One full meal that I can buy with a coupon,  consists of one of the first courses, one of the second courses and two of the cooked vegetables. In the first course meal, there are three choices of pasta and I am only allowed to take one of these. In the second course meal, there are also three choices  whether you want, say beef or fish or vegetarian meal and again I am allowed to take one of these. In the set of vegetable, there are three choices and I am allowed to take two of these three choices.

The question is how many combinations of meal that I can have.

To make it more interesting, let us assume that these following situations can occur.  I am sometimes so hungry so I take the full meal. But there is another time when I do not feel like eating so I do not eat. There is also another time when I want to eat but actually I am not that hungry so I only choose the first course and one of the cooked vegetable.

Anyway, it was a nice experience for me to be able to visit ICTP. I made a lot of friends for different countries there. I gained a lot of information that is really really useful for my research.

Here is a picture taken from my cell phone while I was there.

ps. I am not sure what is the correct answer. I have had already a number in my mind and let us confirm whether my guess is correct.

Logistic map is one of discrete dynamical systems, represented in the following:

.

It is basically a recursive sequence. It looks simple but it is highly not.When the sequence generated will always be inside the interval [0,4].More specifically, when , the sequence converges to a point. When is varied, does the solution still converge to some point?

In fact, when we vary past 3 we have the situation that is best explained through the following figure:

The figure is saying when is just past 3, the sequence is no longer converging to a point. This phenomena is called a period-doubling, as the sequence now tends to alternatingly approaching two points. However, this is the code to create the above figure in Matlab.


clear; clc;
a = 3:0.005:4;  %interval of parameter
x0 = 0.5;       %initial condition
N=500;          %number of iterations
x(1) = x0;      %the first entry is the initial condition
%computation of the orbit
figure(3);hold on;
for i=1:length(a)
for j=2:N
x(j) = a(i)*x(j-1)*(1-x(j-1));
end
y = x(301:end);
plot(a(i),y);
end
hold off
clear

Okay, the fitst task is done. Let us now learn how to make a movie in Matlab. I choose the topic ‘coweb’ as an example. Coweb arises in our topic, which is a discrete dynamical system.

Here is the video

The coweb created in the above movie uses . This exactly where the chaotic dynamic occurs. Therefore, we do not see any pattern there. The iteration that I used is only 90. However, we already see that the sequence makes such a complicated figure.

The code to create such a video is the following.

clc;clear;
myu = 3.9; %initial parameter
maxiter = 90; %maximum iteration
xo = 0.4; %initial condition
aviobj = avifile ( 'logistik2.avi', 'fps', 3 );%create a file logistik2.avi
x = 0:0.01:1; %range of x data
y1 = myu.*x.*(1-x); %range of y data to graph y=myu(x-x^2)
y2 = x; %range for y data to graph y=x
figure(1);hold on; %open figure window and hold on
axis([0 1.1 0 1.1]) % set the y and x axis
plot(x,y1); %plot the graph of y=myu(x-x^2)
plot(x,y2); %plot the graph of y=x
frame = getframe ( gca ); %command 1 to store what is inside figure 1
aviobj = addframe ( aviobj, frame ); %command 2
xbaru = myu.*xo.*(1-xo);
z2 = 0:0.01:xbaru;
z1 = xo;
plot(z1,z2);
frame = getframe ( gca );
aviobj = addframe ( aviobj, frame );
fprintf('the fixed point is %f',(myu-1)/myu);
for i=2:maxiter
if(xbaru > xo)
y1 = xo:0.01:xbaru;
else
y1 = xbaru:0.01:xo;
end
y2 = xbaru;
plot(y1,y2);
frame = getframe ( gca );
aviobj = addframe ( aviobj, frame );
xo = xbaru;
xbaru = myu.*xo.*(1-xo);
if(xbaru > xo)
y1 = xo:0.01:xbaru;
else
y1 = xbaru:0.01:xo;
end
y2 = xo;
plot(y2,y1);
frame = getframe ( gca );
aviobj = addframe ( aviobj, frame );
end
hold off;
H = (myu-1)/myu;
fprintf('please press any key \n');
aviobj = close ( aviobj );
%pause;
%close all;

However, I really really want to learn how to convert this avi video produced by Matlab to a gif image format. So if anyone know how do I do this please let me know.

