Mathematics - 2016-10-09 (page 2 of 3)

DHMO

17:00

I mean, the numbers of 1s usually don't exceed, say, 6

alan2here

they suddenly increase sometimes

Siddharth Venu

hey there guys

alan2here

hey, your right @DHMO

there sort of attracted to 4, 5, 6 range

the concequences of log base 2 vs cubed I guess

Mike Miller

@Ted: You have to if you want to split the bundles!

Even in the complex case.

DHMO

@alan2here sure

Ted Shifrin

17:05

@MikeM: But the principle says you get to just pretend that you've split the bundle on your original base manifold. (That's what Hirzebruch taught me years ago.) Injectivity on cohomology of the pullback you're alluding to justifies that, I think.

Mike Miller

@Ted: Sure, if you have a properly functorial notion of Euler class. I'm not sure if Balarka does.

Ted Shifrin

Now I regret having given away so many of my books :(

Siddharth Venu

hey guys I am having trouble of understanding a question in chapter Straight Lines

DHMO

@SiddharthVenu maybe you should sleep and go back to it tomorrow

it must be very late lol

Siddharth Venu

* Find the values of non-negative real numbers h1,h2,h3,k1,k2,k3 such that the algrbraic sum of the perpendiculars drawn from the points (2 , k1) , (3 , k2) , (7 , k3) , (h1 , 4) , (h2 , 5) , (h3 , -3) on a variable line passing through (2 , 1) is zero *

How can sum of perpendiculars drawn from various points be 0? (my question)

DHMO

17:07

what is algebraic sum?

Ted Shifrin

They either mean slopes or they mean to add specific vectors, @Siddarth.

I have never seen such a thing.

DHMO

exactly

Siddharth Venu

@TedShifrin but they have used the length (they summed up a length and equated it to 0)

in the solution

Ted Shifrin

So when do you get a negative sign on that length?

I guess we arbitrarily decide 'above' the line and that gives a +, 'below' gives a -. Since we're adding to 0, it doesn't matter how we do that.

What a crazy problem.

Salut @DanielC

Siddharth Venu

in the solution, they found the perpendiculars using (ax1+by1+c)/sqrt(a^2+b^2)

Ted Shifrin

17:11

That's the signed distance from the point to the line, yes.

More important is to know how to find such a formula. :)

Siddharth Venu

oh... so can u tell me when will it become negative?

Balarka Sen

@MikeMiller I vaguely do (something something $H^*(X \times X, X \times X - \Delta)$). I can try to reconstruct it.

Ted Shifrin

@Siddharth: Do you know anything about vectors?

@Balarka: That reminds me. Did you ever work out the P.D. of the diagonal?

Siddharth Venu

yes I had a class in physics about it.

DHMO

what's the most interesting group you guys heard of?

Balarka Sen

17:16

@TedShifrin ah, nope. I just needed a sanity-check for a very special case I was working with, so didn't need that. Will write it down somewhere for later.

Ted Shifrin

OK, so the vector $(a,b)$ is perpendicular to the line. So is its negative. The quantity is positive when $(x_1,y_1)$ is gotten by moving from the line "upwards" along the direction $(a,b)$ and negative when you move "downwards" along the direction $-(a,b)$. What's really going on is that you project the vector from a point on the line to your $(x_1,y_1)$ onto the normal vector.

Siddharth Venu

oh! so basically the vectors are just cancelling each of them out?

some negative, some +ve. all of them forming a closed polygon

is that correct?

Ted Shifrin

Right, @Siddharth. :)

Well, no, they're just adding up signed lengths. I don't know that the vectors will actually cancel.

Rrjrjtlokrthjji

Howdy

Mike Miller

@Balarka Again, that's for the tangent bundle, not a general vector bundle. Your approach is hopeless if you don't try to do things in general.

Siddharth Venu

17:20

oh... but distance, it is a scalar. why would it have -ve sign? @Ted

Balarka Sen

@MikeMiller Ah, good point.

Mike Miller

This is one of those times I think you should maybe spend less time trying to build a theory on your own and more trying to understand it from the ground up. Once you have good conceptual understanding you can try to be a theory rebuilder.

Ted Shifrin

signed distance ... I explained how you decide the sign several times :)

Rrjrjtlokrthjji

I've been self-studying Information Theory, mainly from Ash and Shannon-Weaver, Roman. Why we assume for the development of the theory that the probability of error for a single random variable trasmitted is $p < 1/2$ ?

