joigus

Taeto has answered very proficiently your longer-worded question, which has a different scope than the title really. Namely: "How does one prove that a Fourier transform is well defined?" I don't remember the details, nor can I find them on Wikipedia, but a Fourier transform is well-defined when your function is piecewise-continuous. That means it better not have an uncountable number of discontinuities. But the definition is very solid, in the sense that you can even define it for some non-integrable functions or even temperate distributions (strange objects, like the Dirac delta function, that your garden-variety functions can be integrated against.) I hope that adds significantly to your question.

Thank you. I'm sure any comments you may have to add will be most interesting. Here I meant domain and spectrum.

It's an illusion, clearly. It's an illusion, arguably.

It's good to know. It's just that on places like this you're never sure if you're rubbing a person the wrong way.

I don't know if this is necessary, or of any use. I will just have to wait and see. My apologies go specially to Mordred and Markus Hanke, because they know far more than I do about GR, and they've let me play smarty-pants on their turf for a while. My fine instinct tells me I may've p*ssed them off. In case there were any hard feelings (although as I'm writing these lines I've found out that's not the case with Mordred at least) I was just checking both my own knowledge and theirs. If you check your knowledge or solid reason against wise people, you're on your way to understanding more and better. My grievances go to all those people (some OP's and their sock-puppet duplicates or echo chambers who shall remain unnamed) who make me waste time in checking, double-checking and cross-checking and cross-referencing about their silly assumptions or implications, sometimes getting me in more trouble than need be. I don't completely consider that time totally invaluable though as soon as the likes of Mordred and Hanke make an appearance.

My answer would be yes, but as instantiations or "runnings" of a program. My father used to say "We're not starving today!" after supper. It was his way of phrasing something that has been repeated throughout the millennia by humans, and many hominines before us, up to 2.5 million years or more. Every now and then I repeat that sentence after supper. In what sense and to what extent do I know it's not my father again who's saying that? It's his applet running on my hardware, and our atoms are instantiations of a quantum field. So... Dead right!

Should've said reappa or reeppa. Sorry. rip/reap mistake

I'm not taking you seriously 100 %. But I'm interested in your definitions. 1) Your notation is confusing at best. I would first drop the h bar. I don't know what quantum mechanics is doing there. Also, no need for the partial derivative symbol. \[P=-i\frac{d}{dx}\] Your left and right shift operators I have re-named L and R: \[f_{R}\left(x\right)=Rf=f\left(x\right)-f\left(x+1\right)\] \[f_{L}\left(x\right)=Lf=f\left(x\right)-f\left(x-1\right)\] So that, \[G=L\left(XP+PX\right)R\] I'm not sure your operator is compact or admits a compact extension. 2) It seems like you're extending Riemann's zeta function to include a real variable x taking values on the real line. Taking x=1 doesn't take you back to Riemann's zeta function. You should be more explicit about what you're doing there. 3) Your definition of scalar product involves only integration/sum from 1 to positive infinity. That would require some heavy-duty extensions and checks of your definitions (closeness, domains, etc.) 4) X and P are unbounded operators that do not belong to the trace class operators. IOW: Look out for mistakes, especially if you're using anything like traces. Example of simple arguments that miserably fail with them: \[\textrm{tr}\left(XP-PX\right)=0\] \[\textrm{tr}(-iI)=-i\times\infty\] 5) Soundness of passing an operator from left to right in a scalar product depends on subtle questions about domains, not only on real character of its "formal" spectrum. You must check that you're not letting out the accumulation points of your domain in case it's not compact (topological argument.) 6) Related to the previous: Your "left operator" takes you out of the domain of your scalar product. Doesn't that have any bearing on your "proof" of something? I'm just curious. That's all.

That's my lame attempt at sounding British. What part of 欧羅巴 are you from? I gave you +1. Be nice when it's time to come for me.

Linearity is one thing. That's inherited from linearity of the derivative operator. Closure is another thing. That's what Studiot and Taeto are talking about, I think. Because ex, exsin x, excos(x), sin(x), cos(x) are meromorphic (analytic in C ==> analytic in all R) I see no problem with the domains. Closure can be assured by inspection. The most involved cases are exsin(x), excos(x). But, \[\frac{d}{dx}e^{x}\sin x=e^{x}\sin x+e^{x}\cos x\in\textrm{span}\left(A\right)\] \[\frac{d}{dx}e^{x}\cos x=e^{x}\cos x-e^{x}\sin x\in\textrm{span}\left(A\right)\] That would be my answer.

I like the way you say "perisheth," Rippa.

Thanks for the tip. I'm here both to learn and test my knowledge and understanding. Language itself is one motivation too.

