We have 3 positive integers $x$, $a$ and $b$. How can we simplify
$$\left\lceil \left\lceil \left\lceil \frac{xa}{b} \right\rceil \frac{a}{b} \right\rceil \frac{a}{b} \right\rceil$$
?
Thank you.
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$\begingroup$ Doesn't seem easy,what made you to consider this expression? $\endgroup$– kingW3Commented May 1, 2017 at 17:28
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2 Answers
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Consider the case where $b=3$. Then a possible expression for $\left\lceil \left\lceil \left\lceil \frac{xa}{b} \right\rceil \frac{a}{b} \right\rceil \frac{a}{b} \right\rceil$ is $$\left(\left(\frac{xa}{3}+\frac{2}{3}\right)\frac{a}{3}+\frac{2}{3}\right)\frac{a}{3}+\frac{2}{3}\tag{1}$$ and two others are
$$\left(\left(\frac{xa}{3}+\frac{1}{3}\right)\frac{a}{3}+\frac{1}{3}\right)\frac{a}{3}+\frac{1}{3}\tag{2}$$
$$\left(\left(\frac{xa}{3}+\frac{0}{3}\right)\frac{a}{3}+\frac{0}{3}\right)\frac{a}{3}+\frac{0}{3}\tag{3}$$
There are 3 place where one of the fractions $\frac{0}{3},\frac{1}{3},\frac{2}{3}$ can be added. That means there a total of 9 expressions similar to (1), (2), and (3). In general there would be $3b$ such expressions.
To unify these you'd need to make the term added in each of the 3 places be $\frac{Y_i \mod b}{b}$, where $Y_0 = xa$, $Y_1 = \left(\frac{xa}{b}+ \frac{xa\mod b}{b}\right)\cdot a$ and similarly for $Y_2$.
You would end up with a longish expression with 3 modulo operations. I wouldn't consider that simplified, but it does get rid of the ceiling function applications and it does show the complexity the ceiling function adds.
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(Too long for a comment.) All that can be derived is the following pair of inequalities, which don't look amenable to a simple form, unless maybe additional information is known about $x,a,b\,$:
$$ \begin{align} \frac{x a^3}{b^3} \le \left\lceil \left\lceil \left\lceil \frac{xa}{b} \right\rceil \frac{a}{b} \right\rceil \frac{a}{b} \right\rceil &\lt \left\lceil \left\lceil \frac{xa}{b} \right\rceil \frac{a}{b} \right\rceil \frac{a}{b} + 1 \\ &\lt \left( \left\lceil \frac{xa}{b} \right\rceil \frac{a}{b} + 1\right)\frac{a}{b}+1 \\ &\lt \left( \left(\frac{xa}{b} + 1\right) \frac{a}{b} + 1\right)\frac{a}{b}+1 \\ & = \frac{x a^3}{b^3}+\frac{a^2+ab+b^2}{b^2} \end{align} $$
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1$\begingroup$ Minor correction: The inequality should be strict on the right, non-strict on the left. Other than that, it's fine. $\endgroup$– quasiCommented May 1, 2017 at 18:12
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