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Monday Dec 06, 2004

Dec 6, 2004
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so I have this paper to write. I named it "Alan Turing: The Man and The Machine." it's basically a bio of turing, with another section at the end that is more specifically about the development of turing's specialized machine, and then the generalization into the universal turing machine, with discussion of the influences and applications that surround the development.

I had written down that the paper is due thursday, the last day of class. being a devout procastinator, I had not started this paper this afternoon, when I read that it is not due thursday but in fact TOMORROW.

so I had that wave of adrenaline when I realized that I have a ten+ page paper to write and less than a day to finish it, and I have classes and sleep in-between.

right now I'm on page 8. unfortunately, the last two pages will take as long as the first 8, because I could write the first 8 mostly from memory. the last 2 I actually have to research before I can write.

I'm sick of humanities. fuck you, history. I'd rather be working on my math project. which, incidentally, is getting cooler. I think I'll try to work it into a presentation to take to the Texas and Oklahoma Regional Undergraduate Symposium (TORUS). in case anyone is interested, here is the general idea:

Consider subharmonic series M and N. Let C(N) = {real numbers x s.t. there exists a subseries of N converging to x}, i.e. C(N) is the set of all sums of subseries of N.

Define a relation <=. Say N <= M iff C(N) is a subset of C(M). In other words, N <= M iff every real number x that can be expressed as a sum of a subseries of N can also be expressed as a subseries of M.

The problem: determine whether <= is a total ordering on the set of all subharmonic series.
1. reflexivity, i.e. N <= N.
2. transitivity, i.e. N <= M and M <= L implies N <= L.
3. symmetry, i.e. N <= M and M <= N implies M = N
4. comparibility, i.e. N <= M or M <= N for all M, N

1 and 2 are easily proven true, and 4 is easily proven false. 3, though, is tricky. I don't know whether it is true or false. it's easily true if you define equality to mean the sums of N and M equal, or if you definite it to mean C(N) = C(M). but if you define equality to mean termwise equality, which is how it is defined in this problem, then it's not obvious, and may be false.

since we could not make any headway on part c in the general case, we considered the more specific situation where N and M are geometric subharmonic series, and in that situation 3 is provable. however, in this case, I would like to prove a more elegant result, being that C(N) is a subset of C(M) iff a = b^k, where a is the base of N, b is the base of M, and k is some integer.


so if you made it through that, high 5! if you just skimmed that, fair enough. I think I was mostly motivated to write about that project in order to further delay having to continue with my history of science paper. distractions are encouraged.
VIEW 3 of 3 COMMENTS
wyatt:
Congrats on finishing the paper.

Your next assignment is to write a brief essay (20-200 words) on why you should be a member of SG Motorcycle club. The deadline is January 23, but bonus points will be awarded for early completion. Please submit it via my journal.

PS- I'm a history major who's actually made a career out of it. wink I'm also a world-class procrastinator smile
Dec 22, 2004
wyatt:
Thanks for the note. You're in!

Post an introduction, then make yourself at home.
Dec 30, 2004

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