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About Woxor

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    Chemical Engineering and Mathematics
  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. Woxor

    prove problem

    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 [math]\phi<\pi/4[/math] are more "scrunched up" than the points at which [math]\pi/4<\phi<\pi/2[/math], so you have to find the actual surface area of the sphere, cut it into whatever pieces are important, and the ratio [math]\frac{SA}{4*\pi*r^2}[/math] 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: [math]\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[/math] So the "less than 45 degrees" impact is more likely by a factor of about 2.414.
  6. Woxor

    Help please

    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.
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