Mathematics - 2011-11-20 (page 1 of 2)

J. M.

00:00

Why yes, that is a dupe. Let me look around...

Hmm...

Srivatsan

Extra triangle constraint

J. M.

and this is a bit more general.

Srivatsan

the cutting sticks conjecture is still open, right?

@JM How is it general, J.M.?

J. M.

On second thought... cutting sticks isn't an exact fit... I didn't see the n(n+1)/2 at once.

robjohn

01:01

@JM The bounty generated a lot of attempts, though.

They just shouldn't let teenagers integrate around contours without a license.

Henning Makholm

I think my license has expired ... what on earth is going on there?

J. M.

Yeesh...

robjohn

01:22

@HenningMakholm In his question or my answer?

Henning Makholm

Mostly in his question -- but I'm afraid that's only because I haven't really tried to understand your answer yet.

I dimly remember something about evaluating doubly-infinite integrals by completing the real axis into a contour and summing over residues for the enclosed poles. But that depended on the "new" part of the contour being far away where the integrand had faded away. What happens near the 0 end of the integral here?

hmm.. setting xi=exp(2pi/5) I can see that f(xi*z)=xi^2 f(z), so we can integrate outwards along the real axis, around a large arc at some large radius, and come back to 0 along a ray from direction xi. The total of this will be (1-xi^2) times the integral, plus a term for the arc that falls away as the inverse square of the radius. Right.

robjohn

@HenningMakholm The integral is being taken around an infinite "wedge". The incoming piece is z=e^{2\pi i/5}x and the outgoing piece is z=x.

@HenningMakholm almost, there is a factor of e^{2\pi i/5} from the dz as well. so (1-xi^3)

Henning Makholm

@robjohn Yes, saw that in the comments. Got it now.

Benjamin Lim

@HenningMakholm A certain user posted 3 questions in the space of one hour, all in the imperative.....

robjohn

@HenningMakholm My answer sort of anticipated the OP needing to ask questions. That actually helps in that he has to think about the problem a bit more, and in that I don't need to lose his attention in one long monologue.

Benjamin Lim

01:36

@robjohn What do you mean about needing a license to do contous integration?

Henning Makholm

@BenjaminLim The imperative is unproblematic. The problem is asking a question that just consists of an exercise with no thoughts of the OP's own. To criticize the symptom (the imperative mood) rather than the cause (no independent effort) makes it unnecessarily hard for the OP to figure out what he's doing wrong and what he should be doing instead.

robjohn

@BenjaminLim It was just a joke... The question was posed in a much more derogatory way, but then was changed.

Benjamin Lim

You have said more precisely what I wanted to say. It is true that the OP has not put any effort into the questions

robjohn

@BenjaminLim The original title of the question was Does wolframalpha approximate definite integrals badly or did I just calculate badly?

At least he offered the possibility that he was in error.

Henning Makholm

We use the imperative all the time in mathematics. "Let f(x)=x^2+e^x" is an imperative. "Consider the following function" is an imperative. "Differentiate both sides of the equation and then divide by g(y)" is an imperative.

Benjamin Lim

01:39

@robjohn I don't understand what is derogatory about the original title?

@HenningMakholm You're right. I think I wanted to say what the OP posted was as if he was ordering people to do the homework without him trying, like what you said.

robjohn

@BenjaminLim When I saw Does wolframalpha approximate definite integrals badly I felt it derogatory.

Henning Makholm

Right. But I wish people would focus more on "without him trying" than on "ordering" when they leave such comments.

Benjamin Lim

Yeah. I like coming on stackexchange because I know people can correct me when I'm wrong.

robjohn

@HenningMakholm Indeed. I don't see the purpose of criticizing the imperative. That gives me the impression that we are a bunch of prima donnas.

