Theceiling function(also known as theleast integer function) of a real numberx,denoted⌈x⌉,is defined as the smallest integer that is not smaller thanx.
For example,⌈9⌉=9,⌈1.006⌉=2,⌈19⌉=5,⌈π⌉=4,⌈−10.27⌉=−10.
In general,⌈x⌉is the unique integer satisfying⌈x⌉−1<x≤⌈x⌉.
The ceiling is related to thefloor functionby the formula⌈x⌉=−⌊−x⌋.
What is the range ofxthat satisfies
⌈⌈x⌉−1.3⌉=16?
Let⌈x⌉=y, whereyis an integer by the definition of the ceiling function. Then
⌈⌈x⌉−1.3⌉⌈y−1.3⌉15<y−1.316.3<yy=16=16≤16≤17.3=17.(sinceyis an integer)
(1)⌈x+n⌉=⌈x⌉+n,for any integern. (2)⌈x⌉+⌈−x⌉={10ifx∈/Zifx∈Z. (3)⌈x+y⌉=⌈x⌉+⌈y⌉or⌈x⌉+⌈y⌉−1.
These can all be proved from the analogous properties for thefloor function.
Ify=⌊2x+5⌋=3⌊x−4⌋and the product of all the distinct values of⌊3x+y⌋isA, findAmod2016.
Notation:⌊⋅⌋denotes the floor function.
Problem-solving
As with the floor function, it is often easiest to writex=n−r,wheren=⌈x⌉is an integer and0≤r<1.
找到所有的解决方案⌈x⌉⌈2x⌉=15.
Writex=n−ras above. Then⌈2x⌉=2nor2n−1depending onr.In the first case, wherer<21,the equation becomes2n2=15,which has no solution. In the second case, wherer≥21,the equation becomesn(2n−1)=15,so2n2−n−15=0,or(n−3)(2n+5)=0.The only integer solution isn=3.
So the range of solutions is the interval(2,2.5].□
Find a positive integernsuch that⌊1320n⌋+⌈2013n⌉=2013.
This problem is proposed by Ahaan Rungta.
Details and assumptions:
The function⌊x⌋:R→Zrefers to the greatest integer smaller than or equal tox. For example⌊2.3⌋=2and⌊−5⌋=−5.
The function⌈x⌉:R→Zrefers to the smallest integer that is greater than or equal to tox. For example,⌈2.3⌉=3and⌈−5⌉=−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∫−22⌈4−x2⌉dx.
This is clearly2∫02⌈4−x2⌉dx.Now break the interval of integration up into pieces on which⌈4−x2⌉=1,2,3,4.This becomes