Mathematics - 2015-12-27 (page 2 of 2)

Sir Cumference

19:00

Would 3↑↑↑3 be equal to the 3rd pentation of 3?

Khallil

Still here, @BalarkaSen?

Was wondering if you got the definition of simply connected domains and homotopy from a particular book so I can read a bit more on them.

evinda

19:17

Hi @Semiclassical

Semiclassical

hi

evinda

Sir Cumference

Er...anyone know?

evinda

@Semiclassical Are you familiar with the simplex method? I have written my question above...

Semiclassical

in linear programming? not really

evinda

19:21

A ok... no problem.... @Semiclassical

Hi @robjohn
Are you maybe familiar with the simplex method?

robjohn

@evinda I have used it before, but I have not used it much recently.

evinda

Aa... I have written my question above... Did you take a look at it?

Semiclassical

@evinda one thing I do notice is that the first two constraints can be written as $x_1\leq 4-2x_2$, $x_1\leq 1-x_2/2$

Ted Shifrin

19:36

hi @robjohn @Semiclassic (again), guten Abend, @evinda

robjohn

@TedShifrin are you in the land of cotton?

evinda

Guten Abend @TedShifrin

Ted Shifrin

LOL, no, robjohn. Still in San Diego, recovering from a sinus infection. I leave for Ann Arbor Tuesday morning.

I taught the simplex method back in 1982, but I haven't thought about it since. But we didn't throw in extra variables. We just used the reduced echelon form to travel to the right vertex. But I've forgotten the algorithm.

robjohn

@TedShifrin Ah. I had a cold last weekend. I got better in time to do some last minute Christmas shopping.

Ted Shifrin

I've spent the morning cleaning up my place, robjohn, just so it can get dusty in my absence :P

Sir Cumference

19:39

Er...does anyone know Knuth's Up-Arrow Notation?

Ted Shifrin

It means composition of functions that number of times, @SirCumf

Sir Cumference

So would 3↑↑↑3 be equal to the 3rd pentation of 3?

Ted Shifrin

oh, yikes, I have no idea what pentation means

Sir Cumference

it's how many times you do tetration on something

tetration is how many times you raise an exponent

the third tetration of 3 is 3^3^3

Ted Shifrin

Right, so $3\uparrow 3 = 3^{3^3}$. I guess I don't remember what we do with multiple arrows.

Is $2\uparrow\uparrow 3 = \left(2^{2^2}\right)^{2^2}$?

robjohn

19:44

@OFFSHARING We will await your book. Let us know when it is out.

OFFSHARING

@robjohn Who is us? I'll let you know about it privately. Never discuss here my published stuff for avoiding the discussions I had in the past with some users (like Ted, that is an example)

robjohn

@OFFSHARING well, me and I imagine those interested on MSE.

@OFFSHARING You won't let the general populace know?

Ted Shifrin

Of course not.

@robjohn: Perhaps you could privately explain to chris'ssis that most of us respect her mathematical abilities but do not share her taste or her unrelenting what comes across as bragging about her being God's gift (or Ramanujan's gift) to mathematics.

OFFSHARING

@robjohn hehe, I only meant I don't plan to discuss here anything about my published stuff.

Ted Shifrin

And the adversarial tone in here has become unbearable.

OFFSHARING

19:48

Just to avoid other discussions. Moreover I think people already feel better since I don't post anything of my daily creations here.

I'm fine with that. Why to ruin the day of some users? Let them be happy all. ;)

@robjohn btw, did you have to deal with Barnes G-function in some limits, series, integrals? I'm curious if you have some such posts on MSE.

Just to be recorded: I never ever started a fight, but when I was provoked I answered back.

