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**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,905

I found that

!?

IPBLE: Increasing Performance By Lowering Expectations.

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**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,905

I'll try to find function for all x.

Can someone help me?

IPBLE: Increasing Performance By Lowering Expectations.

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**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,905

Here's better function:

*Last edited by krassi_holmz (2005-12-28 07:11:37)*

IPBLE: Increasing Performance By Lowering Expectations.

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**God****Member**- Registered: 2005-08-25
- Posts: 59

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

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**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,905

Here is another:

*Last edited by krassi_holmz (2005-12-28 07:41:38)*

IPBLE: Increasing Performance By Lowering Expectations.

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**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,905

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 (2005-12-28 08:59:49)*

IPBLE: Increasing Performance By Lowering Expectations.

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**ryos****Member**- Registered: 2005-08-04
- Posts: 394

Impressive! How about this:

Floor(x) = x - (x % 1)

Where % is the modulus operator.

El que pega primero pega dos veces.

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**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,905

Picture 1: x-floor{x}

*Last edited by krassi_holmz (2005-12-28 08:55:24)*

IPBLE: Increasing Performance By Lowering Expectations.

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**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,905

Picture 2: be[x]=ArcCos[Sin[x+Pi/2]]

*Last edited by krassi_holmz (2005-12-28 08:56:09)*

IPBLE: Increasing Performance By Lowering Expectations.

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**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,905

Picture 3: sin[x]

*Last edited by krassi_holmz (2005-12-28 08:56:40)*

IPBLE: Increasing Performance By Lowering Expectations.

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**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,905

Picture 4: sinr(x)=sin(Pi x)

*Last edited by krassi_holmz (2005-12-28 08:57:38)*

IPBLE: Increasing Performance By Lowering Expectations.

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**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,905

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

*Last edited by krassi_holmz (2005-12-28 08:58:10)*

IPBLE: Increasing Performance By Lowering Expectations.

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**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,905

Picture 6: The function we've iust assembled.

*Last edited by krassi_holmz (2005-12-28 08:58:47)*

IPBLE: Increasing Performance By Lowering Expectations.

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**John E. Franklin****Member**- Registered: 2005-08-29
- Posts: 3,588

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!

**igloo** **myrtilles** **fourmis**

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**God****Member**- Registered: 2005-08-25
- Posts: 59

Awesome work!

About the square impulse, I'm sure it can somehow be derived from the equation of a circle, since you can control the endpoints by controling the center and radius.

If you allow limits, it becomes even easier (but totally useless for most practical cases)

The corresponding function g(x) would be represented in the form:

g(x) = lim(n->oo)[((x-a)/b)^n]*(x)+(x)

for the interval (a, a+b]

*Last edited by God (2005-12-30 11:58:09)*

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