Farewell, Bandung.

For me, farewell is scary. When someone leaves, there is a slight possibility that you can never see them again. When you leave and somehow you make it to come back, no guarantee everything will stay the same as before. A little fact: it won’t.

I am not a pro at handling farewell. Here’s a little story for you: just last week I left Bandung. My precious four and a half years life in Bandung sure has been very memorable. I met inspiring people there. I met new families there. I gained my precious best friends there. I learned a lot about myself there. I learned a lot about useful life skills there. I graduated there. I did what I like there. So many things to tell, I am getting confused which one I should mention. I was so sad when my train finally departed. I was so sad when I had to say goodbye to my remaining friends in Bandung.

The moment I realised that I will leave Bandung I knew that I will miss this city very much. No, more importantly, I will miss those who have made me miss this city so much. When I’m back later in the future, I know I shouldn’t hope to meet my friends there. They will probably be scattered all around the world by then. Bandung will still be lovable, but it certainly will be different.

Well, I guess it is the people in it that are more valuable to me. I may not be a member of HIMATIKA ITB anymore, but I still come to it when I have spare time. I may not be a member of PSM-ITB anymore, but I still come to FPS-IICC last week. It was because of the people in it. I might forget what I did back then when I was still an active member (in fact, I already forget it by now) but the moment I spent with my friends certainly cannot be forgotten. In fact, I still can laugh if I remember those good moments.

How do you handle farewell? For me.. I think time will heal. I am still in the process of making myself occupied. I do hope that I can accept the fact that I have left Bandung to embrace my awaiting new life ahead. I will make new friends there, I will create memorable moments with them there, i will learn a lot about my calling there, so when my time there has ended, I have many good moments to be cherished.

I think I have learned a new lesson: appreciate everything when you have time to do it. You cannot stay in your current place forever. There must be a time when you have to move, out of your comfort zone. When everything seems to be taken away from you, you will know that the moment you create will still remain in your heart. Your friends may leave to pursue their own dreams, but the moments spent with them will always be in your mind.

Yes, farewell is still scary for me. Some people can really be good at handling farewell. But I am not one of them. Life is somehow fair and unfair. A meeting shall be followed by a farewell. Balance. Action – reaction. Newton’s third law. Well, the unfair part is that I have to go through a painful way to deal with it. I am still coping with that, though. But that doesn’t mean that I won’t meet anyone else ever. I am still going to meet new friends and hopefully when the moment to say Adieu comes again, I won’t be as scared as I am now 🙂

To everyone who knows me and live (or used to live) in Bandung, thank you for the moment I can spend with you.

What’s New?

I’ve been away for such long time. Everytime I want to write something or when a bunch of ideas popped up, either the timing is not good or I have been procrastinating :p

I have been offered to study at UNSW starting from next March. I will take mathematics again (well since I am so faithful), only this time I will undergo a full research instead of taking courses. Sometimes it is called master by research. For two years I will do a research about harmonic analysis on Lie group. For me, that is interesting!

This semester has been quite memorable and different than previous semesters. As a fresh graduate, it means that I do not have to attend class anymore, which is.. quite weird. Sometimes I really want to attend classes, take the exam, and get graded. Although as a lecturer assistant I could get a chance to attend class informally, it still felt different.

Next difference is that I rarely see my friends now. Most of my friends have been graduated and now they have chosen their own path. Most of them are working now. Some of them choose to continue to master study like me. I still meet some of my friends from my batch, sometimes. It is kind of relieving. Being surrounded by juniors makes you feel so old, you know :p

Another difference is that this time I have a chance to be a tutor. You know, in ITB there are Calculus course for first year students. Since they are so many (around 3000 of them), almost all of the lecturer from Department of Mathematics have to teach Calculus. Well, there is 2-hour session per week for tutorial. My job was to assist those first year students to understand the concepts by doing some practices. I even had a chance to teach as a substitute. I have been accustomed to teach informally, but to teach formally in front of 60-70 students is of course something different. Luckily I managed to do it although there may be some minor flaws.

