我也找到了
狼王络波
2021-10-01 21:48:59
( reads)
$$ \Bigg(\bigg(\Big(\big((x)\big)\Big)\bigg)\Bigg) \quad \Bigg[\bigg[\Big[\big[[x]\big]\Big]\bigg]\Bigg] \quad \Bigg\{\bigg\{\Big\{\big\{\{x\}\big\}\Big\}\bigg\}\Bigg\} $$
\[\frac{-b\pm\sqrt{b^2-4ac}}{2a}\]
`a^2+b^2=c^2`
`i(t_b)=A[(1-n)*t/r+b]/[(t_b/r)+b]^(n+1)`