So i have like a very complicated expression involving many variables, and i want to simplify it, the only problem comes from the fact that this complicated expression is itself inside a fractional part $\{...\}$ which we define to be $$\{x\}={\rm fractional\,part\,of\,}x=x-{\rm IntegerPart}(x)$$ in particular this complicated expression takes the form $$\{({\rm complicated\, stuff})^{1/n}\}=({\rm complicated\, stuff})^{1/n}-{\rm IntegerPart}(({\rm complicated\, stuff})^{1/n})$$ I tried raising to the $n$ power and using newton's binomial theorem but we don't have things that get simplified
if anyone knows a way to write that in an alternative way (like getting rid of the sqrt) that would help very much!
TLDR; How to simplify $\{({\rm complicated\, stuff})^{1/n}\}$, or how to write $\{({\rm complicated\, stuff})^{1/n}\}=f({\rm complicated\, stuff})^{1/n}$
if anything is unclear feel free to ask!
the exact expression is $$\left\{\dfrac{1}{\left(1-\dfrac{p}{q}\right)^{1/n}}\right\}$$ p,q, integers
I assume that $0 < p < q$.
$\begin{array}\\ \dfrac{1}{\left(1-\dfrac{p}{q}\right)^{1/n}} &=\dfrac{1}{\left(\dfrac{q-p}{q}\right)^{1/n}}\\ &=\left(\dfrac{q}{q-p}\right)^{1/n}\\ &=\left(\dfrac{q-p+p}{q-p}\right)^{1/n}\\ &=\left(1+\dfrac{p}{q-p}\right)^{1/n}\\ \end{array} $
Since $(1+x)^n \ge 1+nx $, $(1+\frac{x}{n})^n \ge 1+x $ so that $(1+x)^{1/n} \le 1+\frac{x}{n} $.
Therefore $\left(1+\dfrac{p}{q-p}\right)^{1/n} \le 1+\dfrac{p}{n(q-p)} $ or $\dfrac{1}{\left(1-\dfrac{p}{q}\right)^{1/n}} \le 1+\dfrac{p}{n(q-p)} = 1+\dfrac{p/q}{n(1-p/q)} = 1+\dfrac{1}{n(q/p-1)} $.
Does this help?
(added in response to a comment)
If $\dfrac{1}{n(q/p-1)} < 1$, then $\left\lfloor \dfrac{1}{\left(1-\dfrac{p}{q}\right)^{1/n}} \right\rfloor = 1$.
This happens if $n > \dfrac{1}{(q/p-1)} = \dfrac{p}{q-p} $.
Also, by the generalized binomial theorem if $n$ is large, if $r = p/q$, then $(1-r)^{-1/n} =1+\frac{r}{n}+\frac{r^2(1+n)}{2n^2}+... $. This doesn't behave well, since the coefficient of $r^m$ is about $\frac{1}{m!n} $.
It might be better to look at $(1+s)^{1/n} $ where $s = \frac{p}{q-p}$. This is $(1+s)^{1/n} =1+\frac{s}{n}-\frac{s^2(n-1)}{2n^2}+... $. Since this is an enveloping series, $1+\frac{s}{n}-\frac{s^2(n-1)}{2n^2} < (1+s)^{1/n} < 1+\frac{s}{n} $.