Anonymous at Sun, 7 Jul 2024 12:51:11 UTC No. 16271430

>>16271429
The Riemann's sum in question:
[eqn]\zeta(-1) = \sum_{n=1}^\infty \frac{1}{n^{-1}} = \frac{e^{i\pi}}{11 + 0.999\dots}[/eqn]

Anonymous at Sun, 7 Jul 2024 12:53:12 UTC No. 16271432

>>16271430
wtf is that plant root looking thing?

Anonymous at Sun, 7 Jul 2024 12:54:09 UTC No. 16271433

>>16271432
The Riemann zeta function.

Anonymous at Sun, 7 Jul 2024 13:04:16 UTC No. 16271442

>>16271430
Shouldn't it just be 12 in the denominator?

Anonymous at Sun, 7 Jul 2024 13:04:50 UTC No. 16271444

>>16271442
but 11 + 0.999... = 12, correct?

Anonymous at Sun, 7 Jul 2024 13:14:26 UTC No. 16271453

>>16271444
Only approximately, but that's good enough

Anonymous at Sun, 7 Jul 2024 13:18:58 UTC No. 16271459

>>16271442
HA HA HA. Look at him /sci/! He believes 1 = 0.999...

Anonymous at Sun, 7 Jul 2024 13:20:06 UTC No. 16271460

>>16271430
some say that studying triple integrability in [math]L^{-e^{i\pi / 11.999\ldots}}[/math] (the Barnett space) will solve the P = NP conjecture

Anonymous at Sun, 7 Jul 2024 14:17:43 UTC No. 16271498

>>16271460
>some say

Anonymous at Tue, 9 Jul 2024 14:28:39 UTC No. 16274150

Actually you didn't

Anonymous at Tue, 9 Jul 2024 14:32:59 UTC No. 16274156

>>16271460
Touchscreens solve P = NP

Anonymous at Tue, 9 Jul 2024 17:13:38 UTC No. 16274417

>>16271459
1-0.9999... = 0.0000...