Engineering Calculus Notes 440

Engineering Calculus Notes 440 - G at −→ x says that...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
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
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: G at −→ x says that, given ε 2 > 0, for b△ −→ x b < δ 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 )....
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online