Ted Shifrin

@Rrjrjtlokrthjji: Information theory is pretty specialized stuff. It's quite likely no one in this chat knows anything about it. But one never knows ...

Siddharth Venu

17:22

omg ted I got it! Thank you so much for your help :) @TedShifrin

Ted Shifrin

Sure thing, @Siddharth :)

Mike Miller

I'm going to a festival today, but I've got that research itch. we'll see if I can avoid thinking all day.

Ted Shifrin

I'm about to go to dim sum with friends ...

Rrjrjtlokrthjji

A reason I think may be is that the entropy $H(X)$ for the random variables $X_1,X_2, ..,X_n$ has a bound $H(X) \leq logn$, where equality holds for $p_i = 1/n, ~ \forall i=1,2,..,n$ e.g when tossing a fair coin. If the trial of error is $p=1/2$ we get maximum entropy, so there's no reason to work for $p> 1/2$

@TedShifrin Hello Ted

Ted Shifrin

Hello, @Rrjrjtlokrthjji, whoever you be.

Rrjrjtlokrthjji

17:25

@TedShifrin my previous username as Nickolas

Mike Miller

@Ted: We're watching Blue Velvet before heading over. Frank Gehry talks at 1.

Ted Shifrin

Ah, cool, @MikeM. Have fun!

Mike Miller

That's the plan.

Ted Shifrin

Ah, @Rrjrjtlokrthjji. You picked such a lovely name instead.

Rrjrjtlokrthjji

@TedShifrin I recall you are retired now, I've been watching your lectures on multivariable calculus

Balarka Sen

17:26

@MikeMiller Alright, got it.

Rrjrjtlokrthjji

You seem to be a great teacher, although I remember you said many students don't agree on that.

Ted Shifrin

Good way to get some extra nap-time, @Rrjrjtlokrthjji.

Rrjrjtlokrthjji

@TedShifrin we take Information Theory as an undergraduate course on 4th year, along with error-correcting codes.

Ted Shifrin

Thanks for the compliment. I had plenty of students who took 3, 4, 5 classes from me, so I guess they liked it, even though it was hard work. But plenty avoided me at all costs.

That's cool, @Rrjrjt, and cryptography is super important. I'm just saying that most of us know no information theory. Your heuristic makes intuitive sense to me, but I have no background to judge.

Rrjrjtlokrthjji

I guess I'm going to ask the teaching prof for that, although I won't take the course.

Ted Shifrin

17:28

Oh, you made it sound like it was a required course.

Andrew Thompson

I'm now going to Google "Ted Shifrin ratemyprofessor."

Rrjrjtlokrthjji

@TedShifrin no, there are 20 required courses ranging from Linear algebra to complex analysis

then we have to choose other 20 courses from a short list

Ted Shifrin

OK, @Rrjrjtlokrthjji, I just misinterpreted. It's the same most places, although there are always a few that everyone must do.

Have fun, @AndrewT :P

Rrjrjtlokrthjji

@AndrewThompson 4.5 as far as I see

Andrew Thompson

At least no one felt the need to warn about you online.

Yup. The general consensus seems to be "tough but fair."

Rrjrjtlokrthjji

17:30

@AndrewThompson you have any idea on that

Andrew Thompson

On whether it's true? No, I live on the other side of the planet.

Rrjrjtlokrthjji

@AndrewThompson you lean towards pure mathematics?

Ted Shifrin

LOL ... Well, I can't hide reality with the 112 lectures up there for all to see. :P

Andrew Thompson

Yes.

Mike Miller

"Easy but unfair."

I aspire to have ratemyprofessor reviews like that.

Balarka Sen

17:32

lol

Andrew Thompson

"I answered everything correctly and got a C."

I think that sounds worse in America where C is considered a bad grade.

Rrjrjtlokrthjji

@AndrewThompson a prof of us passed only 2 students with 5 out of 10 this semester

Ted Shifrin

At some colleges, the average grade is an A, @AndrewT ... a fact which I find totally abhorrent.

Rrjrjtlokrthjji

What's the equivalent of a C in a scale of ten

Ted Shifrin

A B C D F

Roughly, F is less than 60%, traditionally.