You and your colleagues know that your boss A’s birthday is one of the following 10 dates:

Mar 4, Mar 5, Mar 8
Jun 4, Jun 7
Sep 1, Sep 5
Dec 1, Dec 2, Dec 8
 
A told you only the month of his birthday, and told your colleague C only the day. After that, you first said: “I don’t know A’s birthday; C doesn’t know it either.” After hearing what you said, C replied: “I didn’t know A’s birthday, but now I know it.” You smiled and said: “Now I know it, too.” After looking at the 10 dates and hearing your comments, your administrative assistant wrote down A’s birthday without asking any questions. So what did the assistant write?

When I first read this puzzle, I am confused. Let me just name the reader/myself  “B” in the puzzle above to avoid more confusion. This puzzle is written in some way that the reader is part of the story. Hence the reader, B, was told the month of A’s birthday but we as the reader do not know the month of A’s birthday. Perhaps it is just a part of puzzle that we have to solve, therefore it might not be a big deal.

However, the big deal is the fact that the solution that I found is different from the solution given. I thought I might get wrong, therefore I look for the solution. What happened was I could not understand the solution. I thought it is just me but it turns out there is someone else who has the same experience.

I would like to modify a little bit of the puzzle in order to make this puzzle a bit clearer, well at least for me.

A’s birthday is one of the following 10 dates:

Mar 4, Mar 5, Mar 8
Jun 4, Jun 7
Sep 1, Sep 5
Dec 1, Dec 2, Dec 8
 
We then have the following conversations between A’s friend.
C said: “I don’t know A’s birthday, but I know only the day (the date) of A’s birthday,”
B said: “Well, I know only the month A’s birthday.”
C said: “Now I know it.”
B said: “Now I know it, too.”

From the information given above, can you tell what is A’s birthday?

Perhaps, this version is better than the previous one. What do you think?

1. Suppose is an matrix and it is diagonalizable. Therefore, by definition, there is an invertible matrix and a diagonal matrix . However, I always forget the relationship between and , whether it is or . Generally, those two are no different, but we can choose such that the column vectors of matrix are the eigenvectors of matrix , hence we have to know precisely which one it is, otherwise it gets wrong. Until now, when I forget about this, I always derive it and it takes some of my precious time (haha) but now after I have this, I can just open the Internet and look for this note.
2. Point 1 above talks about diagonalizability of a matrix. But generally, we have a linear operator acting on a general vector field instead. A linear operator is called diagonalizable if there is a basis B such that the matrix representation of this linear operator is a diagonal matrix. It seems for me, diagonalizability of a linear operator and that of a matrix are talking about two different things. One is talking about how to find a basis, while the other is talking about how to find an invertible matrix. However, these two concepts turn out to be the same.

We begin with the following theorem.

Theorem 1 (Coordinate-change matrix)
Let be a standard basis and be another basis for an -dimensional linear space . Then there is an invertible matrix such that for every

.                                                                                                          (1)

Matrix is called the transition or coordinate-change matrix from the basis to the basis . Note that and are the coordinates of vector with respect to basis and respectively. In fact, the column vectors of matrix are the coordinates of the basis vectors of with respect to the standard bases (i.e. if then ).

We won’t prove the theorem as it can be found in any linear algebra textbook. However, using this theorem we can relate the diagonalizability of a linear operator and that of a matrix.Suppose is a diagonalizable linear operator on an -dimensional linear space . Take then let be . Suppose is a standard basis in and is a basis in that consists of independent eigenvectors of .

Then we have the following.

(i) , where is a representation matrix of with respect to standard basis .
(ii) , where is a representation matrix of with respect to basis . We also know that is a diagonal matrix.

We want to show that using the theorem 1, or . From equation (1) we get:

.

Hence, we have . While from point (ii) we have . As a conclusion, we have,

.

We have derived that in fact, is the right one. We also derived that the invertible matrix has such a close relationship with the basis as the column vectors of are the basis vectors of , which are the eigenvectors of the linear operator . This conclude my note.

