Difference between revisions of "Floor function"
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== Properties == | == Properties == | ||
| − | * <math> | + | * <math>\lfloor a+b\rfloor\ge \lfloor a\rfloor+\lfloor b \rfloor</math> for all real <math>(a,b)</math>. |
| − | * [[Hermite's Identity]]: <cmath>\left | + | * [[Hermite's Identity]]: <cmath>\lfloor na\rfloor = \left\lfloor a\right\rfloor+\left\lfloor a+\frac{1}{n}\right\rfloor+\ldots+\left\lfloor a+\frac{n-1}{n}\right\rfloor</cmath> |
==Examples== | ==Examples== | ||
Revision as of 16:25, 24 February 2016
The greatest integer function, also known as the floor function, gives the greatest integer less than or equal to its argument. The floor of 
is usually denoted by 
or ![$[x]$](http://latex.artofproblemsolving.com/b/c/e/bceb7b14e55d33a8bca29b7863ad3cdae95afce4.png)
. The action of this function is the same as "rounding down." On a positive argument, this function is the same as "dropping everything after the decimal point," but this is not true for negative values.
Properties
for all real 
.- Hermite's Identity:
![\[\lfloor na\rfloor = \left\lfloor a\right\rfloor+\left\lfloor a+\frac{1}{n}\right\rfloor+\ldots+\left\lfloor a+\frac{n-1}{n}\right\rfloor\]](//latex.artofproblemsolving.com/8/4/d/84ddeb9332a5aedd6606f0b8711d40bd60fecc9e.png)
for all real 
.![\[\lfloor na\rfloor = \left\lfloor a\right\rfloor+\left\lfloor a+\frac{1}{n}\right\rfloor+\ldots+\left\lfloor a+\frac{n-1}{n}\right\rfloor\]](http://latex.artofproblemsolving.com/8/4/d/84ddeb9332a5aedd6606f0b8711d40bd60fecc9e.png)
Examples
A useful way to use the floor function is to write 
, where y is an integer and k is the leftover stuff after the decimal point. This can greatly simplify many problems.
Alternate Definition
Another common definition of the floor function is
![\[\lfloor x \rfloor = x-\{x\}\]](http://latex.artofproblemsolving.com/f/0/8/f08050f59ce7fe0a9ca89436b6fc4bc457641904.png)
where 
is the fractional part of .
Olympiad Problems
- [1981 USAMO #5] If
is a positive real number, and 
is a positive integer, prove that
is a positive integer, prove that![\[[nx] > \frac{[x]}{1} + \frac{[2x]}{2} + \frac{[3x]}{3} + ... + \frac{[nx]}{n},\]](http://latex.artofproblemsolving.com/e/6/3/e638fb866f4cc5f83251ee225ac311b2dccfe8cd.png)
where ![$[t]$](http://latex.artofproblemsolving.com/8/a/8/8a86702dc60b9db916798b1115a6b6822bf828b9.png)
denotes the greatest integer less than or equal to 
.
- [1968 IMO #6] Let
![$[x]$](//latex.artofproblemsolving.com/b/c/e/bceb7b14e55d33a8bca29b7863ad3cdae95afce4.png)
denote the integer part of , i.e., the greatest integer not exceeding . If is a positive integer, express as a simple function of the sum ![\[\left[\frac{n+1}{2}\right]+\left[\frac{n+2}{4}\right]+...+\left[\frac{n+2^k}{2^{k+1}}\right]+\ldots\]](//latex.artofproblemsolving.com/6/b/a/6ba05e683d271bf4c948c7f9b9420f104be804e1.png)
![\[\left[\frac{n+1}{2}\right]+\left[\frac{n+2}{4}\right]+...+\left[\frac{n+2^k}{2^{k+1}}\right]+\ldots\]](http://latex.artofproblemsolving.com/6/b/a/6ba05e683d271bf4c948c7f9b9420f104be804e1.png)
