Haskell Fibonacci Explanation

Question

I am quite new to Haskell and I\'m trying to wrap my head around how the lazy expression of Fibonacci sequences work.

I know this has been asked before, but none of

Answer 1

This intermediate step is wrong because zipWith has already processed the first pair of items:

fibs = 0 : 1 : 1 : zipWith (+) fibs (tail fibs)

Recall what zipWith does in the general case:

zipWith f (x:xs) (y:ys) = (f x y) : zipWith f xs ys

If you apply the definition directly you get this expansion:

fibs = 0 : 1 : zipWith (+) fibs (tail fibs)                # fibs=[0,1,...]
     = 0 : 1 : zipWith (+) [0,1,...] (tail [0,1,...])      # tail fibs=[1,...]
     = 0 : 1 : zipWith (+) [0,1,...] [1,...]               # apply zipWith
     = 0 : 1 : (0+1 : zipWith (+) [1,0+1,...] [0+1,...])   
     = 0 : 1 : 1 : zipWith (+) [1,1,...] [1,...]           # apply zipWith
     = 0 : 1 : 1 : (1+1 : zipWith (+) [1,1+1,...] [1+1,...])
     = 0 : 1 : 1 : 2 : zipWith (+) [1,2,...] [2,...]       # apply zipWith
     = 0 : 1 : 1 : 2 : (1+2 : zipWith (+) [2,1+2,...] [1+2,...])
     = 0 : 1 : 1 : 2 : 3 : zipWith (+) [2,3...] [3,...]    # apply zipWith
     :

Answer 2

How to visualize what's going on.

  1 1 2 3  5  8 13 21 ...   <----fibs
  1 2 3 5  8 13 21 ...      <----The tail of fibs
+_________________________  <----zipWith (+) function
  2 3 5 8 13 21 34 ...

 Finally, add [1, 1] to the beginning
 1, 1, 2, 3, 5, 8, 13, 21, 34, ...