[gus_massa]

I agree, there are usefull generalization of the derivative. But an important detail is that the operator that is discussed in this post is not linear, i.e. D(x+y) != D(x) + D(y), so it doesn't have all the expected ("intuitive") properties that the usual derivative has.

I don't know the origin of this notation, but I can make up a possible explanation. If you think that every prime is a function of an abstract variable x, so 2 = two(x) = 2 + x, 3 = three(x) = 3 + x, 5 = five(x) = 5 + x, ... evaluated in x=0.

Then, for example, 60 = sixty(x) = two(x) * two(x) * three(x) * five(x) = two(x)^2 * three(x) * five(x)

A number is a function of the primes of the factorizations. You must change the primes into the functions, but you must leave alone the exponents.

Then D is the standard derivative, plus evaluation.

D(60) = sixty'(x) = 2 * two(x) * three(x) * five(x) + two(x)^2 * five(x) + two(x)^2 * three(x) = 2 * 2 * 3 * 5 + 2^2 * 5 + 2^2 * 3 = 60 + 20 + 12 = 92

(I'm mixing the functions and the values when they are evaluated. It's usually not a good idea. If you do that in a Calculus exam the TA will be rightfully angry. But the notation in only text is horrible, so please forgive the technical details.)

With this idea if you have numbers A, B, C such that A = B * C, then A(x) = B (x) * C(x). But if A = B + C then A(x) != B(x) + C(x). This "explains" why the operator D follows the multiplication rule, but not the sum rule.

For example, ten(x) = two(x) * five(x) and six(x) = two(x) * three (x) then sixty(x) = two(x) * two(x) * three(x) * five(x)

But fifteen(x) = three(x) * five(x) and ten(x) = two(x) * five(x), but twenty-five(x) = five(x)^2 and clearly three(x) * five(x) + two(x) * five(x) != five(x)^2.

Edit: Don't take this "explanation" very literally, for example this idea doesn't extend to D^2 directly

D^2 (p^2) = D(p+p) = D(2p) = D(2)p+2D(p) = p+2

D^2(p^2) = D(p+p) != D(p)+D(p) = 2D(p) = 2

D^2(7^2) != (seven(x)^2)'' = (seven(x)+seven(x))' = seven'(x)+seven'(x) = 2seven'(x) = 2

The implicit transformations of numbers into functions and the evaluations in 0 make cause many problems.

[JadeNB]

> But an important detail is that the operator that is discussed in this post is not linear, i.e. D(x+y) != D(x) + D(y), so it doesn't have all the expected ("intuitive") properties that the usual derivative has.

I think some care is appropriate here. The property you quote is additivity, not linearity; for linearity, one would like a ground field (or a ground ring if one is discussing linear maps between modules, I suppose). Since one is not considering any vector space / module structure on the natural numbers, this may hint why linearity (or even its weaker sibling additivity) is not thought to be necessary here.

(EDIT: With that said, I like very much your description of transforming numbers to functions.)