Mathematics - 2018-02-14 (page 1 of 6)

Jake Rose

00:15

(a × b) · (c × d) = (a · c)(b · d) − (a · d)(b · c)

Has anybody got a somewhat easy way of proving this without going into components?

Xander Henderson

00:38

Suppose that $O$ and $P$ are commuting operators. Then $$ O^2P^2 = (OP)^2 = (OP)(OP) = (PO)(OP) = 💩 $$

Semiclassical

2 much OP gives POOP

(truly a deep philosophical insight)

Joe Shmo

Can somebody give some insight/intuition to pullbacks? what did they come to solve?

Ted Shifrin

00:56

@JoeShmo: Pullbacks in what context? Differential forms?

Joe Shmo

yes. I also vaguely remember seeing them in passing in munkres' topology book (along with winding numbers)

Ted Shifrin

@JakeRose: Notice that everything is linear in each variable, so it suffices just to check that it works when $a,b,c,d$ are various ones of the standard $\vec i, \vec j, \vec k$.

Joe Shmo

i could stick them into a pretty diagram. but then what

Xander Henderson

once you've stuck them into a pretty diagram, you are done, no?

Isn't that the goal of life?

Ted Shifrin

I don't think Munkres has anything in there with differential forms. There are fancier answers, but the main point is that you compute integrals by parametrizing the manifold you're integrating over (at least in pieces). When you do that, you pull back the form and write it back in terms of the parameters.

Again, I recommend various of my videos which cover this carefully and with tons of examples.

Joe Shmo

01:00

could you link me?

shameless plug huh? :D)

Ted Shifrin

The link is in my profile. I don't get any money for this — Lots of people have found 'em helpful.

Eric Silva

@Semi my lab was just two labs in a trench coat disguised as one lab and I wanted to cry

Also fuck oscilloscopes

Ted Shifrin

@EricSilva: Was it a Lieutenant Colombo trenchcoat?

Semiclassical

What was the lab on?

Eric Silva

Idk what that is

Ted Shifrin

01:02

Hi Semiclassic

Eric Silva

Faraday's law

Joe Shmo

@TedShifrin I'm only kidding. I'm looking at your videos right now

Semiclassical

ah, that can be fun

Ted Shifrin

Wow ... you never saw the mysteries with Peter Falk as Lt. Colombo :P

Semiclassical

one of my favorite labs in the undergrad sequence is putting a spinning coil inside a magnetic field and seeing the induced current

Ted Shifrin

01:03

@JoeShmo: There are lots of line integrals and surface integrals, but later on there's actually a discussion of homotopy and pullback and some topology theorems.

Eric Silva

@TedShifrin is that a thing from far before I was born

Joe Shmo

@TedShifrin you had me at topology

Cookie Toast

Hey everybody!

Ted Shifrin

@Semiclassic: At MIT they don't tie labs to the usual physics, chemistry, and biology courses, so I never had to do these labs. One must fulfill a lab requirement by taking a lab course and designing one's own experiment. I forget what physics-y thing I did.

Joe Shmo

is it a full course? Do you have stuff on angle forms and i would presume stokes theorem?

Ted Shifrin

01:04

LOL, Eric, everything is a thing from before you were born. You're a baby :)

Eric Silva

@Semiclassical it would've been if it was one lab but we basically did two different labs to confirm Faraday's law that were completely separate and everything broke

Ted Shifrin

Lots on Stokes's Theorem (including proof, Joe Shmo). Angle forms? You mean $d\theta$ in the plane and the analogue in 3-space? Yeah, I talked about the 3-space case in terms more of physics and inverse-square forces, but yes.

MatheinBoulomenos

Sometimes I feel like I'm partially TAing the algebra courses I take

Ted Shifrin

Everything broke?

@Mathein: You mean helping your classmates?

Eric Silva

@TedShifrin like the equipment physically broke down

Ted Shifrin

01:06

Lots of us did that, Mathein.

