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        riddles >> what am i >> Working
        
(Message started by: conehead on May 17^th, 2005, 11:55am)

Title: Working
Post by conehead on May 17^th, 2005, 11:55am

We work on the circle T = R/2piZ i.e. the real line mod 2pi.

Title: Re: Working
Post by Michael_Dagg on May 17^th, 2005, 6:07pm

on 05/17/05 at 11:55:09, conehead wrote:

We work on the circle T = R/2piZ i.e. the real line mod 2pi.

...and such that u in T then u + 2pi = u, you are a Fourier analyst?

Title: Re: Working
Post by conehead on May 17^th, 2005, 10:03pm

on 05/17/05 at 18:07:20, Michael_Dagg wrote:

...and such that u in T then u + 2pi = u, you are a Fourier analyst?

This is very correct. How do you know this?

Title: Re: Working
Post by Michael_Dagg on May 18^th, 2005, 9:03am

on 05/17/05 at 22:03:20, conehead wrote:

This is very correct. How do you know this?

It comes naturally.

What would be more suitable for you: to take the interval [0, 2pi] and bend the ends together a then glue them or wind the real line R into a circle?

Title: Re: Working
Post by StonedAgain on May 18^th, 2005, 8:44pm

on 05/17/05 at 11:55:09, conehead wrote:

We work on the circle T = R/2piZ i.e. the real line mod 2pi.

I have to admit that numbers mod 2pi are completely new to me. I thought you could only do modulo integers - a = b mod 2pi are not integers and besides they it look like they are points on a circle given this conversation.

Title: Re: Working
Post by Another Aggie on May 18^th, 2005, 9:11pm

on 05/18/05 at 20:44:54, StonedAgain wrote:

What you really mean is that it is not the numbers whose generator is mod 2pi are new to you but the generator itself.

Title: Re: Working
Post by towr on May 19^th, 2005, 1:37am

on 05/18/05 at 20:44:54, StonedAgain wrote:

I have to admit that numbers mod 2pi are completely new to me. I thought you could only do modulo integers

Well, now you know. And knowing is half the battle ;)
You can work modulo any number. (And you can also work in non-integer bases, which is loosely related)
x = y modulo r just means that there is an integer n such that x = y + n*r And you can apply that for any real x,y and r, and probably even complex ones.

Quote:

and besides they it look like they are points on a circle given this conversation.

That also holds for when you deal with only integers. But in that case the circumference of the circle is also an integer.
All points of the numberline that, when wrapped around the circle, fall together are the same modulo the circumference of the circle.

Title: Re: Working
Post by Grimbal on May 19^th, 2005, 4:52am

You can even work modulo a polynomial in the space of polynomials. There, the multiple is also a polynomial.
i.e. a = b (mod p) if a = b+pq, where a, b, p, q are all polynomials.

Title: Re: Working
Post by towr on May 19^th, 2005, 6:55am

on 05/19/05 at 04:52:41, Grimbal wrote:

You can even work modulo a polynomial in the space of polynomials. There, the multiple is also a polynomial.
i.e. a = b (mod p) if a = b+pq, where a, b, p, q are all polynomials.

If q can be just any polynomial, wouldn't all a,b be congruent modulo p? Or isn't q = (a-b)/p a polynomial?

Title: Re: Working
Post by musicman on May 19^th, 2005, 8:05am

on 05/18/05 at 09:03:18, Michael_Dagg wrote:

or wind the real line R into a circle?

This is the most suitable and lends more to the idea of unwinding or unrolling a function on T.

Title: Re: Working
Post by Grimbal on May 20^th, 2005, 7:54am

on 05/19/05 at 06:55:47, towr wrote:

If q can be just any polynomial, wouldn't all a,b be congruent modulo p? Or isn't q = (a-b)/p a polynomial?

I don't think so. Especially in cases where (a-b) has a lower order than p.