Absurdism
|
  |
| Joined: 18 Jul 2013 |
| Total Posts: 2568 |
|
|
| 22 Sep 2013 08:28 PM |
Yes, this is related to programming. Let us define a complex primitive Nth root of unity omega:
ω = cos(theta) + i*sin(theta) ω = e^(2*pi/N) By the definition of an nth root of unity, ω is the second solution to the equation x^N = 1.
Can someone explain to me how we can simplify the solutions in exponents via modular arithmetic and whatnot? I sort of understand the generic congruence modulo nonsense:
ω^M where M > N M = Q * N + R where Q is some quotient and R is some remainder. ω^(Q*N + R) = ω^(Q*N) + ω^R = ω^R R is congruent to M(mod N). I don't exactly understand this. The congruence modulo is not only slightly confusing, the process is rather odd. This is how my quantum mechanics professor explained it. I need to input 6 values corresponding to ω (those values are irrelevant to the problem), however I can only use ω and ω^2, even though N = 6. How do I get rid of the other exponents if my problem includes ω^3, ω^6, ω^9, ω^12 and ω^15? |
|
|
| Report Abuse |
|
|
|
| 22 Sep 2013 10:11 PM |
My response:
2+2=4
yay im smart
http://wiki.roblox.com/index.php/User:ElectricBlaze |
|
|
| Report Abuse |
|
|
| |
|
|
| 22 Sep 2013 10:24 PM |
| Sorry, I'm not a arithmaconologist. |
|
|
| Report Abuse |
|
|
Absurdism
|
  |
| Joined: 18 Jul 2013 |
| Total Posts: 2568 |
|
|
| 22 Sep 2013 10:45 PM |
I figured it out.
w^(QN+R)=w^(QN)×w^R=(w^N)Q×w^R=(1)^Q×w^R=w^R
Thank you ElectricBlaze! |
|
|
| Report Abuse |
|
|
|
| 22 Sep 2013 10:46 PM |
No problem, always glad to help people with math!
http://wiki.roblox.com/index.php/User:ElectricBlaze |
|
|
| Report Abuse |
|
|
Absurdism
|
  |
| Joined: 18 Jul 2013 |
| Total Posts: 2568 |
|
| |
|
Dr01d3k4
|
  |
| Joined: 11 Oct 2007 |
| Total Posts: 17916 |
|
|
| 23 Sep 2013 04:39 PM |
| I did this in my last FP2 lesson, yet I have no idea what you are talking about DX |
|
|
| Report Abuse |
|
|
Absurdism
|
  |
| Joined: 18 Jul 2013 |
| Total Posts: 2568 |
|
|
| 23 Sep 2013 04:42 PM |
| I did this in my last quantum mechanics lesson. I barely understand the identities. Complex nth roots of unity are fine self-contained, but manipulating their properties is a vile act of sorcery. |
|
|
| Report Abuse |
|
|
|
| 23 Sep 2013 04:58 PM |
"x^N = 1"
N=0?
Yay, I be smrt! |
|
|
| Report Abuse |
|
|
|
| 23 Sep 2013 05:08 PM |
Screw math. I hate math. Math hates me. |
|
|
| Report Abuse |
|
|
Absurdism
|
  |
| Joined: 18 Jul 2013 |
| Total Posts: 2568 |
|
|
| 23 Sep 2013 05:33 PM |
@Notunknown, by the definition of nth roots of unity, there are n complex solutions. If N were 8, then the following would be the solutions:
1, 1/sqrt(2)*i + 1/sqrt(2), (1/sqrt(2)*i + 1/sqrt(2))^2, (1/sqrt(2)*i + 1/sqrt(2))^3, (1/sqrt(2)*i + 1/sqrt(2))^4, (1/sqrt(2)*i + 1/sqrt(2))^5, (1/sqrt(2)*i + 1/sqrt(2))^6, (1/sqrt(2)*i + 1/sqrt(2))^7
I received 1/sqrt(2)*i + 1/sqrt(2) from the form w = e^((theta*pi*i)/N), which expands and exchanges to cos(pi/4) + i*sin(pi/4). |
|
|
| Report Abuse |
|
|
Absurdism
|
  |
| Joined: 18 Jul 2013 |
| Total Posts: 2568 |
|
|
| 23 Sep 2013 05:34 PM |
| Oh, right: 0 is considered an invalid answer. |
|
|
| Report Abuse |
|
|