# Intriguing Product of Cosines

Just recently I came across a question on Quora Indonesia (so it’s in Indonesian, which is obvious) about simplifying this expression:

\begin{aligned}\prod_{a=1}^{2015} \prod_{b=1}^{2015} \left(1 + \exp\left(\frac{2\pi i ab}{2015}\right)\right).\end{aligned}

I already wrote my answer there, in Indonesian. That is not as simple as I would expect, but it definitely is interesting enough to force me to write this post.

I was thinking, what if I replace that $2015$ by a general positive integer $n$? In other words, what can we say about

\begin{aligned} \alpha(n) = \prod_{a=1}^{n} \prod_{b=1}^{n} \left(1 + \exp\left(\frac{2\pi i ab}{n}\right)\right)?\end{aligned}

As in what I wrote as an answer in that Quora post, let’s denote $1 + \exp\left(\dfrac{2\pi i ab}{n}\right)$ by $\Lambda(a,b)$. Every pair of integers $(a,b) \in \{1,2,\ldots,n\}^2$ will be associated with a complex number $\Lambda(a,b)$.

First, it is quite trivial if $n = 1$ or $n$ is even. If $n = 1$, the above expression just boils down to $2$. Now, if $n = 2k$ for some $k$, then $\Lambda(1,k) = 0$, so the whole product will just be $0$. We shall assume from this point that $n$ is an odd number greater than $1$.

Now, if either $a$ or $b$ is $n$, then $\Lambda(a,b) = 2$. Since there are $2n-1$ pairs of $(a,b)$ where at least one of $a$ or $b$ is $n$, then we can say that

\begin{aligned} \alpha(n) = 2^{2n-1} \prod_{a=1}^{n-1} \prod_{b=1}^{n-1} \Lambda(a,b).\end{aligned}

It is the last product that we should be interested at. Before anything, we shall make a couple of observations about the symmetry of $\Lambda(a,b)$.

1. For any $a$ and $b$, $\Lambda(a,b) = \Lambda(n-a,n-b),$ and
2. For any $a$, $\Lambda(a,b)\Lambda(a,n-b) = 4\cos^2\left(\dfrac{\pi ab}{n}\right)$ for any $b$.

These two observations enable us to write

\begin{aligned} \alpha(n) &= 2^{2n-1} \prod_{a=1}^{\frac{n-1}{2}} \left(\prod_{b=1}^{n-1} \Lambda(a,b)\right)^2 \\ &= 2^{2n-1} \prod_{a=1}^{\frac{n-1}{2}} \left(\prod_{b=1}^{\frac{n-1}{2}} 4\cos^2\left(\frac{\pi ab}{n}\right)\right)^2\\ &= 2^{2n-1} \prod_{a=1}^{\frac{n-1}{2}} \left(\prod_{b=1}^{\frac{n-1}{2}} 2\cos\left(\frac{\pi ab}{n}\right)\right)^4\\ &= 2^{2n-1} \times 2^{(n-1)^2}\prod_{a=1}^{\frac{n-1}{2}} \left(\prod_{b=1}^{\frac{n-1}{2}} \cos\left(\frac{\pi ab}{n}\right)\right)^4 \\ &= 2^{n^2} \prod_{a=1}^{\frac{n-1}{2}} \left(\prod_{b=1}^{\frac{n-1}{2}} \cos\left(\frac{\pi ab}{n}\right)\right)^4. \end{aligned}

By now I hope it is already clear why I wrote ‘product of cosines’ in the title. $\alpha(n)$ is, after all, product of cosines. For simplicity, let us define, for every $a$ where $1 \le a \le \dfrac{n-1}{2},$

\begin{aligned}\mathcal{C}(a) = \prod_{b=1}^{\frac{n-1}{2}} \cos\left(\frac{\pi ab}{n}\right). \end{aligned}

Our function $\alpha(n)$ is therefore

\begin{aligned} \alpha(n) = 2^{n^2} \prod_{a=1}^{\frac{n-1}{2}} \mathcal{C}(a)^4. \end{aligned}

