Understanding Enum.scan

Hi,
It makes sense As you don’t pass initial accumulator, then the first element of the list was taken and the first element on output will be the same as on input - {-1, 3}.
Then, on first iteration

[b: 1, e: 5, x: 3]

b and e are from the second element of the list and x is from accumulator which is the first element of the list.
Then you build new two-element tuple which is {3, 5}, it is used as second value in output list and as accumulator for the next element.
Then on the next iteration

[b: 10, e: 15, x: 5]

b and e are from the last element and x is from accumulator. Newly built tuple is {10, 15}, it is used as third element.
So, we have

[{-1, 3}, {3, 5}, {10, 15}]

Maybe, it would be easier for you if we reimplement Enum.scan using Enum.map_reduce:

def my_scan([], _callback), do: []
def my_scan([head | tail], callback) do
  {map_result, _reduce_result} = Enum.map_reduce(tail, head,  fn x, acc ->
    result = callback.(x, acc)
    {result, result}
  end)
  [head | map_result]
end