  # Woxor

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## Everything posted by Woxor

1. Correct me if I'm wrong, but the end of the spotlight would also not actually travel faster than c, for the same reason that a jet of water (unaffected by gravity) directed similarly would not result in an intersection with the clouds traveling faster than the jet itself.
2. I was totally thinking "future, past, and present." Fits better if you ask me.
3. Yes, that's right. The only real conceptual difference between that and a proper subset is that the trivial subgroup is still a proper subgroup, whereas the null set isn't a proper subset.
4. Very true. For mathematics, I would say, "Don't run before you can walk forwards, backwards, sideways, up and down stairs, in the dark, on ice, and in your sleep. Power walking and moonwalking are also good to know." Math is insanely cumulative.
5. If you assume that all directions have equal probability, then you can't use the circle analogy without allowing for the differences between a sphere and a circle. On a circle, you can treat the arc length of a section over the circumference of the circle as the probability of the random direction falling on that arc, but the points on a sphere (in 3-D polar coordinates) at which $\phi<\pi/4$ are more "scrunched up" than the points at which $\pi/4<\phi<\pi/2$, so you have to find the actual surface area of the sphere, cut it into whatever pieces are important, and the ratio $\frac{SA}{4*\pi*r^2}$ will give the correct probability. Fortunately, the volume can also be used to give an equivalent result. So the ratio of the probability of hitting at less than 45 degrees to the probability of hitting at greater than 45 degrees is given by: $\frac{\int_{0}^{1}\int_{0}^{2\pi}\int_{\pi/4}^{\pi/2}\rho^{2}sin(\phi)d\phi d\theta d\rho}{\int_{0}^{1}\int_{0}^{2\pi}\int_{0}^{\pi/4}\rho^{2}sin(\phi)d\phi d\theta d\rho}=1+\sqrt2$ So the "less than 45 degrees" impact is more likely by a factor of about 2.414.
6. Well, there's no guarantee that you have enough information to find the actual function, but you could try approximating it by first graphing x and y vs. z and then using an appropriate polynomial (or other) approximation. If you don't know about "least squares" approximation, you might want to read up on that. I do, but it's honestly been a while so I would have to double-check myself before I risk explaining anything wrong and leading to even more confusion.
7. That's the whole point -- no countries in the world are or have ever been communist if you hold them to anything beyond their word.
8. How would we know about dark matter if we couldn't detect it? You don't disbelieve in black objects just because you technically can't see them, do you?
9. To the best of my understanding, none of the major considerations of higher dimensions imply that the other (spatial) dimensions are the same as the normal three we experience. Time is its own thing, of course, and yes it behaves like space in many ways, but as far as the other ones go, I don't think it's possible to "get there," really. Some of them are even said to be finite in some sense; that is, once you go so far in the nth dimension, you're back where you started, so you get a looping sort of effect. The closest sort of thing to the scifi you probably want, I would think, would be the concept of "imaginary time" as relates to alternate histories/futures. If you could traverse a distance of imaginary time, you might end up in a world where English muffins were called French muffins, or where Hitler won WWII, or where the sun exploded, etc. But I think Hawking says it's absolutely impossible to travel between these parallel universes, and consideration of their "existence" is not a whole lot more than a mathematical model for our observations of this universe.
10. It actually can be used to prove things about negative numbers. The reason that you learn it with only the natural numbers is that the natural numbers are intuitive and easy to work with. Furthermore, any set where induction can be applied can be thought of as "equivalent" in some way to the natural numbers. If you want to think in terms of "how many" numbers are in the set, the natural numbers are considered to be "countably infinite," which is opposed to, for example, the "uncountable" real numbers. Any set where normal mathematical induction applies, so far as I understand it, must be countable. The reason it must be countable is that induction relies on the enumeration of the elements in the set; that is, you have to take them one at a time and still be able to prove that all of them are accounted for somewhere in the sequence. No one has ever discovered a way to do this with real numbers, and in fact it has been proven impossible given our current formulation of mathematics. Now, as for how you can apply it to negative numbers: you do it mostly the same way. Start with, say, n=0 or something, and prove S(0). Then show that S(k) implies S(k-1) (instead of showing that S(k) implies S(k+1), as you do for natural numbers). By this, you can prove S(n) for all n <= 0. I like to think of induction as the process of finding a starting point in the set of numbers you're working with and then showing that you can "move" all the way through the set with your theorem holding true. Using it on negative numbers is just the process of showing that you can "move" backwards in this way. Regarding your second question: it's pretty tough. There's not a perfectly standardized way to find the formula for a given series (which is what we call the sum of a sequence). Others probably know much more about it than I do, but to the best of my knowledge, we rely heavily on intuition. There are a few theorems that apply generally, but almost all of them check for whether the sum converges (i.e. whether it keeps from shooting off to infinity) rather than what it actually converges to. EDIT: I don't want to be overly confusing, but I don't doubt that there is a form of induction applicable to uncountable sets. For example, you could prove something for the positive real numbers by proving S(0) and then showing that x < y implies that S(x) implies S(y). I don't know how useful this method is, though, and it's probably not called "induction." If this edit is confusing, feel free to ignore it.
