lecture3

# an the point la can be written as the sum j la aj

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

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: tch of Proof. Note that for h �= 0 in Rn , f (a + h) − f (a) − B h → 0 as h → 0 |h| (2.19) f (a + h) − f (a) − B h → 0 as h → 0. (2.20) implies that From this you can conclude that f is continuous at a. Remark. Let L : Rn → Rm be a linear map and a ∈ Rn . The point a can be written � as a sum a = n=1 aj ej = (a1 , . . . , an ). The point La can be written as the sum j � La = aj Lej , and L can be written out in components as L = (L1 , . . . , Lm ), where each Lj : Rn → R is a linear map. Then Lej = (L1 , ej , . . . , Lm ej ), and Li ej = �i,j . The numbers �i,j form an n × n matrix denoted by [�i,j ]. Remark. Let U ⊆ Rn , and let f1 : Rn → Rm1 and f2 : Rn → Rm2 be diﬀerentiable maps. Let m = m1 + m2 , so that Rm1 × Rm2 = Rm . Now, construct a function f : Rn → Rm deﬁned in component form by f = (f1 , f2 ). The derivative of f at a is Df (a) = (Df1 (a), Df2 (a)). (2.21) Remark. Let f : U → Rm be a map. The action of f on input x written out in component form is f (x) = (f1 (x), . . . , fm (x)). So, the map can be represented in component form as f = (f1 , . . . , fm ), where each fi as a map of the form fi : U → R. The derivative of f acting on the standard basis vector ej is Df (a)ej = (Df1 (a)ej , . . . , Dfm (a)ej ) ∂fm ∂f1 (a), . . . , (a)). =( ∂xj ∂xj So, the derivative (Df )(a) can be represented by an m × n matrix � � ∂fi (Df )(a) ∼ Jf (a) = (a) = ∂xj called the Jacobian matrix of f at a, which you probably recognize. 4 (2.22) (2.23)...
View Full Document

## This note was uploaded on 10/02/2010 for the course MAT unknown taught by Professor Unknown during the Fall '04 term at MIT.

Ask a homework question - tutors are online