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Class Sessions
Class Time : 10:30  12:30 , 14:00  16:00
Item

Tentative Date

0 MAXIMA, Review of elementary concepts, *Calculus Useful Formulae 
Aug. 5, 2012 
1 limit of sequences 
Aug. 5, 2012

2 limit of a function 
Aug. 11, 2012

3 Complex Number and Maclaurin series, Derivative Of Simple Functions 
Aug. 19, 2012 
4 Product rule Chain rule and Higher derivatives 
Aug. 19, 2012 
5 Maximum and Minimum of a function 
Aug. 26, 2012 
6 Indefinite Integrals

Sept. 1, 2012

7 Definite Integrals 

8 Application of Definite Integrals


Summary


* Adapted from Erwin Kreyszig's Advanced Engineering Mathematics
Tutorial Materials
Time Slot : Thursday 19:00  20:00 , Sunday 16:30  17:30
Venue : ERB 1009
Instructor's and TAs' contact details
Professor/Lecturer/Instructor:


Name:

Prof. Sidharth Jaggi

Office Location:

SHB Room 706

Telephone:

26094326

Email:

jaggi@ie.cuhk.edu.hk

Other information:

Office Hours: by appointment Calendar

Teaching Assistant/Tutor:


Name:

Sheng Cai Chris TC Wong

Office Location:

SHB Room 803 SHB Room 702

Email:

cs010@ie.cuhk.edu.hk wtc012@ie.cuhk.edu.hk

Tutorial Time and Venue:

TBA

Comments (7)
sidjaggi said
at 7:54 pm on Aug 4, 2010
Hello world!
Welcome to this page. If you have questions/comments/feedback, feel free to write them in here...
ben choi said
at 12:46 am on Aug 9, 2010
Hi SID! I am Ben.
I would like to clarify whether my concept is correct after the course, and the study in these few hours.
"Consider an infinite sequence of numbers a1, a2, . . . , an, . . .. In many realworld situations, it tends to approach a constant number. We discuss rigorously the idea of a sequence converging towards a point called the limit. A number A is the limit of the sequence if the following is true: for any positive ϵ > 0, there exists an integer M (usually depends on ϵ) such that for all n > M,
a n A<ϵ "
here is my guess after reading these page :
for example: here is a sequence"a1,a2......,am,.....,an". when the sequence approach to infin, it converge to A
You give me a "ϵ",a very small number. The relationship between am,A and ϵ is am  A=ϵ
it means whatever any ϵ you give me, there will be a mth number in that sequence may satisfied the condition:
am  A=ϵ
therefore, we should prove that in any terms m in the sequence, there still be n terms such that am<an.
when the condition an  A<ϵ, then we can prove that the limit of this sequence is A.
Am i right?
thx for your answering and bearing my poor english.......
sidjaggi said
at 3:27 am on Aug 9, 2010
Thanks, Ben, for your remarks. And good intuition. Almost correct, but you need to refine it a bit. Here are some questions to think about.
1. Is it always true that in any convergent sequence there is an m such that am  A=ϵ ? why did the definition say a n A<ϵ rather than am  A=ϵ ? can you give an example where the = sign does not hold, but the < sign is correct?
2. does a convergent sequence have to be monotonic? (what is a monotonic function? look it up  http://en.wikipedia.org/wiki/Monotonic_function )
can you give an example of a convergent sequence that is not monotonic? (hint  we studied one in class.)
(just to get some more practice, how about the other way round  can you give an example of a monotonic sequence that is not convergent?)
3. after thinking about 1. and 2., think about this question. is it true that if a_n is a convergent sequence, and a_N A<ϵ for some N, than a_(N+1) A is also always less than ϵ? if so, why? if not, give an example.
4. having thought of 3., do you now see why we say "there exists an integer M (usually depends on ϵ) such that for all n > M, a n A<ϵ "
Apologies if it's frustrating for you to have questions in response to your question, but I think that thinking is the best way to learn something  of course with hints :). Also, if we gave you all the answers, where's the fun? :)
Think about it for a while. If you have answers, feel free to post. Or not  your choice. If you still have questions, feel free to post again. Or ask the tutors. Or meet me.
And to the others, feel free to try to answer the questions above, in this comment box.
Lastly, it'd be nice if you uploaded a photo to your profile, so I could see who's asking :)
Tony said
at 1:28 am on Aug 10, 2010
Hi Jaggi, I am Tony
I would like to clarify something about "limit of sequence"
According to its definition, it claims
n>∞, f(n) > L ... f(n) is a sequence { a_n }
when
for all positive real number ε, ∃N >0, ∀ n > N,  f(n)  L  < ε
=====
For example,
Use the definition to verify lim(n>∞) (sin n)/n = 0 ?
 (sin n)/n  0  < ε
 1/n  < ε
1/ ε < n [ because ∃N >0, ∀ n > N ]
therefore we let N = 1/ε [!!! Q1. Is it a valid way to set the value of N ? ]
Suppose it is okay
 sin (1/ε) / (1/ε)  <  ε   sin (1/ε)  <  ε  = ε
therefore, now we can proof lim(n>∞) (sin n)/n = 0 ...
thanks for your answering ...
sidjaggi said
at 7:08 am on Aug 10, 2010
Hi Tony,
Thanks for your question!
In answer to Question 1, yes! That's exactly what we did in class!
To get some more practice, now try proving that lim(n>∞) n does not exist.
Tony said
at 10:09 am on Aug 10, 2010
 FIRST we suppose lim(n>∞) n exists, denoted by L
that means
for all positive real number ε, ∃N >0, ∀ n > N,  f(n)  L  < ε
when n is large enough, the difference between f(n) and the limit is sufficiently small ...
===
Now we suppose f(n) = n
when n>∞ , f(n)>∞
f(n) is strictly increasing with n and the sequence is said to diverge to infinity ...
Strictly speaking,  f(n)  lim(n>∞) n  is undefined, right ?
the difference between f(n) and L becomes very large ... [ we first suppose L is a fixed and unique real number, and f(n)>∞ when n>∞ ]
and it contradicts the definition of "limit of sequence" [ ! when n is large enough, the difference between f(n) and the limit is sufficiently small ]
therefore we can say lim(n>∞) n does not exist.
sidjaggi said
at 10:34 am on Aug 10, 2010
Good! so now we've seen the power of this technique both for proving the existence of a limit, and for proving the nonexistence of such a limit.
Just to make sure our intuition is good, here are a couple more problems to think about.
1. Does the sequence (1)^n converge (if so, what's the limit/prove) or diverge (if so, prove)?
2. How about the sequence \sum_{i=1}^n 2^{n}, that is, the sequence whose n'th term is the sum 1/2 + 1/4 + 1/8 + ... + 1/(2^n)
Just mental exercises to play with :)
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