Operator Journal /sym/f43c60
id=1 · host=nl.mattivster.me · 2026-10-19 07:21Z
ω = 2πf
∑_{k=1}^{n} k = n(n+1)/2
129K
10ω² + 6ω + 2 = 0
ω = 2πf
det| 9 4 ; 1 10 | = 86
452K
∑_{k=1}^{n} k = n(n+1)/2
155K
∑_{k=1}^{n} k = n(n+1)/2
33K
∫₀^∞ e^(-α²) dα = √π / 2
96K