况下，我希望在实践中，我们将能够使用捷径，但现在我想表明，最坏的情况下，可以在我们的一部分来实现。

I have a application which needs to raise a number to a fractional power. The target platform is an FPGA and I can get estimates on an FPU size for it, but I need an algorithm for raising a number to a fractional power just for a feasibility study. I'm assuming floating point as a worst case, I expect in practice we will be able to use short cuts, but for now I want to show that worst case can be implemented on our part.

想我会问这里，看看是否有，我可以检出任何常用方法。我知道有这样的软件的方法，我想只是一个合理有效的算法开始。我会担心FPGA实现。

Thought I'd ask here and see if there were any common methods that I could checkout. I know there are software methods of doing this, I want just a reasonably efficient algorithm to start with. I'll worry about the FPGA implementation.

推荐答案

是您的输入范围内任意的，在一定范围内已知的？

Is your range of inputs arbitrary, or known within a certain range?

在这两种情况下，X ^M = EXP（M日志x）的，所以如果你能创建功能评价EXP（x）和日志（x）和有多重，你可能可以了，好了。

in either case, x^m = exp(m log x), so if you can create functions to evaluate exp(x) and log(x) and have a multiply, you're probably all set.

你必须搞清楚你想怎么处理x的非正值。

You'll have to figure out how you want to handle nonpositive values of x.

（提示日志（X）：如果这是 IEEE-754浮点，转移尾数如果有必要，直到你结束了一系列的数字2 ^K 2 ^{K + 1} 对于一些价值K.这让你对付一个2：1的范围中是不是太难逼近与多项式。然后你只需少量的可能性来处理指数和移位数目。

(hint for log(x): if this is IEEE-754 floating point, shift the significand if necessary until you end up with a range of numbers between 2^k and 2^k+1 for some value K. This lets you deal with a 2:1 range which isn't too hard to approximate with polynomials. Then you just have a small number of possibilities to handle for the exponent and the shift number.

相应提示为EXP（X）：写X = K + b，其中0℃= B＆LT; 1和k是整数。然后EXP（x）的= EXP（k）的* EXP（二）; b具有在有限的范围内且k具有离散的可能性有限。）

Corresponding hint for exp(x): write x = k+b where 0 <= b < 1 and k is an integer. Then exp(x) = exp(k)*exp(b); b has limited range and k has a limited number of discrete possibilities.)

（提示＃2：数字可能制定出更好的对于x ^M = G（MF（X）），其中f（x）=登录 _{2 X和G （x）= 2 ^X ）}

(hint #2: the numbers probably work out better for x^m = g(m f(x)) where f(x) = log₂x and g(x) = 2^x.)

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