Now, an observation can be made if $n$ is prime. Define

\begin{aligned}\mathcal{S}(a) = \prod_{b=1}^{\frac{n-1}{2}} \sin\left(\frac{\pi ab}{n}\right), \end{aligned}

and observe that

\begin{aligned}\mathcal{C}(a)\mathcal{S}(a) &= \prod_{b=1}^{\frac{n-1}{2}} \cos\left(\frac{\pi ab}{n}\right)\sin\left(\frac{\pi ab}{n}\right) \\ &= \prod_{b=1}^{\frac{n-1}{2}} \frac{1}{2}\sin\left(\frac{2\pi ab}{n}\right) \\ &= \frac{1}{2^{\frac{n-1}{2}}} \times \sin\left(\frac{2\pi a}{n}\right) \times \sin\left(\frac{4\pi a}{n}\right)\times \ldots \times \sin\left(\frac{(n-3)\pi a}{n}\right) \times \sin\left(\frac{(n-1)\pi a}{n}\right) \\ &= \frac{1}{2^{\frac{n-1}{2}}}\times\sin\left(\frac{\pi a}{n}\right)\times\sin\left(\frac{2\pi a}{n}\right)\times\sin\left(\frac{3\pi a}{n}\right)\times\sin\left(\frac{(n-1)\pi a}{2n}\right) \\ &= \frac{1}{2^{\frac{n-1}{2}}}\mathcal{S}(a).\end{aligned}

We then observe that $\mathcal{S}(a) \neq 0$, since none of $\dfrac{\pi ab}{n}$ will be a multiple of $\pi$ if $n$ is prime and both $a$ and $b$ are less than $n$. Hence,

\begin{aligned} \mathcal{C}(a) = \frac{1}{2^{\frac{n-1}{2}}}\end{aligned}

for every $1 \le a \le \dfrac{n-1}{2}$, and thus for $n$ prime we conclude that

\begin{aligned}\alpha(n) = 2^{n^2} \times \left(\frac{1}{2^{\frac{n-1}{2}\times 4}}\right)^{\frac{n-1}{2}} = \frac{2^{n^2}}{2^{(n-1)^2}} = 2^{2n-1}.\end{aligned}

Well, in fact, this is what I initially thought the answer of that Quora question was going to be after trying some small odd integers $n$ to replace $2015$. Until I realised that it is going to be more complicated if $n$ is composite. In this case, there will be $1 \le a,b \le \dfrac{n-1}{2}$ such that $\dfrac{\pi ab}{n}$ is a multiple of $\pi$, and thus implying $\mathcal{S}(a) = 0$. In this case, we cannot use the same argument above to find $\mathcal{C}(a)$: we need to think of another way.

At this stage, I really should try to work on the case of $n = p_1^{k_1}p_2^{k_2}\ldots p_m^{k_m}$ but I think it will be much, much more complicated. So, at least for the time being, I will give you a case where $n = pq$ where $p$ and $q$ are two distinct odd primes.

The result above regarding $\mathcal{C}(a)$ can still be salvaged, though. It does still work for those $1 \le a \le \dfrac{n-1}{2}$ such that $\gcd(a,n) = 1$. So we need to know: how many such $a$ are there? We can use the Principle of Inclusion-Exclusion to find out.

So, out of the $\dfrac{n-1}{2}$ first positive integers, there will be $\left\lfloor\dfrac{n-1}{2p}\right\rfloor$ numbers that are divisible by $p$ and $\left\lfloor\dfrac{n-1}{2q}\right\rfloor$ numbers that are divisible by $q$. There will be no number that is divisible by both $p$ and $q$ in the first $\dfrac{n-1}{2}$ positive integers. Hence, there are

$\dfrac{n-1}{2} - \left\lfloor\dfrac{n-1}{2p}\right\rfloor - \left\lfloor\dfrac{n-1}{2q}\right\rfloor$

such $a$ that are relatively prime to $n = pq$, where $\mathcal{C}(a) = \dfrac{1}{2^{\frac{n-1}{2}}}$.

