Mathematics - 2016-02-10 (page 1 of 3)

Oh, that's the multivariable math book. Also they had web-based homework based on the course. I wrote something like 400 problems that I put up there.

Mike Miller

I managed to sneak in enough time between obligations to answer an interesting question this morning. I guess now's the time for more of that.

Ted Shifrin

12:24 AM

BTW, I glanced at Demailly. His proof of Kodaira vanishing was basically the same.

I'm suspecting it's Kodaira's proof.

Mike Miller

I've come to believe I shouldn't expect an inspirational proof, because $p+q>n$ is sort of an odd condition.

It's gotta come from some formula, and the calculation of $[\Lambda, L]$ is as good a formula as any.

Ted Shifrin

Well, it's not so odd. You're just in a higher level of the Hodge grading than $n$.

Michael

223$ .__.

Mike Miller

Fair enough ... but no inspiration.

PVAL

Anyone know any mathematica?

Ted Shifrin

12:29 AM

I used to know a lot of it, @PVAL, but I'm old and grizzled now.

What are you trying to do?

PVAL

I have a program where the first line is n= some integer. I'd like it to run on a list of integers.

0

Q: Running an entire notebook with different values of an input

I have about a 20 cell notebook which begins at n=3. By replacing 3 by other natural numbers, I can compute what I want using these values, but I do not understand how to automate this process. I'd like it to return all the values when $n$ is in some range (for instance $n\in [3,100]$). The fact...

programming

I'd figure there is a way to do this without tearing up the code.

Ted Shifrin

A random list, or a sequential list, like $n=2,\dots,100$?

PVAL

the second one.

Ted Shifrin

Just a Do[ ..., {n,2,100}] will do it.

PVAL

Over the whole page of code?

Ted Shifrin

12:33 AM

Well, there's a way to turn the code into a procedure that depends on n, and then just do Do[Proc[n],{n,2,100}] if that makes you happier.

You don't know much Mathematica?

PVAL

nope

Ted Shifrin

I wrote my diff geo students a baby primer on mathematica. You might like it for starters.

PVAL

Doing it over the whole code doesn't get an output an error or a running message at the top.

I just want to do one thing at this point...

@TedShifrin You can probably figure out what the matrices I have are.

Ted Shifrin

First of all, you need the whole thing in one cell.

If you want to email me the notebook, I'll look at it while I have my martini soon :)

PVAL

@TedShifrin Alright I sent it. I renamed it something non-descript to see if you can guess what it is.

Ted Shifrin

12:42 AM

LOL, gee thanks.

Michael

what is this sign? $ λ$

Mike Miller

That appears to be a blackboard bold lambda.

Ted Shifrin

you can't do blackboard bold greek

@PVAL: So what output do you want printed for each n?

Michael

How can I just bold it?

Ted Shifrin

you need AMSfonts and then boldsymbol

Mike Miller

12:47 AM

$\bf \lambda$?

Ted Shifrin

I wouldn't worry about it, @Michael.

Mike Miller

guess not

Ah, just like in class, nobody on the internet likes me when I rant about foliations either. :)

PVAL

@TedShifrin The last line.

Count[f, True] - ((2 n - 2) (2 n) - Count[f, True]) the output of this

Ted Shifrin

OK.

Presumably with the value of n.

PVAL

That'd be nice. Though if its in order I can figure it out.

Ted Shifrin

12:52 AM

So it's printing out -16 for your example.

That's correct?

PVAL

For n=3 that should be right

Ted Shifrin

OK, I've got it for you. It'll give you a table with n and your count.

PVAL

Wow thanks a lot.

Ted Shifrin

For large n it may jam up Mathematica. Them is big matrices.

It did n=2,...,5 in seconds, but it's still trying to get to 10.

I just returned your email.

PVAL

n=10 would be a (18*22) by (18*22) matrix.

Ted Shifrin

12:57 AM

Yeah, this is why serious numerical people do Matlab instead of Mathematica.

PVAL

I aint one of those though. I thought modern CAS's wouldn't care about 100 500by500 examples

Ted Shifrin

Mathematica tries to do everything formally rather than numerically, which gets to be a problem with huge problems.

Did you get my email?

PVAL

Ya I got it thanks

Ted Shifrin

You can tell Mathematica to do it numerically. We may have to do that.

