Engineering Calculus Notes 367

Engineering Calculus Notes 367 - x = a : lim x a | f ( x )...

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3.7. HIGHER DERIVATIVES 355 The linear approximation, with −→ a = (1 . 0 , 0 . 5 , 2 . 0), x = 0 . 1, y = 0 . 1 and z = 0 . 2 is T −→ a f (0 . 1 , 0 . 1 , 0 . 2) = 0 . 25 + (0 . 5)(0 . 1) + (1 . 5)( 0 . 1) + (0 . 125)( 0 . 2) = 0 . 125 . The quadratic approximation is T 2 f ( −→ a ) 0 . 1 , 0 . 1 , 0 . 2 = 0 . 25 + (0 . 5)(0 . 1) + (1 . 5)( 0 . 1) + (0 . 125)( 0 . 2) + (0 . 25)(0 . 1) 2 + (3)( 0 . 1) 2 + (3)(0 . 1)( 0 . 1) + (0 . 25)(0 . 1)( 0 . 2) + (0 . 75)( 0 . 1)( 0 . 2) = 0 . 1375 again a better approximation. These examples illustrate that the quadratic approximation, or degree two Taylor polynomial T 2 f ( −→ a ) −→ x , provides a better approximation than the linearization T −→ a f ( −→ x ). This was the expected efect, as we designed T 2 f ( −→ a ) −→ x to have contact oF order two with f ( x ) at −→ x = −→ a . Let us con±rm that this is the case. Proposition 3.7.2 (Taylor’s Theorem For f : R 3 R (degree 2)) . If f : R 3 R is C 2 ( f has continuous second-order partials), then T 2 f ( −→ a ) −→ x and f ( −→ x ) have contact of order two at
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Unformatted text preview: x = a : lim x a | f ( x ) T 2 f ( a ) x | b x a b 2 = 0 . Proof. Equation ( 3.28 ), evaluated at t = 1 and interpreted in terms oF f , says that, xing a R 3 , For any x in the domain oF f , f ( x ) = f ( a ) + d a f ( x ) + 1 2 d 2 vs f ( x ) where vs lies on the line segment From a to x . Thus, f ( x ) T 2 f ( a ) x = 1 2 ( d 2 a f ( x ) d 2 vs f ( x ) ) = 1 2 3 s i =1 3 s j =1 p 2 f x i x j ( a ) 2 f x i x j ( vs ) P x i x j...
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This note was uploaded on 10/20/2011 for the course MAC 2311 taught by Professor All during the Fall '08 term at University of Florida.

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