Now, how about those that share the common factor $p$ or $q$? Let’s have a look at those integers $1 \le a \le \dfrac{n-1}{2}$ where $\gcd(a,n) = p$. We can write $a = pu$ for some integer $u$. Now, in the expansion of

\begin{aligned} \mathcal{C}(a) = \prod_{b=1}^{\frac{n-1}{2}} \cos\left(\frac{\pi kb}{q}\right),\end{aligned}

there will be exactly $\left\lfloor\dfrac{n-1}{2q}\right\rfloor$ terms that are either $1$ or $-1$, coming from those $b$ that are multiples of $q$. How do we compute the product of other $\dfrac{n-1}{2} - \left\lfloor\dfrac{n-1}{2q}\right\rfloor$ terms?

Well, there’s a trick that we can use. We need to split the product into smaller products.

\begin{aligned}\mathcal{C}(a) = \pm\prod_{b=1}^{\frac{q-1}{2}}\cos\left(\frac{\pi kb}{q}\right)\times\prod_{b=\frac{q+1}{2}}^{q-1}\cos\left(\frac{\pi kb}{q}\right)\times\prod_{b=q+1}^{\frac{3q-1}{2}}\cos\left(\frac{\pi kb}{q}\right)\times\cdots\times\prod_{b=\frac{n-q}{2}+1}^{\frac{n-1}{2}}\cos\left(\frac{\pi kb}{q}\right).\end{aligned}

The plus and minus sign there is due to the product of $\cos(jk\pi)$ for all $1 \le j \le \left\lfloor\dfrac{n-1}{2q}\right\rfloor$. It won’t matter which one is the actual sign anyway, since we are interested in $\mathcal{C}(a)^4$.

Now, have a look at the first product. Since $1 \le b \le \dfrac{q-1}{2}$ and $k < q$, none of the $\dfrac{\pi kb}{q}$ will be an integer multiple of $\pi$. Therefore, we can apply the argument involving the product $\mathcal{C}(a)$ and $\mathcal{S}(a)$ as above to deduce that the first product must be $\dfrac{1}{2^{\frac{q-1}{2}}}.$

The rest of the products can be viewed exactly as the first product after you apply a suitable trigonometric identity first. For example, in the second product, you need to use $\cos(\pi k - \theta) = (-1)^k \cos\theta$.

We can then conclude that the product in every group will be $\dfrac{1}{2^{\frac{q-1}{2}}}$. Since there are $\dfrac{2}{q-1}\left(\dfrac{n-1}{2} - \left\lfloor\dfrac{n-1}{2q}\right\rfloor\right)$ smaller products altogether, we conclude that for $1 \le a \le \dfrac{n-1}{2}$ such that $\gcd(a,n) = p$,

\begin{aligned} \mathcal{C}(a) = \left(\frac{1}{2^{\frac{q-1}{2}}}\right)^{\frac{2}{q-1}\left(\frac{n-1}{2} - \left\lfloor\frac{n-1}{2q}\right\rfloor\right)} = \frac{1}{2^{\frac{n-1}{2} - \left\lfloor\frac{n-1}{2q}\right\rfloor}}. \end{aligned}

Using the same reasoning, for $1 \le a \le \dfrac{n-1}{2}$ such that $\gcd(a,n) = q$, we have that

\begin{aligned} \mathcal{C}(a) = \left(\frac{1}{2^{\frac{p-1}{2}}}\right)^{\frac{2}{p-1}\left(\frac{n-1}{2} - \left\lfloor\frac{n-1}{2p}\right\rfloor\right)} = \frac{1}{2^{\frac{n-1}{2} - \left\lfloor\frac{n-1}{2p}\right\rfloor}}. \end{aligned}

We are now ready to compute $\alpha(n)$! From the first $\dfrac{n-1}{2}$ positive integers, there are exactly $\dfrac{n-1}{2} - \left\lfloor\dfrac{n-1}{2p}\right\rfloor - \left\lfloor\dfrac{n-1}{2q}\right\rfloor$ integers whose $\mathcal{C}(a) = \dfrac{1}{2^{\frac{n-1}{2}}}$. There are $\left\lfloor\dfrac{n-1}{2p}\right\rfloor$ integers whose $\mathcal{C}(a) = \dfrac{1}{2^{\frac{n-1}{2} - \left\lfloor\frac{n-1}{2q}\right\rfloor}}$, and finally $\left\lfloor\dfrac{n-1}{2q}\right\rfloor$ integers whose $\mathcal{C}(a) = \dfrac{1}{2^{\frac{n-1}{2} - \left\lfloor\frac{n-1}{2p}\right\rfloor}}.$

