• 沒有找到結果。

Math 2111 Advanced Calculus (I)

N/A
N/A
Protected

Academic year: 2023

Share "Math 2111 Advanced Calculus (I)"

Copied!
1
0
0

加載中.... (立即查看全文)

全文

(1)

Math 2111 Advanced Calculus (I)

Homework 4 Hand in Problems:

2, 3, 4, 6, 7 Lecture Note: 2, 3, 5

1. Let f : [a, b] → R be a continuous function. If f is not a constant function, prove that the range of f is a closed interval.

2. Suppose that S1, S2,· · · , Sn are sets in R and S =S1∪S2∪ · · · ∪Sn. Define Bi = supSi for i= 1,2,· · · , n.

(a) Show that supS= max(B1, B2,· · · , Bn).

(b) IfS is the union of an infinite collection of{Si}, find the relation between supS and {Bi}.

3. LetS1, S2 be sets inR. DefineS ={x |x=x1+x2, wherex1 ∈S2, x2 ∈S2}. Find supS in terms of supS1 and supS2.

4. LetS⊆Dbe a dense subset ofDandf :D→Rbe a continuous function. Suppose that f satisfies the property: for every >0 there existsδ =δ()>0 such that if “ x, y ∈S ” and |x−y|< δ, then

|f(x)−f(y)|< . Prove that f is uniformly continuous on D.

5. (a) Determine whether the function f(x) = sin 1 x

is uniformly continuous on (0,1].

(b) Determine whether the function f(x) =

(

xsin 1 x

, 0< x≤1

0, x= 0

is uniformly continuous on [0,1].

6. Use the Bolzano - Weierstrass Theorem to prove the Extreme Value Theorem.

7. Let f :D⊆R →Rn be a vector valued function. We say that f is uniformly continuous if for every >0 there exists δ =δ()>0 such that

kf(x)−f(y)k< whenever |x−y|< δ.

Denotef = f1,· · · , fn

. Prove thatf is uniformly continuous if and only iffi is uniformly continuous for i= 1,· · · , n.

Lecture Note:

(Page 47)

1. Problem 1.11 2. Problem 1.13 3. Problem 1.14 4. Problem 1.15

5. Problem 1.23(3)(5)(9)

參考文獻

相關文件

I discovered that there is lack of research about the topic that applying Hakkanese teaching into preschool students’learning, so it inspired me to find out Hakka parents’ attitude on

FINAL FOR GEOMETRY Date: Wednesday, June 16, 2004 Instructor: Shu-Yen Pan No credit will be given for an answer without reasoning.. i LetTbe the unit tangent vector