Question

Hy

How do i show the following:
f(n) <= f(n-1) + f(n-2)+ .. + f(1) implies f(n) = O(2^n)

I think we can assume that f is monotonically increasing =>

f(n) <= n*f(n-1)

Answer 1

You were on the right track

f(1) <= f(0)
f(2) <= f(0) + f(1)

|f(2)| <= f(0)|2^2|
|f(1)| <= f(0)|2^1|
|f(0)| <= f(0)|2^0|
Where f(0) is a constant

let x be a constant (equal to f(0))

Suppose that |f(i)| <= x|2^i| for all values of i <= n
Then f(n+1) <= f(n) + (f(n-1) + f(n-2) + … + f(0))
==> |f(n+1)| <= |f(n)| + |(f(n-1) + f(n-2) + … + f(0))|
==> |f(n+1)| <= x|2^n| + x|2^n-1| + x|2^n-2| + .. + x
==> |f(n+1)| <= x|2^(n+1)|
so case n implies case n+1

And by induction the claim holds for all n in N

And from the definition of Big-O notation, f(x) is in O(2^n)