Benjamin Lim

@robjohn Next time I will only focus on the OP not trying. You mean you felt it degraded wolfram alpha?

robjohn

01:44

@BenjaminLim not degraded, just a bit of disrespect. It was only enough to make the joke about driving=contour integration

Henning Makholm

@Benjamin, there is some prior discussion (and links to a longer discussion here) on this matter in the "Imperative questions" room. Unfortunately nothing concrete ever really came of it.

Benjamin Lim

I will take a look

robjohn

@Benjamin: don't get me wrong, I don't think that Wolfram is infallible.

Benjamin Lim

I know what you mean. It is not infallible, sometimes when evaluating integrals it gives complicated answers when the integral is actually simple

robjohn

@BenjaminLim That is very true.

Benjamin Lim

01:46

@robjohn @HenningMakholm I will go for lunch now. See you guys!

robjohn

@BenjaminLim I have had some people ask me why they got the answer wrong because the answer from Mathematica looked so different.

Henning Makholm

Do chatrooms decay if nobody speak in them for some time? Perhaps it would be worth it starting a meta question on the imperative matter just for archiving purposes.

robjohn

@HenningMakholm Oh, no. WolframAlpha particle decay of chatrooms... Film at 11!

Henning Makholm

Um.. random association game?

robjohn

Word association football

I have to take my dog to the park. bbl

Was Film at 11 a foreign phrase?

@Henning: In the US, at least, it is a joke phrase to imply fake importance. It comes from the news stations leading people to watch the late night news. They announce something and then say "Film at 11"

Henning Makholm

01:54

I'm aware of "film at 11", but the alpha particle decay seemed a bit random.

(Though of course, that is how radioactivity works).

robjohn

@HenningMakholm I was just positing a type of decay since you were wondering about chatroom decay.

See you in a bit.

J. M.

@HenningMakholm Nope. As long as there are x number of messages in it, it won't disappear. On the other hand, they do get archived (can't post in it anymore) after y days of inactivity. (where x and y are written somewhere in meta.SO).

Henning Makholm

Okay. Here, I think: meta.stackoverflow.com/questions/104677/chatroom-deleted

Which says that the threshold is 15 messages by at least 2 users. Should be safe, then.

J. M.

Great.

Srivatsan

02:17

I think the OP misunderstood our insistence of manners... =)

Henning Makholm

A propos of the discussion just above ...

Srivatsan

I took back my downvote because I realised I did not understand your answer, @J.M. =)

Will put it back when I get it...

Henning Makholm

I write a nice answer to VVV, and in return he threatens me with armored fighting vehicles...

2

Srivatsan

02:33

Well, I was going to post a comment about blue Rudin, but the name seems somewhat standard. Like baby Rudin. E.g., abebooks.com/Principles-Mathematical-Analysis-1964-blue-Rudin/…

J. M.

03:04

@Srivatsan What? I was out (and terribly distracted.)

(If this keeps up, I might have to take the day off.)

robjohn

@HenningMakholm He threatened me with the same attack force.

I am lame at getting the correct link

Matt

Hey guys can I get your input on a proof?

robjohn

@Matt which proof?

Matt

imgur.com/8wFIb

This one I did

Im not sure my use of modus ponens is valid

nor negation for that matter

Srivatsan

03:23

@JM Sorry, I felt I understood, but on second thought I did not. Sorry about that... =)

Henning Makholm

@Matt, why not just use p|ab => p|a or p|b?

Srivatsan

I too find it hard to concentrate today. (Watching a movie now =)) // I might go through the answer tomorrow...

Matt

@HenningMakholm This is just how I thought to do it

J. M.

@Srivatsan Which one? I'm currently repairing at least two answers I was writing when I got interrupted...

Srivatsan

The Dirichlet convolution one.

I even wrote a comment nitpicking about something. I deleted it realising that I might be the one who got it all wrong, after all =)

J. M.

03:27

Ah. I'm giving the entire gory details just so nobody can complain it was too brief.

Srivatsan

Aw, good for me then.

Henning Makholm

@Matt, is this for a course in formal logic? Otherwise I'd say you're putting too much emphasis on the mechanics of logical reasoning. It dwarfs the actual number theoretic content of the proof.

