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 Re: Juxtaposition in l3fp David Kastrup <[log in to unmask]> Mon, 9 Jun 2014 15:43:46 +0200 text/plain (47 lines) Lars Hellström <[log in to unmask]> writes: > Oh yes. One of the really big mistakes people make when trying to > implement mathematics is believing that the mathematical formula > language is consistent, just because it is precise. Juxtaposition can > denote pretty much *anything* (depending on context), and because > juxtaposition is multiplication, it follows that pretty much anything > can be regarded as a kind of multiplication. ;-) More like "anything can be regarded as a kind of linear operator with respect to the following expression". d/dx is not really something you "multiply" with. Everybody is comfortable writing and reading x y, but when it is x x, it looks suspicious. When you take something like Fourier transforms, linear time-invariant operators _are_ a multiplication after transform. If you take the t->f kind of transform electrical engineers prefer (because you don't need to fiddle with asymmetric or weird scaling factors), you can basically say \nabla -> 2\pi \vec f Which makes little sense on the surface, but a rot written as \nabla \times transforms to 2\pi \vec f \times, a div written as \nabla \cdot transforms to 2\pi \vec f \cdot, a gradient written as \nabla \phi transforms to 2\pi \vec f \Phi, so for a number of purposes, operator juxtaposition works in a rather multiplication-like manner. Of course, like with other mathematical jigglery-pokery in engineering professions, you'll not find anything in the books used to teach the engineers, and the books used to teach mathematicians use completely other terminology and conventions not convenient for the kind of vector analysis an engineer might want to do. And the theoretical physicists do something else entirely in order not to get caught in the crosswinds. You'll be returning to your regular scheduled topic of LaTeX after the break. -- David Kastrup