Square of the sum minus sum of the squares in J (or how to take the train?)

Question

Still in the learning process of J... The problem to solve is now to express the square of the sum minus the sum of the squares of natural integers.

The naive solution is

(*:+/>:i.100) - (+/*:>:i.100)

Now, I want to use a fork to be able to write the list >:i.100 only one time. My fork should like to:

  h
/   \
f   g
|   |
x   x

where f is the square of the sum, g is the sum of the squares, and h is minus. So, naively, I wrote:

((*:+/) - (+/*:)) >:i.100

but it gives me a domain error. Why? I also tried:

(+/ (*: - +/) :*) >: i.100

and this time, it gives me a long list of numbers... I guess it has something to do with the @ conjunction, but I still don't figure out what the At does... Continuing my quest, I finally got

((+/*+/) - +/@:*:) >:i.100

but I don't like the fact I manually compute the squares instead of using the *: operator, and I don't really understand why I need the @: conjunction. Could somebody gives me some light about this problem?

Answer 1

(+/*:) and (*:+/) don't do what you think they do.

Actually, your f is Q (S x) (square of sum of x) and your g is S (Q x) (sum of square of x). You can see that for any f,g it is f (g y) = (f @: g) y.

So, you can write

(Q (S x)) h (S (Q x))

as

((Q @: S) x) h ((S @: Q) X)

which is now equivalent to

(f x) h (g x)

or

(f h g) x

Thus,

((*: @: (+/)) - (+/ @: *:)) >: i.1000

Note also that *: @: (+/) is not the same as *: @: +/, since +/ is not one verb (like *:) but a composite verb from a verb (+) and an adverb (/).