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.

✲ ✻

P Q R O x y

(a) On the straight line from P to Q, y remains constant, so f(x, y) behaves as a function of x only.

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By Taylor’s theorem for one independent variable, f(x + h, y) = f(x, y) + fx(x, y) + h2 2! fxx(x, y) + . . . Notes: (i) fx(x, y) and fxx(x, y) mean ∂f

∂x and ∂2f ∂x2,

respectively (ii) In abbreviated notation, f(Q) = f(P) + hfx(P) + h2 2! fxx(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) + kfx(x + h, y) + k2 2! fxx(x + h, y) + . . . Note: In abbreviated notation, f(R) = f(Q) + kfy(Q) + k2 2! fyy(Q) + . .

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(c) From the result in (a), fy(Q) = fy(P) + hfyx(P) + h2 2! fyxx(P) + . . . and fyy(Q) = fyy(P) + hfyyx(P) + h2 2! fyyxx(Q) + . . . (d) Substituting into (b) gives f(R) = f(P) + hfx(P) + kfy(P)+ 1 2!

• h2fxx(P) + 2hkfyx(P) + k2fyy(P)
• + . .

It may be shown that the complete result can be written as f(x + h, y + k) = f(x, y) +

   h ∂

∂x + k ∂ ∂y

    f(x, y)+

1 2!

   h ∂

∂x + k ∂ ∂y

   

2

f(x, y)+ 1 3!

   h ∂

∂x + k ∂ ∂y

   

3

f(x, y) + . . .

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Notes: (i) The equivalent result for a function of three variables is f(x + h, y + k, z + l) = f(x, y, z) +

   h ∂

∂x + k ∂ ∂y + l ∂ ∂z

    f(x, y, z)+

1 2!

   h ∂

∂x + k ∂ ∂y + l ∂ ∂z

   

2

f(x, y, z)+ 1 3!

   h ∂

∂x + k ∂ ∂y + l ∂ ∂z

   

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f(x, y, z) + . . . (ii) Alternative versions of Taylor’s theorem may be ob- tained by interchanging x, y, z... with h, k, l.... For example, f(x + h, y + k) = f(h, k) +

   x ∂

∂x + y ∂ ∂y

    f(h, k)+

1 2!

   x ∂

∂x + y ∂ ∂y

   

2

f(h, k)+ 1 3!

   x ∂

∂x + y ∂ ∂y

   

3

f(h, k)+ . . .

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(iii) Replacing x with x−h and y with y −k in (ii) gives f(x, y) = f(h, k) +

   (x − h) ∂

∂x + (y − k) ∂ ∂y

    f(h, k)+

1 2!

   (x − h) ∂

∂x + (y − k) ∂ ∂y

   

2

f(h, k)+ 1 3!

   (x − h) ∂

∂x + (y − k) ∂ ∂y

   

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f(h, k) + . . . 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) = f(0, 0)+

   x ∂

∂x + y ∂ ∂y

    f(0, 0)+ 1

2!

   x ∂

∂x + y ∂ ∂y

   

2

f(0, 0)+. . . This is called a “MacLaurin’s series”

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EXAMPLE Determine the Taylor series expansion of the function f

• x + 1, y + π

3

• in ascending powers of x and y when

f(x, y) ≡ sin xy, neglecting terms of degree higher than two. Solution f

 x + 1, y + π

3

  = f  1, π

3

  +    x ∂

∂x + y ∂ ∂y

    f  1, π

3

  + 1

2!

   x ∂

∂x + y ∂ ∂y

   

2

f

 1, π

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; ∂2f ∂x2 ≡ −y2 sin xy = −π2√ 3 18 at x = 1, y = π 3;

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∂2f ∂x∂y ≡ cos xy−xy sin xy = 1 2−π √ 3 6 at x = 1, y = π 3; ∂2f ∂y2 ≡ −x2 sin xy = − √ 3 2 at x = 1, y = π 3. Neglecting terms of degree higher than two, sin xy = √ 3 2 + π 6x+ 1 2y − √ 3π2 36 x2 +

    

1 2 − π √ 3 6

     xy −

√ 3 4 y2 +. . .

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