ℝolliℵg M∀th Thr∑a∂

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Think I did indeed make an error and overlooked some cases and undercounted. Will ponder a while longer.

Help Me, Zond 4 (James Redd and the Blecchs), Saturday, 27 June 2015 23:04 (eight years ago) link

Still tinkering can't quite get the precise adjustment to my lower bound. Sorry to break thread.

Help Me, Zond 4 (James Redd and the Blecchs), Sunday, 28 June 2015 23:22 (eight years ago) link

Okay, now think the recursion formula is
f(n) = 2 f(n-2) + f (n-4) , n even

Help Me, Zond 4 (James Redd and the Blecchs), Monday, 29 June 2015 00:03 (eight years ago) link

Aargh, still missing a few

Help Me, Zond 4 (James Redd and the Blecchs), Monday, 29 June 2015 00:15 (eight years ago) link

Re-reading the circle question, I think I misunderstood it. The arbitrary point (a, b) must be given, otherwise it's too easy. I should have been suspicious that I made some headway on it so quickly.

o. nate, Monday, 29 June 2015 02:19 (eight years ago) link

Gave up and looked up this morning. Interesting. Hopefully can read more carefully tonight.

Help Me, Zond 4 (James Redd and the Blecchs), Monday, 29 June 2015 16:32 (eight years ago) link

this thread is hard to find

anyway, first bunch of results for the 'mustn't go below zero, must end at zero' question
// 2 = 1
// 4 = 2
// 6 = 5
// 8 = 14
// 10 = 42
// 12 = 132
// 14 = 429
// 16 = 1430
// 18 = 4862

it's like a map


/\
/\/\
/\/\/\
/\/\/\/\
/\/\/\/\/\
A -t-> B

how many routes from A to B without backtracking?

koogs, Monday, 29 June 2015 20:43 (eight years ago) link

f(n+1) = f(n) * (2 + (2(n - 1) / (n + 2))

which probably resolves down a lot

koogs, Monday, 29 June 2015 21:08 (eight years ago) link

(oh, i've numbered those 1 - n where they should be 2 - 2n as only the even numbers work)

koogs, Monday, 29 June 2015 21:14 (eight years ago) link

and f(1) = 1

koogs, Monday, 29 June 2015 21:15 (eight years ago) link

scale factors between values are
2 + (0 / 3)
2 + (2 / 4)
2 + (4 / 5)
2 + (6 / 6)
2 + (8 / 7)
2 + (10 / 8)
2 + (12 / 9)
2 + (14 / 10)
etc

koogs, Monday, 29 June 2015 21:20 (eight years ago) link

this thread is hard to find

Searching for "thr" in thread titles works well. Not necessarily so easy to remember, though.

anatol_merklich, Monday, 29 June 2015 21:54 (eight years ago) link

Glad someone is taking a more methodical approach than my inept scribbling. Must have heard this topic mentioned in passing decades ago but obviously I never studied it. Lots of stuff about it on the web, can post link to one nice write-up later if you want.

Help Me, Zond 4 (James Redd and the Blecchs), Tuesday, 30 June 2015 00:19 (eight years ago) link

this thread is hard to find

i just search for "gromov" now

the late great, Tuesday, 30 June 2015 00:26 (eight years ago) link

This morning I was thinking about the simplified version of 2b where the sequence can't go negative but it doesn't have to end on zero, and I had the brainwave that the solution could be written as a recursive function, starting from n=1, and then defining f(n) in terms of f(n-1), since the path dependence means that at the solution at any value of n must build on the solutions at smaller values of n. However, I didn't get much beyond that. Nice to see koogs take it further. I really like the little map drawing.

o. nate, Tuesday, 30 June 2015 02:00 (eight years ago) link

Indeed.

Help Me, Zond 4 (James Redd and the Blecchs), Tuesday, 30 June 2015 02:07 (eight years ago) link

One can see that there has to be a lot of structure there but it is hard to know exactly what it is, or even to enumerate the first few values correctly unless one is very careful or writes a little program to generate them, which I presume is what koogs did. Anyway impressed that he forged so far ahead on his own.

Help Me, Zond 4 (James Redd and the Blecchs), Tuesday, 30 June 2015 02:13 (eight years ago) link

Oops, I was stumbling around related concepts on Wikipedia, starting from Fibonacci numbers and I stumbled on the answer. There is a name for the sequence of numbers that Koog started enumerating above, which apparently solve lots of combinatorics problems, including this one. Now I'm torn on whether I should read the proof, or keep thinking about it.

o. nate, Tuesday, 30 June 2015 03:12 (eight years ago) link

> writes a little program to generate them, which I presume is what koogs did

busted...

was a tiny recursive loop to generate all the possibilities and another method to validate them afterwards. inefficient but was surprisingly quick but i've since (read: in bed this morning) thought of a way of discarding invalid choices whilst creating them, thus cutting down a lot of the work. (although my first attempt got into an infinite loop and i had to kill it, having forgotten to save before running it)...

