3 ms·
Seems to be helpful for physicists, not so much for mathematicians. A physicist will look at an integral and say "oh, yeah, charge is field times area, that ma
by leif 17y ago
Seems to be helpful for physicists, not so much for mathematicians. A physicist will look at an integral and say "oh, yeah, charge is field times area, that makes sense!", while a mathematician isn't going to say the same about measures. I think that's the "revelation" he was going for.
BTW, I always thought of convolution as multiplication. Anyone else feel cheated by this guy for that reason?
- kalid 17y agoThanks -- yep, I'm definitely more in the physics camp vs. the pure math. On convolution, it's an integral and can be thought of as a type of multiplication too, right? :)
- pixcavator 17y agoI feel that integration is more about addition than multiplication, and differentiation is more about subtraction than division. Is it just me?
- pfedor 17y agoNot just you. I seems much more natural to me to think of an integral as "continuous sum." You can often describe the exact same thing using a discrete sum or an integral, e.g., in a classical mechanics class you first consider a system of N connected springs, and a number of properties (like, say lagrangian, energy etc.) of the system will be expressed as sums with N terms. Then you take the continuous limit, the connected springs become one string and voila, all sums become integrals. In quantum mechanics for some systems the set of all energy levels is discrete (e.g., the harmonic oscillator), for some it's continuous (e.g., a free particle), and often it is part-discrete part-continuous (a hydrogen atom.) Many formulas will require you to take a sum for the discrete part and an integral for the continuous part. Some quantum mechanics textbooks introduce a special symbol, a capital sigma overlaying an integral symbol, which means "integrate or add, whichever appropriate."
- fgimenez 17y agoIsn't convolution defined as an integral?