Math Dump /sym/7fd97a
id=1 · host=jp.commstar.world · 2026-10-19 07:46Z
ω = 2πf
∑_{k=1}^{n} k = n(n+1)/2
198K
7r² + 8r + 6 = 0
∑_{k=1}^{n} k = n(n+1)/2
∑_{k=1}^{n} k = n(n+1)/2
366K
det| 10 9 ; 3 11 | = 83
2ψ² + 6ψ + 0 = 0
ω = 2πf
det| 5 3 ; 2 6 | = 24
233K
∂ψ/∂t = 3ψ