But I have found my way of academic recharging. And I give many thanks to ICTP and MIT. Courses that are provided by ICTP and MIT have been recorded and have been available online. I don’t know since when this happens but for me it is great. I can update my knowledge with some new information that I have never learned before.

After downloaded a video, I can watch the course that I am interested. When I don’t understand a thing, I can just pause and google them. I can pause anytime I want when some students knocking my door. I strongly recommended this way of learning to my colleagues and my students as this is much as fun as watching movie.

Here are some references that you might like.

1. http:// www.ictp.tv
2. http://ocw.mit.edu/courses/audio-video-courses/
3. http://www.youtube.com/user/MIT
4. http://www.khanacademy.org/

Here is the puzzle: There are two ropes and one lighter. Each rope has special properties.

1. If we light one end of the rope, it will take exactly one hour to completely burn out.
2. The density of the rope is not uniform, which means that burning half the rope would not take half an hour.
3. Those two ropes are not identical, they aren’t the same density nor the same length nor the same width.

The question is how do we measure exactly 45 minutes using those two ropes.

We shall not discuss the answer here, instead I would like to give some logical consequences of those rules given in the properties of the rope. Those three properties are called axioms in mathematics article. They are not to be proven. We have to believed them, we have to accept it.

The first consequence is summarized in the following lemma (which is a small relatively easy consequence of the axioms).

Lemma 1
If we burn both ends of the rope, the rope will take 30 minutes to be completely burned

This lemma is very useful and it is used in solving the puzzle. It seems obvious but we can’t just take it for granted. Anyway, here is the proof.

Proof:
Let us prove this by a contradiction. Assume it would take minutes by burning both ends of the rope until it is completely burned. Assume . Let us consider the case where .  At first () the rope is burned at both ends and after minutes the rope would be gone. However, after 30 minutes there is still a segment of the rope that hasn’t been burned yet. See the figure below.

And now, let us consider if at first the rope is burned only at one end. It would take then 30 minutes until the segment AB is completely burned and it would take another 30 minutes to completely burn the segment CD. A contradiction, as it would need more than one hour to completely burned out the rope. The case where can be proven similarly. QED

The next consequence is summarized in the theorem below. Theorem usually used to give a significant consequence of the three axioms given above. The theorem tells us that the solution of the puzzle exists.

Theorem 2
There is such a way to measure 45 minutes using only two ropes

Proof: Burn both ends of the first rope and burn one end of the second rope. According to lemma 1, it will take 30 minutes to completely burn out the first rope. The next step is to burn the other end of the second rope, therefore it will take 15 minutes to burn out the second rope if we apply lemma 1 once more to the second rope which has 30 minutes remaining rope.  QED.

The puzzle stops here, but our discussion does not. I would like to discuss if we have ropes then how many minutes we can measure accurately. It turns out this problem is an immediate consequence of the last theorem.

Corollary 3
Suppose there are ropes that satisfy the properties given in the beginning of this section, then there is such a way to measure exactly minutes.

Proof:
This can be shown by inductively applying theorem 2 times to ropes.

The next results given below are dealing with only one rope. As given in the Lemma 1, we can measure exactly 30 minutes with using only one rope. However, there are various ways how we measure exactly 30 minutes. I won’t present the proofs here as it’s just fun to figure it all ourselves.

Lemma 4
Let us take the rope that has special properties given above. Cut the rope into two pieces and burn both ends of the first piece and then burn both ends of the second piece. It takes exactly 30 minutes until the rope is completely burned.

Lemma 5
Let’s take the rope that has special properties given above. Cut the rope into pieces. Burn both ends of the first piece and then burn both ends of the second piece and so on until the last piece. It takes exactly 30 minutes until all the pieces of the rope are completely burned.

All the results above are resulted from burning the end of the rope. We are now asking ourselves what do we get if we burn the rope somewhere in the middle. Logically, the fire will spread in two directions. Depending on the density of the rope, one end of the rope will be caught by the fire first.

Lemma 6
Let’s take the rope that has special properties given above. Burn the rope at one point in the middle. Once one end of the rope is caught by the fire, burn the other end of the rope. It takes exactly 30 minutes until the rope is completely burned.