PVAL

I'd imagine just putting //N after the eigenvalues part would do that.

Ted Shifrin

1:12 AM

Yes, you should definitely do that.

PVAL

Did you figure out what you were computing?

Ted Shifrin

I assume you're computing a signature.

PVAL

Ya but of what..

Ted Shifrin

I didn't pay enough detention.

PVAL

It should be of the Milnor fiber (2,2n-1,2n+1)

Ted Shifrin

1:14 AM

Oh, if I had paid detention, I wouldn't have known that.

Oh, my Mathematica finally finished through n=10.

L33ter

hi @TedShifrin

Ted Shifrin

Hi Karim

PVAL

Well It took like a minute and a half to get to 10.

Ted Shifrin

Whoa, your computer must be a lot more powerful than my IMac, @PVAL.

PVAL

This was after //N

Ted Shifrin

1:15 AM

Ah :)

L33ter

@TedShifrin the edition I have of hatcher defines delta complex than the hatcher book that I borrowed from library

my edition defines it in terms of maps

Ted Shifrin

I should have thought of that while I was reformatting, @PVAL.

L33ter

which I don't understand

PVAL

I can probably make Q sparse. I am not sure if that would help at all.

Ted Shifrin

@PVAL: If you prove something famous, I expect a tiny footnote for helping :D

L33ter

1:16 AM

so what are we doing here

are we breaking up the space X into simplices?

Simeon

Is a coordinate patch always a diffeomorphism?

PVAL

@TedShifrin See I am pretty sure this puts me on the tiny footnote of something famous.

I don't know if you have room.

L33ter

maybe I should collect all of my confusion and post it as post in mathstackexchange

I will do that

the way the first edition of hatcher does delta complexes is much easier to understand

Ted Shifrin

LOL, @PVAL :)

@Simeon: Yes, if you're doing differentiable manifolds.

L33ter

hey @TedShifrin I was marking some question in vector calculus yesterday. Just wanted to be sure my intuition is right since I didn't do this stuff since long time. Suppose we evaluate the limit of the following two variable function $e^{x^2 - 2y}cos(x + 2y)$ with input (2,-1)

so this function is continuos, so we can just input the value.

Ted Shifrin

1:21 AM

Yes, Karim, provided you "know" it's continuous.

L33ter

One student however showed that along the path x = y and y = 0 they have different limits, and said the limit doesn't exist that way.

but, what is wrong in his reasoning that the path x = y doesn't have (2,-1)

so that is why he can't do that right

Ted Shifrin

Yes, Karim, you're breaking $X$ into things more or less homeomorphic to simplices.

L33ter

I see

Ted Shifrin

But he's wrong about that, anyhow, Karim. Those paths intersect at $(0,0)$ and the function is certainly continuous there too.

L33ter

yeah

why do we need the third condition in the definition above ? also, I don't see this remark "Among other things, this last condition rules out trivialities like regarding all the
points of X as individual vertices."

what does he mean by that

1 second @TedShifrin I will just work stuff out in this delta complex definition of the second edition and come fully with all of my questions

to make things clearer in my mind 100 %

Ted Shifrin

1:27 AM

The last condition is saying the topology all works out the way you think it should.

Semiclassical

evening chat

Ted Shifrin

evening, @Semiclassic

user276387

1:47 AM

I'm learning modular arithmetic. If we have $x^2 \equiv 1\mod(pq)$ where $p$ and $q$ are odd primes, I understand that we do $(x-1)(x+1) \equiv 0 \mod(pq)$ which implies $x \equiv 1\mod(pq)$ or $x \equiv -1\mod(pq)$. Okay, so far so good. But could someone please explain why we have extra solutions coming from $\mod{p}$ and $\mod{q}$?

Ted Shifrin

@user276387, your "and" should be "or."

You can only assert that $xy\equiv 0\pmod m$ implies $x\equiv 0\pmod m$ or $y\equiv 0\pmod m$ if you know that $m$ is prime.

For example, $2\cdot 3\equiv 0\pmod 6$, but neither $2$ nor $3$ is divisible by $6$.

user276387

Okay, I understand. So what I said would work for $n^2 \equiv 1\mod{11}$ for example, but it wouldn't work if the argument is composite.