A is 90% or better, traditionally. ... Such traditions are long gone.

Andrew Thompson

17:34

We have an "E" as well, whose description reads: "A performance that meets the minimum criteria, but no more. The candidate demonstrates a very limited degree of judgement and independent thinking."

Here A is 90-92%.

Balarka Sen

What about "T"

for troll answers

Ted Shifrin

In fairness, in the hard classes I taught, I didn't write total joke exams (although Balarka might disagree) or joke homeworks. So I often gave A's down to 80%, B's down to 65% or so. Depended on the class.

Andrew Thompson

If it was a tough exam they go down a bit, but they have no problem giving less than 5% A's.

Alyosha

I'm reading a physics paper where they apply Noether's theorem to the (Weyl) transformation $t \mapsto kt$, $x(t) \mapsto \lambda x(t)$. I've only seen Noether's theorem in the form `this diffeo/symplectomorphism group preserving the Hamiltonian gives rise to this conserved quantitity f:M \to \Bbb{R}', nothing about time translation. Does anyone have a reference for this?

It might be that it's a trivial application of the usual theorem.

Ted Shifrin

@Alyosha: Can there not be time-dependent Hamiltonians?

Sure there can.

Andrew Thompson

17:37

When I TA'd algebra, a student delievered his final exam (which is 100% of the grade) as a joke, writing entirely in a mix of l33t-speak and Python.

Alyosha

@TedShifrin Thanks, for some reason I hadn't thought to include the word `time dependent Hamiltonian' in my searching.

Ted Shifrin

Well, Arnold in his mechanics book talks of the Lagrangian as a function on the tangent bundle, but he only allows diffeomorphisms of the base manifold.

Andrew Thompson

I think I posted this here before, however as the discussion came up I think it's sufficiently funny:

Ted Shifrin

@AndrewT, I had a few students give me joke exams, but not when it affected their grade.

Andrew Thompson

Our Calc2-grades.

"Stryk" = F.

Ted Shifrin

17:40

Even by my standards, that looks like ineffective teaching/learning.

Rrjrjtlokrthjji

@TedShifrin did you read my comment

@TedShifrin only 2 students passed with 5/10

Alyosha

@TedShifrin I think there might be something in the end of Arnold about time-dependent things.

Rrjrjtlokrthjji

hahaha

Ted Shifrin

Oh well ... OK, guys, I'm heading out. Misbehave without me.

Andrew Thompson

Bye!

Rrjrjtlokrthjji

17:42

Bye!

Balarka Sen

@TedShifrin Did you ever give chapter 8.1 as homework, from your book? :)

Mike Miller

@AndrewT How are your things?

Rrjrjtlokrthjji

@MikeMiller some help on information theory

Andrew Thompson

@MikeMiller Quite well, mostly been doing my TA-duties today. If I find the energy I might write some more on the notes for my own stuff, it looks very thin with its one and a half page. How about you?

Mike Miller

@Rrjrjtlokrthjji I can't offer that.

Rrjrjtlokrthjji

17:51

@MikeMiller thank you nonetheless.

Mike Miller

@AndrewT Please do write more. If I've learned anything, it's to write a lot and write often.

I'm dojng good. If you email me I can say more.

Andrew Thompson

done.

Jeff

Help please: If $s=s(t)$ is position and $v=s'$ is velocity how do i show that acceleration $a$ is constant?

oh, and $v=\sqrt{8s+16}$.

Danu

Hi @TedShifrin

Sorry for not replying yesterday---I didn't have time until now.

arctic tern

@Jeff position, velocity, acceleration indicates we might use derivatives. taking derivatives of square roots isn't fun, so square both sides, then take the derivative. what happens?

Jeff

18:01

@arctictern: I get $v^2=8s+16$, then $2v v' = 8s'$, then $2va=8v$ then $a=4$. that was easy (is it right?)

arctic tern

yes

Danu

@TedShifrin Actually, we didn't learn how to compute cohomology of direct products...

We skipped this Kuenneth stuff

Jeff

now the big question is "why didn't i see that?" (but i find myself asking that a lot)

Danu

Next thing: Lefschetz decomposition is about $H^k(X,\Bbb R)$, not $\Bbb C$

All in all, I don't really feel like I can understand your question.

