Engineering Calculus Notes 440

Engineering Calculus Notes 440 - G at x says that, given 2...

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428 CHAPTER 4. MAPPINGS AND TRANSFORMATIONS Now we know that the length of a vector is less than the sum of (the absolute values of) its components, so b L ( −→ x ) b ≤ | ( L ( −→ v )) 1 | + ··· + | ( L ( −→ v )) m | m ( na max v max ) mna max b −→ x b since the length of a vector is at least as large as any of its components. 8 This proves the claim. Now, to prove the theorem, set −→ y = G ( −→ x ) −→ y 0 = G ( −→ x 0 ) and −→ y = G ( −→ x 0 + −→ x ) G ( −→ x 0 ) , that is, G ( −→ x 0 + −→ x ) = −→ y 0 + −→ y . Then the diFerentiability of F at −→ y 0 says that, given ε 1 > 0, we can ±nd δ 1 > 0 such that b△ −→ y b < δ 1 guarantees F ( −→ y ) = F ( −→ y 0 ) + DF −→ y 0 ( −→ y ) + E 1 ( −→ y ) where bE 1 ( −→ y ) b ≤ ε 1 b△ −→ y b . Similarly, the diFerentiability of
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Unformatted text preview: G at x says that, given 2 > 0, for b x b < 2 we can write G ( x + x ) = y + DG x ( x ) + E 2 ( x ) 8 Another way to get at the existence of such a number (without necessarily getting the estimate in terms of entries of [ L ]) is to note that the function f ( x ) = b L ( x ) b is continuous, and so takes its maximum on the (compact) unit sphere in R n . We leave you to work out the details (Exercise 4 )....
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This note was uploaded on 10/20/2011 for the course MAC 2311 taught by Professor All during the Spring '08 term at University of Florida.

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