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This definition does actually have a surprising amount to do with differentiation. The definition works for any unique factorization domain and in particular fo

This definition does actually have a surprising amount to do with differentiation. The definition works for any unique factorization domain and in particular for polynomials. It turns out that the definition here exactly matches the usual derivative for polynomials.

xpe 2y ago
Are you saying that for arithmetic derivatives, the definition (part "a" above) "1 for any prime; everything else calculated via the product rule" has a surprising amount to do with differentiation? If so, can you connect the dots? Or did you mean the properties (part "b" above)?
- eigenket 2y ago
  Yes, > 1 for any prime; everything else calculated via the product rule does indeed have a surprising amount to do with differentiation! If you take the usual polynomial functions in one variable (lets say x is the variable and all our things are complex numbers) then these can be factored: e.g. x^2 + 3x + 2 = (x+1)(x+2). They form a (so called) unique factorization domain, which essentially means that factorization into "primes" works exactly the same as it does for integers. In the example above (x+1) and (x+2) are examples of prime factors which can't be factored any further. If you take the definition "1 for any prime; everything else calculated via the product rule" and apply it to this system where our "numbers" are polynomials and our "primes" are the polynomials we can't factor any further you get a definition of an "arithmetic derivative" for polynomials. The fun fact then is that this arithmetic derivative we just defined is exactly the same as the usual definition of the derivative from calculus: D[(x+1)(x+2)] = (x+1)D[(x+2)] + (x+2)D[(x+1)] = (x+1) + (x+2) = 2x+3 whereas d/dx (x^2 + 3x + 2) = 2x + 3
- gjm11 2y ago
  More the first than the second. There are things other than the integers for which it makes sense to talk about "the primes". One example is: polynomials (with coefficients in, let's say, the complex numbers). In this case it turns out that the "primes" are exactly the linear polynomials (ax+b) where a is nonzero. There's a bit of ambiguity there, just as there is in the integers; 7 and -7 are "the same prime number", and x+3 and 5x+15 are "the same prime polynomial"; if we're going to say D(p)=1 then we need to pick which "version" of p has this property, and the obvious choice is the one of the form (x+a). So, now, if we apply the same definition as for integers to polynomials with these conventions, it says: (1) D(x+a) = 1 and (2) D(fg) = fD(g) + D(f)g when f,g are polynomials. And that turns out to give the exact same result as the "ordinary" derivative for polynomials. Whether "exactly identical to" implies "a surprising amount to do with" depends on how easily surprised you are, I guess. ... I glossed over the sense in which the "primes" are precisely the linear polynomials, so here are a few words about that for anyone who's curious. If we look at polynomials with complex-number coefficients, a beautiful theorem says that they can all be written as A (x-r1) (x-r2) ... (x-rk), and then one polynomial divides another if and only if its set of rj is a subset of the other's (handling repeated roots in the "obvious" way). It's pretty easy to get from this that the linear polynomials are (1) the irreducible ones, i.e., the ones that can't be factored into lower-degree polynomials, and (2) the prime ones, i.e., the ones with the property that if p divides ab then p divides either a or b. (These properties are equivalent for the integers, as well as for polynomials with complex coefficients, but there are other settings in which they come out different, and both of them are useful, so they have different names.) (What happens if we use real rather than complex coefficients? The Wikipedia "Arithmetic derivative" page claims that we still get the usual derivative, but that looks wrong to me, because if we work over the real numbers then x^2+1 is both prime and irreducible, but its derivative isn't 1. Maybe I'm missing something.)
  - eigenket 2y ago
    As far as your point in parentheses goes, I think wikipedia is either wrong or confusingly written (allowing complex factorisations of real polynomials makes what they're written consistent, but is a bit silly). See theorem (20) on page 18 of this pdf for a theorem along these lines https://cs.uwaterloo.ca/journals/JIS/VOL6/Ufnarovski/ufnarovski.pdf https://cs.uwaterloo.ca/journals/JIS/VOL6/Ufnarovski/ufnarov...

xpe 2y ago

Are you saying that for arithmetic derivatives, the definition (part "a" above) "1 for any prime; everything else calculated via the product rule" has a surprising amount to do with differentiation? If so, can you connect the dots? Or did you mean the properties (part "b" above)?

eigenket 2y ago
Yes, > 1 for any prime; everything else calculated via the product rule does indeed have a surprising amount to do with differentiation! If you take the usual polynomial functions in one variable (lets say x is the variable and all our things are complex numbers) then these can be factored: e.g. x^2 + 3x + 2 = (x+1)(x+2). They form a (so called) unique factorization domain, which essentially means that factorization into "primes" works exactly the same as it does for integers. In the example above (x+1) and (x+2) are examples of prime factors which can't be factored any further. If you take the definition "1 for any prime; everything else calculated via the product rule" and apply it to this system where our "numbers" are polynomials and our "primes" are the polynomials we can't factor any further you get a definition of an "arithmetic derivative" for polynomials. The fun fact then is that this arithmetic derivative we just defined is exactly the same as the usual definition of the derivative from calculus: D[(x+1)(x+2)] = (x+1)D[(x+2)] + (x+2)D[(x+1)] = (x+1) + (x+2) = 2x+3 whereas d/dx (x^2 + 3x + 2) = 2x + 3
gjm11 2y ago
More the first than the second. There are things other than the integers for which it makes sense to talk about "the primes". One example is: polynomials (with coefficients in, let's say, the complex numbers). In this case it turns out that the "primes" are exactly the linear polynomials (ax+b) where a is nonzero. There's a bit of ambiguity there, just as there is in the integers; 7 and -7 are "the same prime number", and x+3 and 5x+15 are "the same prime polynomial"; if we're going to say D(p)=1 then we need to pick which "version" of p has this property, and the obvious choice is the one of the form (x+a). So, now, if we apply the same definition as for integers to polynomials with these conventions, it says: (1) D(x+a) = 1 and (2) D(fg) = fD(g) + D(f)g when f,g are polynomials. And that turns out to give the exact same result as the "ordinary" derivative for polynomials. Whether "exactly identical to" implies "a surprising amount to do with" depends on how easily surprised you are, I guess. ... I glossed over the sense in which the "primes" are precisely the linear polynomials, so here are a few words about that for anyone who's curious. If we look at polynomials with complex-number coefficients, a beautiful theorem says that they can all be written as A (x-r1) (x-r2) ... (x-rk), and then one polynomial divides another if and only if its set of rj is a subset of the other's (handling repeated roots in the "obvious" way). It's pretty easy to get from this that the linear polynomials are (1) the irreducible ones, i.e., the ones that can't be factored into lower-degree polynomials, and (2) the prime ones, i.e., the ones with the property that if p divides ab then p divides either a or b. (These properties are equivalent for the integers, as well as for polynomials with complex coefficients, but there are other settings in which they come out different, and both of them are useful, so they have different names.) (What happens if we use real rather than complex coefficients? The Wikipedia "Arithmetic derivative" page claims that we still get the usual derivative, but that looks wrong to me, because if we work over the real numbers then x^2+1 is both prime and irreducible, but its derivative isn't 1. Maybe I'm missing something.)
- eigenket 2y ago
  As far as your point in parentheses goes, I think wikipedia is either wrong or confusingly written (allowing complex factorisations of real polynomials makes what they're written consistent, but is a bit silly). See theorem (20) on page 18 of this pdf for a theorem along these lines https://cs.uwaterloo.ca/journals/JIS/VOL6/Ufnarovski/ufnarovski.pdf https://cs.uwaterloo.ca/journals/JIS/VOL6/Ufnarovski/ufnarov...