OK, it's been ages since you posted this, but I couldn't resist. I'm just refreshing my linear algebra. Then I can pick a basis in which, \[A=\left(\begin{array}{ccccc} a_{1} & 0 & \cdots & 0 & 0\\ 0 & a_{2} & \cdots & 0 & 0\\ \vdots & \vdots & \ddots & \vdots & \vdots\\ 0 & 0 & \cdots & a_{n-1} & 0\\ 0 & 0 & \cdots & 0 & 0 \end{array}\right)\] And, without loss of generality, \[v_{0}=\left(\begin{array}{c} 0\\ 0\\ \vdots\\ 0\\ 1 \end{array}\right)\] Generic n-vector: \[x=\left(\begin{array}{c} x_{1}\\ x_{2}\\ \vdots\\ x_{n-1}\\ x_{n} \end{array}\right)\] Eqs. render as, \[a_{1}x_{1}=0\] \[a_{2}x_{2}=0\] \[\vdots\] \[a_{n-1}x_{n-1}=0\] \[0x_{n}=1\] So no solution for xn.

Yes, exactly. I didn't want to get involved, but I thought it could be useful. False friends are a minefield.

We are all laypeople in some sense or other, aren't we? You're most welcome.

You don't have to apologise. I understand perfectly what you mean, and I was just trying to apply an old Chinese technique which is called "koan" for those kind of difficult questions about "I." It's a well-meaning technique. It's about making you drop our common human need to stick to the "I." In more scientific terms, an electron in my brain is fundamentally indistinguishable from an electron in yours, or another one being kicked off from an atom in the atmosphere. Quantum field theory tells us that elementary particles are just instantiations of one thing called the quantum field. Information is the relevant quantity for describing an "I," or any other physical object. Very recently a very good friend of mine has died. He was younger than me. I lost my parents when I was very young too. To me, all those people are still living in the only sense that I can find physically meaningful: They uploaded software snippets and applets to my brain, so they are still in the world in this particular sense of information processes. Some day I will die too. Hopefully, I will be able to upload my applets --those that prove to be useful, or good in any sense--, to somebody else's brain. That's the only way I can conceive of in which we can perpetuate ourselves. I don't mean to be facetious; only to bring some consolation to you by trying to make you feel more relaxed about the eventual loss of the "I," but I can't think of a better way to finish except with another koan: What is it that makes you John? I hope that helps.

Your English is next to perfect. The only thing I've been able to spot is "I have a doubt." I would have phrased it as "I have a question." More idiomatic. I'm bilingual Spanish-English and have taught both, so if you ever have any nuance, false-friend, etc.-related question, be my guest. The cogujada is a beautiful bird, by the way. Good nickname. Except typos, of course.

Beautiful, and tantalizing. Interesting question. My guess would be no, and that rotation of the accretion disks is mostly driven by local clustering in the accretion area of the newborn star.

Interesting... Tidal effects come to mind at your suggestion, because those are second-order effects, which requires an order-2 tensor.

In other words, ask yourself, is it true that, \[\left[R_{\pi/3}\right]_{\mathcal{B}}\left(\lambda\boldsymbol{u}+\mu\boldsymbol{v}\right)=\lambda\left[R_{\pi/3}\right]_{\mathcal{B}}\boldsymbol{u}+\mu\left[R_{\pi/3}\right]_{\mathcal{B}}\boldsymbol{v}\] for any u, v, lambda and mu?

Here's a starter: Horizons problem (Spatial) flatness problem Cosmological constant problem (vacuum energy) Exact solutions of GR alone could not account for the first two. Part of the reasons why cosmologists moved to inflationary models. More in general (don't forget we're living in the fine-tuning era of physics and cosmology) and how to get over it: Fine tuning Also, and on a different order of questions, but related, what are your predictions for these?: \[\varOmega_{\textrm{matter}}\] \[\varOmega_{\varLambda}\] \[\varOmega_{\textrm{dark matter}}\] \[\varOmega_{\textrm{radiation}}\] all of which can be assessed from observational input.

I agree in a short-winded kind of way.

It depends on your "pervective." 👍

Boy, was that a good explanation! +1 And this was intended as a joke. Coordinates mean nothing, it's the metric tensor contracted with the coordinates, as Markus so brilliantly has explained. Just to clarify...

I'd say a couple of things present themselves as very fundamental and intrinsically quantum (non-classical.) 1) Finiteness of the quantum of action (quantities such as energy x time, momentum x displacement, angular momentum) 2) The need to express probabilities in terms of a more primitive quantity called the probability amplitude, which is complex. But it's impossible to explain quantum mechanics in a few sentences and without mathematics.