For the idea you're using, I would just say something like: By the FTA, let p1^e1 .. pn^en be the prime factorization of a. Then the prime factorization of a² is p1^2e1...pn^2en, so one of the p_i's must be 7 and the exponent 2e_i must be nonzero. Thus e_i is also nonzero. Since 7 occurs in the prime factorization of a with nonzero exponent, 7 divides a. QED.

Matt

03:44

@HenningMakholm yep, its for a discrete math class, im using all my new tools :P

mixedmath

Henning - "Tanks - Battleships!"

that was very funny

Henning Makholm

@Matt Looking a bit further at it, I don't quite think it holds together. It looks like you're keeping some quantifiers on p_i and e_i implicit, and that just won't do when you're trying to be formal about things.

As the proof reads, it looks like you're defining the p_i's and e_i's to be the prime factorization of a. THen you're letting P be the claim a =/=7^e1*p2^e2*..., which is true if and only if p1 happens to be different from 7 -- and Q be a^2=/=7^2e1*p2^2e2*..., which is true if and only if p1 happens to be different from 7.

Matt

ok

So all I know is that 7 definitely divides a^2

J. M.

I hope I haven't forgotten anything...

Matt

So basically what im trying to say by a^2=7^2e1p1^2e1...

is that 7 is a factor of a^2

which must be true if it divides

Henning Makholm

03:56

You probably meant to let P be the statement "there exists e1,e2,... and p2,p3,.... such that a=7^e1*p2^e2" -- but you're not allowed to let the "there exists" be silent. Your argument will be unintelligible then.

Matt

So how would I express this?

Srivatsan

@JM My +1. I certainly did not see the answer at all, so I am glad I pushed you to expand the answer... =)

Thanks for the effort, b.t.w.

J. M.

...and people wonder why number theory gives me headaches... :D

Henning Makholm

@Matt, I don't immediately see a good way to get through if you start by factoring a^2. It seems to end up superfluous no matter how we proceed.

J. M.

(no problem, and thanks for the +1, @Sri.)

Matt

04:00

@HenningMakholm , Hm, ill have to ask the instrutor, thanks for the input, g2g!

Srivatsan

04:17

I am afraid we might drive away a fraction of the new users by our insistence on manners and showing work.

J. M.

"our insistence on manners and showing work." - if it means less crap, so be it...

Srivatsan

One more of those questions: math.stackexchange.com/questions/83835/probability-inequalities

Yes, I am not saying it's a bad thing.

Srivatsan

04:28

@JM I actually agree more with Andrew Stacey's comment below the post.

"If you really don`t want that much low quality questions filter China and India."

I found that comment hilarious... =)

J. M.

That's a bit below the belt, I'd say...

Srivatsan

Yes, it is. But since I am "from India", I have the right to enjoy it nevertheless. =)

J. M.

A programmer and a photographer? Then not likely...

Srivatsan

I want to ask the op why they are solving algebra homework. =)

Henning Makholm

I did.

Srivatsan

04:43

Ok. I am happy as long as I didn't have to ask =) . I even upvoted your comment, Henning...

@J.M. can you correct the "actully" now that you touched the post?

Thanks...

J. M.

Done.

Srivatsan

Thanks, J.M. // I would've done it but I was hoping you might catch up with the 5-min edit window. All is well though...

J. M.

I don't feel like cooking today. I'm eating out. See you guys later.

Srivatsan

Enjoy your lunch, @JM.

Mike Spivey

05:49

@JM Thanks for that link. I hadn't realized that the penalty for downvoting questions had been removed.

yayu

@sri that china and india comment was not by andrew stacey but by some user named zdravko.

a bit sad but true though

Asaf Karagila

07:11

Top of the morrow, idle chat mob.

Matt

07:57

Same to you : P

Zhen Lin

08:44

My laptop failed mysteriously and catastrophically. And I thought I was going to get some Ph.D. application bureaucracy done this weekend...

QED

yep computers

Alexei Averchenko

09:08

hi

i just learned about this monad and adjunction business

it feels weird

Zhen Lin

hah, so did I

for the last 2 weeks or so anyway

Alexei Averchenko

so if i take tangent bundle, zero section and projection from double tangent bundle to the tangent bundle, it's a monad? 0_o

Zhen Lin

It's very algebraic, compared to the other parts of category theory.

Alexei Averchenko

that's the point, the heart of differential geometry is a freaking monad 0_o

Zhen Lin

Did you check the triangular identities?

Alexei Averchenko

09:11

let's see...

looks like the answer is yes

not 100% sure, but if definitely looks like it

Zhen Lin

I don't think it does...

Monads aren't that easy to find, in geometry anyway.

Alexei Averchenko

if \eta is a zero section, T\eta is also a zero section, right?

i can't think of a sensible derivative for which this wouldn't hold

Zhen Lin

Ugh, I can't do that computation in my head, and I'm working on commutative algebra right now so my brain isn't in the right mode anyway...

Alexei Averchenko

now, \eta_{TM} is a zero section from TM to T^2 M, and by projecting it back down we get the identity

wow, this changes a lot in my head :D

Zhen Lin

Yes, that's obvious. But T\eta_M might be something different.

Alexei Averchenko

09:18

i wonder what its algebras are :D

but it's just the pushforward of the zero section!

since the zero section is constant in every chart, it has to be the zero section!

hm

maybe not

Zhen Lin

does x^2 + x y - 1 have any solutions in the integers...?

bleh, it has singularities at infinity... hmmm...

Alexei Averchenko

ok, i messed up about T \eta, it's actually not a pushforward :)

it's a horizontal composition of 1_T and \eta

and it looks right:

T \eta: TM \to T^2 M

(x, X) \mapsto (x, X, 0)

so under the projection we get (x, X) again

i'm still not sure <_<

Zhen Lin

It is the pushforward of \eta_M, no?

Alexei Averchenko

i better ask a question about it

hmmmm

take a curve (x(t), X(t)) (in some contractible chart)

wait

if it's lifted from M by the zero section, it's (x(t), 0)

right?

Zhen Lin

09:34

I guess

Alexei Averchenko

then we lift again to get (x(t), 0, (\dot x(t), 0))

so it's not zero there

so only the coordinate corresponding to the change in tangent vectors are zero

so if we apply the projection here, we won't get 1_{TM} (although i'm confused so we still might :D)

on the other hand, 1_T * \eta_m: TM \to T^2 M is...

you're right, it's the same thing >_<

oh well

Zhen Lin

If you're looking for monads in geometry, consider looking at descent theory. :p hehe...

J. M.

Oh, hell no...

Zhen Lin

verily

Alexei Averchenko

@JM how did she got in stanford? 0_o

can i get there if i play dumb too?

J. M.

09:48

All I know is, if I had to memorize stuff, I'd probably not be doing math as a hobby...

Zhen Lin

I wonder if the user QiL is Qing Liu, hmmm...

J. M.

The gravatars match?

Zhen Lin

No, but QiL does answer a lot of commutative algebra / algebraic geometry questions...

Matt

10:08

What a question and what a user name : (

J. M.

Backwards: [text](url)

Matt

that was my first guess and didn't work, I tried just now. Let me try again

hah! Cool, thanks @JM!

robjohn

@JM It's all memorization; that's how new discoveries are made!

J. M.

:'(

Or maybe "EYYAAAAAAAAA!!!!!1!"

robjohn

Either one works.

:o( sad clown

Zhen Lin

10:49

my fingers hurt, the room is too cold... :-/

J. M.

Do the bathrooms there have hand dryers?

Zhen Lin

Yeah, the ones for visitors. The normal ones don't...

J. M.

In that case, just rub your hands together...

QED

11:17

hot water

t.b.

11:34

I hate it when people delete their own questions (I gave this guy) one standard reference and he doesn't even mention it. Would have been kind of him to mention that this standard reference doesn't contain the result he's looking for, so that people don't have to waste time looking there first.

