Recent content by yoran

I think you misunderstood me. I'm not saying the UK system is similar to the US system, I am just saying that it resembles more the US system than the rest of Europe (Holland, France, Germany, Sweden, etc...). There is a difference between "being closer to ... than ..." and "being equal to ...".

Ok you are partially right :-). So, indeed the UK officially uses a metric system except for some things where the imperial system is still used. Learnt something I didn't know today :-).

Well, the UK is in the European Union but they are kind of an island compared to the rest of Europe (which comes from the fact that the UK is geographically an island). Usage of the imperial system whereas in the rest of Europe the metric system is used, drive on the left side of the road, they...

Hi AronH, Since you are at Ecole Polytechnique I reckon that you have a good level in mathematics. Wouldn't ENS (Ulm) be something for you then? I think they must have one of the strongest programs in Europe... Also, I know the University of Utrecht (that is in Holland) is very good in...

I think this proof is correct. The Turing machine which implements y can see if a proof is correct in the following way. First, you assume you have a formal (that is, a logic expression) statement for "algorithm x always finishes after a finite number of steps". Let's call that expression P...

Well, I don't know the answer to the question myself and thus would be very interested in a proof (which either proves the proposition true or false). I have a feeling that it is connected to Gödel's incompleteness theorem. If you see the sentence "this algorithm finishes after a finite number...

That's an interesting question. Practically, it is impossible to prove (or disproof) termination for all algorithms because there are infinitely many. However, assume you start working by taking algorithms at random. Say you generate the random number x_1 and analyze the termination of...

Well, the part on Wikipedia about termination says this: There is no algorithmic procedure for determining of arbitrary algorithms whether or not they terminate from given starting states. This is exactly the Halting problem. Of course, as you say, it is possible to look at every algorithm...

I agree with that. Computers are so ubiquitous to anything related to science or engineering. You can't do science or engineering research anymore without using a computer. Even some pure mathematicians at my university use mathematical software to test hypotheses and gain insight in problems...

I think mathematics is just as "unnatural" as computer science. All of mathematics is based on axioms and who invented those axioms? Right, people did. Those axioms aren't to be found in nature. In the same way, you can say that the Turing machine (or any equivalent machine) is the basis for...

You could use Arch Linux, which is a minimal system after installation but afterwards you can customize it as you want by installing new packages. As already mentioned, instead of using Gnome or KDE, I would suggest using a more lightweight window manager such as Fluxbox or XFCE.

It's a second order high-pass filter. It looks like this. http://dave.uta.edu/dillon/pspice/images/pst07f.gif but without any concrete values for the components (it's an image I found on the Internet, not mine). I just realized I made a mistake. The current I gave at the resonance...

Ah wait I think I see it. I worked the equations out at the resonance frequency and I get the result that the output voltage is pure imaginary (V(f_r) = \frac{j\omega_r L}{R}V_{in}) while the current is pure real (I(f_r) = \frac{V_{in}}{R}). As you said, then the real power is equal to the...

Isn't it so that when the voltage is high, the current is low and vice-versa?

Oh, I think I see. The output voltage is higher than the input voltage, but then the output current must be a smaller than the input current?