Ted Shifrin

Right :)

But if a number is divisible by $6$, it is divisible by both $2$ and $3$.

BTW, I assume $p$ and $q$ are distinct odd primes.

(Not that the odd matters.)

user276387

How do I show the implication $(x-1)(x+1) \equiv 0 \mod{pq} \implies x=-1 \mod p ~ \text{or} ~x=1 \mod q$

Ted Shifrin

It's what I just said, @user276387: If $pq$ divides a number and $p$ and $q$ are distinct primes, then both $p$ and $q$ must divide the number.

user276387

2:02 AM

Ohhhh yes. :D

Ted Shifrin

Oh, what you wrote isn't right.

There are four options.

Semiclassical

no reason to prefer $(x-1)(x+1)$ to $(x+1)(x-1)$, to put it one way

Ted Shifrin

Well, you need to say that $x\equiv \pm 1 \pmod p$ and $\pmod q$ both.

PVAL

@Ted Well from 2 to 30 has been running for 45 minutes

Ted Shifrin

LOL, those are monstrous matrices, @PVAL :)

Semiclassical

2:04 AM

@PVAL what are the matrices, out of curiousity?

user276387

Many thanks, @TedShifrin.

Ted Shifrin

Sure, @user276387.

PVAL

Ya my rationale was that the nth matrix was of quadratic dimension (which is true), and its m/n times harder to compute the eigenvalues of an$ m \times m$ matrix than an $n \times n$ matrix (which isn't close to true...).So 30 should take around 30 minutes if 10 took 3. That isn't the case.

@Semiclassical Intersection forms of Milnor fibers.

Semiclassical

ah. what kind of a matrix problem is it, though? i'm more curious about the numerical analysis

L33ter

I hate studying with background noise

there some people next to me who are super loud

PVAL

2:08 AM

Computing signature. I did it in the terrible way of taking the number of positive eigenvalues minus the number of negative eigenvalues.

It's probably a lot easier to do Sylvester's Law directly.

Ted Shifrin

I think there's an exponential in there, @PVAL. :)

PVAL

if one properly implements it..

Semiclassical

gotcha

Ted Shifrin

Yes, doing row/column operations is far more efficient.

Semiclassical

what's the form of the matrix?

PVAL

2:10 AM

Does mathematica have sylvesters law built in?

Ted Shifrin

I don't know the answer to that, @PVAL. If you can't google, you can ask on Mathematica.SE.

PVAL

@TedShifrin It seems I already have googled it once...

Ted Shifrin

ah ... then ask on Mathematica.SE. I don't know the answer.

Semiclassical

i just googled and came across your Mathematica answer @pval

and i'm amused to say that i know how to find the exact eigenvalues of that (well, exact as roots of chebyshev polynomials)

PVAL

Probably I should do what's done here mathoverflow.net/questions/107593/computing-signature

Ted Shifrin

2:13 AM

At any rate, @PVAL, I taught you a little of the right way to program this in Mathematica.

Semiclassical

that very matrix, except with 3 being a free parameter, is actually part of the analysis that showed up our last paper

...oh, wait

i missed the extra 1s along the top line :/

Ted Shifrin

See youze guyz.

PVAL

@Semiclassical The matrices are essentially symmetric block matrices with those kind of matrices along the diag and an upper triangular matrix with only 1's in the possible nonzero coordinates on the off diagonal. The one in that mathematica post is not even as a quadratic form and all these are.

Semiclassical

hmm

does it have structure beyond that? block matrices with some overall structure (say, toeplitz or hankel) tend to be a lot more tractable

PVAL

i dont know

Semiclassical

2:24 AM

mmkay

i may play around with the example you gave in the linked question and see if I can point anything out

PVAL

Hmm, I guess the 30th case is around a 3600 by 3600 matrix. I guess that probably means it will take a while. The MO link said a 3000 by 3000 would take an hour on a well optimized machine.

Semiclassical

yeah

PVAL

and a well-optimized program (which I am pretty sure I didnt write..)

