12:03 AM
@robjohn: And now you've got me wondering if some related problems that have somewhat similar recurrences can be solved that way, too. :)

@MikeSpivey I hope some other good comes of it. I'd hate for it to only solve this one problem :-)

@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.

12:23 AM
Hey rob, nice answer! I think yours can take care of the "skipped details" of my answer here.
(The derivation I'm accustomed to uses an integral ansatz to try to find a solution of the difference equation.)

@MikeSpivey Thanks. I skimmed it, but I am in the midst of work right now, so I will have to look a bit later.
@JM does it really apply there? I will have to look closer.

@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.

5 hours later…
5:41 AM
Can anyone help a clueless student with a power series problem? :)

5:57 AM
What up?
@Mr_Crypto: you still here?
Bbl

hi, yes, I am here! :)
@robjohn picture of problem above...I used Maclaurin series of so that x[e^u] = x*Sum(x^(2n) / n!). Not sure if I am on the right path or not?
(where u=x^2)

6:38 AM
Anyone else is welcome to help too if they please...I need all I can get!

1 hour later…
8:00 AM
@Mr_CryptoPrime ...and x\times x^{2n} = x^{2n+1}, right?

hello
what a shame having to answer questions like through a web interface
I wouldn't like that at all

I think it's this misguided attempt to shoehorn pedagogy into technology... or vice versa.

8:19 AM
And now, after I ask it works :)

:D

8:47 AM
My favorite spelling of continuous so far: "continios"

This is interesting - on activity tab there is sometimes a link to question and sometimes link to comment.
E.g. here math.stackexchange.com/users/13854/… - converting fractions and integration help are different.
@robjohn Did you get some reasonable suggestion from anyone?

I don't think my question here is as clear as it could be. Could you guys perhaps suggest some better wordings? math.stackexchange.com/questions/83580/…

@Potato: if you consider arrowheads, one diagonal is easy to sketch, but the others...

9:03 AM
@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

You will not need the jordan curve theorem since it's all straight lines

@MartinSleziak This counts as reasonable for me...

Thanks t.b., I'll have a look.

@Potato, I guess the condition that the polygon does not self-intersect is the key

Yes, indeed. But I'm not sure how to formula that in a precise way that is also helpful for solving the problem.

9:09 AM
huh

I'm far away from my CG books, but I'm pretty sure this was done to death there...
(CG is either "computational geometry" or "computer graphics")

Can someone explain this comment to me?

You've seen Joe O'Rourke's book by any chance, Potato?

@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

@tb Not sure why he wrote that, either...

9:11 AM
@J.M. I haven't, but we are looking for a proof here, not an algorithm, yes?

@QED Actually, there's a nice problem: assume you have an ordered sequence of corners; how do you tell if it's self-intersecting or not?

a working algorithm is a proof

@Potato An algorithm counts as a constructive proof.

@JM A further guy that sometimes could as well talk in Sanskrit, as far as I'm concerned...

@JM I see. I guess I'd prefer a simpler argument, but if it works it works, I suppose.

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)

I'm confused: what is a self-intersecting polygon?
Wouldn't that be a piecewise linear arc?

A polygon where the sides intersect.

@MartinSleziak Yeah, that.

9:16 AM
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.

Unfortunately I can't preview here.

@Potato: Not obviously, I just came up with a counter-example

In my world only the rightmost figure above counts as a polygon.

@t.b. the definition of "polygon" then needs to account for that
that's why I made "non self intersecting" explicit

9:18 AM
Mornin'

the theorem doesn't hold for the red one
hi
so there are n(n-2)/2 lines between the vertices.. and you just want to show at least one is fully inside the polygon
I wonder how many of those lines can be outside the polygon

I can imagine the concave pentagon with only two diagonals...
Conjecture: a (non-intersecting) polygon with an even number of sides n can have up to n/2 internal angles that are reflex (> \pi).