I can make some remarks, at least... Using the Lefschetz decomp. on $H^2(X,\Bbb R)$---together with the bidegree decomposition of $H^k(X,\Bbb C)$ which we can use for $\Bbb R$ by intersecting with $H^k(X,\Bbb R)$ to get what I'll call $H^{p,q}(X,\Bbb R)$---we have $H^2(X,\Bbb R)=H^{0,2}(X,\Bbb R)\oplus H^{2,0}(X,\Bbb R)\oplus H^{1,1}(X,\Bbb R)_\text{p}\oplus \Bbb R[\omega]$.

The $0,2$ and $2,0$ parts are automatically primitive

Adeek

18:16

hi @TedShifrin

Danu

I don't think he's here...

Adeek

so just a quick question the differentiability of $f : M \rightarrow \mathbb{R}$ depends really on charts in some maximal atlas right ? If we use two different charts from two maximal atlas they might not have differentiability property right ?

Danu

Some more points: (i) I never learned how to think about the cohomology of $\Bbb P^n$ in terms of differential forms---my course stuck strictly to singular cohomology (ii) I'm not sure if you were asking for a geometric description type thing, or if you wanted something more dry/algebraic... I don't know much in the direction of the former.

Adeek

Danu

@Adeek You should check that yourself.

Adeek

18:19

they are talking here about compatible charts right ?

Danu

It's a basic but important result: Think about the transition functions between different charts.

Adeek

because $\phi \circ \psi^{-1}$ isn't in general compatible with each other

i.e $\phi \circ \psi^{-1}$ isn't in general $C^{\infty}$

Danu

It's independent of choice of the chart, but of course you have to choose your charts from the same maximal atlas.

Adeek

yeah

that was my question..

Danu

Yeah, okay.

Adeek

18:20

yeah good I am just verifying.

Danu

You have to fix your manifold---which includes specifying a maximal atlas!---and then you can define differentiability on that manifold

Adeek

yeah

ok that is clear thank you @danu

Danu

:)

Glad to help---I don't mind helping out with basic manifolds questions because I'll have to refresh my knowledge of diffgeo soon too, for an exam.

Adeek

yeah I have exam in like 20 days getting ready for it

Danu

okie.

Balarka Sen

18:32

Hi @Danu

Danu

Hi

Jeff

Another help request: If velocity $v=\sqrt{8s+16}$ is a function of time and $s$ is position and $s(0)=6$, then $v(0)=8$. How do I graph velocity vs. time ($v$ as function of $t$)?

arctic tern

@Jeff you know a=4 and v(0)=8, with these initial conditions you can find v(t)

Jeff

@arctic yeah... we haven't done integration yet. but i see how you mean.

Ed_4434

19:09

Hullo, if I were to plot the set $S = \left\{ {z \in \Bbb C: 1 \leq \lvert z\rvert\leq 8, \frac{3\pi}{4} \leq \lvert \arg(z)\rvert \leq \frac{3\pi}{2} }\right\}$ would it look more like a "piece of pizza" or a half circle? Maple plots it as a "piece of pizza" but I think this is because maple uses the principal argument, anyone able to make a comment on that?

arctic tern

@Ed_4434 it should be the sector between 3pi/4 and 3pi/2 of radius 8, but with the same sector of radius 1 eaten out of it.

Ed_4434

i.sstatic.net/RM0Xe.jpg @arctictern

Something like this?

arctic tern

3pi/2 is straight down. that image is from 3pi/4 to 5pi/4.

but otherwise it's correct.

dsillman2000

Morning, guys!

Ed_4434

@arctictern So how come the sector between pi/2 and 3pi/4 isn't filled?

arctic tern

19:21

@Ed_4434 because in the definition you put the angle between 3pi/4 and 3pi/2 no?

how are you defining arg(z) exactly anyway?

@dsillman2000 hello random person

Ed_4434

@arctictern yes but the set has |arg(z)| and I'm defining arg(z) as arg(z) = Arg(z) + 2kpi, k an integer, and Arg(z) in (-pi, pi].

