Timeline for Can we balance $2$-powers?
Current License: CC BY-SA 4.0
19 events
| when toggle format | what | by | license | comment | |
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| Feb 1 at 16:28 | history | edited | LSpice | CC BY-SA 4.0 |
Title of paper
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| S Jan 30 at 6:03 | history | bounty ended | CommunityBot | ||
| S Jan 30 at 6:03 | history | notice removed | CommunityBot | ||
| Jan 29 at 21:29 | answer | added | Max Alekseyev | timeline score: 6 | |
| S Jan 29 at 2:50 | history | suggested | Marco Ripà | CC BY-SA 4.0 |
Minor edits to improve readability (a little bit)
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| Jan 29 at 2:47 | answer | added | Alex Ravsky | timeline score: 1 | |
| Jan 29 at 0:04 | review | Suggested edits | |||
| S Jan 29 at 2:50 | |||||
| Jan 28 at 14:29 | comment | added | Claude Chaunier | @user42355, domotorph has understated it, there is equality between the two sums. The change of sign from $x_i$ to $x_{i+1}$ reflects whether $0$, $-2$ or $2$ has been added to $2x_i$ to yield $x_{i+1}$. While going through $x_1, x_2, \dots, x_k, x_1$ there must be as many flips from positive to negative as from negative to positive. Therefore $\sum_{i=1}^k x_i = $ $\sum_{i=1}^k (2x_i + (0\text{ or }-2\text{ or }2)) = $ $\sum_{i=1}^k 2x_i + \sum_{i=1}^k(0\text{ or }-2\text{ or }2) = $ $\sum_{i=1}^k 2x_i + 0$. | |
| Jan 28 at 10:46 | comment | added | user42355 | @domotorp This only proves $\sum_{i=1}^k x_k \equiv 0 \pmod{2}$, i.e., the sum is an even integer. | |
| Jan 27 at 22:38 | comment | added | domotorp | That is easy because $\sum_{i=1}^k x_i\equiv \sum_{i=1}^k 2x_i$, so it has to be zero. | |
| Jan 25 at 18:28 | comment | added | user42355 | In the linked article, if I understand correctly, the identity $\sum_{i=1}^k x_i = 0$ is derived using a certain monochromatic $k$-gon, whose existence is only conjectural. Is this correct? Or is there some other proof for $\sum_i x_i = 0$? It is easy to see that $\sum_i x_i$ is an even integer, so $\pi$ can only exist if $\sum_i x_i = 0$. | |
| S Jan 22 at 5:03 | history | bounty started | domotorp | ||
| S Jan 22 at 5:03 | history | notice added | domotorp | Draw attention | |
| Jan 20 at 8:52 | comment | added | domotorp | Yes, you are right. | |
| Jan 20 at 7:59 | comment | added | Fedor Petrov | so, $x_i$ is just $2^{i-1}x_1$ modulo 2, where the remainder is chosen in $[-1,1)$, and $(2^k-1)x_1$ is divisible by 2, so the sequence is periodic, right? | |
| Jan 20 at 6:02 | comment | added | domotorp | @Fedor It did, thanks. (Ps. I also left a comment here for you about it, but for some strange reason it disappeared overnight, so let me try again.) | |
| Jan 19 at 22:41 | history | edited | domotorp | CC BY-SA 4.0 |
fixed typo
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| Jan 19 at 22:16 | comment | added | Fedor Petrov | Does the third case in the recurrence contain a typo? | |
| Jan 19 at 21:36 | history | asked | domotorp | CC BY-SA 4.0 |