Hence,

\begin{aligned} \alpha(n) &= 2^{n^2}\prod_{a=1}^{\frac{n-1}{2}} \mathcal{C}(a)^4 \\ &= 2^{n^2} \left(\dfrac{1}{2^{4 \times \frac{n-1}{2}}}\right)^{\dfrac{n-1}{2} - \left\lfloor\dfrac{n-1}{2p}\right\rfloor - \left\lfloor\dfrac{n-1}{2q}\right\rfloor} \left(\frac{1}{2^{4 \times \left(\frac{n-1}{2} - \left\lfloor\frac{n-1}{2q}\right\rfloor\right)}}\right)^{\left\lfloor\dfrac{n-1}{2p}\right\rfloor} \left(\frac{1}{2^{4 \times \left(\frac{n-1}{2} - \left\lfloor\frac{n-1}{2p}\right\rfloor\right)}}\right)^{\left\lfloor\dfrac{n-1}{2q}\right\rfloor} \\ &= 2^{2n-1+8\left\lfloor\dfrac{n-1}{2p}\right\rfloor\left\lfloor\dfrac{n-1}{2q}\right\rfloor}.\end{aligned}

You already see how complicated it is just in the case where $n$ is a product of two distinct primes. In the original Quora question, $n = 2015 = 5 \times 13 \times 31$, so there are more cases to consider: those $1 \le a \le \dfrac{n-1}{2} = 1007$ such that $\gcd(a,2015) = 1, 5, 13, 31, 65, 155, 403$. However, the idea stays the same. In this case, $\alpha(2015) = 2^{13725}$.

I might soon add the case where $n = p^k$ for some $k$. Hopefully I will have time and energy to do the general case 😛

By the way, I am quite certain there is a more elegant way to find $\alpha(n)$. You know, $\Lambda(a,b)$ is too simple for the product to be this complicated. But maybe I am wrong…

# The Power of Simple Things

Well it’s been ages since the last time I updated this blog. To be honest, I did want to write something, only if I haven’t been too busy sleeping. I am currently just doing research, reading literatures, and singing in the choir. Apparently doing research is really time-consuming, since I need to read many literatures and understand them before proceeding with the real research. Also the material I am currently studying is different with what I learned back in ITB, so I need to give more of my time to do the self-learning.

