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Everything posted by DQW
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Maglev is entirely understood in terms of the Maxwellian formulation of electromagnetism. Why should labview learn an entirely new formulation that is known to be wrong ? Naturally, it's his/her choice, but I'd advise against it. The only point I can see in learning Weber's Electrodynamics, is to gain some historical perspective on the evolution of that area of physics in the 19th century.
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Unless I'm completely misunderstanding phaedrus, I don't see where s/he claims to have derived a new definition of a determinant. More on determinants :http://mathworld.wolfram.com/Determinant.html Phaedrus, I don't see any reason why your result should mean anything special. In general, the k-th n-gonal number is linear in n. So a 3X3 determinant of n-gonal numbers will be a cubic in n. What exactly do you find interesting ?
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Thanks matt. I misread the last line. This makes it non-trivial.
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If you are okay woth {NOT(3 < 5)} = {3 >= 5}, why should there be a problem in say, replacing 5 with w ?
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(a) You are given two things. The FIRST thing that would come to my mind is to substitute the first into the second ... (b) Having said (a), I shouldn't say anything more. Sarah, these are nothing more than direct substitution problems.
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Yes, that's correct.
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No, you don't solve for s. s can take any value...which leads you to the next part of the question. Plug in X = 54.74 in the equation for the surface area, A, and thus get an expression for the minimal surface are in terms of s and h. Then you plot A(s,h) for different values of s and h. PS : You can cancel off X/X in all cases EXCEPT when X=0, since "0/0" is not defined. Fortunately, csc(x) is never equal to zero.
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Differential Forms: Who's good at this?
DQW replied to MetaFrizzics's topic in Analysis and Calculus
Those drama queens at PF, eh ? -
Two things : 1. You must state why you are justified in canceling csc(x) - because you can never have csc(x) = 0 for any x. 2. csc/cot = tan/sin = (sin/cos)*(1/sin) = 1/cos = sec(x)
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Didn't know they made woollen blankets with breezes ! Reyveta, did you notice that your electron is also shaped like a pufferfish ? http://employees.csbsju.edu/hjakubowski/classes/ch331/signaltrans/pufferfish.jpg
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Interesting new finds. Sorry if this has been talked about already : http://news.bbc.co.uk/2/hi/science/nature/4708459.stm
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1. Since you haven't defined exactly what you mean by "measured the polarization of the photon at point B", it would be hard to answer your question satisfactorily, 2. Since you've posted this under QM, I'm guessing it's a prelude to an SG question, but that is merely speculation on my part, Please describe how you measure the polarization.
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Oh, I'd imagined that you can only see one "point" on the wave in each frame (ie : that in each frame at time t=t' you can see all values of y(t') ).
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One little detail. I'm pretty sure the last term in the original equation is [imath]3 \sqrt{3} s^2 /2sinX[/imath] , not [imath]3 \sqrt{3 s^2} /2sinX [/imath]. The second option is not even dimensionally correct.
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That's correct ! Notice that all of these are giving you the same result, since [math] \frac {cos(x)}{sin^2(x)} = \frac {cos(x)}{sin(x)} \frac {1}{sin(x)} = cot(x) cosec(x) [/math] Go ahead; you're doing fine. Equate this to 0 and find X.
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Okay, I must be making some really silly mistake here as it looks to me like a one-line proof. Let me lay my head on the guillotine and actually write this down : Assume [imath]a_p = a_q [/imath] for some q>p, then [imath]a_p \equiv a_q~ (mod~ q) [/imath]; but this is not possible since all of the first q terms leave different remainders with q. Hence the assumption was wrong. QED. Feel free to let the blade drop...it won't hurt my feelings !
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Yes, it's 3/2 What's to be confuzzled about? When in doubt, recheck definitions.
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That'll do just fine. And it's a series. Sequence : a1, a2, a3, ... Series : a1 + a2 + a3 + ...
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If the "sum" contained just one term you would have a strict equality [imath]|a_1*a_1| = |a_1|*|a_1| [/imath], but you are told that the series have infinite terms.
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Well ? How do you intend to go about this homework problem ? Any ideas ?
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No Sarah, that proof is incorrect. 1. You are specifically trying to prove a result for a 2-dimensional vector space, instead of for any general vector space 2. You have not used anywhere that a and b actually belong in U/\W. As always, start from the definitions : [math]x~\epsilon~U~ int~ W \implies x ~ \epsilon~ U~and ~x~\epsilon~ W [/math] and conversely. PS : What are the [imath]\LaTeX[/imath] codes for union and intersection, anyone ?
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Sarah, I do not follow what you've done in part (a). What are v1, v2 ? Are they basis vectors of V ? And what are u and w, and why have you defined them to be null-vectors ?
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Not really. Any general plane does not constitute a vector space. But the specific planes chosen by Sarah will work. Please do not mislead students Meta. 1. Circles do not intersect at lines, 2. A circle can not make a vector space - it is closed under neither vector addition not scalar multiplication.
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Differential Forms: Who's good at this?
DQW replied to MetaFrizzics's topic in Analysis and Calculus
A recommendation : http://pzacad.pitzer.edu/~dbachman/forms.pdf