let rec f1 = function 
	| 0 -> 0
	| n -> n + f1(n-1)

let rec f2 n = match n with 
	| 0 -> 0
	| n -> n + f2 n-1 (* here f2 is always given the same n, so OCaml gives a stack overflow error *)

let rec f3 n m = match n with  
	| 0 -> m
	| n -> f3 (n-1) m + n (* here the same m is given all over again to f3. However, n decreases. 
		Check m after calling this function. The result is the same as before calling it. *)

let rec f4 n m k = match n with 
	| 0 -> m + k
	| n -> 1 + f4 (n-1) m+1 k-1 (* OCaml says there is an error while calling f4 function:
	Error: This expression has type int -> 'a
       		but an expression was expected of type int

	What OCaml interpreter knows about the types of the arguments of f4? n is clearly int, m and k 
   are also int because we return m+k. Now for OCaml, f4 is a function int -> int -> int -> 'a. 
   When OCaml parses the expression 1 + f4 (n-1) m+1 k-1, it sees that (f4 (n-1) m) is one of the 
   components of the sum. So, checking the type of (f4 (n-1) m)  OCaml sees that it's actually a 
   function that maps one integer into a resulting value. Hence, it cannot apply the "+" operation 
   over the function (f4 (n-1) m). Expected type is int. *)