∂?t + ε∞ − 1x₁∇·B = 0ℏe^(iπ) + 1 = 0ζ(s)Δx ΔpS¹ ↪ ℝ²Rμν − ½Rgμν∫𝒟φχ(M)P(A|B)c = ℏ = 1∮γ[x, p] = iℏ|ψ|²Γ(½) = √πSO(3)∇²δ(x)⟨x⟩L²(ℝ)ω² = k/mΔS ≥ 0∂Σrank(A)|0⟩ + |1⟩∮ B · dl∫ₘ R√|g| d⁴xiℏ ∂|ψ⟩/∂t = Ĥ|ψ⟩∇×E = −∂B/∂tGμν + Λgμν = 8πGTμν∑ₙ₌₀∞ (−1)ⁿx²ⁿ⁺¹/(2n + 1)!⟨ψ|Â|ψ⟩ / ⟨ψ|ψ⟩d²x/dt² + ω²x = 0limₙ→∞ (1 + 1/n)ⁿ∮ᶜ F · dr = ∬ₛ (∇×F) · dS∂μTᵘᵛ = 0π₁(S¹) ≅ ℤ𝓕{∂f/∂x} = iξf̂(ξ)

∃ caffeine ∈ ℝ⁺: ||Ψ(caffeine) - Ψ*|| < ε

Arnav Mittal

hi, i'm arnav, a student interested in theoretical physics, mathematics, technology, and more!