JUST THE MATHS SLIDES NUMBER 14.9 PARTIAL DIFFERENTIATION 9 - PDF document

“JUST THE MATHS” SLIDES NUMBER 14.9 PARTIAL DIFFERENTIATION 9 (Taylor’s series) for (Functions of several variables) by A.J.Hobson 14.9.1 The theory and formula

UNIT 14.9 PARTIAL DIFFERENTIATION 9 TAYLOR’S SERIES FOR FUNCTIONS OF SEVERAL VARIABLES 14.9.1 THE THEORY AND FORMULA First, we obtain a formula for f ( x + h, y + k ) in terms of f ( x, y ) and its partial derivatives. Let P,Q and R denote the points with cartesian co-ordinates, ( x, y ), ( x + h, y ) and ( x + h, y + k ), respectively. y ✻ R P Q ✲ x O (a) On the straight line from P to Q, y remains constant, so f ( x, y ) behaves as a function of x only. 1

By Taylor’s theorem for one independent variable, f ( x + h, y ) = f ( x, y ) + f x ( x, y ) + h 2 2! f xx ( x, y ) + . . . Notes: ∂x and ∂ 2 f (i) f x ( x, y ) and f xx ( x, y ) mean ∂f ∂x 2 , respectively (ii) In abbreviated notation, f (Q) = f (P) + hf x (P) + h 2 2! f xx (P) + . . (b) On the straight line from Q to R, x remains constant, so f ( x, y ) behaves as a function of y only. Hence, f ( x + h, y + k ) = f ( x + h, y ) + kf x ( x + h, y ) + k 2 2! f xx ( x + h, y ) + . . . Note: In abbreviated notation, f (R) = f (Q) + kf y (Q) + k 2 2! f yy (Q) + . . 2

(c) From the result in (a), f y (Q) = f y (P) + hf yx (P) + h 2 2! f yxx (P) + . . . and f yy (Q) = f yy (P) + hf yyx (P) + h 2 2! f yyxx (Q) + . . . (d) Substituting into (b) gives f (R) = f (P) + hf x (P) + kf y (P)+ 1 h 2 f xx (P) + 2 hkf yx (P) + k 2 f yy (P) � � + . . 2! It may be shown that the complete result can be written as   h ∂ ∂x + k ∂  f ( x + h, y + k ) = f ( x, y ) +  f ( x, y )+     ∂y 2 1   h ∂ ∂x + k ∂  f ( x, y )+     2! ∂y  3 1  h ∂  ∂x + k ∂  f ( x, y ) + . . .     3! ∂y  3

Notes: (i) The equivalent result for a function of three variables is f ( x + h, y + k, z + l ) =   h ∂ ∂x + k ∂ ∂y + l ∂  f ( x, y, z ) +  f ( x, y, z )+     ∂z 2 1  h ∂  ∂x + k ∂ ∂y + l ∂  f ( x, y, z )+     2! ∂z  3 1   h ∂ ∂x + k ∂ ∂y + l ∂  f ( x, y, z ) + . . .     3! ∂z  (ii) Alternative versions of Taylor’s theorem may be ob- tained by interchanging x, y, z... with h, k, l... . For example,  x ∂ ∂x + y ∂   f ( x + h, y + k ) = f ( h, k ) +  f ( h, k )+     ∂y 2 3 1  x ∂  ∂x + y ∂  f ( h, k )+ 1   x ∂ ∂x + y ∂  f ( h, k )+ . . .         2! ∂y 3! ∂y   4

(iii) Replacing x with x − h and y with y − k in (ii) gives  ( x − h ) ∂  ∂x + ( y − k ) ∂  f ( x, y ) = f ( h, k ) +  f ( h, k )+     ∂y 2 1  ( x − h ) ∂ ∂x + ( y − k ) ∂   f ( h, k )+     2! ∂y  3 1   ( x − h ) ∂ ∂x + ( y − k ) ∂  f ( h, k ) + . . .     3! ∂y  This is the “Taylor expansion of f ( x, y ) about the point ( a, b ) ” (iv) A special case of Taylor’s series (for two independent variables) with h = 0 and k = 0 is f ( x, y ) = 2  x ∂ ∂x + y ∂  f (0 , 0)+ 1  x ∂ ∂x + y ∂     f (0 , 0)+ f (0 , 0)+ . . .         ∂y ∂y 2!  This is called a “MacLaurin’s series” 5

EXAMPLE Determine the Taylor series expansion of the function x + 1 , y + π � � f in ascending powers of x and y when 3 f ( x, y ) ≡ sin xy, neglecting terms of degree higher than two. Solution  x + 1 , y + π  1 , π      = f  + f 3 3 2  x ∂  ∂x + y ∂   1 , π  + 1   x ∂ ∂x + y ∂   1 , π      + . . .  f f         ∂y 3 2! ∂y 3  √ The first term on the right has value 3 / 2. The partial derivatives required are as follows: ∂f ∂x ≡ y cos xy = − π 6 at x = 1 , y = π 3; ∂f ∂y ≡ x cos xy = 1 2 at x = 1 , y = π 3; ∂x 2 ≡ − y 2 sin xy = − π 2 √ ∂ 2 f 3 at x = 1 , y = π 3; 18 6

√ ∂ 2 f ∂x∂y ≡ cos xy − xy sin xy = 1 2 − π 3 at x = 1 , y = π 3; 6 √ ∂ 2 f 3 at x = 1 , y = π ∂y 2 ≡ − x 2 sin xy = − 3 . 2 Neglecting terms of degree higher than two, sin xy = √ √ √ √ 3 π 2   2 + π 3 6 x + 1 1 2 − π 3 3 36 x 2 + 4 y 2 + . . . 2 y −  xy −       6  7