11. As far as American conditioning goes, it's not necessarily so much the idea that communism = failure as the fact that true communism is practically impossible and that approximations to it tend to be unstable. Now, I'm not educated on the matter enough to really hold my own, but apparently China has worked out a stable state. That may or may not mean that communism is working over there, but I would wager, despite how conditioned my opinion may seem, that the extent to which it is stable is the extent to which it ignores its true communist ideals. Few in China actually wake up in the morning and think about how they can make the commune better, I assert; human nature doesn't allow it. Where there's smoke, there's fire: an increase in productivity echoes an increase in incentive. Capitalism is nothing more than the economic allowance of this, whereas the incentive for communism appears to come from an abstract knowledge that behaving in thus-and-such a way will make things better for everyone. If a few people manage to think abstractly and consistently long enough to get themselves to do more work than they get credit for, then a few more people might maintain a communist microcosm, but I don't buy the idea that the upper half of China's ability is cheerfully dragging the lower half along. Regardless of culture, most people can see an opportunity to cut one's losses and go for a selfish goal.
12. Looks like it'll be pretty funny. I'm pretty far to the left compared to most people where I live, but these two guys can do no wrong, as far as I'm concerned.
13. Even though I obviously don't know enough about this to justify trying to answer twice, here goes: if by "hardwired" you mean that the calculator simply stores a huge array of ordered pairs corresponding to a sine or cosine function so that the input number is matched with the appropriate output number (rather than it being calculated by the computer upon execution), no, I'm almost positive that it's not done that way. In fact, I believe that sin and cos functions in one of the C math libraries can take a long time to compute (compared to the basic +, -, *, / operators), which would indicate that a calculation (such as a taylor series) is being performed. But when you look at the functions, there's a shockingly small range of values that need be calculated -- all you really need is one quarter of a period (e.g. the part from sin(0) to sin(pi/2)), since the rest of the function, or both, actually, is/are composed of identical curves that have been flipped around. So they may only need a few terms of the Taylor series to get decent accuracy. So yeah, I'd say it's a bunch of simple polynomial operations, like ".4-.4*.4*.4/6+.4*.4*.4*.4*.4/120" or something.
14. (42 choose 6) isn't 57646, but (14 choose 6)/(42 choose 6) = 33/57646, as you've shown. I simply reduced the fraction before writing it out (that is, my calculator reduced it before it printed the answer on its screen). I don't know what you mean by "<>", but I assume that the approximate equality is simply coincidence. You can't find the probability of X after n trials with n*P(X), because otherwise X would have a probability of 1 (or higher!) after a certain number of trials, and that's (somewhat) intuitively not the case. The formula that I used is P(at least one X after n trials) = 1-(1-P(X))^n, which merely approaches 1 as n increases, which makes sense. If you want to see it visually, pick a value of P(X) -- say, 0.4, and graph y = 0.4*n and y = 1-(1-0.4)^n on a graphing calculator, if you have one (or work it out by hand if you're REALLY interested). The two functions obviously don't behave the same way. As for why the formula I used is the correct one, think of it this way: the probability that X will happen at least once is "one minus the probability that X will not happen at all." The probability that X will not happen at all is equal to "the probability that it will not happen the first time" times "the probability that it will not happen the second time," etc. Finally, the probability that it will not happen at any given single time is "one minus the probability that it will happen." Even though the above paragraph is pretty tedious, if you slog through it and write it all down algebraically, you ought to end up with the same thing I did.
15. I think they use taylor series and such, but I'm not that sure. I don't think that strictly numerical methods would work in general since most of them require the calculation of the function value or its derivative (which defeats the point in this case).