Well, since this semester almost ends, it means that I have to move out from Bandung nearly soon. Maybe at the end of this month, maybe the month after. Since I have to start my master study on March, it means I have to depart to Sydney on February. Time does fly. I have to take care of my student visa, prepare the ticket to Sydney, and of course search for apartments in Sydney. Sydney is a very expensive city. The standard rate for student apartment in Sydney is around $250-300 per week, which is around 2.5-3 million rupiahs. It is per week, note that. I have to learn to spend my money wisely :p

That’s about it. Now if you’ll excuse me, I have a Doraemon movie to be watched with my friends at 9.30 pm. Cannot wait to watch it! 😀

The Art of Doing Research

I write this post especially to share my experience in doing research. Now that I have graduated from ITB already, I continue my past thesis project as undergraduate student to a whole new level. I worked in a much more general situation than previous research project and it is kind of abstract, especially to those who are not familiar with my field, harmonic analysis.

So, basically my past thesis project is about strong and weak classical Morrey spaces \mathcal{M}^p_q(\mathbb{R}^n), the set of all measurable function f where \|f\|_{\mathcal{M}^p_q} < \infty, where

\|f\|_{\mathcal{M}^p_q} = \sup\limits_{\substack{a \in \mathbb{R}^n \\ r > 0}} |B(a,r)|^{\frac{1}{p} - \frac{1}{q}} \left(\int_{B(a,r)} |f(x)|^q\ dx\right)^{\frac{1}{q}}.

My main objective was to prove that strong Morrey spaces are contained in its weak spaces w\mathcal{M}^p_q(\mathbb{R}^n), and the inclusion is proper. I studied the structure of the spaces and the behaviour of some nice integral operators defined on it.

To learn something new like that, I spent most of my time reading the papers my supervisor gave me and conducted a little experiment. Do not imagine this experiment conducted in a laboratory. There is no (physical) laboratory at all! My research kit is just a stack of paper and a nice pen. Sometimes I need good food and good music. That’s all. I can just do my research in a restaurant or in a cafe or in my own room or even in the toilet (although that’d be unwise). I spent most of my time imagining, since there is no way I can touch, feel, or see the Morrey spaces. Very cheap yet very abstract.

Turns out, with high effort and lots of luck, I could prove what I was supposed to do. Up until this point, I am still amazed by that miracle.

Now I am working with the generalized version, i.e. generalized Morrey spaces \mathcal{M}^p_{\phi}. You might not want to see the definition. It is rather disgusting lol. But as the name suggests, it is a generalized version of the classical Morrey spaces. With certain \phi, we can have \mathcal{M}^p_{\phi} = \mathcal{M}^p_q. What I want to show now is that two norms in two different Morrey spaces are not equivalent. To prove that, I have to find a certain function that satisfies my hypotheses. I am spending much of my time now just to find a single function.

Now that I have been stuck for at least a month, I learn that

  1. Research needs patience and perseverance. You have to know that you will fail most of the times. At first, it seems frustrating. But hey, successful people must learn how to fail. As long as the number of failure is not infinity, I should be okay.
  2. Not many people will be able to help you. Unlike undergraduate or graduate courses which may be taken by students at the same time, research is way more personal. You should be the one who know the most about your object of study, among with other fellow researchers working in the same field.
  3. Ideas can come at an unexpected time. Sometimes when I just happen to have some ideas or just think of some approach, I immediately jot it down in a piece of paper. I fear that the ideas will never come to me again lol. Some people even say that if you do not dream about it, you haven’t done your research well. Lucky I dreamed about it once.
  4. The feeling of successfully proving something you want to prove is just indescribable. I can smile all day and the day after if that happens. On the contrary, I can be very gloomy if for a month or so, I still couldn’t get the correct approach of solving my problem. Yes, I am referring to what I feel now.
  5. You just cannot stop doing research, once you begin. For someone who is at times can be very curious (to the point where it’s kind of annoying) like me, being able to know more only shows that we actually do not know anything. That is why once you do research, you know what you know and you do not know, and since you do not know that you do not know, you want to know more. That happens all the time.
  6. Good food and good music are good catalysts. Well, since research in pure mathematics only requires paper and pen, of course I have to maintain my mood. That is, by eating a lot of delicious food. And singing. And traveling. And writing like this. This is just my way to regain my mood of doing research. I do hope after writing this and finally going back sleeping, ideas can come to me.

The more I do research, the more I am sure that this is my world. I just love doing research.

Covering Rational Numbers with Arbitrarily Small Intervals

Take any arbitrary \epsilon > 0. There is a collection of intervals \{I_n\}_{n \in \mathbb{N}} so that

\begin{aligned}\mathbb{Q} \subseteq \bigcup_{n = 1}^{\infty} I_n\end{aligned}

and the sum of their length never exceeds \epsilon.

