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, $$\begin{array}{c}&\lceil 9\rceil=9, &\lceil 1.006\rceil =2, &\left\lceil \sqrt{19}\right\rceil=5, &\lceil \pi\rceil = 4, &\lceil -10.27\rceil = -10.\end{array}$$

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

$$\big\lceil \lceil x \rceil - 1.3 \big\rceil = 16 ?$$

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Let $\lceil x \rceil= y$, where $y$ is an integer by the definition of the ceiling function. Then

$$\begin{aligned} \big\lceil \lceil x \rceil - 1.3 \big\rceil &= 16\\ \lceil y - 1.3 \rceil &= 16\\ 15 < y-1.3 &\le 16\\ 16.3 < y &\le 17.3\\ y&=17. \qquad \text{(since }y\text{ is an integer)} \end{aligned}$$

Since $\lceil x \rceil= y,$

$$\begin{aligned} \lceil x \rceil&= 17\\ 16<x&\le 17. \ _\square \end{aligned}$$

(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.

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 $\lceil x \rceil \lceil 2x \rceil = 15.$

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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 $(2,2.5].\ _\square$

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 $\displaystyle \int_{-2}^2 \big\lceil 4-x^2 \big\rceil \, dx.$

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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

$$2\Big((1-0)\cdot 4 + \big(\sqrt{2}-1\big)\cdot 3 + \big(\sqrt{3}-\sqrt{2}\big) \cdot 2 + \big(2-\sqrt{3}\big) \cdot 1\Big) = 2\big(1+\sqrt{2}+\sqrt{3}+2\big) =12.292\ldots. \ _\square$$