so I am trying to write my own function to compose a function n times in Haskell.
so for example, the input
compose (+1) 3
would return f(x) = x+3;
Now my attempt is as follows, but is actually quite naive and currently doesn't work.
compose f 0 = (*1)
compose f n = (compose f n-1).a
In the second case, you are trying to refer to a which I think you meant as f, since then
compose f n = (compose f (n-1)) . f
(note also that compose f n - 1 is parsed as (compose f n) - 1 in your code)
which means you have
compose f 3 = (compose (+1) 2) . f
= ((compose (+1) 1) . f) . f
= (((compose (+1) 0) . f) . f) . f
= ((((*1) . f) . f) . f
Oh and by the way, you can write
compose f 0 = id
which expresses the idea that compose f 0 should do "nothing".
((compose f n) - 1) i.e. it's trying to subtract 1 from compose f n which is not a number.