A = DiagonalMatrix[ConstantArray[-1, 2 n]] +
ConstantArray[-1, {2 n, 2 n}];
B = SparseArray[{{i_, i_} -> 1, {i_, j_} /; i - j > 0 -> 1}, {2 n,
2 n}];
Y = Transpose[B];
W = ConstantArray[0, {2 n, 2 n}];
mat = Normal@
SparseArray[{{i_, i_} -> a, {i_, j_} /; i - j == 1 ->
b, {i_, j_} /; j - i == 1 -> y, {i_, j_} /; Abs[i - j] > 1 ->
z}, {2 n - 2, 2 n - 2}];
Q = SparseArray[
ArrayFlatten[mat /. {a -> A, b -> B, y -> Y, z -> W}]];

There's the mathematica code for the nth matrix.

n starts at 2

Semiclassical

just to check, B is an upper triangular matrix of 1s?

but, actually, a better check

PVAL

Ya thats right

Semiclassical

2:32 AM

is it essentially the same as equation 1.5 of this link?

arxiv.org/pdf/1108.5155.pdf

with A,B as r-by-r matrices

that paper is a bit broader than that, insofar as it's interested in perturbations of that case. but it seemed an obvious reference poitn

PVAL

Ya thats right.

Semiclassical

mmkay

my interest right now is in the spectrum of such matrices, albeit only for the case of $r=2$ for now

PVAL

Ah A is negative definite.

Semiclassical

though it looks like your matrix is $n^2$ dimensional, which is troublesome

PVAL

So they seem to prove things there

They are n(n-2) dimensional

Semiclassical

2:35 AM

ah

main thing is that the blocks are also growing in size rather than just the block matrix itself

which isn't so fun

PVAL

I take it back. A probably isn't negative definite.

It is when its 8 dimensional though..

Semiclassical

yeah, i can see why the numerical analysis is tricky if you tackle it head on

it has a cute matrix plot, at least!

the thing I key into, looking at that plot, is that you can write $A=-B-B^T$

Semiclassical

3:05 AM

@pval I think this is better optimized, at least for the sizes i tried (n=50, for example)

With[{n = 40},
B = SparseArray[{{i_, j_} /; i - j >= 0 -> 1}, {2 n, 2 n}];
mat = Normal@
SparseArray[{{i_, i_} -> a, {i_, j_} /; i - j == 1 ->
b, {i_, j_} /; j - i == 1 -> y}, {2 n - 2, 2 n - 2}];
Q = SparseArray[
ArrayFlatten[
mat /. {a -> -B - B\[Transpose], b -> B, y -> B\[Transpose]}]];]

with the old code, it takes about 3 seconds (according to the AbsoluteTime[] command). the new one takes 0.2 seconds

that's just on generating the matrices, though. haven't done anything spectral yet

Gridley Quayle

Could anyone give some hints for this problem? If f from R to positive R is continuous and periodic period t prove for all a that the integral of f(x)/f(x+a) from 0 to t >= t.

All I've got is that equality occurs for a=kt so far.

PVAL

3:20 AM

@Semiclassical Ya when I was doing it one at a time. The generation (and computing det as a sanity check) was pretty much instant. Ill look at it though.

Maybe you get something pretty nice when you "block" row reduce it (these preserve the signature).

Semiclassical

i'd expect so, yeah

PVAL

I left the computation up to n=30 on tonight. I'll let you know if it returns anything tomorrow (I don't think I am hopeful).

Semiclassical

3:43 AM

@PVAL have you tried using Method->"Banded" for eigenvalue calculations? it appears to help quite a bit

for example, doing

Sign[Eigenvalues[N[Q], Method -> "Banded"]] // Total

takes less than 10 seconds for me

Thomas Rasberry

Hello everyone

PVAL

@Semiclassical I went home and dont have access to mathematica till tomorrow. I'll try that though.

Semiclassical

mmkay

PVAL

@Semiclassical I have done essentially nothing numerical before.

Semiclassical

gotcha. a lot of the stuff i've done over the last few years has been precisely this kind of large matrix stuff in mathematica

PVAL

3:54 AM

@Semiclassical I should mention that this is an interesting question to me when the blocks are size $q-1$ and the block matrices are size $r-1$ (not just when q=2n,r=2n-2) (especially when q and r are coprime and odd which guarantees the determinant is $\pm 1$).

If one interchanges q and r the matrices are actually equivalent as bilinear forms over $\Bbb Z$.