Eric Silva

And then the printer broke so we had to spend an hour copying our data and plots

Ted Shifrin

Yeah, @EricSilva. I had a friend who did a PhD in experimental physical chemistry at Berkeley who took something like 7 years to finish, because first his adviser had to get the lab set up and then he had to get equipment ... and then the equipment didn't work right, etc., etc.

Semiclassical

oh god

print to pdf people

Ted Shifrin

Yeah, what's with copying data and plots?

Eric Silva

@Semiclassical what happened was someone ended up getting it jammed and them physically broke a piece irreparably when they tried to fix it

Semiclassical

01:07

in the printer?

Eric Silva

It was really sad

Ted Shifrin

This is why lab TAs should be watching out for things ...

Eric Silva

@Semiclassical yes

Ted Shifrin

Oh, that was the printer.

Semiclassical

I'm still not seeing why you'd need to copy your data by hand tho

Eric Silva

01:08

Cause we couldn't print out our plots and shit

Semiclassical

what program were you using?

labview or something else?

Eric Silva

Excel

Semiclassical

uh

and you couldn't save the excel files because...

Ted Shifrin

keeps mum

Eric Silva

The TA told us to copy the data by hand and give it to him

Semiclassical

01:10

...

Ted Shifrin

stooopid TA

Eric Silva

Otherwise I would've saved the file and emailed it

Semiclassical

^^

Eric Silva

I was very mad

Ted Shifrin

Justifiably.

TA incompetent.

Semiclassical

01:10

I'll note that in the intro labs here we don't ever do lab printouts

Eric Silva

So a 1 hr lab took 5 hrs

Semiclassical

printing means printing to pdf and emailing stuff out

Ted Shifrin

You need a martini, Eric. Here, let me fix you one.

Semiclassical

anything else is a waste of time and paper

Ted Shifrin

^^^

Eric Silva

01:11

I wish @Ted

Semiclassical

(I mean, you should retain a hard copy in your notebook. but that's as easy as printing it out later and stapling it in)

Eric Silva

@Semiclassical I agree but shrug only a couple more labs

When I go home I'm gonna get a stiff drink and relax after that ordeal

MatheinBoulomenos

@TedShifrin The TA just use my proofs for some exercises (while giving me credit.). Or they ask me "I like your solution, can you present that?". While my TA was busy with his wedding, I actually stepped in for him and hold the 4-hour bonus exam-preperation exercise session. We have the exam for ANT in 2 days, so everyone is studying right now, I've answered around 50 questions on ANT today.

Ted Shifrin

Well, you should take it as a definite compliment. You can put it in your application for PhD work :)

I remember that in our graduate complex analysis course the prof xeroxed the solutions that he liked best and distributed those to the class. A few of us got used a lot.

Semiclassical

Reminds me, there's HW/quiz problem solutions I need to write up

MatheinBoulomenos

01:15

Yes, I enjoy it, but it feels a bit strange, dunno why

Semiclassical

nothing hard but I want to do them good

Ted Shifrin

Besides, Mathein, it means you learn stuff really well, when you have to explain it to others. So it's a win-win for you. As long as you're not obnoxious and don't make the students hate you, I'm sure they appreciate your help and clarity.

@Semiclassic: What's the probability that 50% of your students will study (read?) what you write up?

Semiclassical

Not very high at all.

But the prof wants solutions, so

Ted Shifrin

I remember when I taught abstract algebra the first time (my second year at UGA), I put the standard $R/I\times R/J \cong R/(I\cap J)$ when $I+J=R$ on an exam. Very few got it. I wrote up solutions and said that it would be on the final. Guess how many got it on the final?

Semiclassical

Optimistically, everyone. Realistically, very few.

MatheinBoulomenos

01:18

not that many?

Jake Rose

Hey guys got a vector question

You have two circles

Ted Shifrin

As I recall, precisely the ones who got it the first time. I was livid.

It went on the exam the third quarter, as well.