Asaf Karagila

What's up gang?

QED

hello

J. M.

11:57

@tb We can vote to undelete...

t.b.

@JM Okay, let's do that.

J. M.

Okay, one more needed...

t.b.

@Asaf: could you please vote to undelete here and then vote to close this as a duplicate?

Asaf Karagila

Yes! I could do that!

And I did do that as well!

t.b.

Thanks!

J. M.

12:06

One more jury member and an executioner needed.

Asaf Karagila

13:00

By golly, I got +3 so far, I didn't expect this sort of voting on this very long comment of mine. :-P

Skullpatrol

13:15

When I saw this post the first thing that came to mind was the quote "In mathematics you don't understand anything, you just get used to it" math.stackexchange.com/questions/83900/…

QED

13:31

isn't that physics?

huh! I thought it was freeman dyson on mathematics, actually it's von neumann on mathematics

Jonas Teuwen

13:42

666 rep @QED.

Skullpatrol

@QED Indeed it was von Neumann replying to a physical Chemist who said "I don't understand Mathematics"

@JonasTeuwen What does "666 rep" stand for?

Jonas Teuwen

666 Reputation.

Skullpatrol

@JonasTeuwen Thank you

with 666 being the devil's number?

Jonas Teuwen

13:58

No, 666 with three sixes concatenated.

Skullpatrol

Does anybody find it a bit strange that such a great mathematician would say "In mathematics you don't understand anything, you just get used to it"

QED

I disagree with it

a lot of von neumanns views are a bit wacky

Skullpatrol

I agree

with you QED

QED

John von Neumann

John von Neumann () (December 28, 1903 – February 8, 1957) was a Hungarian-American mathematician and polymath who made major contributions to a vast number of fields, including set theory, functional analysis, quantum mechanics, ergodic theory, geometry, fluid dynamics, economics and game theory, computer science, numerical analysis, hydrodynamics, and statistics, as well as many other mathematical fields. He is generally regarded as one of the greatest mathematicians in modern history. The mathematician Jean Dieudonné called von Neumann "the last of the great mathematicians", while Pe...

that's incredible

Skullpatrol

He is generally regarded as one of the greatest mathematicians in modern history.

@QED So... I had to dig deeper into the context of the quotation and found out that it was made during the research project called "The Manhattan Project" to a physical Chemist who did not understand the transformations needed for shock waves in fluids ...

QED

14:19

interesting

Skullpatrol

@QED I think von Neumann was just trying to tell the Chemist "the quickest way to use these transformations is to memorize them"

QED

yes

Skullpatrol

just like we all had to do with the Pythagorean Theorem or accepting division by zero is not possible

Alexei Averchenko

wow

Marsden, Ratiu, Abraham has all the ingredients I need to make tangent bundle a monad! there's dual tangent rhombic (that can be made into associativity axiom in two ways) and the canonical involution that makes them equivalent! 8)

i gotta work out the details 8)

i need a reference for 3-categories

Skullpatrol

14:42

@QED Do you agree that "Memorization is the beginning of intelligence"

QED

I don't think so

Alexei Averchenko

@QED, how do I say 'equivalence of natural transformations'?

QED

I'm not sure

Skullpatrol

@QED Well ... intelligent life must have some capacity for memory, since humans developed large brains to remember where they stored their food

do you see what I'm getting at?

QED

yes but I disagree with the reason you gave

oh you said it's large brains not memory, that developed for storing food

Skullpatrol

14:53

@QED large brains were needed to remember where the food was stored ...

so a good memory is a more basic form of intelligence

QED

you can have intelligence without memory: memory just increases your ability to make correct decisions

Skullpatrol

@QED I agree this discussion depends on what you call "intelligence"

QED

yes

Skullpatrol

@QED Thank you for your time and attention

"In mathematics you don't understand anything, you just get used to it"

t.b.

