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Everything posted by w=f[z]

  1. Seems like to be invisible, the object in question must be 100% transparent. And then, I suppose it would have to be 100% transparent to all radiation. Oh yeah... and it mustn't emit any radiation. Seems impossible off hand.... Cheers
  2. Ask the audience questions. Make them feel like participants instead of spectators. Cheers
  3. Farsight, Perhaps people would respond better if you were more humble. Cheers
  4. Did Hawking's theory of everything include the fact that I could kick Xerxes' ass in fisticuffs...? Yes... perhaps it was not in good taste. [Edit:] I apologize. I didn't mean to be so insensitive.
  5. Did Hawking's theory of everything include the fact that I could kick his ass in fisticuffs...?
  6. I think we first have to ask "what is a ghost?" and the characteristics of such beings, e.g., they seem intelligent (as opposed to a plant), they pass through walls, etc. Once we have a starting point, ask what it would take to satisfy the basic assumptions, and these criteria must not break the laws of physics. I would bet that we cannot have a system that satisfies our assumptions about ghosts which is logically consistent with the laws of physics. If anyone wants to start a list of basic assumptions on what a ghost is and its characteristics, it might be fun to see how that fits with the laws of physics and go from there. Cheers, w=f[z]
  7. I'm not sure I want to touch this, but I've seen pictures... I've seen pictures of women with dogs. Clearly the dog was "consenting".... Cheers
  8. This question became quite controversial on another forum. I recall that a cyber-fistfight broke out.
  9. Something I'll add that I haven't read here yet... If you are going to teach yourself a subject, it would have to be in your spare time (i.e., after work and on weekends). That's one advantage a Ph.D. has - while they're getting the degree, that is their fulltime job. So I guess the learning process is a bit more efficient for the Ph.D. student. Cheers
  10. Just so we're all on the same page (including myself...) - So the status of this problem is as follows: The original problem (let's call "P1") still remains unsolved as far as the values of the letters (unless river-rat's are correct). If RR's values are correct, then the pattern seems tougher than the second version of this problem. The second (pdf) version (let's call "P2") has been solved for the letters (except j, q, x, & y) by cjohnso0 (and spyman). The pattern between the letters remains unsolved. P2 was solved by linear-algebra techniques (& excel) omitting any of the names (so that we get an augmented matrix-type problem), but P1 requires the omission of only certain names - which seems to require more knowledge in order to determine which ones to omit. Once that is done, then linear-algebra techniques give the values (such as the ones river-rat got). Did I get all that right? River-rat, did you get your values by reducing an augmented matrix omitting only certain names? I'm not sure I caught exactly how you got your values (i.e., the technique). Cheers, w=f[z]
  11. These were the instructions on my version (annoyingly enough... not on the pdf link). I'll type in what I have so forgive any typos. A few comments; This test is no longer on their site, so I thought posting this problem here was fair game now. Also notice their mention of "advanced mathematics" and "reference materials." And here's a minor nitpick, but they encourage "guessing." While guessing might increase your IQ "score" (if you guess correctly)... a correct answer by guessing I don't think is an accurate reflection of one's real IQ (whatever that is...). Okay, so there are the rules that came with the test. The use of computers was fair game. Cheers
  12. Hi doG, Take a look at the other questions on the test (a link to the pdf was provided). Normally I'd agree with you, but this test had no time limit (unlike most internet IQ tests) and the other questions are pretty tough too. My take, is that the folks that would do well on this test would be able to teach themselves the necessary skills needed to answer the questions. Just a thought. Cheers
  13. I might also add an admonition to not confuse mass with matter - an easy mistake to make. Cheers
  14. Hi RR, Thanks, but I think cjohnso0 found the correct values. They seem to work (at least for all the names I tried...). So I probably don't need your mathematica notebook. Thanks though! I think it's a matter of finding the pattern (if there is one...). Nothing jumps out at me yet. Cheers, w=f[z]
  15. Excel... damn, I was trying Maple and Mathematica when good ol' Excel would have done the job. So... if the letters go from 1 to 26, then j, q, x, and y can have possible values of 1, 19, 25, & 26. Which is which is yet to be determined. I am going to give Excel a shot when I get time. Thanks! Edit: It seems like there are only three possibilities; 1. the values were assigned randomly, or 2. there is a pattern in the numbers themselves, or 3. the numbers were mapped to each name/physicist. If 1, then there is no hope of finding y except for guessing (1 out of 4 chance). If 3, then one would need the "key" to decipher the code. My initial guess is that it is 2 - as that would be the easiest to implement (a formula so to speak). The fact that they redid the question (i.e., new values to the names compared to what I originally worked from) makes me even that more confident that it is possibility 2. Cheers