16. The probability of winning once with 14 numbers will be (14 choose 6)/(42 choose 6) = 33/57646. The probability of losing every time after 27 (equally random) trials will be (probability of losing once)^27. Thus, the probability of winning at least once will be 1-(1-P(winning once))^27 = 1-(1-33/57646)^27 = 0.01534 = 1.534%. So actually, it seems like their "investment" may not have been as terrible an idea as it normally is.
17. Yeah, you never said anything about where C and D were, or which angles you were referring to with <1, <2, <3, and <4.
18. This runs into the same sort of problems I had when I tried to figure out a better way to express an organism's taxonomic position: The problem with the current system is that it's non-cladistic (i.e. just because X and Y share a common ancestor Z doesn't mean that the smallest taxon containg X and Y will contain Z, and a taxon containing Z may not contain X or Y), and that it uses very few heirarchical levels of classification (species, genus, order, etc.). Ideally, there would be the same number of levels as there have been speciations in a given species' geneology. Since speciation is never really thought of as happening between more than two species at the same time, you can look at it in a binary way: let each speciation that ever happened be represented by a single binary digit, and put all the digits in chronological order. Thus, 11...01 would be a species that diverged from species 01...00 at the very beginning, but 101...100 and 101...101 would have diverged the most recently. Thus, we have a perfect system for classifying organisms according to strict cladistics and without arbitrary heirarchical levels. EXCEPT: The system is obviously ridiculously impractical. There would be millions if not billions of digits for each species, we would have no idea what most of the digit values should be, there's no account for weird bacterial rejoining and such, and there is little allowance for alteration of the system in light of new evidence -- say you find out that X speciated from Y rather than Z -- then every subsequent digit has to be rethought, there might be an additional number of digits ... you get the idea. The thing doesn't work, and I obviously came up with it only as a thought experiment and without anything like the kind of education I would need to come up with a real, practical system. You run into the same kind of problems here. What if we want to add a new theorem into the system, as we would need to do very often? How do you number the theorems? In chronological order according to when they were proved? That wouldn't be terribly useful, and we would have trouble verifying each historical case anyway. In order of "importance"? How do you decide what's important, and how do you compare the negligible importance of the thousands and thousands of minor lemmas to one another? In groups according to subject? This makes the most sense, but it pretty much eliminates the need for a universal numbering system, leaving each subject to be numbered independently, which is almost what we have now. But you're still left with the above problems on a smaller scale. Furthermore, what about systems with different axioms? Do you really want Euclidean theorems mixed in with those from hyperbolic geometry? If not, how do you determine a way to keep them distinct? How do you denote that each theorem is only true under certain axioms? Well, that's my take on it. I just see too many impracticalities.
19. Yeah, you might be able to somewhat standardize each individual field, but even that's kind of asking for a bigger mess than you started with. Voted "no."
20. I think this belongs in pseudoscience.
21. Try to think of all the possible ways that this minimum=25 event can happen, then try to express each situation in terms of the given probabilities. For example, one way that this can happen is when both are 25. This is the probability of both A and B occuring at once. Another way is for Hirakleio to be 25 and Athens to be hotter. This is the probability of B occuring and C not happening. Continue this until you exhaust the possibilities, and then use things like "The probability of A and B is equal to P(A)*P(B)" to find the total probability.
22. Are you saying it diverges at n=0.9? If so, how are you interpolating the fractional values of n? If you're saying it diverges at x=0.9, then to be on the y=x (or in this case x=n) line, n would have to be 0.9, also, so either way, you need interpolation. How are you doing this? EDIT: Also, I still don't know what you mean by the maximum value, or whether you're saying it diverges assymptotically.
23. It really seems counterintuitive that the values would initially approximate a line. I can definitely believe that they start out "under" it and then cross through it when they rapidly diverge, but perhaps they only seem to lie so close to the line because it's hard to obtain good resolution at that scale. Also, I don't know about your "maximum values" here; there shouldn't be any $n$ at which $x_n$ is undefined or infinite -- in other words, you shouldn't have any vertical assymptotes, if that's what you're talking about. You'll probably just have really rapid divergence (something in $\Theta(e^{mx})$, if I had to take a guess). I don't think your conjecture is true, as I understand it, though I very well may not.
24. Yeah, "such realms" implies that there was some talk of certain "realms" beforehand.
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