That's all.

mathworld.wolfram.com/BarnesG-Function.html

Barnes G-function

In mathematics, the Barnes G-function G(z) is a function that is an extension of superfactorials to the complex numbers. It is related to the Gamma function, the K-function and the Glaisher–Kinkelin constant, and was named after mathematician Ernest William Barnes. Up to elementary factors, it is a special case of the double gamma function. Formally, the Barnes G-function is defined in the following Weierstrass product form: where is the Euler–Mascheroni constant, exp(x) = ex, and ∏ is capital pi notation. == Functional equation and integer arguments == The Barnes G-function satisfies t...

Professor Hari M. Srivastava has some nice stuff in his books about Barnes G-functions, limits, series and some integrals.

I should also mention here Professor Choi Junesang (they both wrote an amazing book).

Simeon

Can a function from $\mathbb{R}^m \to \mathbb{R}^n$ be invertible if $m \neq n$?

evinda

20:04

@Semiclassical I see... But can this be used? I wrote the problem in its canonical form...

Balarka Sen

@Khallil I'm here now. Unfortunately, simply connected domains in $\Bbb C$ are never defined with the minimal amount of machinery as I defined you above (except maybe in good complex analysis books), and homotopy are almost never introduced in complex analysis books. To learn some of these stuff, you need to have a look at algebraic topology textbooks.

Standard reference would be Munkres Topology part II, but you need to know a bit about topological spaces.

(by bit, I mean the basis plus quotient topology and various forms of connectedness, especially path connectedness and local connectedness).

But everything needed can be found in Munkres Topology part I :)

Khallil

I'll start reading it now!

Thanks :-)

Balarka Sen

Note that they won't start with subspaces of $\Bbb C$, but arbitrary topological spaces.

How much of topology are you aware of?

Khallil

^ About that much, @BalarkaSen!

I did an exercise on it when I was taking analysis about 3 weeks ago :-)

Balarka Sen

Hm, so metric spaces?

Khallil

20:16

Yep, I've seen metric spaces and the notion of continuity between normed vector spaces

(The notion of continuity between metric spaces too)

Balarka Sen

You can learn basic notions of homotopy and fundamental group using metric spaces, but it'd be preferable to learn some abstract topology first. Although if you want to I'd encourage you take a peek in Munkres Topology part II.

robjohn

@TedShifrin With a few exceptions, I think that most people are here to help. I cannot deny that there are exceptions, however.

@BalarkaSen it also means wedge product and logical conjunction

Balarka Sen

indeed.

OFFSHARING

20:45

I put a password to the wireless network, I didn't have one before (or better say for a little while). My uncle tried to connect to internet (from a phone) and failed with the password, so I kept it a while like that.

It looks like all moves faster now (or does it just seem to me?).

robjohn

@OFFSHARING I have a password and also allow only certain MAC ID on my wireless. Each time a new device comes in, I need to add its MAC ID to the list.

OFFSHARING

@robjohn Really? Other people connect to your home wireless network?

robjohn

@OFFSHARING Not unless I have added the MAC ID of their device.

OFFSHARING

@robjohn It's nice to have such a neighbour, no payment for internet. :-)

My all neighbours have internet, also wireless connections, but I never connected to them, and I suppose I cannot because they have passwords.

But it's cheap the internet I have here, less than 10 euros/month (TV programmes included).

Mike Miller

@Balarka: contrary to what you say any good complex analysis book defines homotopy and simple connectedness in some form or another.

OFFSHARING

20:54

@robjohn almost the price of 2 cigarette packs per month.

robjohn

@OFFSHARING I have no idea how much that would be

Balarka Sen

ah, then I stand corrected. I haven't seen it defined rigorously in any of the CA books I have read, but not that they are very good.

apparently Khallil's book does not discuss it either

hi @AlexWertheim

Alex Wertheim

Hello @Balarka, how's it going?

OFFSHARING

@robjohn like 8$, the price of 2 cigarette packs.

Mike Miller

Can you name them?

Ted Shifrin

20:56

@AlexW: Happy almost new year!

Alex Wertheim

And a very happy (almost) new year to you too, @Ted. :) And a belated happy Hanukkah!