Speaking about reading literatures, I am currently self-learning one particular structure in mathematics called semisimple algebra. I won’t bother you with all the details, but it is an interesting object to study with, though. In short, an algebra $A$ is a ring which is also a vector space (so we can multiply vector with vector as well as scalar with vector). Now consider a module $U$ over $A$. Modules are basically just like vector space, where now the scalar set is no longer a field but a ring$. We call $U$ as an $A-$module. Now $U$ is called simple if its only submodules are the zero submodule and itself. We call $U$ semisimple if $U$ can be written as a direct sum of simple modules. Finally, we called an algebra $A$ to be semisimple if every non-zero module over $A$ is semisimple. What I am currently investigating is the special case when $A$ is actually a group algebra $\mathbb{C}G$ of $\mathbb{C}-$linear combination of elements in $G$, where $G$ is any finite group. Phew. That was long enough. Algebra is not really my forte, though. (I do miss integral and its friends!!) However, after spending so much time reading this particular thing, I admit I become more fascinated with algebra. To get the idea of this subject, I must have a strong understanding about group theory, but in a fancy level. At first I was quite skeptical about the idea of ‘fancy’ group theory. I didn’t have much experience dealing with groups. I did once learn it back then in my Algebra course in ITB, but what could be made complicated from just a set equipped with one single well-defined operation satisfying associativity, existence of the identity element, and existence of inverse elements? Turns out I was wrong. Like, totally wrong. Up until now I am still amazed by how rich group theory is. I read so many new things I haven’t even heard before, like derived groups, composition series, chief series, solvability; not to mention things I have already heard but I have no idea what they are, e.g. Sylow theorems, general linear groups, group action, representation theory (which has a deep connection with semisimple algebras and module theory). That is just group theory. I haven’t talked about another more complex structures which I believe to possess a richer theory. Before you stop reading because of what you just read, bear with me a little bit longer. What I really want to say from that rather abstract concepts which I believe not all of you would bother to understand is that we shouldn’t underestimate the power of simple things. Simple things can turn out to be not-so-simple and in fact they can probably involve a very serious and deeper understanding to master. Simple things are also the basic building blocks to grasp the more complicated ideas, so speaking that way, they are important. Why do I use such a nonconcrete analogue just to explain that? Well probably because I do feel impressed with all this learning process and I don’t want to wait any longer to write something LOL. Hopefully it won’t take too long time for me to update this blog after this post. Anyway, thanks for reading! # Tentang Kedewasaan Momen ulang tahun kemarin membuat saya berpikir cukup panjang mengenai kedewasaan. Saya sudah 22 tahun sekarang dan seperti harapan-harapan pada umumnya, saya diharapkan untuk makin dewasa. Nah, ini yang membuat saya berpikir. Kedewasaan itu apa ya? Apa yang saya lakukan setahun kemarin belum dewasa? Kalau begitu, kadar kedewasaan itu apa? Jujur, saya tidak tahu apa itu kedewasaan. Bersikap seperti orang dewasa? Sikap yang bagaimana? Orang dewasa juga bisa berbuat salah. Lha, banyak kasus kriminal kan juga rata-rata pelakunya orang dewasa. Jelas kalau berbuat kriminal itu bukan sesuatu yang baik, jadi tidak mungkin dijadikan teladan. Lalu sikap yang seperti apa? Ini membuat saya bingung. Saya cukup lama merenung sampai akhirnya sampai di suatu kesimpulan. Saya tidak perlu capek-capek memikirkan apa itu kedewasaan. Saya tidak perlu tahu juga apa kadar kedewasaan, karena itu bukan sesuatu yang saklek. Yang saya lakukan adalah berusaha menjalani hari ini lebih baik daripada kemarin. No thinking too much about the future. Ya merencanakan masa depan itu boleh, bagus malah. Tapi terlalu terbenam dalamnya, saya rasa tidak baik. Saya berprinsip, lakukan pekerjaan hari ini dengan baik dari kemarin. Kesusahan sehari cukup untuk sehari kok, jadi sebaiknya tidak dibawa-bawa lagi besoknya. Terdengar mudah, namun sulit. Kenapa? Karena.. You know, human. They like to overcomplicate things. Saya pun begitu. Ini yang jadi salah satu resolusi saya untuk tahun ini. Kalau dipikir-pikir lagi.. Saya malah harusnya meniru anak kecil. Mereka selalu bermain, penuh rasa ingin tahu, dan iman mereka sangatlah murni. Tuhan saja berkata kalau Kerajaan Allah itu milik anak-anak kecil (Markus 10:13-16). Ya, kadang dalam beberapa hal, menjadi ‘dewasa’ itu berarti menjadi seperti anak kecil. Membingungkan, bukan? Makanya saya putuskan tidak usah dipikir. Cukup jalani saja hidup seperti hari ini adalah hari terakhir kita hidup. Nah, menjalaninya tentu tidak asal-asalan. Jalani seperti yang Tuhan kehendaki. Sepertinya itu yang memang harus kita lakukan, bukan? Nah, kalau sekarang saya diminta mendefinisikan kedewasaan, saya akan jawab begini. Kedewasaan adalah suatu momen hidup dimana kita bisa menikmati Tuhan seutuhnya. Menjadi dewasa tidak sama dengan menjadi tua. Semakin tua tidak menjamin makin dewasa, meskipun semakin kita dewasa, pastilah kita makin tua 😛 Have a nice day, all! # 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.