tiny-tim Science Advisor Homework Helper Messages 25,837 Reaction score 258 Oct 24, 2012 #32 where's n ?
Daaniyaal Messages 63 Reaction score 0 Oct 24, 2012 #33 tiny-tim said: where's n ? Oh woops θ ε {2nkπ/n}k=0∞
tiny-tim Science Advisor Homework Helper Messages 25,837 Reaction score 258 Oct 24, 2012 #34 now, too many n's !
Daaniyaal Messages 63 Reaction score 0 Oct 24, 2012 #35 AAAAAAH. I swear there's a lot wrong up there. θ ε {2nkπ/k}k=0∞
tiny-tim Science Advisor Homework Helper Messages 25,837 Reaction score 258 Oct 24, 2012 #36 erm … too many k's ?
tiny-tim Science Advisor Homework Helper Messages 25,837 Reaction score 258 Oct 24, 2012 #38 that's the same as your last one (and 2nkπ/k = 2nπ))
tiny-tim Science Advisor Homework Helper Messages 25,837 Reaction score 258 Oct 24, 2012 #40 yes, θ = (2π/n) times k, for any integer k, are the solutions to cos(nθ) = 1 (they're also the solutions for einθ = 1 … they correspond to n equally-spaced positions on the unit circle) is that the answer to the original question?
yes, θ = (2π/n) times k, for any integer k, are the solutions to cos(nθ) = 1 (they're also the solutions for einθ = 1 … they correspond to n equally-spaced positions on the unit circle) is that the answer to the original question?
tiny-tim Science Advisor Homework Helper Messages 25,837 Reaction score 258 Oct 25, 2012 #43 Daaniyaal said: Yes! How would I prove it though? :/ which part of the proof are you not clear about?
Daaniyaal said: Yes! How would I prove it though? :/ which part of the proof are you not clear about?
Daaniyaal Messages 63 Reaction score 0 Oct 28, 2012 #44 I figured it out, if I were to just replace the statement with my conjecture it would be proven, thanks!
I figured it out, if I were to just replace the statement with my conjecture it would be proven, thanks!