Math Dump /sym/7fd97a
id=1 · host=jp.commstar.world · 2026-10-19 07:59Z
ω = 2πf
∑_{k=1}^{n} k = n(n+1)/2
326K
7r² + 8r + 6 = 0
∑_{k=1}^{n} k = n(n+1)/2
∑_{k=1}^{n} k = n(n+1)/2
366K
det| 7 4 ; 4 8 | = 40
201K
det| 10 9 ; 3 11 | = 83
det| 11 9 ; 5 12 | = 87
176K
2ψ² + 6ψ + 0 = 0
det| 4 4 ; 5 5 | = 0
263K
ω = 2πf
∂ψ/∂t = 3ψ