Timeline for Squaring a Leibnitz-like series
Current License: CC BY-SA 4.0
18 events
| when toggle format | what | by | license | comment | |
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| Aug 9, 2021 at 13:29 | audit | First posts | |||
| Aug 9, 2021 at 14:02 | |||||
| Jul 25, 2021 at 14:02 | audit | First posts | |||
| Jul 25, 2021 at 14:05 | |||||
| Jul 20, 2021 at 17:17 | audit | First posts | |||
| Jul 20, 2021 at 17:17 | |||||
| Jul 19, 2021 at 19:57 | vote | accept | Oscar Lanzi | ||
| Jul 19, 2021 at 3:12 | audit | First posts | |||
| Jul 19, 2021 at 3:12 | |||||
| Jul 18, 2021 at 15:02 | audit | First posts | |||
| Jul 18, 2021 at 15:19 | |||||
| Jul 18, 2021 at 1:55 | answer | added | No-one Important | timeline score: 5 | |
| Jul 17, 2021 at 9:49 | history | edited | Oscar Lanzi | CC BY-SA 4.0 |
added 38 characters in body
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| Jul 16, 2021 at 22:25 | history | edited | Oscar Lanzi | CC BY-SA 4.0 |
Described some own research.
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| Jul 16, 2021 at 12:58 | comment | added | Jack D'Aurizio | Cancellation is tricky here, since the serie appearing in (2) is not absolutely convergent. Anyway, have a look at Knopp's lemma at page 70 here: drive.google.com/file/d/0BxKdOVsjsuEwdjBEM1dpRkhMa2s/view | |
| Jul 15, 2021 at 20:06 | comment | added | Oscar Lanzi | Criss products mean terms like $2ab$ in $(a+b)^2=a^2+2ab+b^2$. In this case there are infinitely many and somehow the positive ones cancel the negative ones exactly leaving just the Squares. Why does that cancelation work for just this series? | |
| Jul 15, 2021 at 19:39 | comment | added | Vishu | It’s not about bravery, but hard work. | |
| Jul 15, 2021 at 19:24 | comment | added | Oscar Lanzi | Glory comes to the brave ... . | |
| Jul 15, 2021 at 19:24 | comment | added | Vishu | I’m pretty sure that will work, but too lazy to. :) | |
| Jul 15, 2021 at 19:23 | comment | added | Oscar Lanzi | Try it and see, maybe? | |
| Jul 15, 2021 at 19:13 | comment | added | Vishu | The cross terms can be written as $$\sum_{0\le a\lt b\lt\infty} \frac{(16a+4)(16b+4)}{(8a+1)(8a+3)(8b+1)(8b+3)} - \frac{(16a+12)(16b+12)}{(8a+5)(8a+7)(8b+1)(8b+7)} $$ I suppose one way of showing this equals zero is by using partial fractions and the closed -form for $\sum_{n=1}^{\infty}\left( \frac{1}{n+p} -\frac{1}{n+q} \right)$. | |
| Jul 15, 2021 at 18:32 | history | edited | Andrew Chin | CC BY-SA 4.0 |
added 3 characters in body
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| Jul 15, 2021 at 18:19 | history | asked | Oscar Lanzi | CC BY-SA 4.0 |