Approximate and accurate proof of time complexity of the Fibonacci series

The Fibonacci series can be derived from many applications. We know that the time complexity of the Fibonacci series is exponential. Now let's roughly prove it:

Fibonacci SeriesRecurrence:

F (n) = f (n-1) + f (n-2)

F (1) = F (2) = 1

It is roughly proved that decision_tree can be used. For more intuitive purposes, I reference another constant function f (x) = 0; X = 1, 2, 4, 5 ,............

SoFibonacci SeriesRecursive deformation is as follows:

F (n) = f (n-1) + f (n-2)+ F (N)

F (1) = F (2) = 1

Draw the decision_tree

// F (n) = f (n) /// F (n-1) f (n-2) /////// F (n-2) F (n-3) f (n-3f )//.... //... F (2) /// \\// F (1) F (2) 1 ///\/// 1 1 1 1

Because F (n) is always equal to 0, F (n) is equal to the sum of leaf nodes multiplied by theta (1). The longest link is the leftmost subtree, and the shortest link is the rightmost subtree.

So there are: 2 ^ (n/2) <F (n) <2 ^ N, that is, F (n) = theta (2 ^ N ).

Exact proof: