Mathematics

12:08 AM

DogAteMy: That's the "standard" proof (using induction) that many of us use, and I put it in my book. There are group theory proofs, too.

J. M.

@Owatch single-shift QR doesn't really play nice with Wilkinson's strategy in the unsymmetric case, as you've by now seen. Double-shift is only marginally more expensive; why not use that instead? And yes, "any nonzero number with modulus equal to unity" is mathematically rendered as $|z|=1$, or what @Semiclassical said earlier.

Ted Shifrin

hi @J.M.

J. M.

@Owatch The idea is that it's easy to find the zero eigenvalues of a matrix via the QR iteration. Thus, it boils down to finding $\lambda$ in $\mathbf A-\lambda\mathbf I$ so that it has a zero eigenvalue, in which case $\lambda$ is now a good approximation to the eigenvalue you were looking for!

Hey @Ted, just needed to do a little linear algebra lecture; I see I was pinged in the midst of my slumber.

Ted Shifrin

Chatters are never supposed to slumber.

J. M.

@TedShifrin tss, guess I ain't one after all. ;P

Ted Shifrin

12:19 AM

Sometimes I wished I'd learned more numerical analysis. Oh well.

Mike Miller

Hi folks.

Ted Shifrin

G'night, @MikeM.

Mike Miller

Waiting on the instructor to put out his exam review guide so I can write a review session out of it :P

Ted Shifrin

It's for the kidlets to ask you questions at a review session, no?

Mike Miller

I don't trust them to have questions.

Ted Shifrin

12:27 AM

Then I guess you don't help them ...

This is one thing I've always been insistent about, myself.

Of course, I've never provided exam review sheets, either. And certainly not "practice exams" that would turn out to be the actual exam. The way some professors curry favor.

J. M.

@TedShifrin It's not that late... come to the applied side. ;)

Ted Shifrin

@J.M.: It's past that late.

J. M.

Okay, but I've seen 60-year olds do something completely different after 30+ years...

Mike Miller

What about 20 year olds?

Can't teach a young dog any tricks?

Ted Shifrin

@J.M.: You're assuming I don't actually want to be retired.

J. M.

12:34 AM

@MikeMiller Kids always have plenty of options. That's why I don't have any sympathy for "I'm BOOOOREEED!!!1!"

@TedShifrin well, sitting on the porch has its charms. I'll grant that.

Ted Shifrin

Well, I spent this afternoon playing bridge. A bit more work than rocking in a rocking chair.

(Not to mention the 2+ hours of driving in traffic. ugh.)

J. M.

@TedShifrin Couldn't you play bridge somewhere nearer?

Ted Shifrin

I usually do. I gave in this time. Never again.

J. M.

Yeah, unless the other guy is an exceptionally entertaining player, no point in wasting the time, gas, and emotional stuff. ;)

Ted Shifrin

well, at least I'm not the only sexist person in here ... :) the other person is a woman. We have a good time, but ... not worth the traffic.

J. M.

12:37 AM

fine, "gal". :P

Ted Shifrin

Oh, much better.

J. M.

statistically tho, "guy" is more likely than "gal" when I hear "bridge partner".

Ted Shifrin

Most people at the duplicate games are old women ... then old men. I look young in that crowd.

J. M.

I'd say it's something to be proud of. :D

Ted Shifrin

Yeah. We tied for first.

Akiva Weinberger

12:44 AM

@TedShifrin Really? Huh, I guess I'm the weird one for not knowing it

I just knew the group-theory one

Ted Shifrin

Well, DogAteMy, you're just starting :P

You excited about Math Camp?

Akiva Weinberger

The group-theory one seems more natural, since it doesn't rely on binomial coefficients or anything, only the definition of a prime

@TedShifrin YEAH

Leaving tomorrow

Ted Shifrin

So we won't see much of you for a few weeks :P

But it's really cool that $\binom pk \equiv 0\pmod p$ for $k=1,\dots,p-1$. I love that.

Akiva Weinberger

I'll still have my phone on me, so maybe — though I suppose I should be socializing with fellow campers (something I didn't do enough of last year)

Ted Shifrin

Yes, you should.

Akiva Weinberger

12:48 AM

Hm. I never thought about that identity combinatorially. (Re: binom coeffs)

Mike Miller

There are great combinatorialnproofs of most standard number theory theorems in george andrews' book

Ted Shifrin

I don't know how to do it combinatorially, I guess. It's just a consequence of the usual $p|ab\implies p|a \text{ or } p|b$.

Akiva Weinberger

I should be able to partition the 3-subsets of 17 into 17 equinumerous sets.

@TedShifrin Yeah; that and the formula for them gives it automatically.

Ted Shifrin

So if you have a combinatorial way to see it, please tell me.

Akiva Weinberger

So I have two tasks right now:

divide the 3-subsets of 17 into 17 equinumerous sets,

and dinner.

Ted Shifrin

12:50 AM

LOL ... very late for dinner.

Akiva Weinberger

I was out biking.

Ted Shifrin

Good for you!

Go eat first.

Akiva Weinberger

I went to the library, where I found the thing about Fermat's Little Theorem.

(Side note: If the Last theorem is "FLT", should the Little theorem be "flt"?)

('Cause it's little.)

Ted Shifrin

Yup :P

Akiva Weinberger

I didn't actually intend to go to the library. I went to a large park ("Prospect Park"), and there was a library right next to it.

Ted Shifrin

12:52 AM

I'm familiar with Prospect Park.

Akiva Weinberger

It distracted me.

Ah, you are?

Ted Shifrin

Yup. My dad grew up in Brooklyn and I have good friends in Brooklyn Heights.

Mike Miller

Prospect park is beautiful.

Akiva Weinberger

@TedShifrin I suppose I'm one of those "good friends", then!

(I hope.)

Ted Shifrin

Well, if we ever meet ... perhaps :P

Akiva Weinberger

12:54 AM

Now I'm wondering if there's someone in Brooklyn Heights we both coincidentally know. But I should eat.

Ted Shifrin

Go eat.

It's conceivable you know each other from temple, but I dunno.

Akiva Weinberger

Would you happen to know of Congregation Bnai Avraham? On Remsen?

Ted Shifrin

Hmm, you're probably a lot more orthodox than they are.

They're on Hicks and they go to something down Hicks north of Clark, I think.

Akiva Weinberger

Oh, "north of Clark" is on the exact opposite side of the neighborhood, in fact

Mike Miller

I worked at the USAO in Brooklyn Heights one summer. But I probably wouldn't know anyone you do.

Ted Shifrin

1:00 AM

I figured, DogAteMy.

But you can still be my friend, DogAteMy :P ... if you want to be.

Akiva Weinberger

@MikeMiller Don't know what that is

Mike Miller

us attorney office

Akiva Weinberger

Ah

Semiclassical

1:18 AM

evening chat

Akiva Weinberger

Yes

@Semiclassical $\binom pk$ is a multiple of $p$ for $0<k<p$ and $p$ prime, right? Do you think there's a combinatorial proof of that?

Like, for example, a way to split the 3-subsets of 17 into 17 equinumerous subsets.

Semiclassical

dunno

Samuel Handwich

can somebody help me interpret an algebraic topology question? I'm not sure what it's asking.

question is this: "Does every function $f(x)=x^2$ admit a continuous extension $\hat{f}:\beta \mathbb{R}\rightarrow \mathbb{R}$ to the Stone-Cech compactification?"

the wording is what confuses me. my reading is that $f(x)=x^2$ is one particular function. I'm not sure what the every means.

shai horowitz

1:43 AM

perhaps its trying to admit things like $f(cos(x))=cos(x)^2$? its weird wording

does the kid in the sixth sense have the shining

Ted Shifrin

@SamuelH: Sounds like a bad translation from some other language?

Samuel Handwich

@TedShifrin i suspect maybe it should have been "the function f(x)=x^2" or just "every function"

it was on my phd topology exam, but i failed that, so i was missing something

Ted Shifrin

I guess the point is that Stone-Cech allows every bounded real-valued continuous function to be extended, not every function? I've forgotten all this stuff.

Samuel Handwich

1:58 AM

so have i, evidently :-(

Ted Shifrin

So, I'm guessing $f(x)=x^2$ has no continuous extension.

shai horowitz

but guys I found an upper bound of the trajectory of the collatz function thats true for the first trillion numbers. do publications accept conjectures about conjectures

Ted Shifrin

Not in my experience, @shai.

shai horowitz

darn i'll have to continue my work actualy riguoursly proving the max length a f(x)=(3x+1/2), collatz trajectory of n is clearly bounded by f(x)^n

Semiclassical

first trillion numbers isn't necessarily that big

i mean, Wikipedia indicates that it's been tested for terms starting as large as 2^60

i.e. about a million trillion

Axoren

2:59 AM

I'm looking for something that looks like an approximation method but isn't. I forgot what it was. It was function $f(x)$ in terms of $f'(x)$ and $f''(x)$. It was like the Taylor's Theorem for 3 terms, but it wasn't in terms of a known value $a, f(a)$.

Can't for the life of me remember the book or paper I saw it in.

@Huy You broke Wolfram Alpha. Now we're all doomed.

Semiclassical

@Axoren dooooomed

Axoren

This isn't fixing itself nicely at all...

@Huy
[0 to 1454](http://tinyurl.com/hbw9vb4)

[1454 to 1455](http://tinyurl.com/hvcev82)

[0 to 1455](http://tinyurl.com/zuc5mk8)

$-1.09358 \approx 22.7858 + 0.02147$

I just can't link at ALL, for some reason. Thanks @SevenSidedDie for fixing the messed up post.

0celo7

3:52 AM

Good evening

l' - ' - ' - ' - 'l

Hello

Samuel Handwich

hi

Forever Mozart

4:17 AM

they cut my hair :)

:(

now I feel sad

l' - ' - ' - ' - 'l

4:45 AM

why would they cut your hair

you can keep your hair if you want, who would force this upon you?!

Huy

5:45 AM

@Axoren: 22 is way too large though, the result should be less than 1 because it's a probability

Axoren

6:22 AM

@Huy That's the beauty of it. -1.09358 is too low, 22.7858 is too high, yet this sum approximated both of them.

Axoren

6:33 AM

Thinking about it, doesn't the $\approx$ relation just work on damn-near everything? In sacks of rices, $962\,593 \approx 1\,000\,000$. With such a massive difference, you'd have approximately $37\,407$ grains when you had none at all.

Tobias Kildetoft

@Axoren it is not actually a well-defined relation usually

Axoren

@TobiasKildetoft Not at all.

It's quite a shame, given how often it's used.

Maybe there is a proper interpretation of it that doesn't violate reason.

Like $\approx$ not being symmetric or transitive.

You wouldn't say that 1,000,000 is approximately 935,326, for example. But you would say the reverse.

Maybe it would be transitive, but trivially so.

Every standard unit would be approximate to itself, but it wouldn't be approximate to any other unit. So anything that approximates to it, won't approximate to anything else.

idonutunderstand

8:09 AM

anyone want to tell me wtf type theory is?

Tobias Kildetoft

@idonutunderstand There are several type theories

idonutunderstand

how about intuitionistic type theory?

Tobias Kildetoft

Not sure about specifics for any of them. As I recall, the general idea is to separate sets into different "types" to avoid things like Russell's paradox

idonutunderstand

I've heard something like that.

I've also heard something like, for example, a function is no longer defined as a relation.

Rather a function is more primitive than a relation.

I don't know which type theory that would refer to.

I've been listening to type theory lectures and trying to absorb the material.

But the idea of it still eludes me.

Transcript for

$Mathematics$ Mathematics