Julian Rachman

Can anyone check my analogy and see if they understand it or if it is a good analogy pertaining to this paper?: sharelatex.com/project/569c7839de5631c80dafbfad/output/…

Thomas Rasberry

4:10 AM

Semiclassical

@pval neat

Thomas Rasberry

I saw Category Theory way too many years ago, Julian. I probably would be able to help you in a year but I'm busy recovering what I lost. I'm in my last grad course (a second course in abstract algebra) and I'm struggling with module tensors via the way Dummitt/Foote introduces them without categories.

Julian Rachman

@ThomasRasberry well, I don't think you need any category theory for the analogy I made in the first paragraph of my paper.

That first paragraph is what I would like people to take a look at and see if the understand.

Thomas Rasberry

I would use "is related" instead of "is relate," but yeah, that's a great paragraph.

Julian Rachman

Thank you! I will change it right now.

Thomas Rasberry

4:20 AM

I have never seen the empty-letter concept before. Is that so each k-dictionary A^k can include all dictionaries A^n, n<k?

Julian Rachman

yep

@ThomasRasberry So if you take a look at Theorem 1.1, could you say that by reading my analogy and then reading the theorem, you could grasp a better understanding of the meaning and importance of the said theorem?

Thomas Rasberry

Yes, speaking from the perspective of never encountering quasi-orderings myself, I can understand Theorem 1.1 with that analogy.

Julian Rachman

@ThomasRasberry Ok. Thanks for the help!

Actually, since I have already gotten you into this (sorry...), do you think my abstract is strong or needs work? Suggestions?

(@ThomasRasberry)

Thomas Rasberry

//The problem that arises is that these
formulations do not provide any applicability to other fields of mathematics and are only
focused based on a computational point-of-view.//

Perhaps rephrase concisely as "These computational formulations lack applications in other fields of mathematics." Otherwise, it's perfectly readable to me!

Julian Rachman

4:36 AM

Ok. Thanks! I just fixed it so everything should be right.

Thomas Rasberry

On page 3 line 3, do you mean a_j \not \sqsubset a_i instead of using x there?

Julian Rachman

Ah yes. Thank you for the catch.

If you reload the page it should be fixed.

I appreciate it that you are "attempting" to read my paper. It is still in the works but I have the final proof already done. I just need to put it in words.

Thomas Rasberry

No problem! It gives me practice. I'm doing my dissertation now and having to learn tons of new things on the fly that I haven't encountered in a classroom.

Julian Rachman

Ah. I wish you luck and I hope that once you are done with your paper I may read it and you can read mine :)

Thomas Rasberry

Page three, definition for "good" infinite sequence isn't clear to me, it has two consecutive "if and only ifs." Do you mean to say a "good" infinite sequence is not an infinite antichain and not incomparable?

Otherwise it is perfectly readable. I can't go into Section 3 since I am too rusty on categories, but I understood it up until then just fine.

Julian Rachman

4:57 AM

Ya. I do mean that an infinite sequence is good iff it is not an infinite antichain nor imcomparable.

How should I re-word it?

Wait. Are we talking about "good" or well-foundedness? @ThomasRasberry

Julian Rachman

5:51 AM

@ThomasRasberry Just made a huge update to the paper (sharelatex.com/project/569c7839de5631c80dafbfad/output/…). Let me know what you think.

Huy

7:06 AM

@PVAL yes, I'm actually going through the proofs in Farb and Margalit's text but some of the proofs skip details that I'm not perfectly familiar with yet, so I'm trying to fill those.

usukidoll

7:16 AM

hiiiiiiiiii

Tobias Kildetoft

7:53 AM

@Huy Proof of what?

Huy

@TobiasKildetoft: the bigon criterion

Tobias Kildetoft

@Huy Doesn't ring a bell (other than it sounds like "big one")

Huy

@TobiasKildetoft: let $\alpha, \beta$ be two transverse simple closed curves in a surface $S$. we say they form a bigon if there is an embedded disk in $S$ whose boundary is the union of an arc of $\alpha$ and an arc of $\beta$ intersecting in exactly two points. the bigon criterion states that $\alpha$ and $\beta$ are in minimal position (i.e. the intersection number cannot be lowered by free homotopy) if and only if they do not form a bigon.

