« first day (3877 days earlier)      last day (1131 days later) » 

12:00 AM
@leslietownes great, so i never have to pay attention to pi day ever again
(i don't mind Pi Day as an excuse to eat pie, but I find the focus on Pi a bit silly)
 
What is $\text{GL}_2(\Bbb R)/\text{GL}_2(\Bbb Z)$? It's the unit tangent bundle to the modular curve $X(1) = \Bbb H^2/\text{PSL}_2(\Bbb Z)$, I think.
Fun fact, $\text{GL}_2(\Bbb Q_p)/\text{GL}_2(\Bbb Z_p)$ (p-adics) is the $p+1$-regular tree. This is the "unit tangent bundle" on the vertices of the $p+1$-regular tree, I guess.
OK, I'm gone
 
i really hated pi day when i was an academic. it seemed like it only really began to catch on after 2000. drove me crazy. now i am one degree removed. it's silly to focus on pi. i have identified more interesting constants in my published work.
 
TREE(3) day
 
12:19 AM
Some pi trivia. 22/7 is an ok approximation, but it's even better if you subtract 0.04%. That is, (22/7)(1-4/10000). Of course, 355/113 is much better, but that denominator isn't much fun when you're doing mental arithmetic.
 
when i'm doing what, now? :)
i was on a call the other day where three people spent about ten minutes computing the volume of a very small chamber. the stumbling block seemed to be once you've cubed everything, you're getting thousands of cubic meters, which aren't liters. i remained silent and hoped that nobody remembered my background. it was a success.
i may have opened this window and chatted about something else while it was playing itself out.
 
I must admit that my mental arithmetic skills aren't as sharp as they were a few decades ago. But I still like to do some stuff mentally, to keep those brain circuits functional.
 
and when i say about ten minutes, i mean, at least ten minutes. it was a long time. i floated the possibility of suggesting that people compute a volume offline, but abandoned it because i was one of five or six conference participants and didn't want to disrupt the flow of the meeting.
i think i have only gotten better with mental arithmetic over time, but i still suck.
i did a finance interview where they asked me to estimate the number of gallons of gas sold in the US in a day. my estimate was right, to the number of digits. they didn't hire me anyway. it's all a game.
 
One good thing about going to school during the era of slide rules & log tables: you got pretty good at keeping track of powers of 10.
 
i did mention this. i said "we all agree it's 1.32 times something and we just need to pin down the power of 10." it was ignored.
 
12:29 AM
I'm slightly disappointed that I got no responses to this:
Mar 8 at 20:46, by PM 2Ring
Quick Fermi problem: What's the square root of the ocean? That is, what volume is $v$, such that $n$ is the number of molecules of water in $v$ and $n^2$ is the number of water molecules in the ocean, IOW, the volume of the ocean is $nv$.
 
also it's not going to be a large power of 10, or a small power of 10. same as when i used to go to the grocery store and buy apples, i know if i buy five apples it's not going to be $25 or $0.50.
i hate fermi problems. they take me back to finance interviews, which i generally performed well in until the non-fermi part of the interview. and finance guys are d*cks anyway.
and by guys i do mean guys. i did not have a single female interviewer.
 
@leslietownes Exactly. Kids who depend on calculators don't seem to see the benefit in doing rough estimates just to make sure you haven't screwed up a decimal place somewhere.
 
it's really that simple. i'm not buying a new car for $2000 or $200,000. i might buy a car for $20,000.
oh, mathjax, where would i be without you.
 
$ ds^2=c^2dt^2-dx^2-dy^2-dz^2 $ anybody know how to write this in null coordinates?
 
12:33 AM
geocalc, no time travelling. we go positive t only. none of this goofy business.
 
@PM2Ring 5 mL
or a teaspoon
 
one of my officemates in grad school was notoriously bad at practical math. to the point where he had a collection agency going after him for missed cell phone payments. another officemate, completely useless at practical calculation, off by powers of a thousand. but she could spot errors in arguments. she had a superhuman ability.
 
