You can define the order recursively as well. For instance, let's say you have a function f. To calculate f(n) takes k steps. Now you want to calculate f(n+1). Lets say f(n+1) calls f(n) once, then f(n+1) takes k + some constant steps. Each invocation will take some constant steps extra, so this method is O(n).
Now look at another example. Lets say you implement fibonacci naively by adding the two previous results:
fib(n) = { return fib(n-1) + fib(n-2) }
Now lets say you can calculate fib(n-2) and fib(n-1) both in about k steps. To calculate fib(n) you need k+k = 2*k steps. Now lets say you want to calculate fib(n+1). So you need twice as much steps as for fib(n-1). So this seems to be O(2^N)
Admittedly, this is not very formal, but hopefully this way you can get a bit of a feel.
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