1

There is some sense in which the derivative of a function $\frac{df}{dx}$ can be written as a "product" $Df$. And while solving, treat $D$ as a "number".

What is this process called, if it even has a name?

Sofviic
  • 33
  • You're examples are wildly different, but you give them as example of the same thing. Very unclear question. – Jens Renders Mar 30 '21 at 15:44
  • @JensRenders In what sense are they different, if you don't mind me asking? Is it because $exp$ is a function, while $\frac{d}{dx}$ is an operator? – Sofviic Mar 30 '21 at 15:46
  • regarding the application of a linear operator as product has nothing to do with defining a shorthand exp for the function $e^x$ – Jens Renders Mar 30 '21 at 15:50
  • @JensRenders I see, I'll remove the mention of $exp$ then. – Sofviic Mar 30 '21 at 15:55
  • I don't think this has a name though – Jens Renders Mar 30 '21 at 16:01
  • I guess the process you are looking for is that the linear maps (say) $C^{\infty}(\mathbb{R}) \to C^{\infty}(\mathbb{R})$ form a vector space. – Qi Zhu Mar 30 '21 at 16:01
  • @QiZhu except that a vectorspace forgets that you can apply the elements.This application is the product here. So this answers "treat as a number" but not "written as a product" – Jens Renders Mar 30 '21 at 16:04
  • @JensRenders What product are you referring to? If $D$ is in the vector space, then it can be applied to $f$, giving $Df$, as $D$ is a map $C^{\infty}(\mathbb{R}) \to C^{\infty}(\mathbb{R})$. If you further want composition of such linear maps, then just take the structure of a ring/algebra instead of a vector space. – Qi Zhu Mar 30 '21 at 16:08
  • @QiZhu Indeed, but OP seems to be asking about regarding $Df$ as a product. Like it is done in the finite dimensional case with matrix vector products (this has some properties like distributivity and associativity with the vector space/algebra operations). And my comment was about the fact that a vector space (or algebra for that matter) is agnostic to the fact that its elements can be applied, whether you regard that as product or not – Jens Renders Mar 30 '21 at 16:11
  • @JensRenders As I've already said, this is not any vector space but rather a vector space of maps. But if you wish, you can just let it be an abelian group and take the corresponding canonical group action. – Qi Zhu Mar 30 '21 at 16:15
  • @QiZhu I have not ignored anything you said, I just comment that "vector space" cannot answer this question as it does not covers the actual operation the question is about. The specific vector space you mention has elements that can be applied and I never contradicted that. The vector space structure is just agnostic to that. That group action is the structure that carries that application structure, and indeed group actions are often seen as product, so perhaps that is the correct answer to this question. (except that we can scale the operator, so it's more like a vector space action) – Jens Renders Mar 30 '21 at 16:25
  • There's a very general term "[by] abuse of notation", for when you treat objects of type $X$ as if they follow the rules for objects of type $Y$, without having established that they obey those rules. An example might be writing an infinite series of terms without having defined convergence. – JonathanZ Jun 11 '21 at 18:06
  • @JonathanZsupportsMonicaC Oh right! That actually describes it really well. "[by] analogy" could also be used in a similar vain. If you post your comment as an answer, I'll accept it. – Sofviic Jun 16 '21 at 12:20

1 Answers1

0

There's a very general term "[by] abuse of notation", for when you treat objects of type X as if they follow the rules for objects of type Y, without having established that they obey those rules. An example might be writing an infinite series of terms without having defined convergence.

And as noted by the OP, one could also describe this as "[by] analogy with Y".

Whichever description one uses, you've admitted that you've left the path of rigorous deduction, and what follow may not be true/valid. Usually one follows up with either

1 - Going back and rigorously proving that X's do behave as you presumed, or

2 - If all you wanted was to find a solution to some problem, you prove that the answer you found does in fact provide a solution, and the way that you found it is considered to just be a trick or heuristic or ansatz.

JonathanZ
  • 10,615