Defining pure function @ vs /@

Question

Learning how to program I dont fully get the difference of /@ vs @

l = Table[n (n + 1) (n + 2) (n + 3), {n, 1, 10}]
{24, 120, 360, 840, 1680, 3024, 5040, 7920, 11880, 17160}

Sqrt[l + 1]
{5, 11, 19, 29, 41, 55, 71, 89, 109, 131}

Trying to obtain the same result

a = (#^(1/2)) &@ (# + 1) & /@ l
{5, 11, 19, 29, 41, 55, 71, 89, 109, 131}

b = (#^(1/2)) & /@ (# + 1) & /@ l
{25, 121, 361, 841, 1681, 3025, 5041, 7921, 11881, 17161}

c = (#^(1/2)) & /@ (# + 1) & @ l
{5, 11, 19, 29, 41, 55, 71, 89, 109, 131}

The way I understand /@ is to apply the function to every item of the list, so not sure why I get different results on a, b, c. Particularly why in b the function (#^(1/2)) & is not evaluated at all

Answer 1

There are three things at play here:

• The precedence of @,/@ and & (note the parentheses):

  HoldForm[#&@#&/@l]
  HoldForm[#&/@#&@l]
  HoldForm[#&/@#&/@l]
  (* ((#1&)[#1]&)/@l *)
  (* ((#1&)/@#1&)[l] *)
  (* ((#1&)/@#1&)/@l *)
  

• The Listable attribute of functions like Plus and Power:

  1+#&@{x,y}
  #^2&@{x,y}
  (* {1+x,1+y} *)
  (* {x^2,y^2} *)
  

• The behavior of Map applied to atomic expressions: (note: does not apply to all types: e.g. Association and SparseArray are handled differently - thanks @Alan for pointing this out)

  f /@ {x, y}
  f /@ x
  (* {f[x], f[y]} *)
  (* x *)
  

Putting everything together, you can easily explain what you're seeing (I'll leave it as an exercise to explain the individual cases)