What is f(x) = 1 + (x/3) + (x/3)^2 + (x/3)^3 + (x/3)^4 + ...

Question

I am trying to derive the formula for the nth Fibonacci number by using infinite-degree polynomial.

Answer 1

1/(1-x)=(1-x)^-1=1+x+x^2(-1)(-2)/2!-x^3(-3)(-2)(-1)/3!...=1+x+x^2+x^3+x^4+...

Therefore, 1/(1-(x/3))=(1-(x/3))^-1=1+(x/3)+(x/3)^2+...=f(x).

Here's a way to find Fibonacci numbers: x=1/(1+x). Start with x=0; this gives another value for x=1; put this value in and get another x=1/2; put this in and we get 2/3; then 3/5, 5/8, 8/13, 13/21,... Fibonacci series is [0,]1,2,3,5,8,13,21,... The quotient of any two adjacent terms approaches the Golden Ratio as the series progresses. Hope you find this interesting. The solution for x in the equation is in fact the Golden Ratio itself: x^2+x-1=0, x=1/2(sqrt(5)-1) (positive root)=0.61803..., so that 1/0.61803=1.61803=1/2(sqrt(5)+1).