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Recall Taylor’s Theorem from single variable calculus: If y = f(x) is a function with continuous derivatives up to order k + 1 (that is, f is differentiable up to order k + 1) at x 0, then f(x 0 + h) = f(x 0 ) +f (x 0 ) h + f (x 0 ) ——— h 2 +... 2! f (k) (x 0 ) + ——— h k + k! x 0 + h (x 0 + h – t) k ————— f (k+1) (t) dt k! x 0 R k (x 0, h) where R k (x 0, h) = Consider a function z = f(x,y) with continuous first and second partial derivatives at x 0 = (x 0, y 0 ). Then, Theorem 9 (page 137) tells us f is differentiable at x 0, which implies that the tangent plane is a “good approximation” of f near x 0 : f(x 0 + h) = f(x 0 ) +f x (x 0 ) f y (x 0 ) h1h2h1h2 + R 1 (x 0, h) tangent plane approximation This is Theorem 2 on page 196 with n = 2. linear approximation

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Consider a function z = f(x,y) with continuous first, second, and third partial derivatives at x 0 = (x 0, y 0 ). Then, by defining g(t) = f(x 0 + th) and applying the second order Taylor polynomial from single variable calculus (by using the chain rule), we get Theorem 3 on page 196 with n = 2 : f(x 0 + h) = f(x 0 ) +f x (x 0 ) f y (x 0 ) h1h2h1h2 + R 2 (x 0, h) linear approximation f xx (x 0 ) f yx (x 0 ) f xy (x 0 ) f yy (x 0 ) + h 1 h 2 h1h2h1h2 quadratic approximation 1 — 2 This matrix of second derivatives is called the Hessian matrix.

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Consider a function z = f(x,y) with continuous first, second, and third partial derivatives at x 0 = (x 0, y 0 ). Then, by defining g(t) = f(x 0 + th) and applying the second order Taylor polynomial from single variable calculus (by using the chain rule), we get Theorem 3 on page 196 with n = 2 : f(x 0 + h) = f(x 0 ) +f x (x 0 ) f y (x 0 ) h1h2h1h2 + R 2 (x 0, h) f xx (x 0 ) f yx (x 0 ) f xy (x 0 ) f yy (x 0 ) + h 1 h 2 h1h2h1h2 1 — 2 f x (x 0 ) h 1 + f y (x 0 ) h 2 f xx (x 0 ) h 1 2 + f xy (x 0 ) h 1 h 2 + f yy (x 0 ) h 2 2 1 — 2 1 — 2

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Find the second-order Taylor polynomial for f(x,y) = e x cos y at the point x 0 = (x 0, y 0 ) = (0,0). f(0,0) =f x =f x (0,0) = f y =f y (0,0) = f xx =f xx (0,0) = f yy =f yy (0,0) = f xy =f xy (0,0) = 1e x cos y1 – e x sin y0 e x cos y1 – e x cos y– 1 – e x sin y 0 f(x 0 + h) = f(h) = f(h 1, h 2 ) = (1)(h 1 ) +(0)(h 2 ) + 1 — (1)(h 1 ) 2 + 2 (0)(h 1 )(h 2 ) + 1 + 1 —(–1)(h 2 ) 2 2 + R 2 (0, h) 1 1 = 1 + h 1 + — h 1 2 – — h 2 2 + R 2 (0, h) 2 2 x – x 0 y – y 0 xx2x2 y2y2

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Find the second-order Taylor polynomial for f(x,y) = sin(x + 2y) at the point x 0 = (x 0, y 0 ) = (0,0). f(0,0) =f x =f x (0,0) = f y =f y (0,0) = f xx =f xx (0,0) = f yy =f yy (0,0) = f xy =f xy (0,0) = 0cos(x + 2y)1 2 cos(x + 2y)2 – sin(x + 2y)0 – 4 sin(x + 2y)0 – 2 sin(x + 2y) 0 f(x 0 + h) = f(h) = f(h 1, h 2 ) = (1)(h 1 ) +(2)(h 2 ) + 1 —(0)(h 1 ) 2 + 2 (0)(h 1 )(h 2 ) + 0 + 1 —(0)(h 2 ) 2 2 + R 2 (0, h) = h 1 + 2h 2 + R 2 (0, h) Note that the first-order and second-order Taylor polynomials are identical!

