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Everything posted by John

  1. Though I agree that it's drifting pretty far off-topic, on Einstein and special relativity, the well-cited Wikipedia article on the history of special relativity makes for good reading. If the entire article is too long, then section 3.1, on Einstein in particular, may be sufficient and includes a telling quote from the man himself. I wouldn't say it's impossible for unknown outsiders to make meaningful contributions. However, as others have noted, it's extremely unlikely and will probably become increasingly so as science continues to advance. In the Speculations forum, at least in my experience, ideas aren't generally dismissed automatically. What usually seems to happen is that an idea is presented, questions are raised (sometimes, admittedly, with at least a hint of condescension), and the OP either answers those questions or fails to answer those questions. In the latter case, the OP either gracefully accepts that his idea is flawed or makes unfounded claims that the scientific establishment is wrong or conspiring against outsiders or whatever. The latter ties into what's been mentioned elsewhere, that part of science education is learning to accept being wrong and having one's ideas shot down in front of others. To me, the most frustrating instances (again of the latter scenario mentioned above) are those in which the OP is clearly intelligent but simply has no knowledge or understanding of what science has already established. Maybe the lack of education is due to youth, maybe it's due to laziness, or maybe it's due to pride (as if having to build on what's already known somehow lessens any breakthroughs the person might have), but in any case, it's a shame to see.
  2. In case there's any confusion, the answer to OP's question is that [math]\sqrt{16}=4[/math]. The notation [math]\sqrt{x}[/math], when x is positive, refers to the principal square root, which is also positive. If you want to refer to both square roots of 16, then you should write [math]\pm \sqrt{16}[/math]. As studiot mentioned, "the square root" usually refers to the principal square root, also.
  3. As someone else who returned to studying mathematics after a fairly lengthy absence, I feel your pain. After nearly two years of being back in, my careless errors have become less frequent (being subject to having my work graded has helped), but they still happen from time to time. I guess just keep working at it, and of course feel free to post here with any questions or concerns, even if it's just to get a second opinion on work you're pretty sure is correct. For at least some textbooks, you might be able to find detailed solutions to some or all exercises online, as well. In any case, take care, and good luck with your studies.
  4. I think there's some confusion here between the elements of each set and the cardinality (number of elements) of each set. C - P is the set containing the students who are in C and are not in P (i.e. the set of students who only play soccer, which you recognize is also the result of (P U C) - P). What you've done is subtract the cardinality of P from the cardinality of C, which is |C| - |P| = 13 - 7 = 6, which is different. Three of the students in P are not in C to begin with, so they don't factor into C - P. The reason subtracting the cardinality of P from the cardinality of P U C worked is that P is a subset of P U C, and thus all 7 students in P factor into the set difference. Edit: I should also add that in the example at the end of your post, you have (P U C) - P = {a, b}. However, P U C = {a, b, c, d, e, f} and P = {a, b, c}, so (P U C) - P = {d, e, f}, which you can see is the same as C - P.
  5. John

    Mean Value

    That's what I got. You may want to change the denominator from 2^(2n) to 4^n and arrange things such that the answer is a bit "simpler," but I wouldn't expect that to matter too much.
  6. John

    Mean Value

    That matches my answer when n=3, which may mean you're on the right track (or that my answer's wrong, but I think it's correct ). How did you arrive at that result?
  7. John

    Mean Value

    We start with an initial property value c. Do you know how to calculate the expected property value after the first game? Once you do that, you can use the new value as the input for the second game, and then use the result of that as the input for the third game, and so on. After a few steps you may notice a pattern, and that pattern will allow you to express the expected property value after n games. I'm sorry if I'm being a bit unclear. Trying not to give too much away. Also, this is assuming I'm understanding the question correctly. I take the "mean value" in this case to be the expected value of the property, but maybe you mean something different.
  8. John

