I'll try to find function for all x.
Can someone help me?

cool... how did you come up with that?

How did I Came up with that?
I'll explain.

The function x - floor[x] = {x} is periodic with period 1. (picture 1)
I wanted to represent {x} with trigonometric functions.
First, ArcCos[Sin[x+Pi/2]]= |x| for x ∈ [-Pi,Pi] and this function is peridic with period 2Pi.
Let ArcCos[Sin[x+Pi/2]]=be[x]. (picture 2)
I reduced be[x] to have period 1:
ber[x]=be[Pi x].
Now you can notice that if x ∈(2n,2n+1) ber[x] ≡ {x} and if x∈(2m-1,2m) 1-ber[x]≡{x} ; n,m∈N. This is ecvivalent to:
Now i have to find function g[x]:
{|g[x]-ber[x]|}≡{x}, so
g[x]=
1. 0, if x ∈(2n,2n+1)
2. 1, if x ∈(2m-1,2m)
If I find function that changes its sign in period of 1, I'll be able to reduce it to g[x].
I can take a periodic function, for example sin[x]. (picture 3)
It has period Pi. We need period 1. So we'll use
sinr[x]=sin[Pi x] (picture 4)
The new function has period 1.
Now we construct function that gives us the sign of sinr[x]:
ss[x]=sinr[x]/(|sinr[x]|) (picture 5)
If x∈(2m-1,2m) ss[x]=-1
If x∈(2n,2n+1) ss[x]=1.
Now the only thing we need is a function q(x) for such
q[-1]=1
q[1]=0
This is linear function:
-a+b=1 && a+b=0 => 2a+1=0 => a=-1/2;b=1/2
so
q[x]=-1/2x+1/2
and q[ss[x]]=g[x] and
(|q[ss[x]]-ber[x]|)={x}
and
Floor[x]=x-(|q[ss[x]]|)
And when you simplify and substitute modulus with sqr(x^2) you should get the second example.
The final result depends on what periodic shall we take.

Last edited by krassi_holmz (2005-12-29 07:59:49)

Picture 1: x-floor{x}

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Last edited by krassi_holmz (2005-12-29 07:55:24)

Picture 3: sin[x]

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Last edited by krassi_holmz (2005-12-29 07:56:40)

Picture 5: ss[x]=sinr[x]/(|sinr[x]|)

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Last edited by krassi_holmz (2005-12-29 07:58:10)

Totally rad! You must be breaking new ground.
I'll go over this more closely soon.  This is terrific!
How about an impulse function, a square wave that
occurs only once.  This could be used as a pass filter
to partition out certain parts of a function within a
certain range of x.  So instead of saying y=5 for 10 < x <20,
you could multiply by the square impulse.
Awesome work, krassi!