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wu :: forums    riddles    what am i (Moderators: Eigenray, Icarus, Grimbal, towr, ThudnBlunder, william wu, SMQ)    Working « Previous topic | Next topic » Pages: 1  Reply Notify of replies Send Topic Print    Author  Topic: Working  (Read 1606 times) conehead Guest Working   « on: May 17th, 2005, 11:55am » Quote Modify Remove We work on the circle T = R/2piZ i.e. the real line mod 2pi. IP Logged Michael Dagg Senior Riddler     Gender: Posts: 500 Re: Working   « Reply #1 on: May 17th, 2005, 6:07pm » Quote Modify on May 17th, 2005, 11:55am, 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? IP Logged Regards, Michael Dagg conehead Guest Re: Working   « Reply #2 on: May 17th, 2005, 10:03pm » Quote Modify Remove on May 17th, 2005, 6:07pm, 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? IP Logged Michael Dagg Senior Riddler     Gender: Posts: 500 Re: Working   « Reply #3 on: May 18th, 2005, 9:03am » Quote Modify on May 17th, 2005, 10:03pm, 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? IP Logged Regards, Michael Dagg StonedAgain Guest Re: Working   « Reply #4 on: May 18th, 2005, 8:44pm » Quote Modify Remove on May 17th, 2005, 11:55am, 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. IP Logged Another Aggie Guest Re: Working   « Reply #5 on: May 18th, 2005, 9:11pm » Quote Modify Remove on May 18th, 2005, 8:44pm, StonedAgain wrote:   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.   What you really mean is that it is not the numbers whose generator is mod 2pi are new to you but the generator itself.     IP Logged towr wu::riddles Moderator Uberpuzzler Some people are average, some are just mean.     Gender: Posts: 13730 Re: Working   « Reply #6 on: May 19th, 2005, 1:37am » Quote Modify on May 18th, 2005, 8:44pm, 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. « Last Edit: May 19th, 2005, 1:48am by towr » IP Logged Wikipedia, Google, Mathworld, Integer sequence DB Grimbal wu::riddles Moderator Uberpuzzler     Gender: Posts: 7527 Re: Working   « Reply #7 on: May 19th, 2005, 4:52am » Quote Modify 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. IP Logged towr wu::riddles Moderator Uberpuzzler Some people are average, some are just mean.     Gender: Posts: 13730 Re: Working   « Reply #8 on: May 19th, 2005, 6:55am » Quote Modify on May 19th, 2005, 4:52am, 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? IP Logged Wikipedia, Google, Mathworld, Integer sequence DB musicman Guest Re: Working   « Reply #9 on: May 19th, 2005, 8:05am » Quote Modify Remove on May 18th, 2005, 9:03am, 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. IP Logged Grimbal wu::riddles Moderator Uberpuzzler     Gender: Posts: 7527 Re: Working   « Reply #10 on: May 20th, 2005, 7:54am » Quote Modify on May 19th, 2005, 6:55am, 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. IP Logged Pages: 1  Reply Notify of replies Send Topic Print Forum Jump: ----------------------------- riddles -----------------------------  - easy   - medium   - hard => what am i   - what happened   - microsoft   - cs   - putnam exam (pure math)   - suggestions, help, and FAQ   - general problem-solving / chatting / whatever ----------------------------- general -----------------------------  - guestbook   - truth   - complex analysis   - wanted   - psychology   - chinese « Previous topic | Next topic »

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