Engineering Calculus Notes 368

Engineering Calculus Notes 368 - 356 CHAPTER 3. REAL-VALUED...

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Unformatted text preview: 356 CHAPTER 3. REAL-VALUED FUNCTIONS: DIFFERENTIATION and so → →→ |f (− ) − T2 f (− ) − | x ax →→ − −− 2 x a 3 3 1 ≤ 2 ∂2f − ∂2f |△xi △xj | (→) − a (s) → ∂xi ∂xj ∂xi ∂xj △− 2 x i=1 j =1 ≤ |△xi △xj | n2 ∂2f − ∂2f max (→) − a (s) max → . (3.29) i,j 2 i,j ∂xi ∂xj ∂xi ∂xj △− 2 x By an argument analogous to that giving Equation (2.20) on p. 159 (Exercise 7), we can say that max i,j |△xi △xj | → 2 ≤1 △− x and by continuity of the second-order partials, for each i and j lim −− →→ x→a ∂2f − ∂2f − (→) = x (→) . a ∂xi ∂xj ∂xi ∂xj Together, these arguments show that the right-hand side of Equation (3.29) → → → goes to zero as − → − (since also s → − ), proving our claim. x a a We note in passing that higher-order “total” derivatives, and the corresponding higher-degree Taylor polynomials, can also be defined and shown to satisfy higher-order contact conditions. However, the formulation of these quantities involves more complicated multi-index formulas, and since we shall not use derivatives beyond order two in our theory, we leave these constructions and proofs to your imagination. Exercises for § 3.7 Practice problems: 1. Find ∂2f ∂2f ∂2x , ∂2y , and ∂2f ∂y∂x for each function below: (a) f (x, y ) = x2 y (b) f (x, y ) = sin x + cos y (c) f (x, y ) = x3 y + 3xy 2 (d) f (x, y ) = sin(x2 y ) (e) f (x, y ) = sin(x2 + 2y ) (f) f (x, y ) = ln(x2 y ) x+y (g) f (x, y ) = ln (x2 y + xy 2 ) (h) f (x, y ) = 2 x + y2 ...
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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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