To see this, for any \epsilon > 0, take any positive sequence (a_n) converging to \epsilon. For example, a_n = \dfrac{\epsilon}{2^n} will do. Since rational numbers are countably many, we can cover every rational number r_n with open interval I_n centered at r_n with \ell(I_n) = a_n.

This is used to show that rational numbers, in the universe of real numbers, have measure zero, since we can control how small our \epsilon is going to be.

Hampir Tidak Ada Bilangan Rasional, Tapi…

Sebelum mulai, saya mau minta kalian pilih satu buah bilangan apa saja yang terletak di antara 0 dan 1 (ini berarti 0 dan 1 tidak boleh dipilih ya). Kalau sudah, diingat-ingat saja. Gak akan saya apa-apain kok.

Kenapa saya minta seperti itu? Kalau saya minta sebutkan satu bilangan di antara 0 dan 1, pasti kebanyakan akan jawab \frac{1}{2}. Mungkin ada beberapa yang kreatif jawabnya \frac{5}{6}. Lebih kreatif lagi kalau jawabnya \frac{354566}{583948}. Tapi jauh lebih kreatif orang yang jawab \frac{1}{2}\sqrt{2} atau \frac{\pi}{4} atau \log(2). Nah, seberapa kreatifkah kalian? 😛

Faktanya, hampir setiap orang yang diminta untuk berikan satu angka di antara 0 dan 1, pasti akan memberikan jawaban berupa bilangan rasional. Mungkin karena bilangan-bilangan itu yang dekat dengan mereka. Mungkin cuma anak matematika yang akan jawab \frac{\pi}{4} atau \ln(2), karena mereka lebih sering berurusan dengan bilangan-bilangan semacam itu ketimbang anak-anak lainnya.

Sekarang, saya mau berikan suatu fakta yang mungkin agak ironis: di kumpulan bilangan real, hampir tidak ada bilangan rasional. Artinya, jumlah bilangan irasional jauh, jauh, jauh lebih banyak daripada bilangan rasional. Lebih mengejutkan lagi, dengan menganggap semua bilangan di antara 0 dan 1 terdistribusi secara merata, kalau kita harus mengambil satu angka dari antara 0 dan 1, katakanlah a \in (0,1), maka peluang bahwa a bilangan rasional adalah 0. Dengan kata lain, peluang bahwa a bilangan irasional adalah 1.

Buktinya tidaklah sederhana. Bahkan tampak aneh dan tidak wajar ya? Jelas-jelas dari interval (0,1) bilangan rasional ada banyak. Ambil saja \frac{1}{2}, \frac{1}{4}, \frac{1}{6}, \ldots yang belum lagi ditambah \frac{1}{3}, \frac{1}{5}, \frac{1}{7}, \ldots. Itu baru yang bentuknya \frac{1}{n}, belum ditambah dengan yang bentuknya lebih ‘tidak karuan’ seperti \frac{11}{5476} dan lain sebagainya. Mau ditulis semua juga tidak mungkin saking banyaknya, kan? Bagaimana mungkin kalau kita mengambil satu bilangan dari antara 0 dan 1, peluang terambil bilangan rasional adalah 0?

Penjelasan di bawah ini sudah masuk ke ranah yang lebih matematis. Coba dibaca saja, tidak harus terlalu dipahami. Buat yang memang tidak bisa paham, percaya sajalah :D. Mungkin bisa langsung lompat ke dua paragraf terakhir.

Untuk menjelaskan ini, kita butuh suatu konsep yang disebut ukuran (measure). Secara kasar, namanya juga ukuran, pasti ia dibutuhkan untuk memberikan gambaran seberapa besar suatu objek. Nah dalam konteks ini, predikat ukuran melekat pada himpunan. Jadi, kita bisa mengukur seberapa besar suatu himpunan.

Secara intuitif, kalau kita punya selang I = (0,1), ukuran himpunan ini adalah panjangnya kan? Jadi ukurannya 1. Sekarang, kalau kita punya persegi panjang di bidang berdimensi dua R = \{(x,y) \in \mathbb{R}^2 \mid 0 \le x \le 2, 0 \le y \le 3\}, ukuran himpunan ini secara intuitif adalah luasnya, yaitu 6. Kalau kita punya kubus di ruang berdimensi tiga Q = \{(x,y,z) \in \mathbb{R}^3 \mid 0 \le x \le 2, 0 \le y \le 2, 0 \le z \le 2\}, ukuran himpunan ini secara intuitif adalah volumenya, yaitu 8. Ukuran memang dipakai untuk memperumum konsep panjang (di ruang berdimensi satu), luas (di ruang berdimensi dua), dan volume (di ruang berdimensi tiga). Memperumum di sini maksudnya apa? Maksudnya adalah kalau kita ingin bekerja di ruang berdimensi n, kita tetap bisa punya suatu alat untuk mengukur seberapa besar suatu himpunan.