it seems, because of the duality of inside/outside it's no easier to count the lines that are outside than it is those that are inside
how about this stronger statement: You can always connect it by vertices i and i+2 (for some i)
any counter examples?
aw
(the "stronger statement" is equivalent to the notion of a "strictly convex vertex" in the paper)
so I was on the right track
@Martin, why do you keep deleting it? anyway I think that answers this question perfectly

9:40 AM
Hopefully this qualifies as a fair use. - just 4 pages from book.

and the key to proving that is looking at the angles, as J.M. was getting at

I think it does. Now if somebody could summarize O'Rourke, that'd be neat... :)

would we have solved it, I wonder? :)

@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've given google books link above, but J.M. said it's not viewable for him. chat.stackexchange.com/transcript/message/2474854#2474854

yeah google books is a horrible "resource"

9:43 AM
@Martin, yeah, there should be some text before the URLs so that the software doesn't render the pics.

it's really irritating that it's become popular to link to it on wikipedia

J.M. I tried that but when I enter text than upload button does not work.
QED I find google books a good source.

I've taken to cheating: I use the upload facility in the question/answer panes.

it's just imgur.com

@MartinSleziak I agree it's good, the only caveat is that the access is spotty.

9:45 AM
QED As you mention it, do you edit wikipedia?

no

Sometimes, Google is cruel: you have this one unavailable page in between available pages...
...and that unavailable page is the page that has the theorem/proof you need.

@JM That was simple - I should have figure that out.... :-)

Is there a name for systems like.. say a triple of natural numbers and a set of arithmetic operations you can do it?

9:48 AM
BTW it seems that many wikipedians were again using them - see this discussion en.wikipedia.org/wiki/Wikipedia:Templates_for_deletion/Log/…

That's surprising, but I guess it's too late now anyway
e.g. consider the system where you can do (a,b,c) -> (a+2,b-1,c-1) as well as permutations
and we want to answer questions like "when is it possible to reach (0,0,n)"
[N.B. this is easily solved but I have a harder one I can't solve]

If you consider those triples as points... you're asking if some piecewise linear path will hit a certain point?

interesting! Had not thought of viewing it geometrically

The "permutations" become "reflections" in the geometric viewpoint.

this problem was phrased about snakes on an island of three different colors: when two of different colors meet they become the third color

9:54 AM

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))

The idea is very similar to what J. J. outlines in his answer to potato's question.

ah, that (2b,b-a,c) would be a bit too elaborate to do geometrically...

J.J.s i think answer is exactly what Martin posted
oops
I should have written (2a,b-a,c) - corrected now.

@QED: another possibility is to treat them as vectors that you're multiplying by simple matrices...

9:57 AM
thanks for helping me notice that - that's important because it means we always stay on the surface {(a,b,c) in N^3 | a+b+c=n}
I'm scared of word problems regarding matrices.. :)
the tricky thing is you can't just multiply the matrices together however you want: negative numbers aren't allowed

@QED For example, the matrix you need is {{2,0,0},{-1,1,0},{0,0,1}}

so even if you could factor M (where M(a,b,c) = (0,x,y) with a+b+c=x+y) into the matrices, you need to check at each stage they never go below zero
also I asked the computer to graph this system for some numbers and it has multiple connected components
so it's a tough problem :)

Wow - Prof. O'Rourke has an account here: math.stackexchange.com/users/237/joseph-orourke

I know; that's exactly why I brought him up. :)
I'm hoping he'll see it.

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
but I made no progress in the last few days :(

10:27 AM
@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.

Ah, I must have missed that while I was typing... I don't have the SE modifications installed.

Perhaps the comments in activity tab of profile pages might help sometims.
As I've mentioned.
But it only works sometimes - I don't know why.

It sort of makes me sad that this question will probably get more than a "yes, obviously" as an answer.

@MartinSleziak That is how I must have gotten the link when I got it. I think it was from an activity tab.

10:32 AM
Even with an upvote.

@tb Perhaps it is not as obvious to everyone. Those people might find such an answer beneficial.
I sometimes have trouble remembering that things are not as obvious to everyone.
Obviously ;-)

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."

10:49 AM
hopeless

@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.

