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Topic: Floor Functions
Replies: 2   Last Post: Apr 28, 2009 4:09 PM

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No-one

Posts: 1
From: :)
Registered: 1/17/09
Floor Functions
Posted: Jan 17, 2009 12:17 PM
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I am/was/still doing a research [ not the only 1] on these types of functions [ check wikipedia if you don't know what they are ]. I was a university student and I liked math alot. However things in real life has made me loose interest lately [ lately is not quite right, 4 years maybe?], and I had to withdraw [not cause of laziness].

I really wish to share some stuff with you :). Tell me if you like it or not :). Most of my work is nothing special, but math has been my favorite hobby since I was young and sharing stuff with people makes me happy :). I have never been on online forums about math before so not sure if I am posting this in the right place.

I discovered floor functions before I even learned them. I was proud that I managed to find stuff before I even learned it, and it was right. Of course I managed to also find some stuff thats interesting :).

check this formula

I think you may like this :) .

I will use the symbol [ .. ] for floor functions rather then the normal symbol because its easier to type.


A) a floor function to convert numbers from base 10 to other bases.

let assume you want to convert base 10 to base X.

then { the n in the dn should be a subscript }

dn = [ U/ b^(n-1)] - b[ U/b^n]

where, U = value of number in base 10.
, b = the base you want to convert the number
n = the digit position of the number in BASE X.[ n=1 is left to radix point, n=0 is right to radix point ]
dn = the value of the digit at digit position n in BASE X.

where , U 10 = ....{ d3 }{ d2 }{ d1 }.{ d0 }{ d-1 }.... x ( by those brackets I mean a digit )

[10 and x should be small letters]

Example: 123 base 10 to base 4.

using formula.

its a known fact that the number in base 4 will not have any decimals or fraction, so its useless trying n= 0 or less.

let n = 1,

d1 = [ 123/4^(1-1)] - 4[ 123/4^1]
d1 = [123] - 4[30.75] = 123 - 4(30) = 123 -120 = 3

let n = 2,

d2 = [123/4^(2-1)] - 4[ 123/4^2] = [ 30.75] - 4[7.6875] = (30) -4 (7) = 2

let n = 3,

d3 = [ 123/ 4^(3-1)] - 4[ 123/4^3] = [ 7.6875] - 4[1.921875] = (7) - (4) = 3

let n = 4,

d4 = [ 123/ 4^(4-1)] - 4[ 123/4^4] = [1.921875] -4[0] = 1

putting n= 5 or more, always shows 0., putting n = 0 or less, always shows 0, because there is no decimal.

this means the number in base 4 is, 1323

It works with all numbers , as long as b is a whole number. It also converts fractions. It still faster to use algorithym, but mathematicians may prefer formulas :).

Nothing important but its still interesting :). { if you interested in proof reply bk }

[ I am doing a research about floor function integrals. If you are interested I will tell you about it :).]

I will keep posting similar curiouse stuff from time to time if you want :).


Date Subject Author
1/17/09
Read Floor Functions
No-one
2/14/09
Read Re: Floor Functions
No-one1
4/28/09
Read Re: Floor Functions
Hecman Gun

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