Ceiling Function
The ceiling function (also known as the least integer function) of a real number \(x,\) denoted \(\lceil x\rceil,\) is defined as the smallest integer that is not smaller than \(x.\)
For example, \[\begin{array} &\lceil 9\rceil=9, &\lceil 1.006\rceil =2, &\left\lceil \sqrt{19}\right\rceil=5, &\lceil \pi\rceil = 4, &\lceil -10.27\rceil = -10.\end{array}\]
In general, \( \lceil x \rceil \) is the unique integer satisfying \( \lceil x \rceil - 1 < x \le \lceil x \rceil.\)
The ceiling is related to the floor function by the formula \[ \lceil x \rceil = -\lfloor -x \rfloor. \]
What is the range of \(x\) that satisfies
\[ \big\lceil \lceil x \rceil - 1.3 \big\rceil = 16 ?\]
Let \(\lceil x \rceil= y\), where \(y\) is an integer by the definition of the ceiling function. Then
\[\begin{align} \big\lceil \lceil x \rceil - 1.3 \big\rceil &= 16\\ \lceil y - 1.3 \rceil &= 16\\ 15 < y-1.3 &\le 16\\ 16.3 < y &\le 17.3\\ y&=17. \qquad \text{(since }y\text{ is an integer)} \end{align}\]
Since \(\lceil x \rceil= y,\)
\[\begin{align} \lceil x \rceil&= 17\\ 16<x&\le 17. \ _\square \end{align}\]
Contents
Properties
(1) \( \lceil x+n \rceil = \lceil x \rceil + n,\) for any integer \( n.\)
(2) \( \lceil x \rceil + \lceil -x \rceil = \begin{cases} 1 & \text{if } x \notin {\mathbb Z} \\ 0 & \text{if } x \in {\mathbb Z}. \end{cases}\)
(3) \( \lceil x+y \rceil = \lceil x \rceil + \lceil y \rceil\) or \( \lceil x \rceil + \lceil y \rceil - 1. \)
These can all be proved from the analogous properties for the floor function.
If \(y=\left\lfloor 2x+5 \right\rfloor =3\left\lfloor x-4 \right\rfloor \) and the product of all the distinct values of \(\left\lfloor 3x+y \right\rfloor \) is \(A\), find \(A \bmod{2016} \).
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Notation: \(\lfloor \cdot \rfloor \) denotes the floor function.
Problem-solving
As with the floor function, it is often easiest to write \( x = n-r,\) where \( n=\lceil x \rceil \) is an integer and \( 0 \le r < 1.\)
Find all solutions to \( \lceil x \rceil \lceil 2x \rceil = 15. \)
Write \( x= n-r \) as above. Then \( \lceil 2x \rceil = 2n \) or \( 2n-1 \) depending on \( r. \) In the first case, where \( r < \frac12, \) the equation becomes \( 2n^2 = 15, \) which has no solution. In the second case, where \( r \ge \frac12,\) the equation becomes \( n(2n-1) = 15,\) so \( 2n^2-n-15 = 0,\) or \( (n-3)(2n+5) = 0.\) The only integer solution is \( n=3 \).
So the range of solutions is the interval \( (2,2.5].\ _\square \)
Find a positive integer \(n\) such that \( \left\lfloor \frac {20n}{13} \right\rfloor + \left\lceil \frac {13n}{20} \right\rceil = 2013 \).
This problem is proposed by Ahaan Rungta.
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Details and assumptions:
The function \(\lfloor x \rfloor: \mathbb{R} \rightarrow \mathbb{Z}\) refers to the greatest integer smaller than or equal to \(x\). For example \(\lfloor 2.3 \rfloor = 2\) and \(\lfloor -5 \rfloor = -5\).
The function \( \lceil x \rceil : \mathbb{R} \rightarrow \mathbb{Z} \) refers to the smallest integer that is greater than or equal to to \(x\). For example, \( \lceil 2.3 \rceil = 3 \) and \( \lceil -5 \rceil = -5 \).
As with floor functions, the best strategy with integrals or sums involving the ceiling function is to break up the interval of integration (or summation) into pieces on which the ceiling function is constant.
Find \(\displaystyle \int_{-2}^2 \big\lceil 4-x^2 \big\rceil \, dx.\)
This is clearly \( 2\displaystyle \int_{0}^2 \big\lceil 4-x^2 \big\rceil \, dx.\) Now break the interval of integration up into pieces on which \( \big\lceil 4-x^2 \big\rceil = 1,2,3,4.\) This becomes
\[ 2\Big((1-0)\cdot 4 + \big(\sqrt{2}-1\big)\cdot 3 + \big(\sqrt{3}-\sqrt{2}\big) \cdot 2 + \big(2-\sqrt{3}\big) \cdot 1\Big) = 2\big(1+\sqrt{2}+\sqrt{3}+2\big) =12.292\ldots. \ _\square \]
Calculate \[ 2 \int_{2017}^{1} \big( x - \lceil x \rceil \big) dx. \]
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Hint: Sketch out the graph.
If you like this problem, check out this problem as well.
\[ \large \sum_{k=1}^{202} \left \lceil \sqrt k \ \right \rceil = \ ? \]