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

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