Mathematical Methods for Physics and Engineering - Matematica.NET

PARTIAL DIFFERENTIATIONThus, from (5.17), we may write∂∂x =cosφ ∂∂ρ − sin φ ∂ρ ∂φ ,Now it is only a matter of writing∂ 2 f∂x = ∂ ( ) ∂f= ∂ ( ∂2 ∂x ∂x ∂x ∂x(= cos φ ∂∂ρ − sin φ ∂ρ ∂φ(= cos φ ∂ ∂∂ρ − sin φρ ∂φ)f∂∂y =sinφ ∂∂ρ + cos φρ)(cos φ ∂)(cos φ ∂g=cos 2 φ ∂2 g 2cosφ sin φ ∂g+∂ρ2 ρ 2 ∂φ+ sin2 φ ∂gρ ∂ρ + sin2 φ ∂ 2 gρ 2 ∂φ 2∂ρ − sin φ ∂ρ ∂φ∂ρ − sin φ )∂gρ ∂φ−2cosφ sin φρand a similar expression for ∂ 2 f/∂y 2 ,(∂ 2 f∂y = sin φ ∂2 ∂ρ + cos φ )(∂sin φ ∂ρ ∂φ ∂ρ + cos φρ=sin 2 φ ∂2 g 2cosφ sin φ ∂g−∂ρ2 ρ 2 ∂φ+ cos2 φ ∂gρ ∂ρ + cos2 φ ∂ 2 gρ 2 ∂φ . 2+2cosφ sin φρ∂∂φ .)g∂ 2 g∂φ∂ρ∂∂φ)g∂ 2 g∂φ∂ρWhen these two expressions are added together the change of variables is complete andwe obtain∂ 2 f∂x + ∂2 f2 ∂y = ∂2 g2 ∂ρ + 1 ∂g2 ρ ∂ρ + 1 ∂ 2 gρ 2 ∂φ . ◭ 25.7 Taylor’s theorem for many-variable functionsWe have already introduced Taylor’s theorem for a function f(x) of one variable,in section 4.6. In an analogous way, the Taylor expansion of a function f(x, y) oftwo variables is given byf(x, y) =f(x 0 ,y 0 )+ ∂f ∂f∆x +∂x ∂y ∆y+ 1 [ ∂ 2 ]f2! ∂x 2 (∆x)2 +2 ∂2 f∂x∂y ∆x∆y + ∂2 f∂y 2 (∆y)2 + ··· , (5.18)where ∆x = x − x 0 and ∆y = y − y 0 , and all the derivatives are to be evaluatedat (x 0 ,y 0 ).160