Ted Shifrin

You too :) So, you still standing, despite mean Mike? :)

Alex Wertheim

glares at Mike

Haha, more like thanks to Mike. Yep, good first semester. :)

Ted Shifrin

Quarter :)

Glad you're smiling :)

Mike Miller

Aw... I have to write a lesson plan for the first class.

Alex Wertheim

21:00

Haha, right you are. I always forget, and I have two to go.

Ted Shifrin

Lesson plan sounds so ... High school.

Alex Wertheim

We're both glad for that :) how's San Diego been?

@Mike: you're not just going to talk about Chern-Simons theory? =P

Ted Shifrin

Pretty good ..

Heading east in two days.

Alex Wertheim

Oh really? Where to?

Mike Miller

@AlexW: Oh, thanks, you wrote it for me.

Ted Shifrin

21:01

Ann Arbor, then ATL and Athens

Balarka Sen

@MikeMiller Hmm, I've read a bit of Titchmarsh, but I don't remember simple connectedness defined in a formal way. Also, a book full of computations Lipschitz-Spiegel just defines it hand-wavingly ("every loop can be shrunk to a point")

Alex Wertheim

Ooh, Ann Arbor is nice. Are you visiting family or math people (or both)?

Ted Shifrin

Friends ...

Alex Wertheim

(Sorry, that probably made it sound like I think Athens isn't nice. I just have never been)

Mike Miller

I haven't even heard of those books.

Ted Shifrin

21:03

@MikeM: I actually answered a real diff geo question this morning.

Titchmarsh is a classic.

Mike Miller

@Ted: :S

I just looked at the answer.

Balarka Sen

The first one is good, but old, and the second one is not really for math people.

Ted Shifrin

Alex, no apology needed. You're fine.

Alex Wertheim

:) always good to ring in the new year with good friends.

math.stackexchange.com/questions/1590279/…

Mike Miller

I stand by my statement, perhaps with the modification "Any good book that covers much ground"

Alex Wertheim

21:06

Eric Wofsey's answer is probably the best way to go here, but is there a way to make the Frobenius endomorphism work here?

Balarka Sen

I believe you, and I stand corrected. :)

Ted Shifrin

@AlexW: Is there only the obvious ring structure on $\Bbb F_q$?

Alex Wertheim

With the standard field structure on a field of characteristic $p$ which isn't equal to its prime subfield, I believe I have an argument that Frobenius can't be represented by a multiplication map. But I know very little about the different ring structures on fields of char $p$...

Mambo

It would be really helpful if anyone suggest places where harmonic analysis is done in more of a functional analysis approach rather than measure theoretric completely!!

Ted Shifrin

Aren't there lots of discontinuous ring structures on $\Bbb R$?

Alex Wertheim

21:11

@Ted: that's what I don't know, I suppose. :) If that's the case, then there's another example. For small enough values of $q$, if you require rings have unity, one can march through multiplication tables, but I don't know many stronger methods.

Khallil

@MikeMiller Could you possibly recommend one?

Mike Miller

eg one might want to know "Is every harmonic function the real part of a holomorphic function?" or "can I find a branch of log?" or...

anon

@AlexWertheim what do you mean "represented by a multiplication map"?

Balarka Sen

@anon Reminder: We have a discussion on ambiguity cohomology to do :P If you don't care anymore, it's fine though.

Hi, by the way.

Mike Miller

@Khallil: I could, but I'm not sure if others would agree.

anon

21:14

if you mean x->ax, well, then that doesn't send 1 to 1 unless a=1

@BalarkaSen yeah, I'm good now

hit me a link again

Alex Wertheim

@anon: yes. But this is with the standard ring structure on $F_q$.

Mambo

$\Huge\text{Help :}$ It is really tough to guess what a professor does from his publications.

Khallil

Long time no see, @anon. Hope you're doing well! :-)

anon

@AlexWertheim as opposed to what?

Balarka Sen

@anon.

