uncool

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uncool last won the day on February 7 2016

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  1. Your reasoning is almost correct - but you miss that there is a switch there. Remember, the group acts on x, not on the argument of the function. In other words: (g1 . (g2 . f)) (x) = (g1 . h)(x), where h(x) = f(g2^-1 x). Then by the definition, to apply the action of g1, we apply it to x, and get h(g1^-1 x) = f(g2^-1 g1^-1 x).
  2. It is correct; the interesting part comes with the question of why. Can you think of both a conceptual (dealing with the concept of limits) and a formal (dealin with epsilon-delta definition) reason why?
  3. Angular momentum is also conserved when there is a change in radius, as in the example of an object moving past a point with no force acting on the object; the only requirement is that any force be central (that is, in a direction parallel to the radius from the point angular momentum is being measured from).
  4. This seems trivially falsified if we take the case of an object moving with no force affecting it.
  5. As a note, without having read the poem myself, "drop of blood on the seafoam" makes me think of the birth of Aphrodite, not Christian mythology.
  6. Flat things can still be finite, though. For example, a 2-torus can be made flat (though not in 3-dimensional space).
  7. There are infinitely many numbers of the form 10^m, but only finitely many possible remainders/equivalence classes modulo k, so by Pigeonhole Principle, at least two (in fact, infinitely many) distinct numbers of the form 10^m must have the same remainder/be in the same equivalence class. Then their difference is divisible by k. This is a palindrome because it's of the form 99.....9900...00, which can be turned into 00...0099...9900..00 with exactly as many leading as trailing 0s. For example: 14. The remainders modulo 14 of the first few powers of 10 are: 1, 10, 2, 6, 4, 12, 8, 10, 2, 6, 4, ... So 10^1 and 10^7 have the same remainders, so 10^7 - 10^1 is divisible by 7. 10^7 - 10^1 = 9999990 = 09999990, which is a palindrome with that leading 0.
  8. There is no "z1" or "z2" in the equation 1 = 0. Not really. Again, use it to analyze something and get a result in an easier way. Groups, for example, are useful because they tell us about the symmetries of an object. That is one of the major reasons mathematicians study them in the first place. That goes to basic logic. You can't have that a statement is both true and false at the same time, under the usual assumptions of logic.
  9. It means you don't have to conclude that 1 = 0, which is quite useful. It also allows you to relax the other axioms when applied to this "secondary" form of multiplication, which would be necessary for your axiom to make sense. The same is true if you tangibly have 4 apples and you multiply by 3.14; you still end up holding 4 apples. All you're saying is that multiplication doesn't apply in that way. So no, I still don't understand how what you are doing is in any way useful.
  10. How can they both be true, but "not at the same time"? The only way this can work is if you are trying to define a new operation, call it "z-multiplication", above and beyond the standard 2 structures, that does not satisfy some of the usual field axioms for multiplication. Only then can you have that 0 *z a is not 0. And more importantly: If you're proposing an axiom, you should show some examples of things that satisfy that axiom, and (more importantly) how the extra axiom makes them more useful.
  11. No. The universe could be finite, but not have an edge. Think, for example, of a 2-D "universe" that exists on the surface of a sphere. There is no edge - and so no distance-to-edge.
  12. It's true for any positive integer. More specifically, every integer is a factor of some number of the form 10^m - 10^n, which is automatically a palindrome if you allow leading 0s. Proof isn't hard through the Pigeonhole Principle.
  13. 1. Re3+ Possibilities: 1... Re4 2. Qd5# 1...Ne4 2. Bf4# 1... Be4 2. Bf4# 1... Kd6 2. Qxe7#
  14. Perhaps up to some dimensionless constants, e.g. 2 pi or 4 pi or e or ln(2). 2 pi for angular frequency vs frequency, 4 pi if someone focuses on flux per unit area rather than total flux (as the area of a sphere is 4 pi r^2), e or ln(2) in case they decide to deal with exponentiation slightly differently than we do.
  15. Remember some assumptions on the commutation relations of position and momentum - position and momentum operators with different indices (e.g. [math]\hat{Y},\hat{P_z}[/math]) commute, while ones with the same index (e.g. [math]\hat{Z},\hat{P_z}[/math]) do not. In your expression, the only terms with common indices deal with Z, so you can feel free to move the others in any way you want - including, as here, to the front. So when you do that for the first term, you get [math]\hat{Y}\hat{P_z}\hat{Z}\hat{P_x} = \hat{Y}\hat{P_x}\hat{P_z}\hat{Z}[/math]. However, you can't then switch [math]\hat{P_z}[/math] and [math]\hat{Z}[/math], as they do not commute - so this is the simplest you can get. Similarly, the second term becomes [math]\hat{P_x}\hat{Y}\hat{Z}\hat{P_z}[/math]. Does that help?