A somewhat unfortunate abbreviation...

J. M.

15:05

Tsk, tsk...

Zhen Lin

15:23

I know who this story is about!

J. M.

You know "AF" personally?

Zhen Lin

No, but I do know someone who matches the description perfectly.

I have no idea how plausible the story is though.

robjohn

15:41

@Henning: high five

We both commented about the same thing on the same question.

@JM: and to you, too :-)

J. M.

It's a pet peeve... :D

Quite a number of the notation questions have the OP neglecting to mention where s/he saw that confusing bit...

t.b.

The Schwarz of inequality / And lemma too, he has no T. / The "distribution" Schwartz, you see / Is French, and so he has a T., yet again :)

5

Alexei Averchenko

hi guys

do you know anything about that tangent bundle being a monad business?

t.b.

hi, no.

Alexei Averchenko

i have a rather obvious idea

t.b.

15:53

shoot

Alexei Averchenko

if for some bizzarre reason nobody thought about it yet, i'll have my first research paper :D

open marsden et. al.

there're two diagrams there

dual rhombic

t.b.

there are many Marsden et al. which one?

Alexei Averchenko

manifolds, tensors and applications

t.b.

page?

Alexei Averchenko

146 of my not-yet-published copy

there's dual rhombic which turns either \pi T or T \pi into the multiplication (satisfying the associativity axiom)

and one of the exercises is about the canonical involution

which is exactly modification (3-cell) between the two!

i strongly suspect that the zero section will complete the monad, but i haven't worked the details yet

t.b.

15:58

The problem is that I don't have the exact copy you have, could you please provide a link or give me the exact page here? Otherwise it's impossible to follow you.

Alexei Averchenko

p. 163 for the dual tangent rhombic

p. 166 for the canonical involution

(the diagram is incomplete, there are two arrows down to TM, one is Tt_M and one is t_{TM})

monads aside, this is still a very interesting pair of identities

one could become the multiplication axiom (after horizontal multiplication by T either from the left or from the right)

t.b.

Unfortunately, I can only see page 163, 166 is not currently displayed for me.

Alexei Averchenko

166 has this:

Zhen Lin

and I can't see 166, so between us we see all of it!

Alexei Averchenko

T^2M <------ s_M --------> T^2M

\ /
t_{TM} T t_M
\ /
TM == TM

hm

you get the idea, two morphisms from T^2M to TM

and s_M is invertible, in a chart s_M(u, e, e_1, e_2) = (u, e_1, e, e_2)

t.b.

16:05

robjohn

@tb :-D I once gave a talk at Princeton about Muckenhoupt weights. They are called A_p weights and in my talk I was talking about the quality of the weights and when I said something about their "A_p-ness" I felt very exposed, but no one else even giggled.

t.b.

Alexei Averchenko

yep :)

Zhen Lin

The rhombus is simply the assertion that \tau is a natural transformation, no?

Alexei Averchenko

no

J. M.

16:06

Alexei, is your book digital or physical? If digital, you might want to take screenshots instead of torturing yourself with ASCII art...

Alexei Averchenko

digital

theo just gave the screens

t.b.

@robjohn :D This sort of reminds me of the Italian saying in a talk (with strong accent): "Now we see with surprise that we have a wonderful R-action"

Alexei Averchenko

so my idea is to check out the zero section, basically

and my last ditch hope if it fails is to use the modification s to obtain some weaker notion of monad :)

does it sound plausible?

Zhen Lin

aaaaagh dimension theory is killing me

@Alexei: The involution s appears to be a commutativity condition.

robjohn

@tb Ah, embarrassing moments in front of groups when you wish you could commute.

Alexei Averchenko

16:13

it is

Zhen Lin

You need to check the unit laws (which we discussed earlier) and associativity laws (which involve triple tangent bundles... :-|)

Alexei Averchenko

the rhombic guarantees the associativity for T t and t T

i have a suspicion that the canonical involution will help with build the unit :)

i'm about to work it out

Zhen Lin

What's your proposed multiplication?

