# Finding a formula for this data

• Oct 16th 2009, 09:25 AM
andy450
Finding a formula for this data
I'm trying to work out a formula (i dont even know if it possibly exists) but this is the information is have:
The 1st term - 1
The 2nd term - 1
3rd - 1
4th - 2
5th - 2
6th - 2
7th - 2
8th - 2
9th - 3
10th - 3
etc, etc, but basically i realise that the pattern is you square root the term and round down, eg, if u wanted to find the 29th term, u squeare root, so u get 5.39 and round down to 5, so the 29th term is 5.
What i need to know is there a formula for this, not matter how complicated???
• Oct 16th 2009, 10:09 AM
aidan
Quote:

Originally Posted by andy450
I'm trying to work out a formula (i dont even know if it possibly exists) but this is the information is have:
The 1st term - 1
The 2nd term - 1
3rd - 1
4th - 2
5th - 2
6th - 2
7th - 2
8th - 2
9th - 3
10th - 3
etc, etc, but basically i realise that the pattern is you square root the term and round down, eg, if u wanted to find the 29th term, u squeare root, so u get 5.39 and round down to 5, so the 29th term is 5.
What i need to know is there a formula for this, not matter how complicated???

Quote:

is there a formula for this
Yes.
A floor function or greatest integer function is needed (for rounding down).

$\displaystyle f(x) = INT \left( \sqrt{x} \, \right )$

.
• Oct 17th 2009, 07:58 AM
CaptainBlack
Quote:

Originally Posted by aidan
Yes.
A floor function or greatest integer function is needed (for rounding down).

$\displaystyle f(x) = INT \left( \sqrt{x} \, \right )$

.

The floor function even has its own notation:

$\displaystyle \lfloor \sqrt{x} \rfloor$ denotes the greatest integer less than $\displaystyle \sqrt{x}$

There is also the similar ceiling function - the least integer greater than ..

so for $\displaystyle \sqrt{x}$ this would be writtem $\displaystyle \lceil \sqrt{x} \,\rceil$.

CB