Recent content by bobibomi

ω = √- \stackrel{(D((2*cos(ak)- 2))}{(m)} ω = √\stackrel{(-D((2*cos(ak)- 2))}{(m)} -(A-B) = (-A - -B) = (-A+B) = (B-A) ω = √\stackrel{(-D((2*cos(ak)- 2))}{(m)} = √\stackrel{(D((-2*cos(ak)- - 2))}{(m)} = √\stackrel{(D((-2*cos(ak) + 2))}{(+m)} = √\stackrel{(D((2-2*cos(ak)))}{(m)} Is...

ω = √- \stackrel{(D((2*cos(ak)- 2))}{(m)} ω = √\stackrel{(-D((2*cos(ak)- 2))}{(m)} -(A-B) = (-A - -B) = (-A+B) = (B-A) ω = √-\stackrel{(D((2*cos(ak)- 2))}{(-m)} = √\stackrel{(-D((2*cos(ak)- 2))}{(--m)} = √\stackrel{(-D((2*cos(ak)- 2))}{(+m)} = √\stackrel{(-D((2*cos(ak)- 2))}{(m)} Is...

ω = √- \stackrel{D((2*cos(ak)- 2)}{m} I don´t know how do you mean that. Maybe you can show it

ω = √- \stackrel{D((2*cos(ak)- 2)}{m} Is this right?

m * -1*ω2 = D((2*cos(ak)- 2) omega as a function of D,m,a,k maybe this: -1*ω2 = \stackrel{D((2*cos(ak)- 2)}{m} -ω = √ \stackrel{D((2*cos(ak)- 2)}{m} ω = - √ \stackrel{D((2*cos(ak)- 2)}{m} Where is my mistake?

m * -1*ω2 = D((2*cos(ak)- 2) omega as a function of D,m,a,k maybe this: ω = -√\stackrel{D((2*cos(ak)- 2)}{m} is this right?

B*(-A) = 2A+2A*C -AB = 2A + 2AC divide by A: -B = 2 + 2C so in the real equation: m * -sin(aik)*ω2 = D((2*sin(aki)cos(ak)- 2*(sin(aik)) divide by sin(aik): m * -1*ω2 = D((2*cos(ak)- 2) Is this right?

B*(-A) = 2A+2A*C -AB = 2A + 2ACSorry bu I don´t know how to go on.

m * -sin(aik)*ω2 = D((2*sin(aki)cos(ak)- 2*(sin(aik)) I think: sin(aik) we can canceled out: m *ω2 = D(2*sin(aki)cos(ak)- (sin(aik)) Is this right? and then?

m * -sin(aik)sin(ωt)*ω2 = D(sin(ωt)(2*sin(aki)cos(ak)- 2*(sin(aik)sin(ωt)) Ok, this was the last right step, but I don´t know how I should go on

m * -sin(aik)sin(ωt)*ω2 = D(sin(ωt)(sin(a (i+1) k) + sin(a (i-1) k)) - 2*(sin(a (i) k) sin(ωt)) 1) a(i+1)k = aki+ak 2) a(i-1)k = aki - ak m * -sin(aik)sin(ωt)*ω2 = D(sin(ωt)(sin(aki+ak) + sin(aki - ak)) - 2*(sin(aik)sin(ωt)) Sin(A+B) = sin(A)cos(B)+cos(A)sin(B) m *...

1)a(i+1)k = aki+ak 2)a(i-1)k = aki- ak is the first step right?

m * -sin(aik)*ω2 = D(sin(a (i+1) k) + sin(a (i-1) k)) - 2*(sin(a (i) k) Sin(A+B) = sin(A)cos(B)+cos(A)sin(B) m * -sin(aik)*ω2 = D(sin(a (sin(i)cos(1)+cos(i)sin(1)) k) + sin(a (sin(i)cos(1)+cos(i)+sin(1)) k)) - 2*(sin(a (i) k) Is this right?

m * -sin(aik)sin(ωt)*ω2 = D(sin(ωt)(sin(a (i+1) k) + sin(a (i-1) k)) - 2*(sin(a (i) k) sin(ωt)) m * -sin(aik)*ω2 = D(sin(a (i+1) k) + sin(a (i-1) k)) - 2*(sin(a (i) k) I´m sorry but I don´t know a formula for expanding sin(A+B).

m * -sin(aik)sin(ωt)*ω2 = D(sin(a (i+1) k) sin(ωt) + a (i+1) + sin(a (i-1) k) sin(ωt) + a (i-1) - 2*(sin(a (i) k) sin(ωt) + a (i)) m * -sin(aik)sin(ωt)*ω2 = D(sin(ωt)(sin(a (i+1) k) + sin(a (i-1) k)) + a (i+1) + a (i-1) - 2*(sin(a (i) k) sin(ωt) + a (i)) m * -sin(aik)sin(ωt)*ω2 =...