Of all results presented above, with using only one rope, we can measure exactly 30 and 60 minutes. But I would like to ask whether there is any other time that we can measure using only one rope.

Saya selalu merasa kagum atas kehebatan sebuah jendela dalam layar komputer kita, apapun nama jendelanya (entah itu internet explorer, firefox, safari, google chrome, opera, ect.). Jendela ini telah membawa kita (setidaknya saya) ke sebuah dunia yang luar biasa dimana informasi dapat ditransfer dengan hitungan milidetik. Salah seorang supervisor saya pernah berkata tentang kekagumannya terhadap internet. “Dengan internet, siapapun dapat menjadi guru sekarang ini,” katanya. Setuju, mengingat banyak sekali informasi tersedia di dalam jendela tersebut, tergantung bagaimana seseorang tersebut memanfaatkannya.

Permasalahan yang akan saya bicarakan adalah di dunia pendidikan, khususnya di dunia matematika, dan ini juga berlaku di dunia akademik manapun tentunya. Memanfaatkan jendela dalam layar komputer, saya menemukan beberapa cara mengakses informasi tentang matematika, apakah itu buku pengajaran, catatan kuliah, dan bahkan kuliahnya sekalipun dan juga diskusi gratis tersedia. Lewat tulisan ini, saya ingin mensharingkan pengalaman saya yang saya yakin cukup berharga untuk kita semua. Lewat tulisan ini juga saya ingin menyimpan list-list situs ini secara permanen, karena jika tidak, situs-situs ini hanya akan menjadi tulisan di post-it notes yang tertempel di meja saya, yang akan terbuang ketika saya membereskan meja saya akhir bulan ini. Mohon maaf jika ada kesalahan dan silahkan untuk mengomentari atau mengupdate item-item yang telah saya tuliskan.

1. Itunes

Saya harap semua orang kenal dengan program ini. Itunes adalah sebuah software yang didesain untuk mendengarkan musik, menonton film ataupun video klip. Tak hanya itu, software ini dapat digunakan ‘belanja’ lagu ataupun film, tentunya dengan pembayaran credit card. Salah satu fiture dalam itunes yang membuat saya terbantu adalah fiture iTunes U. Tidak sulit untuk mengakses fiture ini, tingga kita klik iTunes Store dan arahkan kursor anda ke pilihan iTunes U.

Isi dari fitur ini adalah kumpulah kuliah kuliah dari berbagai bidang, yang telah diunggah oleh lecturernya sendiri ataupun oleh sebuah organisasi yang biasanya merupakan sebuah universitas terbuka. Melalui fitur ini, saya bisa belajar tentang berbagai kuliah matematika yang dibawakan secara menarik oleh tiap tiap kontributornya. Saya juga dapat menikmati kuliah yang dibawakan oleh Gilbert Strang, profesor terkemuka dari MIT, dan juga menikmati seminar tentang Einstein yang dibawakan oleh Terence Tao, seorang profesor termuda yang mendapat Field Medal.

2. Khan Academy

Sharing saya yang kedua masih berupa kuliah matematika yang tersedia secara gratis di jendela komputer kita. Khan academy adalah sebuah organisasi non-profit yang bertujuan sangat mulia yaitu mencerdaskan semua orang. Pencetusnya adalah Salman Khan yang sampai saat ini telah mempunyai koleksi 2100 video pengajaran khusunya di bidang matematika.

Khan academy, selain mempunyai website sendiri, juga memanfaatkan situs youtube untuk menyebarluaskan pengajarannya. Saat ini siapa yang tidak kenal youtube. Tujuannya tak lain adalah agar semua orang dapat merasakan manfaatnya.

Sudah beberapa kali saya menikmati kuliah kuliah yang tersedia di situs ini. Saat ini kuliah kuliah yang ada mungkin lebih cenderung erlementer, namun saya yakin seiring bertambahnya waktu, akan semakin banyak kuliah-kuliah dari yang paling dasar sampe yang paling advance akan tersedia disini. Lebih lagi, kita diundang untuk menjadi salah satu kontributor dari situs ini.