Joe Shmo

@TedShifrin huge segway but sometime over the winter break Jordan's theorem finally clicked. I had one last issue i needed to straighten out at the time, hence i was asking stupid questions (if i remember correctly). But basically a linear space can be decomposed into a direct sum of generalized eigenspaces, each of which can be decomposed into a direct sum of T-invariant (important!) cyclic subspaces.One would choose v not in ker(T^(m-1)) where m is the degree of T on a given eigenspace.

Jake Rose

$ \mod (r-a) = p $

Ted Shifrin

Do you mean $T$ or $T-\lambda I$, @JoeShmo?

Jake Rose

01:20

Dont know how to do mod signs

Semiclassical

on that note, I was making a stoopid linear algebra mistake a few days ago (not an algebra mistake)

Jake Rose

Actually nevermind

Ted Shifrin

Just absolute value will work, @JakeRose, or \|

Jake Rose

$\|x$

Ted Shifrin

Never mind what?

Jake Rose

01:20

Ill write it out in full

Joe Shmo

@TedShifrin yes

...to construct these cyclic subspaces, and iterate it w.r.t T-lambda*I

Semiclassical

one of the facts we love in physics is that $[A,B]=0$ with $A,B$ diagnolizable $\implies$ $A,B$ have a common eigenbasis

Joe Shmo

Sorry. T-lambda*I

Ted Shifrin

Yes, I love it too, Semiclassic.

Right, JoeShmo.

Oh, @Cookie is lurking.

MatheinBoulomenos

@TedShifrin when I TA'ed LA, the prof promised that he would give everyone (nb over 500 people) candy if nobody incorrectly tries to use "Sarrus's rule" on a $4\times 4$ or higher matrix.

Semiclassical

01:21

but I was being dumb and had myself convinced that: "Oh, $\psi$ is an eigenvector of $A$? Then it must also be an eigenvector of $B$!"

MatheinBoulomenos

He didn't have to buy the candy ...

Ted Shifrin

@Semiclassic: Try $A$ the identity :P

Semiclassical

yeah

Ted Shifrin

I don't know what that rule is (by name), but I can sure guess, @Mathein.

0celo7

Germans love naming things after people

Semiclassical

01:23

The Leibniz formula is the right version of that, isn't it?

Ted Shifrin

With 500 (!!) students, it's not surprising that at least 1 would screw up.

What in the world is that, Semiclassic?

MatheinBoulomenos

It's just the Leibniz rule for $n=3$, but in this case there's a nice way to remember it, which doesn't work for $n>3$

Semiclassical

Leibniz formula for determinants

In algebra, the Leibniz formula, named in honor of Gottfried Leibniz, expresses the determinant of a square matrix in terms of permutations of the matrix elements. If A is an n×n matrix, where ai,j is the entry in the ith row and jth column of A, the formula is det ( A ) = ∑ σ ∈ S n sgn ⁡ ( σ ) ∏ i ...

Ted Shifrin

I've never heard that name, either. yes, it's the criss-cross rule for $3\times 3$.

MatheinBoulomenos

Rule of Sarrus

Sarrus' rule or Sarrus' scheme is a method and a memorization scheme to compute the determinant of a 3×3 matrix. It is named after the French mathematician Pierre Frédéric Sarrus. Consider a 3×3 matrix M = ( a 11 a 12 ...

Ted Shifrin

01:24

Right, that.

Semiclassical

Not a lot of permutations on 3 elements

Jake Rose

Two spheres
$\r-a|=p and \r-b|=p $

Show the plane of intersection is $2(b − a) · r = p^2 − q^2 + b^2 − a^2$

I dont know how to add vector signs to the letters so in the first equation a and b are both vectors and the 'r' is different for each equation. Also the a and b in the third equation is $=\a|$ and same for b

Ted Shifrin

Never saw the name before ... in my 50+ years of knowing this.

Semiclassical

I've seen it, but I don't remember it

0celo7

@MatheinBoulomenos Well you just do the Sarrus rule 4x for a 4x4

so what is the issue

Jake Rose

01:24

*$\a|$

0celo7

how did people manage to do that wrong

Semiclassical

...but then, I also remember the Leibniz formula so there

0celo7

what did they even do?

