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Oct 29, 2015 at 15:38 history edited Dr. Wolfgang Hintze CC BY-SA 3.0
EDIT #1 changed: wrong statement found, proof not conclusive
Oct 29, 2015 at 15:31 history edited Dr. Wolfgang Hintze CC BY-SA 3.0
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Oct 28, 2015 at 13:05 comment added Dr. Wolfgang Hintze @ seno44: (28.10.15 14:04) unfortunately, I have just discovered a flaw in the "proof" of the recent comment. I try to mend it and return.
Oct 27, 2015 at 20:22 history edited Dr. Wolfgang Hintze CC BY-SA 3.0
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Oct 27, 2015 at 20:17 history edited Dr. Wolfgang Hintze CC BY-SA 3.0
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Oct 27, 2015 at 20:09 comment added Dr. Wolfgang Hintze @ seno44: If the generalization I proposed in my last comment is correct, the Taylor method is easily generalized. As far as I can see it leads to an infinite set of equations of the form $(vx^k, vv) = 0 (k=0,1,2,...)$ where $vv = (v_{10}, ..., v_{N0})$ , $x^k = (x_{10}^k, ..., x_{N0}^k)$, and $(a,b)$ is the scalar product of the vectors $a$ and $b$. As the equations must hold for all k, the only solution for $vv$ is the trivial one. QED.
Oct 27, 2015 at 19:37 comment added Dr. Wolfgang Hintze @ seno44: Ok. Please be more specific.Do you wish to generalize 0 = y_{10} + y_{20} to 0 = y_{10} + y_{20} + ... + y_{n0} ?
Oct 27, 2015 at 15:07 comment added user45183 The problem is that I want to generalize this consideration to $N$ particles rather than $N=2$. In that respect the Taylor expansion approach does not generalize nicely in my opinion.
Oct 27, 2015 at 13:35 comment added Dr. Wolfgang Hintze @ seno44: I agree, if you consider the Taylor Expansion "very dificult" which, honestly, I wouldn't. But, you are right, there should be some simple symmetry argument to give the proof (even simpler than in the answer of Fritz Veeman) because the contrary holds for a symmetric potential, i.e. you can have $d=0$ iff $x_{10}+x_{20}=0$ and $y_{10}+y_{20}=0$
Oct 27, 2015 at 13:30 history edited Dr. Wolfgang Hintze CC BY-SA 3.0
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Oct 27, 2015 at 12:32 comment added user45183 I think you have chosen the route that I was explicitly not interested in as I mentioned in the question.
Oct 27, 2015 at 9:53 comment added Dr. Wolfgang Hintze @seno 44: Thanks for the hint to my trivial error. I have given a complete solution of the OP in the meantime.
Oct 26, 2015 at 15:08 history edited Dr. Wolfgang Hintze CC BY-SA 3.0
Beautifying of subscripts, oberservation added
Oct 26, 2015 at 14:57 history edited Dr. Wolfgang Hintze CC BY-SA 3.0
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Oct 26, 2015 at 14:52 history edited Dr. Wolfgang Hintze CC BY-SA 3.0
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Oct 26, 2015 at 14:43 history edited Dr. Wolfgang Hintze CC BY-SA 3.0
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Oct 26, 2015 at 11:46 history edited Dr. Wolfgang Hintze CC BY-SA 3.0
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Oct 26, 2015 at 11:31 history edited Dr. Wolfgang Hintze CC BY-SA 3.0
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Oct 26, 2015 at 11:05 history edited Dr. Wolfgang Hintze CC BY-SA 3.0
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Oct 26, 2015 at 10:56 history edited Dr. Wolfgang Hintze CC BY-SA 3.0
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Oct 26, 2015 at 10:27 history edited Dr. Wolfgang Hintze CC BY-SA 3.0
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Oct 26, 2015 at 10:16 history edited Dr. Wolfgang Hintze CC BY-SA 3.0
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Oct 26, 2015 at 10:09 history edited Dr. Wolfgang Hintze CC BY-SA 3.0
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Oct 26, 2015 at 10:07 history undeleted Dr. Wolfgang Hintze
Oct 24, 2015 at 20:36 history deleted Dr. Wolfgang Hintze via Vote
Oct 24, 2015 at 20:28 history edited Dr. Wolfgang Hintze CC BY-SA 3.0
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Oct 24, 2015 at 20:25 comment added user45183 the starting point is not correct. It should be $x^2(t)$ in the $\dot{y}$ equation!
Oct 24, 2015 at 20:18 history edited Dr. Wolfgang Hintze CC BY-SA 3.0
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Oct 24, 2015 at 20:11 history edited Dr. Wolfgang Hintze CC BY-SA 3.0
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Oct 24, 2015 at 20:03 review First posts
Oct 24, 2015 at 20:10
Oct 24, 2015 at 20:03 history answered Dr. Wolfgang Hintze CC BY-SA 3.0