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taeto

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taeto last won the day on March 2 2020

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About taeto

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    Hong Kong
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    Mathematics

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  1. I documented that he authored a book that describes his experiences with being transferred from his home near Windsor, ON, to become a subject of Nova Scotia's mental health system. So maybe you are right to say that I "infer" something about his mental health. To say that I "offer diagnosis" is a little far fetched. A diagnosis can assume to have been established already before the act of confining a person to a psychiatric institution. Would you be happy if I suggested that you seem to offer the opposite diagnosis, probably not. Yes, the moderation is excellent in this case, as always.
  2. I hope in a similar way that some damning comments about the OP have so far not been considered irrelevant. He clearly has a record of mental illness, and he may honestly firmly believe that he has a scientific approach to the world somehow. He obviously has no leg to stand on in a science forum, and moderation has to intervene. Still a little bit of empathy instead of only cusswords is also in order.
  3. As a possibly not entirely irrelevant side note, I got to thinking back at One Flew Over the Cuckoo's Nest, which for me installs the question in my mind, how would I deal with being put in a mental institution and how might my personality and powers of straight thinking have changed, if and when I got to leave it again to return to 'normal' society. Now the way that Jack Nicholson's character got entry into the system isn't likely the way that I would manage to do it. But you can think of scientists who in earlier times got close, such as Alan Turing, and others certainly did, like Cantor and
  4. You are right, I apologize for being so bitchy. I care a little bit, because I wanted to find a good argument for the integral, and I could not readily do that. Your 'pedestrian way' ought to provide the correct answer. But it does not work as a formal proof, unless you can show that the integrals converge as Lebesgue integrals. Something like the Henstock-Kurzweil integral may work even if Lebesgue does not.
  5. Please do not patronize. The theorem that I mentioned specifically concerns the Lebesgue integral. Before that I already mentioned that we cannot be talking about the Riemann integral, because that is not even defined for this integrand with the given integration domain. Now you should justify your "this reduces to". Typically by appealing to the monotone convergence theorem or to the dominated convergence theorem for Lebesgue integrals. Or to something else?
  6. Thanks joigus for following up on this! It is a mistake when you say "this reduces to". Because it will not be difficult to come up with examples like \[ \int_a^b f(x)+g(x) dx\] where \(f\) and \(g\) are both not defined in the entire interval \([a,b]\) and where \[\int_a^b f(x)+g(x) dx \neq \int_a^b f(x)dx + \int_a^b g(x)dx\] holds, even when all those three integrals converge. The Riemann integral is not defined unless the integrand is defined throughout the integration interval, and this is not the case here. In such a case when it is not defined, you can resort to the theorem tha
  7. Zodiac has gone full circle.
  8. Then the system that you are talking about, in which an electron spontaneously disintegrates, might be neither created nor established. And that is the only reason why we never see it happen, an electron going off like a firecracker?
  9. I think the point the physicists are making is that the electron on its own is not unstable. it does not decay.
  10. Will that be in this thread or in a new thread? In a thread which explicitly discusses gravity, it should be quite interesting and novel to combine it with a discussion of the strong force. Let me guess: at a certain distance between nucleons, they are attracted to each other because of the existence of vacuum in between them. But as they approach each other too closely, they get repulsed, because they physically bump into each other in a similar fashion as billiard balls?
  11. Maybe nowadays. In my time it was just a trick that we learned in a first course on topology. Actually the textbook for the course was probably on mathematical analysis. I would not have expected that this topic would come to any notoriety in the mean time. If a student would come to me and suggest to write a PhD on compactification I would try to dissuade, since it does not seem a serious research topic. Maybe I am just too out of touch.
  12. You are doing too much work. To show that \(x^2-2\) and \(x-2\) are linearly independent polynomials, you have to show that the equation \(c_1 (x^2-2) + c_2 (x-2) = 0,\) as a polynomial equation, has \(c_1=c_2=0\) as the only solution. This is very easy, because \(c_1 (x^2-2) + c_2 (x-2) = c_1x^2 + c_2 x - 2(c_1 + c_2) = 0x^2+ 0x+0\) immediately tells you that \(c_1=0\) and \(c_2=0,\) just from looking at the coefficients of the powers of \(x\). All that is left is to argue why \(W^\perp\) does not have additional polynomials in its basis other than these two.
  13. The elements of the basis are polynomials. You correctly found that the polynomials in \(W^\perp\) have the form \(ax^2+bx+c,\) where \(2a+2b+c=0.\) The basis of \(W^\perp\) has to be a set of two polynomials which satisfy this. You could choose \(\{x^2-2,x-2\}.\) You said this already. Do you know how to prove that the set of these two polynomials make a basis for \(W^\perp\)?
  14. It looks like you are doing more work than you need to. And some mistakes. Towards the end of your long calculation of \(\langle p, x+1\rangle\) you assume \(p(0)=0?\) But if \(p(x)=ax^2+bx+c,\) then you have \(p(0)=c.\) I guess you just forgot to type it in. So your conclusion that \(p(x)=ax^2+bx+c \in W^\perp\) if and only if \(2a+2b+c=0\) is correct. And you correctly found two linearly independent solutions to this equation. Your solution is correct.
  15. You can answer question (a) easily, I assume? And for any \(q(x) \in P_2\) you know the condition for determining whether \(q(x)\) is in \(W^{\perp}\)?
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