@AkivaWeinberger Very good. I think it's closer to 8 mL, but the water volume I used includes all the water in the atmosphere, and groundwater in the crust. And it's probably a fair bit more if we include the water "dissolved" in the mantle.
 
Yeah I'll take it as being within the margin of error
 
@geocalc33 Mr Wiki knows.
In special relativity, light-cone coordinates is a special coordinate system where two of the coordinates, x+ and x− are null coordinates and all the other coordinates are spatial. Call them x ⊥ {\displaystyle x_{\perp }} . Assume we are working with a (d,1) Lorentzian signature. Instead of the standard coordinate system (using Einstein notation) d s 2 = − d t...
 
12:39 AM
1-2 tsp
 
@AkivaWeinberger Sure. If you're within an order of magnitude or two, you're fine.
 
Another way to phrase your question is, what's the geometric mean of the oceans and a molecule of water
$\sqrt{ab}$
 
Yep
We had a similar sort of question recently on Astronomy.SE. If you could bring all of the stars from the Milky Way together into one giant ball, without changing their density, how big would that ball be. There's somewhere between 100 & 400 billion stars in the galaxy.
 
$\delta_{ij}dx^idx^j$ what does this expression mean?
 
it would seem to not care about the quantity unless i = j, in which case it does care. maybe an integral involving coordinates labeled with superscripts.
 
12:49 AM
1
Q: How big would the Milky Way be if all the stars were emptied into one sphere?

Yevgeny SimkinNeglecting the fact that this ball of gas would just collapse on itself - I'm curious if there's an agreed-upon measurement that takes into account the volume of all the ~100 billion stars to predict the diameter of a milky way where all the gas is in one orb. If not, is it safe to just use the v...

 
the sun is a mass of incandescent gas. some wise sage said that
 
@geocalc33 What leslie townes said.
In mathematics, the Kronecker delta (named after Leopold Kronecker) is a function of two variables, usually just non-negative integers. The function is 1 if the variables are equal, and 0 otherwise: δ i j = { 0 if i ≠ j ,...
 
@PM2Ring yup!
 
so $ds^2=dudv+dr^2+dw^2$?
 
12:55 AM
One of my science teachers had a telescopic cylindrical slide rule, with helical scales. It only had a few scales on but the scale length was 60".
 
actually I think it's $ds^2=dudv-dr^2-dw^2$
 
my high school calculus teacher had a box full of damaged or otherwise inoperable slide rules. he let me keep one. it had something that let it do trig functions, or would have let it do that, but the indicator had slid away and it was just two marked pieces of plastic. it needed the third piece of plastic.
 
@geocalc33 That looks better. Assuming the u & v are the null coords, and r & v are the spacelike ones.
Standard slide rules had sin & tan scales, usually on the back of the slide. Having cos as well would've been a waste of space.
Oops. I meant *and r & w are the spacelike ones. Sorry about that.
I still have a small 5" slide rule. But it's packed away in storage, under the house.
 
1:15 AM
our storage space is above our garage. we did not inspect when we evaluated the house. when we moved in it was clear that somebody had been growing weed there. at least until it became legal. there were newspapers from the 1980s. who knows what happened.
there were all these housings for lights and fans and homemade cardboard ductwork to put skunky aromas out of the garage.
it's legal now, you can walk literally 5 minutes away and get anything. probably ruined the point of all the ductwork.
 
@PM2Ring I had a prized circular slide rule I bought in high school. I bequeathed it to the UGA Math Club when I retired.
 
1:31 AM
I had a circular one too, as well as a standard one. The circular slide rules were great, due to the continuous scale. But they were a bit too big to fit into a pocket. :)
 
i want a circular one.
also want a pocket antikythera mechanism. and as long as we're wanting, a pony. for my daughter.
someone should investigate the invisible link between children and horses.
 
There were tiny circular ones (3” diameter), but mine was respectable!
 
Yeah, I've got a 3" circular one, embeded into a large rectangular thing, but it's practically useless, so I didn't mention it earlier.
 
ebay is overpricing these things. it's marked plastic.
no sale.
 