The rhombus is really just asserting that \tau is a natural transformation. You need a different diagram, and a different natural transformation, to get a monad multiplication \mu.

Alexei Averchenko

i think you're mistaken

tau is the canonical projection TM \to M

Zhen Lin

And a monad multiplication is a natural transformation T^2 \to T

Alexei Averchenko

16:20

the naturality square for tau is this:

TM - Tf -> TN

Zhen Lin

Set N = TM :p

Alexei Averchenko

M - f -> N

Henning Makholm

@robjohn It's not as if there seemed to be much else to say there (except for J.M.s superior solution).

Alexei Averchenko

you mean M = TN?

Zhen Lin

Oh yes, that one.

Alexei Averchenko

16:22

TTN - Tt -> TN

you're right

ok

multiply the rhombic horizontally by T

from left or from the right

in either case you'll get the associativity axiom for either T t or for t T

robjohn

@HenningMakholm I was just amused that we commented within 25 seconds of each other.

Zhen Lin

Yes. Now you should check the unit law.

Alexei Averchenko

ok

now i gotta remember how to compute second derivatives in this setting :D

Matt

@t.b.: Thank you. Reading what you wrote about the writing made me happy.

t.b.

@Matt As I said, I was seriously impressed. (by the way: ping me using "@tb" in chat, otherwise it doesn't seem to work).

robjohn

16:29

@tb: so using @t.b. doesn't ping?

t.b.

@robjohn No, for some reason it doesn't.

J. M.

So pings here work differently from pings on main...

It boggles...

robjohn

@tb: So we should believe the type ahead box that pops up and not the pseudonym shown on the chat.

Zhen Lin

@JM Different software. shrug

J. M.

You'd think they'd reuse components, no? ;)

Zhen Lin

16:34

"Show that X is smooth at P if and only if R_m is a regular local ring." — isn't that the definition...? I'm so confused.

t.b.

@JM: diagonal matrices are a polynomial of every matrix (according to the convention A^0 = identity matrix).

J. M.

Oh, I was thinking of two different diagonals with wholly different entries...

t.b.

Ah, sorry. Didn't read carefully enough...

J. M.

e.g. diag(2,3,1) and diag(11,19,17).

They'd commute, but I can't see at once how to express one in terms of the other.

QED

as a polynomial?

J. M.

16:39

Yeah. Is there? (I don't know, really.)

t.b.

@JM: That should be easy, though.

QED

are we looking for P(A) = B or P(A,B) = 0?

not that I know how to do either..

t.b.

Just find a polynomial such that p(2) = 11, p(3) = 19, p(1) = 17. This definitely exists.

J. M.

@QED This is what we're considering.

Ah, okay. So nix on those diagonals... :)

t.b.

make it simpler diag(1,1,0) diag(1,0,0), then you're in trouble

Matt

16:44

I've never seen as many unanswered questions on SE at once. Do they usually peak on Sunday evening?

QED

I wonder if there's any I can answer

J. M.

Yeah. Since you can't construct an interpolating polynomial with repeated abscissas. Neat, t.b.! :)

@Matt The "unanswered" is because people like to enjoy their weekends, I reckon. :)

t.b.

Plus, there's been an avalanche of questions over the last hour.

Matt

@JM: By hanging out together in this chat room?

Silly homework with deadlines was my first thought but only two of them are tagged homework.

J. M.

@Matt Well, I've been out of the house at least thrice today... :D

Matt

16:48

@JM: I haven't. I spent a major part of my day fighting my homework.

Zhen Lin

math.stackexchange.com/questions/81846/… - It looks to me like OP doesn't like Theorema Egregium as an answer and wants something from the theory of PDEs instead. :-|

Matt

Too many symbols.

J. M.

Maybe he doesn't understand how TE fits into it...

...or he hasn't played around with a rubber ball, for that matter.

Zhen Lin

On the other hand, there are surprising facts, e.g. the existence of a flat metric on the torus...

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