3. Gigapedia

Sharing saya yang ketiga adalah tentang buku pengajaran. Sering saya mendapatkan kesulitan untuk mengakses sebuah buku matematika. Hal ini dikarenakan saya kebetulan bekerja di sebuah universitas yang budgetnya sedikit untuk matematika. Alhasil, buku pengajaran untuk matematika sangatlah jarang. Solusinya adalah mencari buku elektronik dimana kita harus mencarinya lewat situs pencari seperti Google ataupun Yahoo.

Namun kini ada berita bagus untuk kita semua pemburu buku elektronik. Silahkan anda masuk ke situs gigapedia dimana disitu kita dapat menemukan lebih dari 10 ribu buku elektronik dari berbagai bidang. WOW. Sungguh sebuah hadiah yang bagus, karena dengan adanya buku elektronik tersebut, kita dapat belajar dimana pun bahkan ketika kita sedang bepergian, tentunya jika kita membawa laptop ataupun ebook reader.

4. Physics Forum

Apakah anda familiar dengan forum internet. Banyak sekali forum internet di dunia maya ini, ada yang membahas olah raga, politik, komputer, teknologi. Nah, ini adalah sebuah forum akademik yang unik di mana disini kita dapat belajar sambil berbagi ilmu. Dari namanya, tentunya kita menyangka forum ini berbicara tentang fisika, namun ternyata tidak. Selain fisika, terdapat matematika, engineering, dan ilmu sains lainnya.

Keunikan dari forum ini adalah anda dapat bertanya tentang homework anda khusunya untuk mahasiswa. Mahasiswa modern adalah mahasiswa yang dapat mencari informasi sebanyak mungkin yang dapat membantu dirinya dalam kuliahnya. Walaupun anda dapat bertanya tentang peer kuliah anda, tetapi anda harus berusaha terlebih dahulu barulah anda akan mendapatkan response dari pertanyaan anda. Dan anda tidak akan mendapatkan jawaban 100%, karena anda hanya akan mendapatkan petunjuk untuk menjawab pertanyaan anda.

Selain pr, kita juga dapat berdiskusi tentang masalah masalah dalam bidang akademik kita. Menurut pengalaman saya memakai forum ini, setiap saya mendapatkan suatu masalah, uneg-uneg, pasti ada setidaknya satu orang di forum ini yang mengalami uneg-uneg yang sama.

5. Mathoverflow

Sama seperti situs internet sebelumnya, situs ini merupakan situs yang dapat kita gunakan untuk berdiskusi. Perbedaannya disini hanya topik-topik di sekitar matematika yang berada di level research. Sering kali ketika kita melakukan research dan harus membaca buku-buku di level ‘graduate’ yang tidak pernah kita buka sebelumnya, kita mempunyai pertanyaan dan disinilah yang akan membantu kita. Bahkan, topik-topik yang didiskusikan disini kadang-kadang merupakan sebuah ‘open question’ yang belum terjawab.

6. Wolfram, PlanetMath dan Scholarpedia

Walaupun wikipedia mempunyai banyak sekali informasi yang dapat kita baca, namun sayangnya kita tidak bisa menjadikan wikipedia sebagai rujukan informasi. Tetapi terdapat situs laiinya yang dapat dipakai untuk rujukan. Ada tiga website yang menjadi situs tersering yang sering saya kunjungi untuk mengakses informasi matematika secara cepat dan akurat.

Masih banyak situs-situs internet lainnya yang bisa membantu kita entah dalam perkuliahan, penelitian ataupun menyiapkan kuliah. Keenam situs internet diatas hanyalah sebagai contoh bahwa banyak yang dapat kita temukan dan manfaatkan dalam jendela dalam komputer kita.

Sekali lagi saya mohon maaf jika ada kesalahan dalam penulisan artikel ini. Ucapan terima kasih sebanyak-banyaknya jika ada komentar dan masukannya, atau jika anda dapat menambah situs-situs yang lain yang akan tentunya sangat berguna.