Tobias Kildetoft

8:15 AM

@Huy Hmm, I think I understood most of that, apart from the fact that I don't quite remember what the intersection number is, nor do I know what free homotopy is

Huy

@TobiasKildetoft: intersection number in this case is just $|\alpha \cap \beta|$

and free homotopy is just a homotopy without a fixed point

Tobias Kildetoft

ahh, ok

Huy

it's actually very intuitive but not quite that easy to prove

Tobias Kildetoft

so the idea is that any bigon can be moved to make the two intersections into one via a homotopy?

Huy

or even zero

one direction is really easy

if they form a bigon, then they are not in minimal position

the other direction is the one that requires work

Tobias Kildetoft

8:19 AM

So, you pick an intersection that can vanish, and you need to find another one where the curves form a bigon. Yeah, I can see that that becomes tricky

Huy

@TobiasKildetoft: the proofs are actually very "different". there's a topological and one that uses hyperbolic geometry that I know. they take advantage of a lemma that states that if two transverse single closed curves do not form any bigons, then any lifts of them to their universal cover intersects at the most once

Idle001

9:24 AM

is there sthing wrong in my answer here ?

Tobias Kildetoft

@Idle001 It lacks explanation I would say

Idle001

oh look that i just added an example+summary

Tobias Kildetoft

@Idle001 The explanation for what you are doing should come first. And there is no point of an example if the rest is explained properly

also, I have no idea what the second line says

Idle001

i thought it can be got in mind for some1 habitual with recurrence relations

second line ?

Tobias Kildetoft

second line of your answer. Apart from using the word "indicate" to mean "equal" which confused me at first, it seems to set $2a_{n-1}$ to two different values

Idle001

9:32 AM

nvm , i did my effort

keep downvoting, i m living this fact with a big smirk

Tobias Kildetoft

@Idle001 I didn't bother to downvote it, I just tried to explain what was wrong with it

I assumed you were genuinely interested since you asked here, rather than just expecting to be told that the answer was perfect and people were fools for downvoting it.

Balarka Sen

@L33ter Tell me your intuition.

@Huy Have you tried out the thing with stability I pinged you with? I am not sure if it works, but I think that's worth giving a thought.

Huy

@BalarkaSen: no but I'd rather not introduce even more things that I've never heard of to be honest

Balarka Sen

Internet died out yesterday, sorry for leaving so abruptly.

Huy

no problem

Balarka Sen

9:44 AM

Fair enough. Stability of transversality is something you can take as a blackbox though.

Huy

@BalarkaSen: if you have no other things you have to do first and are interested at all in the topic, you can check out the proof and what comes before in the PDF available online. it's pretty much at the beginning of the text.

Balarka Sen

The proof of what, precisely?

Huy

maybe I've overlooked some important argument or so

@BalarkaSen: the bigon criterion

in the second topological proof they state "thus, $H^{-1}(\beta)$ in the annulus is a $1$-submanifold"

I think you're onto something with the stability thing though, just with different words than I'm used to

Balarka Sen

@Huy Oh, so they do prove that claim?

Idle001

@TobiasKildetoft sorry i really still want to know why did i recieve negative reaction, for correcting myself as first reason

Newb

9:51 AM

anyone up for a good probability problem?

I've made substantial headway but am looking for someone to help me finish it off

Huy

@BalarkaSen: they do mention that wlog they assume $H$ to be transverse to $\beta$

that's what you meant by stability?

Balarka Sen

The prospect of reading and learning something new is definitely very exciting, @Huy, but unfortunately I have a lot to do before. I need to get started on those.

@Huy Uh, I thought $H$ is just a homotopy of $\alpha$ which reduces the intersection number to minimal? What is true is that since $H(-, 0) = \alpha$ is transverse to $\beta$, so is $H(-, t)$ for all $t < \epsilon$ for some $\epsilon$. This is the stability theorem (transverse things remain transverse under small perturbations)

Huy

ok

Tobias Kildetoft

@Idle001 The answer is still poor, so I don't see why you would expect otherwise.

Huy

@BalarkaSen: it's hard to explain a proof that I haven't understood yet

Balarka Sen

9:54 AM

Oh, but if you assume $H$ is transverse to $\beta$, then certainly $H^{-1}(\beta)$ is a 1-fold. This is a consequence of transversality theory.