I bought a couple of planimeters on ebay. One ended up photographed for my multivariable book.
 
1:44 AM
I've just been looking at slide rules on Ebay. They have a few circular ones, but they're mostly pretty pricey. I did see one for 4 bucks, though.
 
i'm afraid if i buy the four buck one i get a color printout of the ebay listing.
 
I'm almost tempted to write an SVG file for printing a slide rule...
 
my daughter is absolutely losing her mind right now. it's mealtime. she's throwing dishes around the kitchen. terrible twos.
 
Maybe you should be in there and not here.
 
yeah, probably. you remind me of my wife's therapist. that's a positive thing. she seems nice. she argues my point of view and i'm not even there.
 
1:49 AM
Go.
 
i'm on it. thanks.
 
Bye :)
 
2:03 AM
@robjohn Please justify the construction :-)
 
2:13 AM
daughter is now in my home office, yelling at the cat. this counts as parenting.
my daughter was asked to explain her actions and said: "it's a mystery." flashbacks to A Serious Man.
another ready-baked answer she has for why she did something is: "for some reason." as long as you did it for some reason, it's unassailable. check mate.
 
in the UK, they punish their kids for being terrible when they were two, by putting them in preschool when they turn three
 
i've investigated when my daughter becomes the problem of the state. it seems like no earlier than 5.
in the meantime it's just all of this nonsense. i secretly love it and will miss it when it's gone. but it's a pain in the ass.
we spent 20 minutes after she got home from (paid, expensive) day care feeding a stuffed mouse to a bird puppet.
she really loves the idea of animals eating other animals. again, at some point this becomes the state's problem.
 
2:39 AM
 
3:20 AM
The Tangent Secant Theorem says that $PT^2=PM^2=PA\cdot PB$
and the Right Triangle Altitude Theorem says that $PD^2=PC\cdot PB$
since $\triangle CDB$ is a right triangle
 
In mathematics, a pair of pants is a surface which is homeomorphic to the three-holed sphere. The name comes from considering one of the removed disks as the waist and the two others as the cuffs of a pair of pants. Pairs of pants are used as building blocks for compact surfaces in various theories. Two important applications are to hyperbolic geometry, where decompositions of closed surfaces into pairs of pants are used to construct the Fenchel-Nielsen coordinates on Teichmüller space, and in topological quantum field theory where they are the simplest non-trivial cobordisms between 1-dimensional...
 
4:06 AM
@robjohn Thanks :-)
It was really an interesting construction!
@robjohn O_o How did it become a xkcd styled figure?
 
5:03 AM
@Astyx Do you remember you had a question about the sequence $0 \to \operatorname{Br} k \to \bigoplus_{v} \operatorname{Br} k_v \to \Bbb Q/\Bbb Z \to 0$ ? What was that question? lol
 
5:15 AM
i missed the slide rule convo. still have my faber castell.
 
 
2 hours later…
6:46 AM
Is there some relation between pullback in differential geometry and categroy theory?
I just saw the word pullback in my analysis on manifold class
 
7:12 AM
@love_sodam I'd say that pullback in differential geometry corresponds more to contravariant functor in category theory
in DG you're pulling back something on Y to something on X via a map X -> Y
in CT terms this gives you a contravariant functor from manifolds
 
7:35 AM
one question. If $f,g\in C^\infty(M)$ where $M$ is a smooth mfd, then $d(f/g) = (gdf-fdg)/g^2 on the set where $g\neq 0$. The definition of $df$ (differential) is for $f\in C^\infty(M)$, df_p(v) = vf for $v\in T_pM$.
I proved this using the coordinate representation of $df$
How can I prove it only from the definition of differential
?
 