Ted Shifrin

Right, @Jake. Best thing to do is remember that $\|x\|^2 = x\cdot x$.

Expand out your two equations of your spheres and note that $\|r\|^2$ will cancel out.

It isn't a valid formula for the determinant, 0celo. Note that you need $4!$ terms and you won't get nearly that many.

Jake Rose

What do you mean expand it out?

Xander Henderson

01:26

@TedShifrin Just one more question, sir..

0celo7

@TedShifrin Can't you pick a top term, then do the 3x3, then the next top term, a 3x3, etc.?

Semiclassical

not unless you do it in a bit more complicated way than the usual Sarrus rule at any rate

Ted Shifrin

@JakeRose: Write $\|r-a\|^2 = (r-a)\cdot (r-a)$.

Semiclassical

That's not the rule of Sarrus, tho

Ted Shifrin

That's cofactors. That's totally different.

MatheinBoulomenos

01:26

you only get 8 terms, when you do the same thing for a 4x4 matrix

Jake Rose

But its not squared originially?

0celo7

I must not understand this Sarrus then

Semiclassical

Rule of Sarrus is specifically the thing about writing the first two columns again in parallel and drawing the diagonals

0celo7

let's see

Ted Shifrin

Well, square it, silly :) @JakeRose

Semiclassical

01:27

which works for 3-by-3

Jake Rose

Is there a way to do it via equation of plane?

0celo7

Oh, I see

I never learned this, I just do cofactors

Ted Shifrin

No. You're trying to find points that satisfy both sphere equations, @JakeRose.

Semiclassical

that's usually the right way

Leibniz is nice because it makes doing determinants of sparse matrices easy

Ted Shifrin

Cofactors are a computational nightmare unless there are lots of 0s.

0celo7

01:28

@Semiclassical Impossible to get wrong modulo signs and other errors!

MatheinBoulomenos

It's not necessary to know it, as long as you can compute the determinant somehow

Jake Rose

But surely you can find it by somehow doing the (b-a).r? and then doing something else after that

0celo7

Since linear algebra I haven't had to compute a determinant by hand, so...

Semiclassical

There's also dodgson condensation but who the hell does that

0celo7

@Semiclassical lol

Ted Shifrin

01:29

@JakeRose: Ignore me if you wish. I'm done.

0celo7

so does your series look nice finally?

Semiclassical

Series?

0celo7

fourier

MatheinBoulomenos

I heard that once a numerical analysist taught LA here and he barely taught determinants, because they are "computionally expensive and numerically instable"

Semiclassical

oh

yeah, it's not bad

Ted Shifrin

01:30

That's irresponsible, @Mathein.

Secret

Sometimes I am wondering whether the individual terms in a determinant have geometric meaning

0celo7

@Semiclassical have you tried summing like 1,000 terms?

Secret

e.g. for a 3x3 matrix each term of the determinant is a product of 3 entries

Jake Rose

Could you explain how your method gets the equation of the plane? I dont understand it fully

Ted Shifrin

No, @Secret, because they're completely basis-dependent.

Semiclassical

01:31

@TedShifrin I actually had an email exchange last night whose resolution relied on the determinant of a matrix being the volume of a parallelpiped :)

Ted Shifrin

Sit down and work it out, @JakeRose.

Signed volume, @Semiclassic.

Semiclassical

yes, fair. (it was (det M)^2 in this case so I was being careless)

user193319

Claim: Let $A$ be a two point set in $\Bbb{R}$. Show that $A$ is not the retract of $\Bbb{R}^2$. Proof: Since the subspace topology on a finite set in a Hausdorff space is discrete, $A$ is totally disconnected. But the continuous image of $\Bbb{R}^2$, a path-connected set, which is absurd since the image equals $A$.

Ted Shifrin

It's actually one of the ways I explain that for a surface in $\Bbb R^3$ the determinant of the first fundamental form is the fudge factor for the surface area integral.