2

Khallil

21:15

Any recommendation is better than a random one I find on the web, @MikeMiller! :-)

Mike Miller

Ahlfors.

Alex Wertheim

@anon: the questioned linked allows for different ring structures, including ones without unity. I don't have an argument for those cases.

Ted Shifrin

Hi @anon. Happy almost new year!

Balarka Sen

I believe that thing is some version of Cech cohomology. And merry Christmas by the way.

Khallil

Thanks, @MikeMiller!

anon

21:18

yeah, I looked up Cech cohomology. the nerve idea I also had when trying to capture the idea of local-possibility vs global-impossibility, but I was trying to define a fundamental group that way.

Mambo

I have a soft question !

Ted Shifrin

@Khallil: Gamelin at UCLA wrote a good book.

Balarka Sen

@anon If I am not wrong, you can define a fundamental group.

anon

also, yes, hi, merry xmas, happy new year, etc.

Khallil

Wicked! I'll try to see if my library has a copy of that as well, @TedShifrin. Thanks ^_^

Balarka Sen

21:19

It's called Cech fundamental group.

I haven't sketched out how it could be used here though.

It's the same idea in the end. You construct the nerve from your covering, and then you take fundamental group of the whole thing.

Mike Miller

@Ted: What does he do in it?

Balarka Sen

@anon By the way, one thing: Penrose's remark on $d_{ij}$ not depending on the choice of $A_{ij}$'s is actually correct and important, contrary to what you (and admittedly I) believed :)

anon

my idea was to define the fundamental group to be the homotopies of an embedding of the disk into the picture which at all times during the homotopy has "possible" image

I'll have to reread my question to remember what the dijs even were

go ahead and start writing an answer if you haven't already

Ted Shifrin

All the standard stuff, @MikeM, a bit more intuitive discussion of Riemann surfaces than most.

Balarka Sen

@anon I am not sure if I understand that. Elaborate?

Mike Miller

21:24

@Ted: Harmonic conjugates? Riemann mapping? Logs?

Ted Shifrin

Of course.

Balarka Sen

@anon Cool, I'll start writing one, will let you know when I finish (it'll take time since I'll do it in pieces).

Mike Miller

I'm happy.

Balarka Sen

Also, your interpretation of the ambiguity group is wrong.

Penrose meant something else in his paper (I do not blame you: it's so hand-wavy). At least, it's the only way it becomes Cech cohomology.

anon

@BalarkaSen Say the picture is a space X, and we have an open cover {U} of all open sets which, when you restrict your attention to only the picture as it exists inside U, are "possible" (rather than impossible like the whole picture). Let *:D->X be an embedding of the open disk U into X whose image is "possible" set U. A "path" in the fundamental group of X should be a homotopy H:*->* such that H(t,-):D->X has image which is possible for all t.

basically, you're shining a light on the picture, and moving around your light but are restricted to making sure you're only shining light on a region which depicts something possible

Balarka Sen

21:28

You see, it's the group of symmetries of the figure which makes changes to your pieces $Q_i$, but the figure remains the same. In the tribar example, one can enlarge one piece by $\lambda$, enlarge (i.e., shrink) the other piece by $1/\lambda$ and leave the third as it is. This leaves the picture invariant. So collection of all such symmetries exactly forms the ambiguity group!

You have a natural composition. First do one symmetry operation, then do another.

So it's precisely the set $\{\lambda \in \Bbb R: \lambda > 0\}$ under multiplication. Namely, $\Bbb R^+$ under multiplication.

anon

ah, I hadn't thought of them as symmetries of the space of logically possible interpretations

Balarka Sen

Ultimately, the group's meant to detect "ambiguity", not impossibility.

In the Necker cube, note that ambiguity is in the choice of a linking point/vertex to point outwards or inwards. So two choices. Choosing one vertex to be pointing outwards makes the figure "unambigous", as all others are automatically pointing outwards. Choosing it to be pointing inwards makes the figure unambigous again. So choice is in that single vertex. Reflection is our symmetry here.