  16. If there is a definite answer, then we would have to account for the letters j, q, x, and y. And we would also have to determine if the values start at 1 and go to 26, or start at 2 and go to 27. My guess would be that they start at 1. Seems like in order to maximize our confidence that we have figured out the correct value for y, we should also figure out the values for j, q, & x. My thought is that there should be a pattern to how the letters are assigned values. Determining what that pattern is should be the next step. I'm so excited I'm about to pee my pants! Cheers
  17. !!!!! Your values indeed work! (At least for the named I tried.) There is still the matter of the missing 4 letters/values, j, q, x, & y. I got so excited that I had to post a message before thinking about how y might fit in. How exactly did you solve it? Good old linear algebra? If so, which names did you exclude/include? Great job! Cheers, w=f[z]
  18. I just double checked with my original printed version. What I have matches the OP. But I would still suggest working this with the new values. Yes, I saw that too. They have taken this problem off their website so I figured it was fair game now. Plus, I've had this problem in the back of my brain for so long now, that for my own "well being," I really want to see the solution. Cheers, w=f[z]
  19. Not bad CPL. I see you refrained from converting to SI. There's still room for someone to verify CPL's result.... Good fun! Cheers
  20. Son of a gun... I'll have to double check with my printed version (which I left home today ). I could have swore what I had matched my printed version.... Cheers Edit: Okay... still have my printed version at home, but I am pretty darn sure "Heisenberg" and "Schrodinger" had the same value (122) because I tried to use that fact before. Just going from memory here... but something is very fishy about this d*mn problem now. That really pisses me off! Hell... maybe with the "new" values, it works out nicer.... Edit #2: It seems these guys were using the same values I originally posted too... http://episteme.arstechnica.com/eve/forums/a/tpc/f/6330927813/m/6740938424/p/1 Edit #3: Looking at the date-stamp on the pdf link I gave for the puzzle, I see that it is 11/22/05 - which is about 3 years newer than what I was originally working from. In light of that, it might be a good idea to work form those numbers in case there indeed was a mistake that the puzzle makers corrected. Sheesh! Sorry guys for the values I supplied, but I was only posting what I had in front of me and what I originally used....
  21. Apparently the "fastest 1/4 mile" question is even better. It has yet to be solved.... Cheers
  22. Hi Jim, I see from your post history that you don't get too many responses. This is probably because you know more about this stuff than most (if not all) of us. It's been a few years since I had to think about relativity in the sense of actually working through tensor equations and such. And then I would have to wonder how much of it I understood in the first place. So I am beyond rusty. But... I have a couple friends on another forum that seem to enjoy getting their hands dirty with the stuff you ask about. I'll send the word for them to come over here and specifically have a look at your posts and see if it's something they would like to get into. That's the best I can do for now. Nice TeXing by the way. Cheers, w=f[z]
  23. If the top quark is indeed pointlike, then wouldn't the density go to infinity? Afterall, if we're talking about mass density, we'd use the formula [math]\rho=m/V[/math]. Cheers
  24. Hi Jordan, The mass I think you mean to be [math]GeV/c^2[/math]. (Unless you defined c to be 1....) Can you shed some more light on this problem for us? By density, do you mean mass density (there is also number density, etc.)? For starters, where does this information/problem come from? Cheers, w=f[z]
  25. It may be helpful in the future to know that the fundamental theorem of calculus has a few 3-dimensional analogs: Fundamental theorem for gradients: [math]\int^{\vec{b}}_{\vec{a}}(\vec{\nabla}F)\cdot d\vec{l}=F(\vec{b})-F(\vec{a})[/math] where the integral is a line integral. Fundamental theorem for divergences (divergence theorem, Green's theorem, or Gauss' theorem): [math]\int_{\upsilon}(\vec{\nabla} \cdot \vec{v}) d\tau=\oint_{\partial\upsilon}\vec{v}\cdot d\vec{a}[/math] where the LHS is over a volume and the RHS is over the surface. Fundamental theorem for curls (Stoke's theorem): [math]\int_{\sigma}(\vec{\nabla} \times \vec{v}) \cdot d\vec{a}=\oint_{\partial\sigma}\vec{v}\cdot d\vec{l}[/math] where the LHS is over the area and the RHS is over the perimeter. I use the [math]\partial[/math] symbol in the sense to mean the boundary. Compare those to the FTC: [math]\int^b_a\frac{df}{dx}dx=f(b)-f(a)[/math] Do you see the similarity? The key feature these have in common with the FTC is that the RHS is over the boundary of that of the LHS - just like the FTC. Cheers, w=f[z]
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