Timeline for Make $n$ numbers equal using pairwise averages
Current License: CC BY-SA 4.0
9 events
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| May 7, 2022 at 6:58 | comment | added | jh w | Thank you for your answer. The trouble is ''a wrong step on can destroy the solvability'' by @YCor. I have tried to solve this problem in this approach, but it doesn't seem to go any further. | |
| May 7, 2022 at 6:42 | history | edited | Brendan McKay | CC BY-SA 4.0 |
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| May 7, 2022 at 6:30 | comment | added | Brendan McKay | @FedorPetrov I take it back. To show that only a finite number of steps is needed, one has to know that the sequence is bounded below, but that is obvious for both functions. I just chose one that reaches zero rather than some multiple of the square of the mean when all the numbers are equal, on a whim if you like. Our functions have an affine relationship anyway so the proof is equivalent. | |
| May 7, 2022 at 6:25 | history | edited | Brendan McKay | CC BY-SA 4.0 |
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| May 7, 2022 at 6:05 | comment | added | Fedor Petrov | I do not understand. The minimum of what? | |
| May 7, 2022 at 5:48 | comment | added | Brendan McKay | @FedorPetrov First part now incorporated. The sum of squares is fine but then one has to argue what the minimum is (trivial, yes, but one extra step). | |
| May 7, 2022 at 5:47 | history | edited | Brendan McKay | CC BY-SA 4.0 |
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| May 7, 2022 at 4:47 | comment | added | Fedor Petrov | The numbers are rational, so you should start with multiplying them by a common denominator. Also, for $V$ we alternatively may take the sum of squares of all numbers. | |
| May 7, 2022 at 4:13 | history | answered | Brendan McKay | CC BY-SA 4.0 |