So, ambiguity group is $\Bbb Z/2$.

L33ter

hi

Balarka Sen

@anon Ah. Also, I think you mean isotopy instead of homotopy there.

anon

probably

Ted Shifrin

21:35

Been to the Grand Canyon yet, Karim? :)

Balarka Sen

Oops, just noticed you already said it.

By isotopy, I mean homotopy through "possible" maps D^2 --> X here, of course.

L33ter

yeah @TedShifrin was super fun

Ted Shifrin

Ha ha

L33ter

today @TedShifrin we ate in some burger place

Balarka Sen

@L33ter So... you did not fall into it?

:(

L33ter

21:37

the burger was like 10x size

its crazy

Ted Shifrin

Balarka: There isn't time for them to have gone.

L33ter

we didn't go @BalarkaSen

I was just kidding

we will not have time to go today, but today we will go to circus de alley

Ted Shifrin

Karim: if you're going to be a tourist, you should actually do homework beforehand and figure out good things to do :)

L33ter

yeah

well, we are going to california we are planning to go to Universal studios, sea world, and disney land

Ted Shifrin

None of that is cheap, but you'll have fun. I've never done any of it. :)

L33ter

21:40

my uncle lives in california, so he said he will invite us to that :D

so I don't mind

Ted Shifrin

Very nice.

L33ter

btw I am marking for vector calculus next semester

I forgot all of that stuff

I don't know how I will be marking for it

Balarka Sen

@L33ter Or maybe you have gone, you're just kidding from your grave.

Ted Shifrin

You've heard me say how important it is ... and the analysis understanding of it.

L33ter

yeah @BalarkaSen :D I am just ghost here talking to you from my grave

Balarka Sen

21:41

There's always a hope.

L33ter

yeah I will attend his lectures @TedShifrin and also review this stuff again

I took calculus 1,2,3 and differential equations all in my first year

Ted Shifrin

Calc 3 is badly taught, in general :(

L33ter

so, most of calculus aside from integration I forgot it

no, my professor was quite good

and I remember I used to read stewart calculus chapter before doing the assignments

and understanding those stuff

Ted Shifrin

Most calculus book authors don't understand multivariable deeply :(

Balarka Sen

@anon I think $\tau$ should be some sort of 1-st betti number or something.

Ted Shifrin

21:43

And the organization is sort of universally not ideal in many places ... But I don't have opinions.

L33ter

yeah

@TedShifrin I agree

I want to visit Berkeley when I go to CA @TedShifrin

Ted Shifrin

It's about 400 miles from LA.

L33ter

oh

Ted Shifrin

Hmm, maybe 350. I've forgotten.

You really need to open a map on your computer.

L33ter

yeah

it is 200 miles

Ted Shifrin

21:47

No, that's not right.

L33ter

3 h 19 min (200.7 mi) via CA-99

Ted Shifrin

No, you've got something wrong.

Oh, LA = Los Angeles. Are you still screwing that up?

L33ter

yeah

lol

I don't know the notation of the cities here haha

Ted Shifrin

Well, how LA could stand for Vegas is beyond me.

Huy

^

LAs vegas

Ted Shifrin

21:49

rolls eyes

L33ter

haha

Huy

well it is CAli fornia

right ???

Ted Shifrin

no, Vegas is not

Huy

:P

L33ter

I am very excited to visit Berekley

that would be cool

Ted Shifrin

21:51

Visiting over holidays isn't so good. Things are all closed up and no students or faculty are around.

L33ter

oh

Balarka Sen

@anon Here's how I think it's related to Cech cohomology. Let $X$ be your ambiguous figure. $\{Q_i\}$ be a good cover (good cover means every open set and all intersections are contractible). Declare $C_0(X)$ to be collection of functions $\{Q_i\} \to \{q_i\}$ (these are the 0-cochains). Declare $C_1(X)$ to be collection of functions $\{Q_i \cap Q_j\} \to \{d_{ij}\}$ such that $d_{ij} = d_{ji}^{-1}$.

