The function n -> 2^n takes input of size O(log(n)) and returns output of size O(n).
Of course, n = 2^(log n), so our function takes exponential time/space in input size to simply write the output.
The Fibonacci function is like that.
Of course, discussions of complexity become cumbersome if you don't specify variables; that's why we talk about complexity in n of computing Fib(n).
Potato, potato anyway.
The function n -> 2^n takes input of size O(log(n)) and returns output of size O(n).
Of course, n = 2^(log n), so our function takes exponential time/space in input size to simply write the output.
The Fibonacci function is like that.
Of course, discussions of complexity become cumbersome if you don't specify variables; that's why we talk about complexity in n of computing Fib(n).
Potato, potato anyway.