Ada banyak jenis ukuran, namun yang populer salah satunya adalah ukuran Lebesgue. Sekarang, sedikit lebih formal. Misalkan m adalah ukuran Lebesgue, maka m adalah suatu pemetaan yang membawa himpunan tak kosong ke suatu bilangan real non-negatif yang diperluas \mathbb{R}_+^{\star} = \mathbb{R} \cup \{\infty\}, yang kelak dinamakan ukuran himpunan ini. Dari penjelasan sebelumnya, kita memperbolehkan ukuran suatu himpunan tak berhingga.

Sekarang, mari bicara dalam domain bilangan real \mathbb{R} saja. Jika I = (a,b), definisikan m(I) = b-a. Sekarang, bagaimana untuk kelas himpunan yang lebih besar? Sekarang, coba pandang himpunan (0,1) \cup (2,3). Masuk akal kan kalau kita mendefinisikan m((0,1) \cup (2,3)) = m((0,1)) + m((2,3)) = 1 + 1 = 2? Ini sudah membuat lebih banyak himpunan bisa diukur. Sekarang, berhingga banyaknya himpunan-himpunan yang saling disjoin (yaitu tidak memiliki irisan) bisa dihitung ukurannya.

Secara umum, jika a_1 < b_1 < a_2 < b_2 < \ldots < a_n < b_n, maka

m\left(\displaystyle\bigcup_{i = 1}^n (a_i, b_i)\right) = \displaystyle\sum_{i = 1}^n m((a_i, b_i)).

Bahkan, ini kita bisa perumum lagi untuk kasus terhitung banyaknya himpunan-himpunan yang saling disjoin, yaitu jika a_1 < b_1 < a_2 < b_2 < \ldots < a_n < b_n < \ldots, maka

m\left(\displaystyle\bigcup_{i = 1}^{\infty} (a_i, b_i)\right) = \displaystyle\sum_{i = 1}^{\infty} m((a_i, b_i)).

Ekspresi terakhir tidaklah masalah karena ukuran suatu himpunan boleh tak berhingga.

Sekarang, coba kita formalkan. Jika E suatu himpunan terukur (yaitu yang ukurannya terdefinisi dengan baik), maka

  1. m(E) \ge 0.
  2. Jika E_1, E_2, \ldots adalah subhimpunan dari E yang saling disjoin, maka m\left(\displaystyle\bigcup_{i = 1}^{\infty} E_i\right) = \displaystyle\sum_{i = 1}^{\infty} m(E_i).

Dari dua sifat di atas kita bisa dapatkan fakta ini.

Fakta: Jika A dan B adalah dua himpunan terukur dan A \subset B, maka m(A) \le m(B). Ini bisa didapat dengan menggunakan fakta bahwa B = A \cup (B\setminus A), dan karena A dan B\setminus A adalah dua himpunan yang saling disjoin, maka m(B) = m(A) + m(B\setminus A) \ge m(A).

Sekarang, kita ingin menghitung ukuran suatu singleton, yaitu himpunan yang terdiri dari satu anggota. Misalkan A = \{a\} dan B = (a - \frac{\epsilon}{2}, a + \frac{\epsilon}{2}), untuk suatu \epsilon > 0. Jelas A \subset B dan kita tahu bahwa m(B) = \epsilon.

Karenanya, 0 \le m(A) \le m(B) = \epsilon. Tapi kita bisa pilih \epsilon yang sekecil-kecilnya, sehingga ini memberikan m(A) = 0. Jadi ukuran dari suatu himpunan singleton adalah 0.

Sekarang, perhatikan himpunan bilangan rasional \mathbb{Q}. Karena \mathbb{Q} terhitung, maka kita bisa enumerasi \mathbb{Q} = \{q_i \mid i \in \mathbb{N}\}. Karenanya,

m(\mathbb{Q}) = m\left(\displaystyle\bigcup_{i = 1}^{\infty} \{q_i\}\right) = \displaystyle\sum_{i = 1}^{\infty} m(\{q_i\}) = \displaystyle\sum_{i=1}^{\infty} 0 = 0.

