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Also, the abstract algebra definition is an actual generalization, because the (analysis) derivative acts as a linear operator in the vector space of (suitably
by codeflo 2y ago
Also, the abstract algebra definition is an actual generalization, because the (analysis) derivative acts as a linear operator in the vector space of (suitably smooth) funtions. It blew my mind when I first encountered this way of looking at derivatives. And from there, it makes complete sense to look at operators with similar properties in other vector spaces.
The definition presented here is a loose analogy to derivatives rather than an actual generalization, which doesn't fully justify using the name IMO.
- deleted 2y ago[deleted]
- JadeNB 2y ago> And from there, it makes complete sense to look at operators with similar properties in other vector spaces. It does, but this one's on a rig (= ring - negatives), not a vector space over any field.
- codeflo 2y agoI don't see how you would define linearity in a ring (I don't mean a module, just a ring). I.e. D(af) = aD(f) doesn't make sense if you don't have scalar multiplication.
- gizmo686 2y agoThe minimum would be to ask for D(a+b) = D(a) + D(b). EDIT: Actually, according to wikipedia, that is exactly what is done in differential algebra https://en.wikipedia.org/wiki/Differential_algebra https://en.wikipedia.org/wiki/Differential_algebra
- JadeNB 2y ago> I don't see how you would define linearity in a ring (I don't mean a module, just a ring). I.e. D(af) = aD(f) doesn't make sense if you don't have scalar multiplication. Every ring is a module over itself. But you wouldn't want the definition you propose; instead, you'd want `D(ab) = aD(b) + D(a)b`. If you really like some sort of linearity to be present, you could observe that this property forces every derivation to be linear as a transformation of `R_0`-modules, where `R_0` is the subring `ker(D)` of "constants".