Just one question missing to 2k questions tagged calculus: math.stackexchange.com/questions/tagged/calculus

11:05 AM
@MartinSleziak what question is missing? ;-)

"Hai all what is derivate of e^x?"
"Plzz don't use so many formulas."

lol

12:07 PM
Wasted.

hm?

You went through your whole hooch supply, Jonas?

No, luckily not.

12:37 PM
Ello.
@JonasTeuwen: Still wasted from yesterday or newly wasted from this morning?

From yesterday.

1:07 PM
@JonasTeuwen They say that youth is wasted on the young. It appears that the young are simply wasted.
2

1:21 PM
Joke's old, but still funny... :D

Heh.

1 hour later…
2:26 PM

It boggles...

The profile says it all...

I noticed only now that MaX has a whole pile of questions...

I'm surprised I'm not in that list...

2:45 PM
@t.b.: Is this the list of idiots?

"Idiots" is too much... it's alright to ask a lot of questions, as long as they're good questions.
Joe O'Rourke on MO asks a lot of questions for instance. He's certainly no dummy...

Well "the profile says it all" -> read the profile. Then post a link of people with tons of questions. Am I reading too much into this?

No, I brought MaX up separately from the other guy who was asking on Hilbert spaces...

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.

3:06 PM

No.

3:21 PM
It's a mixture of English humour and trying to troll me. The next suggestion was 'daikon', which is Japanese for radish.

What sort of puppy is this?

Hmm, "radish" would have been appropriate if the little one was all white...

I think we should give her the name "ai" (Japanese for "love") and then call her "ai-chan".
The argument against that is that they can't pronounce it properly : )
Do you have a pet yourself?

3:36 PM
Had. My current living arrangement doesn't allow for dogs or cats...

We brought a second cat today. We're trying it for a bit before we decide.

The opposite gender from the previous one?

Nope.
Ours is about a year old female, this is a two-three months old female.
There is a lot of hissing from the older one. The young one just seems confused and wants to play.

Mooornin

Hi Matt : )

3:51 PM
This is no good. Two Matts on the same channel.
@Matt: Oh no, I am pinging two people at once!!!

Lol

: )

@AsafKaragila Gives "catfight" a whole new meaning...

I ain't that fond of number theory either... :D

4:07 PM
I don't know yet. I bought a book 2 years ago and been meaning to read it.

Well, it's a "different strokes" thing. You can see on this site that number theory is a rather popular subject here.
Most of the dudes with high rep are those with number theory expertise.

Anyone wanna help me with some mathematical induction?

I can try, go on...

Eveything should make sense
Just not seeing any algebra that I could do
So from my understanding of induction I just have to make the p(k) equation = the p(k+1) equation
And that is sufficient proof

Give me a sec, I'm doing it on paper.

4:24 PM
Thabks! :)

hm. I got my sums wrong somewhere.

algebra error?

Is this thing homework?

@JM That is a Great Dane, if I am not mistaken

No, study review

4:35 PM
@AsafKaragila From experience, both see a highlighted name, but only one gets pinged.

@rob: It is; Matt had linked to the wiki article before the pic... :D

@JM Ah, I just saw the pic as I was scrolling to here.

A docile dog, for the most part.

Aha! the other Rob (the one who is not a troll) must have not been here recently. That @rob actually pinged me.
@JM Yes, my wife had one when I first started dating her. They are mostly gentle.

hello

4:37 PM
@QED hi there

@robjohn He changed handles; that may have something to do with it...

any "rules of thumb" that could help someone determine whether or not they could succeed at a PhD?

@QED I don't have you on my clocks, so I don't know what time it is there. 'hi there' is as good as I can do ;-)

I don't think rob was trolling, just philosophical
I am using UTC+0

@JM Ah, who is he now?
@QED That's why I said the one who is not a troll :-D

4:39 PM
Skull something... last I checked.

which implies...

oh YOU are the other Rob

Indeed :-)
Ah, I see his avatar by skullpatrol

I think I might just give up on the buckets puzzle
I've hit it with everything I've got

Have you posted a question on the main site?