Semiclassical

Just some nice facts about the Gram matrix of a set of unit vectors, really.

user193319

01:33

Does this sound right? If so, couldn't $A$ be replaced by any finite set with cardinality greater than $2$?

Ted Shifrin

@user193319: Connected is all you need. You don't need to mention path-connected.

Jake Rose

Okay I've worked it out

Im just trying to understand why that gives the solution

Ted Shifrin

Yes, @Semiclassic, but the way to prove the Gram determinant is what it is uses this, of course.

user193319

@TedShifrin You're right. Thanks!

0celo7

@user193319 are your real thingies supposed to have different dimensions or typo?

Ted Shifrin

01:34

If you do the algebra with the two equations, @JakeRose, it immediately turns into that plane equation.

Jake Rose

So you assume that r is equal for both

Semiclassical

Right, I was using the oriented volume formula as a way to justify those properties @TedShifrin

Ted Shifrin

Yes, it's a vector for a point in both spheres, @JakeRose.

user193319

@0celo7 Yeah...they should all be $\Bbb{R}^2$.

Ted Shifrin

Yes, sure, @Semiclassic. Extend to an orthogonal basis ...

Jake Rose

01:34

But theres only two points that exist so how does it map the rest of the plane?

Ted Shifrin

That makes no sense to me, @JakeRose. I have no idea what you're thinking.

There's no mapping.

Semiclassical

e.g. if $M=AA^T$ then $x^T M x = \|Ax\|^2\geq 0$

Jake Rose

Sorry bad terminology

Ted Shifrin

By the way, the intersection of two spheres is NOT a plane. It's contained in a plane. That's the equation you wrote down.

Semiclassical

Which translates in the case of interest into a nice QM bound that people have known about

Ted Shifrin

01:36

Actually, you need $M=A^\top A$ for that, @Semiclassic.

Or else it's $\|A^\top x\|^2$.

Jake Rose

If you assume that r is equal for both equations, shouldnt that only work for the circle where that intersetion is? Why does it give the equation for the full plane?

Semiclassical

ah, right. should be $A^Tx$ based on what I wrote

Ted Shifrin

@JakeRose: Did you read what I just wrote? You only get points lying in the plane, not all the points of the plane.

The statement should be that any vector $r$ lying in both spheres also lies in that plane. Not vice versa.

Cookie Toast

Just lurking a tad :) I just finished studying for an exam and I'm procrastinating going home

Semiclassical

I should probably be writing it as $M=A^T A$ anyways since I want to write $A=[\hat{a}\, \hat{b}\, \hat{c}]$ for unit column vectors in RR^3

Ted Shifrin

01:38

Well, fine, @Cookie. Don't even say hello, then.

Cookie Toast

Heya :P

Jake Rose

The question which i lifted this from states ", show that the plane in which the circle of intersection lies is given by"

The by being the equation we proved

Ted Shifrin

That is correctly stated.

Jake Rose

So shouldn't we have a plane that extends beyond the intersection of the two spheres?

Ted Shifrin

We do. You really need to read the things I type to you.

I've said it twice.

Semiclassical

01:40

With the choice x=[1 1 1]^T corresponding to the one people have been interested in. that's minimized exactly when $\hat{a}+\hat{b}+\hat{c}=0$, and the only way to get that with unit vectors is to have them at 120 angles in a common plane.

Jake Rose

Sorry I must have misinterpreted what you said

Semiclassical

Anyways. Some nice simple geometry in here.

Ted Shifrin

Anyhow, I need to leave. Dinner and then duplicate bridge for the evening. Bye, all.

Semiclassical

(I count my main contribution to the work as being to get rid of most of the trigonometry the person was having to do initially :P)

Jake Rose

Can anybody else explain why that solution works for the entire plane?

We have ssaid that r is the same for both. And so shouldnt r only be on the circle of intersection and not anywhere else?

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$Mathematics$ Mathematics