Engineering Calculus Notes 367

# Engineering Calculus Notes 367 - −→ x = −→ a lim...

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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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