SEQUENCES u1=real number, un+1=un²-1, given u2=u1 find two values of u1?

[ashleya3]

Well im given the recursive definition, but i don't how to find the first term.The answer is ½(1±√5 )but i dont know how to get there, looking for the logic behind the answer

[gudspeling]

u[n+1] = u[n]^2 - 1u[2] = u[1]^2 - 1u[2]=u[1]Lt u[1]=xx=x^2-1x^2 - x - 1=0You can solve this using the quadratic formulax= [1±sqrt(1+4)]/2x= (1/2)(1±sqrt(5))u[1] = ½(1±√5)

[whatagowk]

the logic is there. u just have to discover it. like a gold mine. until it is discovered it is useless.

[Dr D]

You have all the info you need. U2 = U1. Now apply for formula for n = 1.U2 = U1 = U1^2 - 1So U1^2 - U1 - 1 = 0Now apply the quadratic formula to get the above answer.

[Kemmy]

U(n+1) = [U(n)]^2 - 1 and since U2=U1, let n=1,U2 = [U1]^2 - 1U1 = [U1]^2 - 10 = [U1]^2 - U1 - 1{Quadratic equation, x = [-b±√(b^2-4ac)] / 2a}U1 = [1±√(1-4(1)(-1))] / 2U1 = [1±√5]/2U1 = ½(1±√5 )

[Imran]

All these things they just go above my head...