Huy

that's why I keep telling you to look for yourself. :P

Balarka Sen

I will, but first I need to do a few things :)

Gridley Quayle

If we have a set of n (n distinct numbers such that the sum of all n are equal to 0, the amount of possible walks that always stay positive I've conjectured to be 1/n. This works for cases n=1,2,3,4 and Matalb supports my conjecture for larger cases but I haven't yet found a proof. Any help

? Want to make sure it is not trivial before posting on the main site

Huy

ok, I'll see if I know what the transversality theorem claims

Balarka Sen

oops, I mean transversality theory. have a look at Guillemin-Pollack.

Tobias Kildetoft

9:56 AM

@GridleyQuayle What is a walk in this context?

Huy

@BalarkaSen: I'll finish some lunch first, then I'll do that :)

Balarka Sen

if $f : X \to Y$ is transverse to a subfold $Z \subset Y$, then $f^{-1}(Z \cap Y)$ is a subfold of $X$.

@Huy good luck!

Gridley Quayle

The sum of the numbers so the ith position is the sum of a_1 to a_i

Total number of walks is trivially n!

Tobias Kildetoft

@GridleyQuayle Are the number ordered beforehand?

Gridley Quayle

I thought if I assume the numbers are integers, then It would work for all rationals if we multiply by the largest common denominator. No, the numbers aren't ordered.

Idle001

10:00 AM

@TobiasKildetoft now ?

Gridley Quayle

As the rationals are dense in the reals, if proven for the rationals it should also extend into the reals So I'm considering integers for the moment as I've run out of tools to use

Tobias Kildetoft

@GridleyQuayle So a walk is basically picking an ordering, and by "staying positive" you just mean except at the last step where it becomes $0$? Or is it allowed to also be $0$ along the way?

Gridley Quayle

It is allowed to become zero along the way yes.

Tobias Kildetoft

@GridleyQuayle And by $1/n$ you mean that this should be the fraction of the total number of walks with this property?

Gridley Quayle

YEs

Tobias Kildetoft

10:05 AM

@GridleyQuayle Hmm, I would definitely say that it does not appear to be entirely trivial, though there may well be some neat counting trick that does it in an easy way

Gridley Quayle

I first thought it was a probability problem as it is similar to the ballot problem but am sure it is combinatorial now

Tobias Kildetoft

@GridleyQuayle So we can certainly assume that $0$ is not one of the numbers, though this does not seem to help much

@GridleyQuayle Hmm, what if we extend the problem at first, to consider sets of numbers summing to $m$ and walks staying above $m$?

Then if we remove an element, we get a smaller set where we should be able to solve the problem recursively

The issue is that we would need an element that can be inserted anywhere in such a walk

Gridley Quayle

I see at the moment how helpful that would be, I think I'll ask on the main site. Would this come under the random walk and combinatorics tags?

Tobias Kildetoft

@GridleyQuayle random walks are usually something done on graphs or grids, so it has a different meaning than walk here

Aayush Agrawal

10:23 AM

is there any software to help you write the mathjax?

Huy

@BalarkaSen: does the transversality theory also take care of why the preimage can't be of "Y-shape"?

Newb

10:39 AM

@TobiasKildetoft do you want a good probability problem?

I've made some progress but I'm stuck

Tobias Kildetoft

@Newb Not really

Newb

11:46 AM

@ThomasAndrews if you're around, do you think you can help me on a probability problem?

It's a good problem, and I've made some headway, but I am stuck

Idle001

12:01 PM

@robjohn sorry to bother you are you familiar with recurrence

Idle001

12:31 PM

@TobiasKildetoft i really dont understand ur couple of discouraging comments there

Tobias Kildetoft

@Idle001 I don't understand why you refuse to accept that your answers are bad

Idle001

why should i explain a blatant recurrent relation that gives me correct sums ?

Tobias Kildetoft

@Idle001 Because that is what math is about. Explaining why something is indeed correct

Idle001

what is more illustrating that 2 examples and step-by-step definition ?

knock knock, hello, any logic there ?

seems that some people have a serious issue with ME, not anything relating to my posts

Tobias Kildetoft

@Idle001 I am getting tired of trying to reason with you. Your answers are bad because they contain no explanations of where anything comes from or why they are correct. There is nothing more to it, and I really don't feel like repeating it any more now.

Transcript for

$Mathematics$ Mathematics