Gilad Reti

[Gilad Reti @gilad_reti]

@linusgroh @GitHubCopilot @code https://t.co/Oxjks8gAfw

[Gilad Reti @gilad_reti]

@74WTungsteno Since cos(x)≈1-x²/2, it reduces to the planar Pythagorean theorem in the small angle limit

[Gilad Reti @gilad_reti]

@ilarrosac @diegorattaggi @potetoichiro Nice! I think that the last part may be solved without using the general argument but rather that 5^n grows much quickly than 2^(n+1).

[Gilad Reti @gilad_reti]

@ixardul @raymondh Make its name dotted. For example:

from types import SimpleNamespace

matches = SimpleNamspace(your_var=your_var)

match x:
case matches.your_var:
...

[Gilad Reti @gilad_reti]

@AnalysisFact Let t=x-y, then the above is equivalent to ||(t+y)^n-y^n|≤(|t|+|y|)^n-|y|^n which easily follows from the triangle inequality

[Gilad Reti @gilad_reti]

Took a while, but our blog post regarding CVE-2020-27352 is finally out.

https://t.co/IX5hpCVbxy

[Gilad Reti @gilad_reti]

@linuxtoday eBPF!

[Gilad Reti @gilad_reti]

@pstrandmark @gvanrossum Here are some of Guido's thoughts:
https://t.co/FwmqOsWXYr

[Gilad Reti @gilad_reti]

@angeris @InertialObservr Something similar to this?
https://t.co/wRUeOdBwsc

[Gilad Reti @gilad_reti]

@TopologyFact This generalizes to higher dimensions too, replacing "connected" with "n-connected"

[Gilad Reti @gilad_reti]

@diegorattaggi @SrinivasR1729 The magic fades out when you actually calculate the integrand and it turns out to be \frac{\sqrt{5}-1}{2} \sin(x) :(

[Gilad Reti @gilad_reti]

@maths_language @AlgebraFact nCk is the coeeficient of x^ky^n. To see that, note that \sum_n \sum_k x^ky^n=\sum_n y^n \sum_k x^k = \sum_n y^n(1+x)^n = 1/(1-y(1+x)).