    Mean Value

    I don't know what specific course you're taking or how advanced it is, but if you know how to calculate the expected value at each step, you can find a recurrence relation, then use that to figure out the sequence itself. Based on that sequence, you can find a pattern and figure out the answer from there.
  9. Linear algebra develops from techniques used to solve systems of linear equations. Building on these methods, linear algebra gets into the study of vector spaces, which are sets of vectors combined with two operations such that certain requirements (which we call axioms) are met. Abstract algebra is the study of algebraic structures (including vector spaces, but also groups, rings and others) and thus serves as a generalization of all this, i.e. it deals with the properties of various sets of elements combined with various operations following various axioms.
  10. You would fix the specific numbers in the first however many slots and then work out the number of combinations for the remaining slots. In your initial example of finding all groups, you're calculating [math]50\choose4[/math] to find the number of possible combinations of four integers between 1 and 50 inclusive. If you wanted to find all combinations containing 1 and 2, then you would assume 1 and 2 occupy two slots, and then calculate how many ways to choose two numbers from the remaining 48 possibilities, i.e. you'd calculate [math]{{48}\choose{2}} = \frac{48!}{2!(48-2)!} = \frac{48\times47}{2}[/math].
  11. I don't think Feynman was criticizing the rules themselves, but rather criticizing the tendency to get caught up in symbol pushing without really understanding what's going on. The fact that we can solve for x in an equation like x + 3 = 7 by subtracting 3 from both sides is useful, but immediately answering, "Well, x must be 4," rather than saying, "Well, x + 3 = 7, so x + 3 - 3 = 7 - 3, so x = 4," is fine (unless you're in a class and the teacher insists you show every step). As overtone mentioned (and as I'm sure Feynman recognized), the specific rules and methods taught in class are taught for a reason. Even for seemingly trivial exercises, the use of a new technique can provide insight into what's actually going on at a fundamental level. But we shouldn't get so caught up in any particular algorithm for finding a solution that we lose sight of the general idea.
  12. Take this all with a grain of salt, as at the end of the day it's just a single example, and I'm not exactly an expert in neuroscience education: There's a guy at my school in a similar situation. He's got a bachelor's in psychology and wants to go to graduate school for neuroscience. He wasn't accepted into grad school, and he says his lack of natural science in undergrad was the main problem. Now (or at least, as of the last time I saw him) he's doing a second bachelor's, in chemistry. I don't know what classes he took during his psych program or where he applied for grad school, so perhaps your preparation is better than what his was, the grad programs you're looking at are more forgiving and/or willing to let you take required science courses with them. However, if you can't get more natural science in before you graduate, you may have to at least take a few post-bacc courses before graduate programs will accept you.
  13. 1. I'm not sure there are other *common* uses, but really you can use the same notation to describe just about anything you want (though for at least some definitions, other people might look at you funny). Wikipedia gives an example of using the notation for sums of vector spaces. Keep in mind I'm a student, not a professional mathematician. There may be very common uses of the notation that I haven't seen yet. 2. I don't see why not. For instance, [math]\mathbb{R}^n[/math] is commonly used to denote an n-dimensional Euclidean space, [math]\mathbb{C}^n[/math] for complex coordinate spaces, etc.
  14. 1. That notation usually refers to the Cartesian product, yes. 2. Yes.
  15. You're welcome. Glad I could help.
  16. Yes, almost. You'll also need to subtract the probability of getting zero red balls (which, of course, is a bit simpler to calculate).
  17. What have you tried so far? Although this may be giving too much away, have you covered Bernoulli trials yet?
  18. That only includes one digit. I like that, though. If f(3) is the largest number that can be expressed using 3 digits, though (in combination with f() itself), then f(333) must be less than or equal to f(3), f(3...3) (containing 333 3's) must be less than or equal to f(333), and so on. For arbitrarily large x, then, f(x) must be less than or equal to f(3), which suggests f(3) being infinite.
  19. I thought we were only ruling out division by 0, not infinity itself. The set of hyperreals *R includes infinities and infinitesimals, and we can divide a finite number by an infinitesimal to yield infinity. But, no matter. So now the rules are: 1. Use three digits and at most one operator (of any convenient arity--and is this one operator per digit, or one operator for the entirety of whatever is constructed by those digits?). 2. We cannot define new operations, though operations defined in the places mentioned in your post are fine. 3. The result must be a real number. Anything I'm missing?
  20. Might need to be even more precise than that, else I'll go with [math]D(999)[/math] where [math]D:\mathbb{^*R}\to\mathbb{^*R}[/math] is a unary operation that divides its input by the infinitesimal [math]x[/math].
  21. Yeah, I slightly misread imatfaal's post. The reason I bring up primes larger than 6 is that, previously, you claimed we shouldn't need to consider 2R/(6p) for p > 7, or (later) primes greater than sqrt(p_2). That's what I was addressing here. The entire point to the twin prime conjecture is the lower limit does go to infinity, i.e. there are infinitely many twin primes. If the possibility exists that there aren't infinitely many, then the conjecture isn't proven. Indeed, though I'm not sure how easy it is to find said highest prime for very large ranges.
  22. This is a different definition for R from what you've told us previously. You defined R as the difference between p_1^2 and p_2^2, which in the case of 31 and 37 works out to be 408, which means 287 is in R. Furthermore, imatfaal used the same definition for R, and now you've said his math is correct aside from the 3p vs. 6p thing you mentioned. However, assuming this new definition is what you meant, there are still relevant multiples of primes greater than 6 in the range, for example 979, which aside from 1 and 979 only has factors 11 and 89 and is equal to 6(163) + 1. No worries. Though I'd imagine that if the proof were based on a simple idea, it would have already been found, there's a chance you're right and we're all wrong. But it doesn't seem that way so far. I think I've heard before of statistical arguments in favor of the twin prime conjecture. However, obviously that sort of argument doesn't constitute proof. I'm not sure what you mean here. If the lower limit is never infinity, then for arbitrarily large ranges, the possibility still exists that only a finite number of twin primes exist. Even if the lower limit in some range is Graham's number or something, that means there could possibly be Graham's number plus two twin primes, and no more.
  23. I see your point about 205. However, 287's only factors (besides 1 and 287) are 7 and 41, both of which are greater than sqrt(37). For my part, I'm not certain the argument holds water, and I think imatfaal's example of 31, 37 has been shown to be a counterexample, unless there's still some part of the argument we're missing. No need to get upset. If the twin prime conjecture were easy to prove, everyone would be doing it. Even if our objections are actually silly, any proposed proof of the twin prime conjecture will be put through the ringer by the mathematical community, so be prepared if you take it that far. As for Dr. A. Mathematician, he's one guy. Maybe the argument is actually correct, or maybe he's missing something too, especially if it was presented to him less rigorously than it is in this thread. Regardless of all that, you must show that, assuming your argument is correct, the lower limit eventually reaches infinity. Simply getting larger isn't enough, as convergent infinite series show.
  24. But the square root of p_2 in our go-to example (p_1 = 31, p_2 = 37) is a little over six, and clearly there are primes larger than six that have multiples of the form 6n +/- 1 in R. Even primes larger than p_2 itself have such multiples (e.g. if you look at 41, it has multiples 205 = 6(34) + 1 and 287 = 6(48) - 1).
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