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Unformatted text preview: 356 CHAPTER 3. REAL-VALUED FUNCTIONS: DIFFERENTIATION
|f (− ) − T2 f (− ) − |
− −− 2
2 ∂2f −
|△xi △xj |
x i=1 j =1 ≤ |△xi △xj |
→ . (3.29)
2 i,j ∂xi ∂xj
x By an argument analogous to that giving Equation (2.20) on p. 159
(Exercise 7), we can say that
i,j |△xi △xj |
→ 2 ≤1
x and by continuity of the second-order partials, for each i and j
x→a ∂2f −
∂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.
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
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 )
(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.
- Fall '08