analiza wektorowa seymour: wie ktoś jak udowodnić, że rot(rotA)=grad(divA)−ΔA?

Seymour: Ponawiam pytanie

piotr: (−f^(0,0,2)(x,y,z)−f^(0,2,0)(x,y,z)+g^(1,1,0)(x,y,z)+h^(1,0,1)(x,y,z), f^(1,1,0)(x,y,z)−g^(0,0,2)(x,y,z)−g^(2,0,0)(x,y,z)+h^(0,1,1)(x,y,z), f^(1,0,1)(x,y,z)+g^(0,1,1)(x,y,z)−h^(0,2,0)(x,y,z)−h^(2,0,0)(x,y,z)}

piotr: rot(rot(f(x,y,z), g(x,y,z), h(x,y,z))) = (−f^(0,0,2)(x,y,z)−f^(0,2,0)(x,y,z)+g^(1,1,0)(x,y,z)+h^(1,0,1)(x,y,z), f^(1,1,0)(x,y,z)−g^(0,0,2)(x,y,z)−g^(2,0,0)(x,y,z)+h^(0,1,1)(x,y,z), f^(1,0,1)(x,y,z)+g^(0,1,1)(x,y,z)−h^(0,2,0)(x,y,z)−h^(2,0,0)(x,y,z))

piotr: grad(div((f(x,y,z), g(x,y,z), h(x,y,z)))−Δ((f(x,y,z), g(x,y,z), h(x,y,z)) = ( −f^(0,0,2)(x,y,z)−f^(0,2,0)(x,y,z)+g^(1,1,0)(x,y,z)+h^(1,0,1)(x,y,z), f^(1,1,0)(x,y,z)−g^(0,0,2)(x,y,z)−g^(2,0,0)(x,y,z)+h^(0,1,1)(x,y,z), f^(1,0,1)(x,y,z)+g^(0,1,1)(x,y,z)−h^(0,2,0)(x,y,z)−h^(2,0,0)(x,y,z) )