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Find the first-order and second-order Taylor approximations to f(x,y) = sin(xy) at the point x 0 = (x 0, y 0 ) = (1, /2). f(1, /2) =f x =f x (1, /2) = f y =f y (1, /2) = f xx =f xx (1, /2) = f yy =f yy (1, /2) = f xy =f xy (1, /2) = 1y cos(xy)0 x cos(xy)0 – y 2 sin(xy) – 2 /4 – x 2 sin(xy)– 1 cos(xy) – xy sin(xy) – /2 f(x 0 + h) = f(1 + h 1, /2 + h 2 ) = f(x,y) = (0)(x – 1) + (0)(y – /2) + 1 + + R 2 ((1, /2), h) 1 —(– 2 /4)(x – 1) 2 + 2 (– /2)(x – 1)(y – /2) + 1 —(–1)(y – /2) 2 2 The first-order (linear) approximation is sin(xy) The second-order (quadratic) approximation is sin(xy) 1

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Find the first-order and second-order Taylor approximations to f(x,y) = sin(xy) at the point x 0 = (x 0, y 0 ) = (1, /2). f(x 0 + h) = f(1 + h 1, /2 + h 2 ) = f(x,y) = (0)(x – 1) + (0)(y – /2) + 1 + + R 2 (0, h) 1 —(– 2 /4)(x – 1) 2 + 2 (– /2)(x – 1)(y – /2) + 1 —(–1)(y – /2) 2 2 The first-order (linear) approximation is sin(xy) The second-order (quadratic) approximation is sin(xy) 1 2 1 1 – —(x – 1) 2 – —(x – 1)(y – /2) – —(y – /2) 2 8 2 2

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Find the linear and quadratic Taylor approximations to the expression (3.98 – 1) 2 / (5.97 – 3) 2. Compare these approximations with the exact value. We shall find the linear and quadratic Taylor approximations to f(x,y) = at the point x 0 = (x 0, y 0 ) = (, ). (x – 1) 2 / (y – 3) 2 4 6 f(4,6) =f x =f x (4,6) = f y =f y (4,6) = f xx =f xx (4,6) = f yy =f yy (4,6) = f xy =f xy (4,6) = 2(x – 1) / (y – 3) 2 2/3 –2(x –1) 2 / (y –3) 3 – 2/3 2 / (y – 3) 2 2/9 6(x – 1) 2 / (y – 3) 4 2/3 –4(x – 1) / (y – 3) 3 – 4/9 f(x 0 + h) = f(4 + h 1, 6 + h 2 ) = f(x,y) = (2/3)(x – 4) +(–2/3)(y – 6) + 1 + + R 2 ((4,6), h) 1 —(2/9)(x – 4) 2 + 2 (–4/9)(x – 4)(y – 6) + 1 —(2/3)(y – 6) 2 2 1

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The linear approximation of (3.98 – 1) 2 / (5.97 – 3) 2 is 1 + (2/3)(–0.02) + (–2/3)(–0.03) = 1.00666 Find the linear and quadratic Taylor approximations to the expression (3.98 – 1) 2 / (5.97 – 3) 2. Compare these approximations with the exact value. We shall find the linear and quadratic Taylor approximations to f(x,y) = at the point x 0 = (x 0, y 0 ) = (, ). (x – 1) 2 / (y – 3) 2 4 6 f(x 0 + h) = f(4 + h 1, 6 + h 2 ) = f(x,y) = (2/3)(x – 4) +(–2/3)(y – 6) + 1 + + R 2 ((4,6), h) 1 —(2/9)(x – 4) 2 + 2 (–4/9)(x – 4)(y – 6) + 1 —(2/3)(y – 6) 2 2 The quadratic approximation of (3.98 – 1) 2 / (5.97 – 3) 2 is 1 + (2/3)(–0.02) + (–2/3)(–0.03) + (1/9)(–0.02) 2 + (–4/9)(–0.02)(–0.03) + (1/3)(–0.03) 2 = 1.00674 The “exact” value is 1.00675.

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The Second Order Taylor Formula is really just the terms in a power series expansion up to degree two. If we know the power series expansion, then we can extract the terms that would be in the Second Order (or any order!) Taylor Formula. For example, we know that for –1 < r < 1, 1 + r + r 2 + r 3 + r 4 +... = 1 —— 1 – r

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Consider the function 1 f(x,y) =———— =———— 1 – x – y 2 1 – (x + y 2 ) We can obtain the linear and quadratic approximations (or any other order approximations!) from the fact that 1 f(x,y) = ———— = 1 – (x + y 2 ) for –1 < x + y 2 < 1 The first order Taylor formula is f(x,y) 1 + x The second order Taylor formula is f(x,y) 1 + x + x 2 + y 2 The third order Taylor formula is f(x,y) 1 + x + x 2 + x 3 + y 2 + 2xy 2 1 + (x + y 2 ) + (x + y 2 ) 2 + (x + y 2 ) 3 + ….

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