Coboundary map $C_0(X) \to C_1(X)$ is just takes a 0-cochain assigning $q_i$ to $Q_i$ to the 1-cochain assigning $q_i/q_j$ to $Q_i \cap Q_j$.

OFFSHARING

@robjohn the thing I like a lot at this notebook is that most of the time is pretty cold, and that's very confortable.

No heat at all, no noise, just much silence.

Balarka Sen

@anon Note that if you had triple or quadruple or so on intersections, you could continue similarly.

I bet the "right" coboundary map is going to satisfy $\partial \partial = 0$.

And then you can compute cohomology of that.

Of course, it measures how close you are for $(1)$ in your question to hold. That measures ambiguity/impossibility.

And higher cohomologies would exist too :)

The reason it doesn't exist for the tribar is that $Q_1 \cap Q_2 \cap Q_3$ is null.

L33ter

@TedShifrin I am quite excited for my last semester

and motivated 2

Superstar Monica

22:01

It looks like I have a new nickname now

Jasper deleted his account again?

Balarka Sen

@L33ter BTW, you were asking for smash product above. Here's what it is. Take $S^1 \times S^1$, where $S^1$ is the circle. This is a torus. Now a torus has a "meridian" (the circle which wraps around it and is tight around the handle) and a "longitude" (the circle going around the donut hole). These two intersect at a point. Now collapse (i.e., quotient) this subspace consisting of two circles touching at a point. You get $S^2$.

Thus, we say "smash" of two circles is $S^2$. Similar construction works for arbitrary spaces.

Superstar Monica

I'm Superstar Monica now. That's a cool nickname, isn't it?

L33ter

oh cool

that is so cool @BalarkaSen

Balarka Sen

Why is it cool?

L33ter

it is is geometric construction

I like such stuff

Superstar Monica

22:04

20

Q: Evaluating $\int_{0}^{\pi/4} \log(\sin(x)) \log(\cos(x)) \log(\cos(2x)) \,dx$

What tools would you recommend me for evaluating this integral? $$\int_{0}^{\pi/4} \log(\sin(x)) \log(\cos(x)) \log(\cos(2x)) \,dx$$ My first thought was to use the beta function, but it's hard to get such a form because of $\cos(2x)$. What other options do I have?

That's worth trying.

However, it is a deep integral, a profound approach is required.

Mary Star

22:22

Suppose that a unit-speed curve $\gamma$ with curvature $\kappa>0$ and principal normal $n$ is a parametrization of the intersection of two oriented surfaces $S_1$ and $S_2$ with unit normals $N_1$ and $N_2$.

Suppose that $\alpha$ is the angle between the two surfaces.

How can we calculate $\|(N_1 \times N_2)\times n\|^2$ ?

Khallil

22:49

Where does your question arise from, @MaryStar?

(I may be imagining it incorrectly, but I reckon $n$ will point either in the same or opposite direction to the cross product of $N_1$ and $N_2$, so it'd just be $0$?)

Mary Star

@Khallil I am looking at an exercise in differential geometry, at which I have to calculate this.

So $n$ is parallel to $N_1 \times N_2$ ? @Khallil

Khallil

That's what I would've thought.

I imagined something like that. Of course the positive curvature would mean there'd be bending of some kind in there.

I may well be wrong, however. I haven't studied much differential geometry, @MaryStar.

Mary Star

What do you mean by "bending in there" ? @Khallil

Khallil

There are intersecting planes (of $0$ curvature) in the image, but the surfaces will not be planes because $\kappa > 0$, @MaryStar.

Mary Star

Ok...

I haven't really understood why $n$ is parallel to $N_1 \times N_2$.
Could you explain it further to me? @Khallil

Khallil

22:59

Oh wait, I used the wrong word. I meant they would be perpendicular so the cross product of $N_1$ and $N_2$ would be parallel to the tangent to the intersection of the surfaces, @MaryStar. ^_^"

So in the picture, the cross product of $N_1$ and $N_2$ would point along the red line.

