@robjohn I'll let you know if I can make it work on some other problem. I just tried it on the successions recurrence (see here) but couldn't quite get it to work. There are a couple of other similar problems I may try it on.
You might be interested in that post for its own sake, though, since the number of permutations with no successions has a simple expression in terms of derangements.
@Srivatsan He's a bit of a mess. It's like God spilled a person...
@robjohn The numerators and denominators of the continued fraction there are essentially subfactorials in disguise, so methinks only a little tweaking is necessary.
@J.M. Well, I'm thinking in order to do this rigorously, you need to show that given a list of ordered pairs of real numbers, where the first and last element of the list are the same, the set of points lying on the line between them lies in the inside of the polygon. But I'm not sure how to stipulate "non-selfintersecting polygon" or formula an "interiorness" condition in coordinate geometry
@Potato: Define a polygon as a list of vertices, a self intersection as two lines crossing and a non-self intersecting polygon as with no two lines crossing
Does this stronger theorem hold: every vertex of a non-self intersecting polygon can be joined by a line to some other vertex (where the line is inside the polygon)
but QED, isn't that obviously false? Take an "arrowhead" quadrilateral. The 2 of the vertices cannot be joined to any others by a line lying inside the figure.
@QED I wanted to post the pages without the annoying big previews - but I did not know another way. (Always when I pressed upload, the chat posted a new picture.)
I forgot about this thread where I've linked to an article containing an O(n) algorithm to compute the convex core of a (simple) n-gon after a discussion in the comments with Mike Spivey.
The tough one is permutations and (a,b,c) -> (2a,b-a,c) [note they are natural numbers including zero so you can't always do the subtraction] - it was phrased about three buckets with pebbles in it
the question is whether you can always empty one bucket (reach (0,x,y))
I was trying to solve it in a similar way you prove the simple typed lambda calculus reduces: finding a reduction strategy and a decreasing measure.. but I think I can rule that out (I don't think it's possible to solve it that way)
I could come up with loops for all the simple strategies I thought of
as for positive results: you can eliminate common factors.. reducing (ka,kb,kc) is the same as reducing (a,b,c)
so that leaves us with only infinitely many cases left to solve
2
I looked up "arithmetical dynamical systems" but that's not what this is
There's probably no theory about this stuff, seeing as how quickly it becomes Turing equivalent
@MartinSleziak I haven't heard anything about the links to comments. I know that at some point I was given a link to one, but could not figure out how to do it at any given time I wanted.
Yes, perhaps there's nothing wrong with answering them for the benefit of the others. But upvoting? "This question shows research effort, it is useful and clear."
@tb I know it is in the hover message for the upvote arrow, but some questions get upvoted for the answers they inspire (whether that be right or wrong).
@tb I always have problem with this. Sometimes a question shows no effort, but I find such question very useful to by on a Q&A site. Sometimes the other way round - the question seems extremely easy to me (but probably not to everyone), but I see that OP has put lots of work into it, so he deserves a reward.
I think that people have various reasons for upvoting/downvoting. (But it's probably more relevant for answers.)
@MartinSleziak I submit that a lot of questions satisfy only one of "This question shows research effort, it is useful and clear.". And lot more satisfy neither.
Completely unrelated: We're getting a puppy and now we've been arguing about the name. My partner wants to call her Aleph Naught, I don't think that's even mildly amusing.
A Nordic mathematical competition asked to prove that for all $x$ we have $x^8-x^7+2x^6-2x^5+3x^4-3x^3+4x^2-4x+\frac{5}{2}\geq 0$ for all real $x$. I heard that it follows from Hilbert's problem that one can prove this by writing the polynomial as sum of squares. How can I find such a representat...
I was missing a paren that was messing everything up
@QED 1) a) Prove that if a^2 is divisible by 7 then a is divisible by 7. b) Use a) and the quotient remainder theorem and case reasoning to prove that the square root of 7 is irrational.