@arctictern The confusion is stemming from the absolute value bars around arg(z). I thought it should be the union of $S_1 = \left\{ {z \in \Bbb C: 1 \leq \lvert z\rvert\leq 8, \frac{3\pi}{4} \leq \arg(z) \leq \frac{3\pi}{2} }\right\}$ and $S_2 = \left\{ {z \in \Bbb C: 1 \leq \lvert z\rvert\leq 8, \frac{-3\pi}{2} \leq \arg(z) \leq \frac{-3\pi}{4} }\right\}$

arctic tern

bleh

Ed_4434

@arctictern Apologies for the bombardment haha

Jeff

@arctictern TY for help. I set it up as "need a function whose derivative is a constant, so v=4t+C" to avoid integration.

arctic tern

@Jeff what makes you think integration is something to be avoided here?

Jeff

19:28

@arctictern class hasn't done integration yet.

arctic tern

what kind of class? if it's physics, might it expect some calc as a prereq?

Wenqin Ye

why is "the number of ways to have a 5 card hand with no clubs" not 52c5 - 13c5

why is it 39c5?

Jeff

@arctictern I'm referring to the problem like an hour ago. You may be confusing with someone else's question. It's calculus 1.

arctic tern

@WenqinYe 52c5 - 13c5 counts the 5-card hands, and subtracts out all hands which all 5 of which are clubs.

Wenqin Ye

ohhhhh

arctic tern

19:30

a 5-card hand with no clubs is created by picking 5 cards out of the 39 non-club cards

Wenqin Ye

thank you

arctic tern

@Jeff I know what the problem we're talking about. That doesn't mean I automatically know it's from a calc I class and not a physics class.

Jeff

@arctictern ok. anyway, thanks again.

Ed_4434

@arctictern Should I ask the question on the main site? :P

Mike Miller

19:47

yo

Rrjrjtlokrthjji

Any help on information theory?

JeSuis

20:01

hi

Balarka Sen

Hi @Mike.

Andrew Thompson

Ok, I really need to start printing papers. I found one online now, and I was all excited about it, and as I was about to favourite it I found that I had already done so.

Mike Miller

Printing and note taking is good!

Ed_4434

20:17

Hullo, if I were to plot the set $S = \left\{ {z \in \Bbb C: 1 \leq \lvert z\rvert\leq 8, \frac{3\pi}{4} \leq \lvert \arg(z)\rvert \leq \frac{3\pi}{2} }\right\}$ would it look more like a "piece of pizza" or a half circle? Maple plots it as a "piece of pizza" but I think this is because maple uses the principal argument, anyone able to make a comment on that? Note the absolute value bars around the argument.

JeSuis

Given $p:A'\to A$ an epimorphism from free abelian groups of finite rank, how can I prove that $\ker{p|_B'}=\ker p$? Where $B'=p^{-1}(B)$ and $B$ is a subgroup of $A.$

There is an obvious inclusion but for the other one, I don't get it.

Balarka Sen

Funny, I placed my notebook so that the enter button would not get accidentally pressed while I am writing. I suspect there's some key combinations I am not aware of which is equivalent to "enter".

Ed_4434

@BalarkaSen hehe

heather

So, I'm teaching myself calculus with *Calculus and Analytic Geometry* by Thomas, and I'm sadly already kind of stuck. The exercise is "plot the given point $P$ and such of the following points as may apply [...] (d) The point $T$ such that $PT$ is perpendicular to and is bisected by the 45 degree line $L$ through the origin bisecting the first and third quadrants. Give the coordinates of $T$ assuming equal units on the axes." For problem one, it gives point $P$ as (1, -2).

It also says in a note that $T$ and $P$ are symmetric with respect to $L$. I had no clue how to figure this out (plea…

Lozansky

20:34

Man, Frodenius method is time consuming

I wish I was better at sums

JeSuis

up ^^

Ted Shifrin

20:56

@Danu: You can easily do the (co)homology of $S^2\times S^2$ by Mayer-Vietoris, not even knowing Künneth. ;)

Adeek

hey @TedShifrin

Ted Shifrin

@heather: You have to draw pictures. You get two congruent right triangles, one with $x$ and $y$ as legs, the other with $y$ and $x$ as legs.

hi Karim

@Danu: The point is that the Kähler form $\omega$, which is the (positive) generator of $H^2(\Bbb P^n,\Bbb Z)$ generates all the cohomology. Wedging $\omega$ with itself $k$ times gives the generator of $H^{2k}(\Bbb P^n,\Bbb Z)$. I was encouraging you to understand this via Poincaré duality and thinking about the linear subspaces of $\Bbb P^n$.

