How to represent $x^y$ = [Integer] + [Remainder]?

Question

For example: $5^{\frac{1}{2}} = 2.23606\ldots = 2 + 0.23606\ldots$ Can we do this for $x^y$ in general?

Motivation: a way of expressing the floor, $ \lfloor x^y\rfloor $, of an exponential $x^y$ using algebra.

Note: $x,y \in \mathbb{R} > 0$

I'm thinking that maybe the trick may involve generating a power series of $x^y$ and somehow extracting terms that represent the integer and remainder parts, but otherwise I'm stuck.

Any comments/answers/suggestions/edits would be appreciated.

Many thanks,

-McMath.

Answer 1

Directly using the binomial theorem: we can write (for $x,y \geq 0$) $$x=k+r \qquad k \in \mathbb{N}, r \in [0,1)$$ $$y=\ell+s \qquad \ell \in \mathbb{N}, s \in [0,1),$$ we see that $$x^y= (k+r)^s(k+r)^{\ell} = k^s(\underbrace{1+\frac{r}{k}}_{\in [1,2)})^s \sum_{i=0}^{\ell} \binom{\ell}{i} k^{\ell-i}r^i.$$ Thus, $\lfloor x^y \rfloor$ ulimately depends on $r=\{x\}$ and $s=\{y\}.$

My bet is there is no formula that does not involve all four of $\lfloor x \rfloor , \lfloor y \rfloor, \{x\}, \{y\}.$

Answer 2

I don't think you will find anything. Let $5^x=6.9999, 5^y=7.00001$. $x$ and $y$ are very similar, but the integer part of the power is different. Powers are nicely continuous but floor functions are not, so they don't want to play together nicely.