Thursday 10 March 2016

Updates on life

Who knew, that I would one day write a blog post just to give an "update on life"? It's almost like I think that someone will read this, apart from me. Actually, it's pretty impressive that I haven't completely given up posting things yet, its been at least a couple of months (I think). Anywhoo, life has been hella tiring. Seriously. Imagine running a marathon, whilst solving the Schrodinger equation for a N-particle system, whilst head juggling a soccer ball for ten hours straight. That's what life is like right now, and I DON'T EVEN KNOW WHY. I don't even have that much to do, but somehow I still feel so tired. Perhaps it's related to the fact that I am constantly surrounded by people for 80% of my waking hours, so I don't get any lonely time. Who knows.

Anyhow, here's two small things that I've done (ish) :

Ah, the classic block on a plane. Just as classic as foggy glasses whilst eating two minute noodles at 5 am, or getting wet because I was walking in the rain. Absolute classic. A friend of mine however (a man named Y) suggested a variation of the problem. In the variation, the triangular plane (the black object in the picture) exists on a frictionless table, i.e. as the red block slides down the plane, the plane will (probably) slide to the left. The question is, will the red block reach the bottom of the plane faster if the plane is on a frictionless table, or if the plane is fixed in place?

To solve this problem, I introduced two variables, $x$ and $y$, and just Euler-Lagranged it like WHAM WHAM WHAM.. DOWN FOR THE COUNT! The answer I got is:

$$T' = T \sqrt{\dfrac{M + m \text{sin}^2 \theta}{M+m}}$$

Where $T'$ is the time it would take the red plane to get to the bottom if the plane was on a frictionless surface, and $T$ is the time it would take if the plane is fixed. Checking the boundary case of $\theta = \pi / 2$, we see that $T' = T$ as required.

Anywhoo, there's only one other thing of any slight interest that I've done:

A man sits in a room. Every second, he says a random number between $1$ and $N$. How many seconds on average pass before he has said every number between $1$ and $N$ at least once?

This has real implications, in that originally I was trying to estimate the time it would take for a system to explore all of it's 'accessible' microstates. The rough way I sort of maybe solved it was to consider the expected number of 'duds' (i.e. number of numbers that the man says that he has already said at least once) as a function of the number of unique numbers that he has said so far, and sum that up. (by linearity of expectednesstation)(I think). Anywhoo, the answer I get involves some harmonics (sums of $\frac{1}{N}$), but crudely (for large $N$), it's probably:

$$T = N \text{ln}(N) + N$$

(I say probably, because I'm running a simulation right now to see if I'm right or just completely off whack. I'm probably right. Probably.)


The above plot, for $N$ up to $100,000$, shows the 'actual' rough value in blue, and the guessed value in green. As can be seen, it seems an OK guess

Just another note on life: My sleep pattern is completely off whack, it's getting close to 3:00 AM right now, which is pretty bad. But I'm running an experiment, recording all my sleep times, and whether or not I have a dream, and whether I remember it or not, to see if dreams have any relation to sleep times.
Anyway, Ima pop ma clockers, see you laterz

And this is for $N$ up to a million.