@Adeek: Since you don't reply without a ping ... I just saw your question. No, you're wrong. Any two charts in an atlas that overlap by definition give a compatible notion of smoothness.

Adeek

I see

Ted Shifrin

You need to move past these first-week definitions and learn some real stuff !

Hi @quid pro quo.

@Balarka: Oh, with regard to your 8.1 question, I always assigned those as challenge problems on the first homework in the chapter. Not worth too many points, but ...

quid

21:13

Hello @TedShifrin!

Ted Shifrin

Don't think I've ever seen you around "in person" before, @quid :)

quid

I am sorry I am not sure what you mean.

Mike Miller

Frank Gehry's talk was quite good.

Ted Shifrin

@quid: I've seen lots of your answers to questions, comments, etc., but I don't recall ever seeing you in chat ...

Oh, cool, @MikeM ... what did he say?

Mike Miller

It was an interview. He talked about his creative process, the violence of angles, why there aren't so many women in top levels in architecture.

And a lot else.

He said a few things about the building of his he likes least and the building in LA he hates most that I'm not allowed to repeat.

Ted Shifrin

21:23

Ah, very interesting. I assume it was not a Trumpist discussion.

Mike Miller

Not from Frank Gehry, no. There were some jokes about Trump.

Ted Shifrin

I hadn't thought of architecture as fitting with the physical sciences and math in terms of discouraging women, but I guess it seems right. I (used to) know a lot of names of famous architects, all men :(

Balarka Sen

@TedShifrin Heh.

Ted Shifrin

Actually, @Balarka, many students over the years were captivated by the "why does a mirror reverse left and right but not up and down?" and would email me about it years later. ... I have to confess I first "met" that question when it was given to me at a math party after some colloquium. It stumped me for a while.

Mike Miller

The point he and the moderator came to is that part of success is in walking away from projects, from saying no to the person who commissioned the building. And that's a culture that women are systematically biased against in.

Maybe if you say no to someone you don't get the next job.

quid

21:26

@TedShifrin Ah I see. I am here rarely only, that's true. But IIRC we even talked some time ago. You told me you planned to retire.

Are you enjoying your retirement by now?

Mike Miller

Now we're eating vegan ramen and banh mi poutine.

He's retired. The jury is out on enjoying it.

Ted Shifrin

I'd forgotten, @quid, I apologize. Yeah, more or less ... still working on settling in in some ways.

Poutine never sounds vegan to me, @MikeM.

@MikeM: WRT the bias, I would assume the same worry might apply to men, but once again it's primarily a man's world in that arena, so ...

quid

No worries.

Balarka Sen

@TedShifrin I don't see why that should be hard. If you have a picture of something asymmetric and connected, if you place it front-facing to the mirror, the image is exactly a translated copy of the picture as seen from someone with eyes in the back of the piece of paper the picture is on. That's precisely the reflection of the object along the perpendicular axis.

I guess it'd be hard to rigorously write all of that down.

Ted Shifrin

No, you know all the words to make everything precise. Reflection along an axis? I don't think so.

The question (which is natural) is why does it appear to affect one axis in the plane of the mirror and not the other?

Balarka Sen

21:35

Because distance from object to mirror is exactly "distance" from mirror to image?

Ted Shifrin

That didn't address my query.

Odd that I told the OP the question was wrong, and yet, a day later, there it still sits as is.

Balarka Sen

@TedShifrin I am not sure what you want me to say. If I have a stick, top painted red, and bottom painted blue, and if the image had top painted blue and bottom painted red then (perpendicular) distance from eg blue tip to mirror (call the point the perp hits the mirror x) would not be the same as the distance from x to image of the blue tip.

Is that insufficient?

Ted Shifrin

Why does this argument not apply equally well to a stick placed horizontally rather than vertically?

Balarka Sen

Well, if you place the stick horizontally the image would exactly be the same. The tips do not switch places.

Ted Shifrin

Huh?

They don't in the vertical case either.

I actually don't follow your argument.

Balarka Sen

21:45

I think we're misinterpreting each other by having two different meaning about horizontal and vertical in mind. Fix a coordinate system with origin on the mirror. By horizontal, I think you mean along the x-axis.

Whereas in my last statement, I meant y-axis.

Ted Shifrin

Shrug. xy axes in the plane, z axis perpendicular to plane?