Do you see what I mean, @MaryStar?

Mary Star

But why are $n$ and $N_1\times N_2$ perpendicular? I got stuck right now... @Khallil

Khallil

Is $n$ not supposed to be perpendicular to the tangent along the curve of intersection, @MaryStar?

Mary Star

Yes. @Khallil

And why is $N_1 \times N_2$ parallel to the tangent along the curve of intersection? @Khallil

$N_1$ is perpedicular to the surface $S_1$, $N_2$ is perpedicular to the surface $S_2$.
We have that $N_1\times N_2$ is perpendicular to both surfaces, right?

Khallil

23:17

Yep, I agree with that. I'm not entirely sure why. Pictorially, it just seems to make sense.

I can't see why it isn't true.

Mary Star

Ok...

Khallil

I can understand your dissatisfaction, @MaryStar.
I've not thought about it before but now that I have, I feel bad.

L33ter

Hey @BalarkaSen $R^w$ under the box topology is locally compact right

but not under the product topology right ?

Mary Star

@Khallil You don't have to feel bad...

Is it as follows?

@Khallil

Superstar Monica

@robjohn how things happened here

?

8

A: Why is this definite integral antisymmetric in $s\mapsto s^{-1}$?

HINT: Your integral can be brought to this form $$\int_{-1}^1 \frac{\log \left((1-x)^s+(1+x)^s\right)-s \log (2)}{x+1} \, dx$$ and then you can split the interval and calculate each integral separately. Use on the positive side that $\frac{1-x}{1+x}\mapsto x$ and $\frac{1+x}{1-x}\mapsto x$ on the...

@robjohn I don't need the bounty but I'm curious to know why the other answer received the bounty (half of the bounty).

Khallil

23:31

Yep, that's how I see it.

Superstar Monica

Then I'm shocked some prefer to give the bounty to no one although it's about points that are definitely lost.

Khallil

$\implies \| \left( N_1 \times N_2 \right) \times n \|^2 = \| \| N_1 \times N_2 \| \| n \| \sin \left( 90^{\circ} \right) \hat{n} \|^2 = \| N_1 \times N_2 \|^2 \| n \|^2 $

Superstar Monica

@DanielFischer maybe you have more time. I'm curious to know how that bounty was granted.

Khallil

Then I guess that we'd need $\| N_1 \times N_2 \| = \| N_1 \| \| N_2 \| | \sin (\alpha) | \ \| \hat{n} \|$ where $\| \hat{n} \| = 1$.

Do you agree, @MaryStar?

Daniel Fischer

@SuperstarMonica Your answer was posted before the bounty was placed, and for an automatic award of bounties, only answers that were posted during the bounty period are eligible.

Superstar Monica

23:38

@DanielFischer OK, thanks.

Daniel Fischer

@SuperstarMonica You're welcome.

Superstar Monica

Bad for those 25 points that were lost. I would have been happier to see the other user takes 50 points rather than half.

Mary Star

Khallil

It'd be $| \sin (\alpha) |$ but the final result seems fine.

Does the principal normal have unit magnitude, @MaryStar?

Mary Star

Oh yes... So the result is $ \sin^2 (\alpha) $, right? @Khallil

Khallil

23:42

I think so.

Very cool question.

May I ask where you got it from, @MaryStar?

Mary Star

I am looking at the second part of the following exercise of the book "Elementary Differential Geometry" of Andrew Pressley :

@Khallil

Khallil

Which page is it on, @MaryStar?

Mary Star

@Khallil 170

Btw... Thank you very much for your help!! :-) @Khallil

Khallil

No problem! Glad to have been of help :-)

I'm out. Good night!

Mary Star

23:58

Good night!! @Khallil

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$Mathematics$ Mathematics