Page 58 - Vector Analysis

54 CHAPTER 2. Differentiation of Functions of Several Variables

Example 2.57. Let f : R Ñ R2 be given by f (x) = (cos x, sin x). Then f (2π) ´ f (0) =
(0, 0); however, f 1(x) = (´ sin x, cos x) which cannot be a zero vector.

Example 2.58. Let f be given in Example 2.32, and U be a small neighborhood of the

curve

C = ␣(x, y) ˇ x2 + y2 = 1, x ď 0( Y ␣(x, ˘1) | 0 ď x ď 1( .
ˇ

Then

f (1, ´1) ´ f (1, 1) = 3π .
2

On the other hand,

[ ´y x ] [] 2x
(Df )(x, y)(0, ´2) = x2 + y2 + 0
x2 y2 ´2 = ´ x2 + y2

which can never be 3π since ˇ 2x ˇ ď 3 if (x, y) P U while 3π ą 3. Therefore, no point
2 ˇ x2 + y2 ˇ 2

(x, y) in U validates

()
(Df )(x, y) (1, ´1) ´ (1, 1) = f (1, ´1) ´ f (1, 1) .

Example 2.59. Suppose that U Ď Rn is an open convex set, and f : U Ñ Rm is differen-
tiable and Df (x) = 0 for all x P U. Then f is a constant; that is, for some α P Rm we have
f (x) = α for all x P U.
Reason: Since U is convex, then the Mean Value Theorem can be applied to any x, y P U
such that fi(x)´fi(y) = Dfi(ci)(x´y) = 0 (7 Dfi = 0) for i = 1, 2, ¨ ¨ ¨ , m; thus f (x) = f (y)
for any x, y P U. Let α = f (x) P Rm, then we reach the conclusion.

2.6 The Inverse Function Theorem（反函數定理）

反函數定理是用來探討一個函數的反函數是否存在的問題。只要一個函數不是一對一的，
一般來說都不能定義其反函數，例如三角函數中，正弦、餘弦及正切函數都是周期函
數，所以全域的反函數不存在。但是我們也知道有所謂的反三角函數 sin´1 （或 arcsin）,
cos´1 （或 arctan）及 tan´1（或 arctan），這是因為我們限制了原三角函數的定義域使其
在新的定義域上是一對一的（因此反函數存在）。因此，要討論一個定在某一個（大範圍
的）定義域的函數的反函數，常常我們最多只能說反函數只在某一小塊區域上存在。

如何知道一個函數在一小塊區域上的反函數存在，我們首先該問的是在定義域是一維
（或是指單變數函數）的情況下發生什麼事？由一維的反函數定理 (Theorem A.10) 我們知