Balarka Sen

Yep.

Ted Shifrin

Then I'm asking how the mirror should distinguish at all between x and y directions.

Lozansky

Wouldn't the mirror flip along the z-axis then?

O'''''''''''''''''''-

I'm having trouble with introductory measure theory :C

heather

21:47

what would be a good book to learn trigonometry from?

Ted Shifrin

Any precalculus book is fine, @heather. Or a Schaum's outline. I'm sure you can find plenty on-line with Khan Academy, etc.

heather

@TedShifrin, okay

Balarka Sen

@TedShifrin If you have a stick along the x-axis (red dot in the front, blue dot in the back) pointing towards the mirror, the image would have blue dot far ahead of the red dot for precisely the reason I said- the distance from the blue dot is larger, hence it has to be further "inside the mirror". Whereas for something along the y-axis, both have the same distance from the mirror, so nothing happens.

Mike Miller

Don't you have homework?

Ted Shifrin

@Lozansky: Sorry, missed your comment. I wouldn't quite say it that way, but, yes, I think we agree.

Balarka Sen

21:51

I guess I do, whoops.

Ted Shifrin

@Balarka: Your purported argument is not consistent with symmetry. The mirror cannot tell the difference between x and y.

You seem to be having a stick that goes along the z-axis, not x.

LOL @MikeM for repeated chiding.

Balarka Sen

Nope, I do have x. I guess ~~we're~~ I'm miscommunicating. Oh well.

O'''''''''''''''''''-

Any recommendations for a measure theory book with good exercises? Stein and Shakarchi has helpful exposition but I think a lot of the exercises are too hard for me right now. Are Folland's easier?

Ted Shifrin

Royden is the standard source from many years ago, @O'''''''''''''''''''- ... have you tried that?

O'''''''''''''''''''-

I've also been opening tabs for questions tagged measure-theory here and quite a few of those seem to be at an appropriate level :)

Ted Shifrin

21:53

Folland is not much easier, I don't think.

Mike Miller

@Balarka Someone I know is at Bonn right now, having a great time. They apparently couldn't resist a Bengali pun: they named the photo album "Bonnbash".

O'''''''''''''''''''-

@TedShifrin I'll take a look, thanks!

Mike Miller

O-man: You should really try to work through SS with the help of faculty. You will learn a lot, even if you find the exercises difficult.

Balarka Sen

@TedShifrin Oh, my coordinate system has z-axis on the mirror, not xy-plane on the mirror. lol.

Ted Shifrin

@MikeM: I was grossly underwhelmed by and unhappy with SS for the complex analysis course. I started out excited to use it, then ended up regretting it miserably.

Balarka Sen

21:54

@MikeMiller Hahaha. Excellent.

Ted Shifrin

@BalarkaSen Completely contradicting your answer to me when I queried 15 minutes ago.

Balarka Sen

Sorry for misunderstanding.

Ted Shifrin

Up and down is in the plane of the mirror, not out of the mirror.

O'''''''''''''''''''-

@MikeMiller Hm I would love to do that, but I took too many classes this semester to really get into it. How about I revisit S&S next semester having gotten the basic background now?

Mike Miller

@Ted: I have looked at their series and I agree with your claim on complex and stand by mine on real.

Ted Shifrin

21:56

Are you another over-eager person overdoing it, @O'''''''''''''''''''-? Take it easier and take the time to take only 2 courses at a time and really learn them. :P

@MikeM: Yeah, my colleague at UGA who taught real analysis a lot used SS and was generally happy, although he used Folland for a good portion of the course.

Balarka Sen

I like the exercises in the complex book.

O'''''''''''''''''''-

@TedShifrin Yes, I mean I haven't been doing that each semester. But next one I will definitely take much less--if I survive now.

Ted Shifrin

Aside from the emphasis on number theory (which I wasn't going to treat), I was stunned at how late in SS's complex analysis the logarithm and LFTs appear. But oh well ...

@Balarka: I liked a lot of them, but, as usual, I added a lot of my own.

Over the years of teaching out of Ahlfors, I developed a number of different exercises, some stolen from Polya-Szëgo (which I gave to Pedro when he visited me).

OK, I went over 40K. Now I really can retire. End of MSE.

Balarka Sen

Just don't retire from the chat.

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$Mathematics$ Mathematics