Difference between revisions of "Floor function"

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*<math>\lfloor -3.2 \rfloor = -4</math>
 
*<math>\lfloor -3.2 \rfloor = -4</math>
  
A useful way to use the floor function is to write <math>\lfloor x \rfloor=\lfloor y+k \rfloor</math>, where y is an integer and k is the leftover stuff after the decimal point. This can greatly simplify many problems.  
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A useful way to use the floor function is to write <math>\lfloor x \rfloor=\lfloor y+k \rfloor</math>, where y is an integer and k is the leftover stuff after the decimal point. This can greatly simplify many problems.
  
 
==Alternate Definition==
 
==Alternate Definition==

Revision as of 18:08, 19 September 2020

The greatest integer function, also known as the floor function, gives the greatest integer less than or equal to its argument. The floor of $x$ is usually denoted by $\lfloor x \rfloor$ or $[x]$. 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

Examples

  • $\lfloor 3.14 \rfloor = 3$
  • $\lfloor 5 \rfloor = 5$
  • $\lfloor -3.2 \rfloor = -4$

A useful way to use the floor function is to write $\lfloor x \rfloor=\lfloor y+k \rfloor$, 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

\[\lfloor x \rfloor = x-\{x\}\]

where $\{x\}$ is the fractional part of $x$.

Olympiad Problems

  • [1981 USAMO #5] If $x$ is a positive real number, and $n$ is a positive integer, prove that

\[[nx] \geq \frac{[x]}{1} + \frac{[2x]}{2} + \frac{[3x]}{3} + ... + \frac{[nx]}{n},\] where $[t]$ denotes the greatest integer less than or equal to $t$.

AoPS discussion 1

AoPS discussion 2

  • [1968 IMO #6] Let $[x]$ denote the integer part of $x$, i.e., the greatest integer not exceeding $x$. If $n$ is a positive integer, express as a simple function of $n$ the sum \[\left[\frac{n+1}{2}\right]+\left[\frac{n+2}{4}\right]+...+\left[\frac{n+2^k}{2^{k+1}}\right]+\ldots\]

See Also