4:48 PM
that's what I mean by give up
since I can't solve it

@Matt: OK. So all you need to do to finish your induction is expand the two numerators.
i.e. multiply 2(k+1)(k+2)(2k+3) and 2k(2k^2+3k+1)+6(k+1)^2
If you get the sums right they should be the same.

what a shame you have to prove these things by induction
no wonder everyone finds induction so hard to learn, given how it's taught

@matt, ah ok, let me try that

@QED: It's shoehorning, I tell you...

@Matt: they should be 2(2n^3 + 9n^2 + 13n + 6)

4:56 PM
hmm
from 2k(2k^2+3k+1)+6(k+1)^2 I get (4k^3+6k^2+2k)(6k^2+12k+6)
Which then, according to wolfram expands out to 24 k^5+84 k^4+108 k^3+60 k^2+12 k
ah wait

And according to wolfram, 2k(2k^2+3k+1)+6(k+1)^2 expands to 4 k^3 + 12 k^2 + 14 k+ 6 : )

(4k^3+6k^2+2k)(6k^2+12k+6) should be (4k^3+6k^2+2k)+(6k^2+12k+6)

yes
@Matt: Let me know if it worked.

1

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...

is that Hilberts "theorem 90"?

will do, still working it out

5:07 PM

I want to work some mathematics but I can't think what :S

5:24 PM
@QED, Im studying some discrete questions you could work them with me :P

yes
I would help if I can

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.
Thats what im starting on now

Have you proved that prime factorization is unique?

Not yet, just getting started

You can prove that every natural number has a prime factorization by "strong" induction.
Then you can prove that it is unique by induction on the number of prime factors.
This would be excellent practice if you are learning how to use induction, and as a bonus you can solve (a) with it

5:33 PM
Is that the standard method of proving this or something fancy? lol
My inital thought for a is that if n is divisible by 7 then n = 7k

I know two ways to prove it and this is the simplest and most straight foward

Matt: Well done! : )

@Matt I feel like its incomplete, any recommendations on a therefore at the bottom?

can I give a style tip
Instead of proving that 2A = 2B prove that A=B

@Matt: Therefore "For all integers ... " (what you have on line one)

5:39 PM
@QED what do you mean?

You proved \sum_{i=1}^n 2i = 2 n(n+1)(2n+1)/3
If you instead proved \sum_{i=1}^n i = n(n+1)(2n+1)/3 you wouldn't have to carry around these useless '2's for the entire proof

@QED: That's how I got my sums right in the end : P
First I did it with the factor of 2 twice and messed up twice.

Really? is it like a constant in an integral for sums?

something as simple as: \sum_{i=1}^n i = n(n+1)(2n+1)/3 implies \sum_{i=1}^n 2i = 2 n(n+1)(2n+1)/3 -- can be left implicit in proofs-for-human

You can just factor it out?

5:41 PM
just think about it this way: You can prove 2A = 2B by proving A = B
do note that this is just stylistic, it doesn't have any impact on the mathematics

I'm coming up with recipes again. Huzzah.

ah, ok

all you actually need to do to prove this is check it for 4 different values of n
using induction is very very weird
(above I was writing i inside the sum when I should have written i^2)
something like unique prime factorization makes use of induction for the right reasons

What do you think of my proof for 1a?

That is not a valid proof

5:48 PM
It isnt?

You need to use the unique prime factorization thing I've been going on about (or at least Euclids lemma)

hmm

Stylistically, there is a problem because you didn't introduce 'k'
if you try to fix the proof by introducing 'k' properly you'll notice where the mathematical error is

yes
What do I need to do regarding unique prime factorization

sorry what do you mean?

5:53 PM
How would I begin the proof using that method

do you know the theorem?
it says that a natural number n is a product of primes in exactly one way
so if a^2 = 7k, take the prime factorization of both sides
is it not making sense?

Slightly
The prime factorization of the right side would be..

for example, 2541^2 = 7*922383

7(k)?

no
k is not necessarily prime
let's consider a concrete example

5:58 PM
ok

the prime factorization of both sides of the above is: 3^2*7^2*11^4