Solve for n: x_n=(phi^n+(1-phi)^n)/sqrt (5)

Question

Can you solve for exponents n in equations likethe fibonacci formula

Answer 1

x_n=(φⁿ+(1-φ)ⁿ)/√5

The closed from of the Fibonacci number is  x_n=(φⁿ-(1-φ)ⁿ)/√5, where

φ=(1+√5)/2 (golden ratio). 

x_n=(φⁿ-(1-φ)ⁿ)/√5+2(1-φ)ⁿ)/√5=F_n-1+2(1-φ)ⁿ)/√5=

                                                 =F_n-1+(2/√5)*[(1-√5)/2]ⁿ

 

Alternatively, if we look at  the difference x_n-x_n-1, then

x_n-x_n-1=(φⁿ+(1-φ)ⁿ-φ^n-1-(1-φ)^n-1)/√5)=

           = [(φ-1)φ^n-1-φ(1-φ)^n-1]/√5)=

           = [φ(φ^n-1-(1-φ)^n-1)-φ^n-1]/√5)=φF_n-1-φ^n-1/√5. 

So, 

x_n=φF_n-1-φ_n-1/√5+x_n-1=φF_n-1-φ^n-1/√5+(φ^n-1+(1-φ)^n-1)/√5=

    =φF_n-1+(1-φ)^n-1/√5 and assuming φ=(1+√5)/2, we can write as

x_n=φF_n-1+(1-√5)^n-1/(2^n-1√5).