then it was just a case of trying to spot a pattern in the results. started by dividing the one by the previous, got lucky.

would like to see the proper answer - it feels so far like those calculus lessons where they get you to derive everything from first principles and then point out that it was just n * a^(n - 1) all along

koogs, Tuesday, 30 June 2015 08:33 (eight years ago) link

I would definitely read up on it. Isn't this how the learning process is supposed to work: you try to do it yourself first, you do some calculations, you look for some relationships, patterns and connections, you beat your head against the wall and then, unless you are very very smart, diligent and lucky, you throw in the towel and look at the literature. Then it makes much more sense: I knew there was a relation something like that, I knew I needed an extra gimmick I couldn't come up with in the time allotted

Help Me, Zond 4 (James Redd and the Blecchs), Tuesday, 30 June 2015 11:04 (eight years ago) link

Aka Dyck paths to watch out for

Help Me, Zond 4 (James Redd and the Blecchs), Tuesday, 30 June 2015 11:04 (eight years ago) link

flopson to thread, to comment on our progress and assign new problems

I Want My LLTV (James Redd and the Blecchs), Tuesday, 30 June 2015 23:11 (eight years ago) link

flopson has given up on us

I Want My LLTV (James Redd and the Blecchs), Wednesday, 1 July 2015 09:11 (eight years ago) link

is a single step along a Dyck Path called a Dyck Move?

koogs, Wednesday, 1 July 2015 10:04 (eight years ago) link

Dunno.

Two things I was able to observe during my initial fumblings:

1) The number of paths that are strictly positive until they return to zero at 2n is the same as the number of paths that are nonnegative that return to zero at 2n-2.

2) The recurrence formula seems to depend on more than a handful of prior terms. As we now know, it depends on all of them.

Should have been able to use 1) to derive the recurrence relationship if I had thought about it long enough. I think somebody posted a similar sentiment a few posts ago.

I Want My LLTV (James Redd and the Blecchs), Wednesday, 1 July 2015 10:37 (eight years ago) link

Been resisting the temptation to say:
flopson has gone off the net because of koogs

I Want My LLTV (James Redd and the Blecchs), Wednesday, 1 July 2015 10:45 (eight years ago) link

Do not peek if you are still calculating: http://www.math.ucla.edu/~pak/lectures/Cat/pakcat.htm

How I Wrote Matchstick Men (James Redd and the Blecchs), Sunday, 5 July 2015 21:58 (eight years ago) link

sorry i've been slacking on serving up with the answer, i'm trying to get a detail right and the friend who told me these riddles isn't responding. be back very soon with a fully satisfying proof

flopson, Sunday, 5 July 2015 22:52 (eight years ago) link

I liked this particular write-up: http://www.math.ku.edu/~jmartin/courses/math724-F13/count-dyck.pdf

How I Wrote Matchstick Men (James Redd and the Blecchs), Sunday, 5 July 2015 23:12 (eight years ago) link

oh hah! once you see the pun it all follows (well, 2b does). here is something that i think is not too terrible a hint. write "i get a dollar" as "(" (open paren) and "i give a dollar" as ")" (closed paren). now consider what "allowable" sequences look like syntactically in such a translation :-)

got bent (mild cheezed off vibes) (s.clover), Sunday, 5 July 2015 23:42 (eight years ago) link

my dad is asking for a good introductory group theory text. nothing too dry.

he is working on this, but i think he finds it dry:

http://www.amazon.com/gp/product/0521312493/

any suggestions?

the late great, Monday, 13 July 2015 01:45 (eight years ago) link

he specifically asked for group theory, so my sense is maybe some of the stuff in that book is over his head and he wants something a little more basic?

the late great, Monday, 13 July 2015 01:48 (eight years ago) link

Kind of an interesting article about the Navier-Stokes equations, which tormented me in my days as a Mechanical Engineering student:

https://www.quantamagazine.org/20150721-famous-fluid-equations-are-incomplete/

o. nate, Thursday, 23 July 2015 02:30 (eight years ago) link

still looking for a group theory book

the late great, Thursday, 23 July 2015 02:40 (eight years ago) link

Try Springer undergrad book by Armstrong.

(Xp)Interesting. Will read later hopefully

Archaic Buster Poindexter, Live At The Apollo (James Redd and the Blecchs), Thursday, 23 July 2015 02:44 (eight years ago) link

Μ. Α. Armstrong, Groups and Symmetry. His topology book seems to be pretty good as well.

