Martin Geisler Online

I’ve just been to our local Føtex where I bought some chocolade for my exam tomorrow in Abstract Algebra. I’ll also take some bottles of CocaCola with me so that I can get enough energy for the brain :-)

I’ve also been at the University of Aarhus today to hear the last words of wisdom from our lecturer and to talk with our instructor. There I found out that a Rasmus Villemoes has published a full solution to the exam questions from the previous years. And a very good solution that is. I’ve been training with these questions myself for the last couple of days, and there has (unfortunately) been a couple of questions that I couldn’t answer. So it’s interesting to see how simple and elegantly you can solve most questions — I hope it’ll help me tomorrow. –Martin Geisler

No more exams! I had my last exam last Tuesday: Calculus. It went fantastic — I got the grade 13 which is the top grade in the Danish school-system!

The exam was an aural exam with 30 minutes preparation. I was so lucky that I got “The Complex Exponential Function” as my question. That was probably the easiest question among the 15 that I had to choose from. The other questions were about things like the length of a curve in Rⁿ, Fourier Series and so on — much harder questions. I don’t think I would have gotten a 13 if I had gotten one of the other questions… perhaps 10 or 11 instead (there’s no 12 in the Danish grade system).

But that doesn’t matter: I got the question and after waiting 30 minutes I told them everything they wanted to know about it. They then asked me some questions about other things, such a Fourier Series. That also went well — It’s not that I bad at handling Fourier Series, it’s just that most proofs are terribly long and boring. In the end the censor got curious: he wanted to know if the Exponential Function had an inverse function, e.g. he wanted me to talk about the complex logarithm. We haven’t seen this in our books, but I did manage to find some of the formula for z given w = e^z:

The formula is found by taking advantage of, that |e^z| = |w| and arg(e^z) = arg(w). Here we have that |e^z| = e^x and that arg(e^z) = y and the result follows easily from that.

I’ve just received a preliminary score for my written test in Calculus — I got 92 points out of 100. This is just a quick count done by my lecturer, the censors have yet to see it, so the score might shift a little. I’ll know my final score after my aural test this Tuesday — that’s just five days away! But so far it looks good :-)

I’ve been rehearsing with Jérémy since the written test six days ago — we’ve been training from 14:00 to ~19:00 each day. There’s 31 questions and only 10 days to train, so it’s been a busy week. But we’re getting there, we now have four days to discuss the last 12 subjects.

Yes!! Today I had my last written exam in this round. That means that I’ve had three of my four exams — the last one is aural Calculus (mathematical analysis).

Todays test was written Calculus. I believe it went pretty well — I answered all the questions, and with the help of my TI89 graphics calculator I was able to check most of my answers. It’s a fantastic help to have a calculator that can do symbolic manipulation — many of the calculations were quite long and boring, and the risk of making a little mistake was high. I cought myself saying “2 × 3 = 5″…

I now have about ten days to prepare for the next and final exam. I have to make an outline for 31 subjects in those ten days, so there’ll be plenty to do. But we get 30 minutes to prepare ourselves at the exam, so I don’t think it’ll be that bad — or perhaps I’m just being optimistic :-)

Manitou and I met yesterday and made a couple of Java assignments which were overdue. We implemented a Heap using the Locator pattern and a Vocabulary using a Trie — it was a very good learning-experience to actually implement these ADTs instead of just talking about them.

We also talked about other things — we met at 14:00 and went home again at 23:30 so there was plenty of time :-) One of the things we discussed extensively was, that he had a book, which said, that the set of all subsets of the empty set, which is denoted 2^∅, is {∅, {∅}}. I don’t believe that this is true — the only subset of ∅ is ∅ itself, so I would say that 2^∅ = {∅}. If 2^∅ = {∅, {∅}}, then that would imply, that ∅ ∈ ∅ which is clearly false, as the empty set is empty. We’ll ask Jørgen Hoffman-Jørgensen about it — he should know as he’s our teacher in Probablility Theory.