Jadi, ukuran bilangan rasional adalah nol! Kalau kita batasi diri bekerja pada bilangan rasional yang terletak di antara 0 dan 1, yaitu \mathbb{Q} \cap (0,1), maka dengan menggunakan fakta di atas bisa didapatkan m(\mathbb{Q} \cap (0,1)) = 0.

Apa artinya ukuran suatu himpunan 0? Mengingat bahwa ukuran adalah perumuman konsep panjang, luas, atau volume, kita bisa bilang kalau himpunan yang berukuran 0 pada dasarnya dapat diabaikan; hampir tidak ada. Kenapa hampir? Ya karena sebenarnya mereka ada. Kita bekerja dengan bilangan rasional dari SD, kok. 😛

Kembali ke permasalahan di atas. Dari penjelasan secara matematis di atas (semoga tidak terlalu sulit untuk dipahami ya 😦 ) dapat disimpulkan bahwa bilangan rasional hampir tidak ada. Kalau kita kaitkan dengan peluang (yang sebenarnya juga adalah ukuran), kita bisa bilang bahwa peluang mengambil bilangan rasional dari antara 0 dan 1 adalah 0, karena ukuran bilangan rasional adalah 0. Artinya, kalau kita harus mengambil satu bilangan yang terdistribusi merata dari antara 0 dan 1, maka pasti bilangan tersebut adalah bilangan irasional.

Mengingat hampir setiap orang yang diminta untuk mengambil suatu bilangan dari antara 0 dan 1 akan menjawab dengan bilangan rasional, maka sebenarnya ini sangat ironis, kan?

‘Paradoks’ Diagonal

Banyak orang yang saya kenal cenderung memanfaatkan gambar untuk membuktikan suatu pernyataan matematika tertentu. Padahal dalam membuktikan suatu kebenaran matematika, gambar tidak bisa digunakan sebagai acuan utama. Gambar hanyalah alat bantu. Apa yang tampak di gambar belumlah tentu seperti yang terjadi sebenarnya.

Untuk mengilustrasikan hal ini, saya mau menulis tentang apa yang orang sebut sebagai Paradoks Diagonal. Sebenarnya ini bukan suatu paradoks, karena tidak ada pernyataan yang saling bertolak belakang sebagai implikasi dari pernyataan ini. Saya lebih suka menyebutnya ‘Paradoks’ Diagonal. *ketahuan gak kreatif ya* :p

Di sebuah persegi satuan, kita tahu panjang diagonalnya adalah \sqrt{2}. Namun, kalau kita buat ‘tangga’ sepanjang diagonal, yaitu kumpulan garis horizontal dan vertikal sepanjang diagonal, lama kelamaan ia akan semakin mirip dengan diagonal. Namun, panjang total tangga ini adalah panjang total garis horizontal + panjang total garis vertikal yang adalah 2. Jadi, apakah ini berarti 2 = \sqrt{2}?

Ilustrasi:

diagonal paradoxKita harus berhati-hati ketika mengambil limit di sini. Fungsi l \colon A \rightarrow \mathbb{R} di mana A adalah himpunan semua kurva dengan panjang berhingga dan l(C) menyatakan panjang kurva C bukan fungsi yang kontinu. Dengan kata lain, jika kita punya barisan kurva \{C_n\} yang konvergen ke kurva C, maka belum tentu l(C_n) juga konvergen ke l(C).

Untuk melihat fakta tersebut, misalkan pada selang (a,b) kita punya barisan fungsi \{f_n\} yang konvergen ke g, yaitu

f_n(x) \rightarrow g(x) saat n \rightarrow \infty, \forall x \in (a,b).

Nah, untuk setiap n \in \mathbb{N}, dari pelajaran Kalkulus kita tahu bahwa

l(f_n) = \displaystyle\int_a^b \sqrt{f_n'(x)^2 + 1}\ dx dan l(g) = \displaystyle\int_a^b \sqrt{g'(x)^2 + 1}\ dx.

Dalam kasus anak tangga ini, f_n'(x) hanya bernilai 0, \infty, dan bahkan tidak terdefinisi di titik tertentu, padahal g'(x) = 1. Jadi, dapat disimpulkan l(f_n) tidak konvergen ke l(g).