Archaic Buster Poindexter, Live At The Apollo (James Redd and the Blecchs), Thursday, 23 July 2015 03:04 (eight years ago) link

^that's probably a good rec. i read his topo book, very fun. iirc there's a good chapter in coxeter's geometry textbook. group theory can't be learned without reference to geometry/symmetry imo. i learnt groups from a number theorist, took years to undo the damage

flopson, Thursday, 23 July 2015 03:44 (eight years ago) link

Lol. Yes, he does the basic symmetry ideas early on and then leads to stuff like orbits and stabilizers in the second half.

Archaic Buster Poindexter, Live At The Apollo (James Redd and the Blecchs), Thursday, 23 July 2015 03:50 (eight years ago) link

Book we used in college was Jacobsen, probably a little too abstract for most tastes. Teacher was awesome though, Jonathan Rogawski (RIP), who was a student of Langlands himself.

Archaic Buster Poindexter, Live At The Apollo (James Redd and the Blecchs), Thursday, 23 July 2015 03:56 (eight years ago) link

Just glanced at Armstrong again. Really touches on a lot of good, interesting stuff in a nice way, such as the kind of group theory and linear algebra that physicists use, or the free group- as seen in topology! - as opposed to a Hard Algebra approach, say, in which you might spend a lot more time on rings and monoids. His topology book comes highly recommended in The Princeton Companion by this guy: http://www.math.ucla.edu/~totaro/

Archaic Buster Poindexter, Live At The Apollo (James Redd and the Blecchs), Thursday, 23 July 2015 10:56 (eight years ago) link

thx for the recommend ... just grabbed the .djvu of "groups and symmetry" and also his (?) "basic topology" too

the late great, Thursday, 23 July 2015 15:45 (eight years ago) link

http://web.math.princeton.edu/generals/examiner.html

princeton oral math exams

some nice questions in there

If f_n is a seqeunce of integrable functions, when is int(f_n)
convergent? Can you give an example where this fails?

Set a(n) = 1/n + ... + 1/2n. Compute lim a(n) as n --> infinity.

Let En be a sequence of measurable sets in [0,1] with m(En) --> 1. Does
there exist a subsequence whose intersections all have measure > 1/2?

flopson, Saturday, 1 August 2015 23:24 (eight years ago) link

sorry for dropping the ball on those combo coin flipping problems. i wanted to write up a nice proof for 2a that everyone could understand but i'm too busy writing a dumbass masters thesis right now. anyways 2a uses reflection principle and for 2b you can find a bijection between unique maximums and multiple maximums by reflecting about the peak /\ in a clever way. the proof of 2b my friend showed me was actually an original one that he's getting published so i prob shouldn't write it here. but i'll link to it later

bonus (easy) analysis question with multiple answers:

find a sequence X_n in R such that lim_{n->infty} X_n - X_{n-1} = 0 but lim_{n->infty} X_n doesn't exist

flopson, Saturday, 1 August 2015 23:31 (eight years ago) link

Let En be a sequence of measurable sets in [0,1] with m(En) --> 1. Does
there exist a subsequence whose intersections all have measure > 1/2?

i think this is 'yes' but cannot prove it

dead (Lamp), Tuesday, 11 August 2015 01:45 (eight years ago) link

after posting and thinking about those for 5 mins i realized none of those questions make sense, i think there's some missing info

like, En = [0,1] for all n is an answer to that one you mention Lamp. m(En) -> 1 and m(Ei int Ej) = 1 > 1/2 for all i, j (taking the sequence itself as subsequence)

flopson, Tuesday, 11 August 2015 01:59 (eight years ago) link

either that or princeton oral exams are a scam

flopson, Tuesday, 11 August 2015 02:00 (eight years ago) link

i need to revisit rudin someday, never got a good handle on measure they/real analysis and ended up doing actuary stuff so having that basis for probability would be valuable if only for exams

art, Tuesday, 11 August 2015 02:06 (eight years ago) link

ilx poster eteaoe convinced me to take measure theory in this very thread. it's ok, a little too dry for me but i didn't pursue stochastic calculus or any of the fancy applications of it, so it mostly just looked like an overly fussy way of doing integration to me at the same. this is my favourite result from measure theory https://en.wikipedia.org/wiki/Borel%E2%80%93Cantelli_lemma which states that

if the sum of probabilities of a random sequence of events is finite, the probability that infinitely many of them occur is zero

flopson, Tuesday, 11 August 2015 02:20 (eight years ago) link

my advisor was super psyched to teach a real analysis class (he was an algebraic topology/knot theory guy) bc he said he felt like he really "got" the subject, but the course was still, as you say, pretty dry. your synopsis is about where i landed, but having gotten into stats since i feel like i shd take another stab.

art, Tuesday, 11 August 2015 02:34 (eight years ago) link

like, En = [0,1] for all n is an answer to that one you mention Lamp.

I took it to mean "is it the case that, for EVERY sequence E_n....."

Guayaquil (eephus!), Tuesday, 11 August 2015 06:06 (eight years ago) link


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