Rujukan

1. http://www.apple.com/education/itunes-u/
2. http://www.khanacademy.org/
3. http://library.nu/
4. http://www.physicsforums.com/
5. http://mathworld.wolfram.com/
6. http://planetmath.org/
7. http://www.scholarpedia.org/

The problem
A 200-galon tank is half full of distilled water. At time , a solution containing 0.5 pound/galon of salt enters the tank at the rate of 5 gal/minutes, and the well-stirred mixture is withdrawn at the rate of 3 gal/minutes.
a. Give a mathematical model representation of the above problem.
b. At what time, will the tank be full?
c. At the time the tank is full, how many pounds of salt will it contain?

The mathematical model
The model describing the above process is based on the following formula:

rate of change of the amount of salt  = (rate of salt arrived) – (rate of salt departed)

Suppose is the volume of the liquid in the tank at time . The volume of liquid inside the tank will increase as the inflow (5 gal/minutes) is bigger than the outflow (3 galon/minutes).  Thus, we have

100 galon + (5 gal/min – 3 gal/min) min,
gal,

as we know that initially the tank is half full (100 gal).

Suppose is the amount of salt inside the tank at time . The rate in of salt is as follows,

rate in (pound/min) (0.5 pound/gal) (5 gal/min) 2.5 pound/min,

while the rate out is
rate out (pound/min) outflow (gal/min)
(pound/min)
(pound/min).

Therefore, the mathematica model is given by,

.          (1)

The solution
First we determine time at which the tank will be full. We use the following equation,

,

We solve for when the volume is 200 galon,  then we obtain .

To solve the differential equation (1) we need the initial condition
as follow,

.

We are going to use Maple to solve the problem

> model := diff(y(t),t) = 2.5 - (3*y(t))/(100+2*t);
> init := y(0)=0;
> dsolve({model,init});

We obtain the solution that is given by

When , the amount of salt inside the tank will be

.

where is an matrix and are vectors.

To mention a few, the followings are the solvability conditions (equivalent conditions) such that the linear system above has a solution:

1. determinant of is not zero
2. rank of is full
3. all eigenvalues of are non zero
4. the vector columns (also rows) are independent
5. has an inverse
6. et cetera

However, I would like to discuss the case when either (#) is not a square matrix or
(##) the rank of is not full (< $latex  n$).

There is actually a condition such that the linear system has a solution whenever we have either (#) or (##). The solvability condition is

.

The thing is this condition is not applicable to most problems I have ever had. For instances, I bumped into the following problem.

Let be the following

,  and . I want to seek a condition such that the linear system is solvable for . I want to find a condition for and such that there is a solution of .

The statement did not help me at that time and I look for another implication for this statement that is applicable enough to my problem.

Fundamental Theorem of Orthogonality (see [1])
Let be any matrices.  The row space of is orthogonal to the null space of .

Proof: Suppose is a vector in the nullspace. Then . This equation can be considered as rows of multiplying :

= = .

We can clearly see that a dot product of row 1 and is zero, a dot product of row 2 and is zero, and so on. This means that the dot product of every row of and is zero, which means every row of is orthogonal to , which completes the proof.

Corrolary 1
The column space of is orthogonal to the left null space of .

Proof: The proof is easy, we just need to consider the transpose of the matrix .

Going back to our original problem, when the vector is in the column space of , the vector will be orthogonal to any vector in the left null space of .

Thus, we have the following solvability condition:

Solvability condition of for any matrices
The linear system has a solution if the dot product of and is zero, where is the base vector of the left null space of .

Remark, the above solvability condition is closely related to the linear version of Fredholm solvability condition.

Reference
[1] Strang, G., “Linear Algebra and Its Applications,” Thomson Brooks/Cole

A. If the package in the sty format

1. create a new directory under tex tree

–> sudo mkdir /usr/share/texmf-texlive/tex/latex/<packagename>

2. copy the package inside the new directory

3, run the following command

–> sudo mktexlsr

B. if the package is not the sty format

1. Download <packagename>.ins and <packagename>.dtx (you need both)

2. Run the following command

–> latex  <packagename>.ins

3. You will get the package in the sty format and you can follow instructions in A.

I hope this helps. Thanks to this page.