8:06 AM
@Wolgwang I've incorporated the Theorems into the image, too ;-)
 
 
2 hours later…
9:43 AM
@leslietownes this sounds all very reasonable and I hope the state won't make her stop
Especially yelling at the cat. Cats deserve it. Elegant, mischievous b*stards
 
10:11 AM
@EdwardEvans IIRC, I was told the arrow $\bigoplus \operatorname{Br} k_v \to \Bbb Q/\Bbb Z$ was just the sum $\Bbb Z/2\Bbb Z\oplus \bigoplus \Bbb Q/\Bbb Z \to \Bbb Q/\Bbb Z$
I was wondering whether $\Bbb Z/2\Bbb Z$ was indentified to ${1\over 2}\Bbb Z/\Bbb Z \subset \Bbb Q/\Bbb Z$ in that sum
(since $\operatorname{Br}k_v \cong \Bbb Q/\Bbb Z$ for $v\ne \infty$)
 
10:40 AM
1
Q: Exercise 11.17 on John Lee's Introduction to smooth manifolds [Proof verification]

barista Let $f(x,y) = x^2$ on $\Bbb R^2$ and let $X$ be the vector field $X = \text{grad}\ f = 2x\frac{\partial}{\partial x}$. Compute the coordinate expression for $X$ in polar coordinates (on some open subset on which they are defined) and show that it is not equal to $\frac{\partial f}{\partial r}\fr...

Can anyone check my proof is correct?
 
10:54 AM
Is it possible to find a mentor on this website, who is interested in "keeping track" of my progress in a certain field? Do not want to ask this on the meta, because it will be closed anyways, but maybe someone with the necessary expertise is reading. :)
 
@Astyx right then the answer is yes since Br k_v is 1/2 Z/Z for v a real prime and 0 for v a complex prime lol, I was just reading about it last night and the sequence came up
I don't know the deets tho
 
11:18 AM
What do you mean by "complex prime"?
I've seen this sequence when $k=\Bbb Q$, I don't know about general $k$
 
11:34 AM
Hello everyone, I would like to invite you all to help me clear some doubts in math.stackexchange.com/questions/4062893/… Any help will be really appreciated. I don't have enough resources to get a maths teacher. Kindly help.
 
 
1 hour later…
12:34 PM
@Astyx places corresponding to pairs of complex embeddings of your number field
 
1:05 PM
Ah ok
that makes sense
 
hello
 
Hold up, aren't all places coming either form $\Bbb R$ or from a p-adic norm ?
Oh that's only on $\Bbb Q$
What are you reading ? @EdwardEvans
 
Hey, say I want to see if there is a video lecture online for a particular research paper has anyone got any search tips? Other than just the name of the article in youtube...
 
1:54 PM
@DanielAdams that's probably extremely rare. It would be one of the authors of the paper presenting it, and probably a recording from a conference. You can search the authors' names in youtube and see what comes up, and as a last attempt you can email one of the authors and see if they ever recorded a video of a talk on the paper (unlikely to occur given how many talks get recorded and how many papers released).
 
2:42 PM
@Astyx for number fields the valuations are p-adic for a prime ideal p, or a real/complex embedding
 
Right, I figured it would be something like this
@EdwardEvans There's a very nice MSE post on Brauer groups somewhere
I'll try to find it
 
3:43 PM
5
A: Which number fields can appear as subfields of a finite-dimensional division algebra over Q with center Q?

Infinity$\def\Q{\mathbf{Q}}$ $\def\Z{\mathbf{Z}}$ $\def\Br{\mathrm{Br}}$ $\def\inv{\mathrm{inv}}$ $\def\Gal{\mathrm{Gal}}$ The questions you ask are essentially straightforward enough, but require a little theory. In fact, I once asked a starting graduate student to answer your question as preparation f...

 
4:29 PM
Elementary kinematics question
If an object at rest falls from $10\mathrm{m}$ with acceleration $9.8\mathrm{m}/\mathrm{s}^2$, what will be its velocity when it touches the ground?

If upwards direction is considered positive, then $s = 10\mathrm{m}$, $u = 0\mathrm{m/s}$, $a = 9.8\mathrm{m/s^2}$. I should be able to get $v$ from applying $v^2=u^2+2as$, however that results in a non real value.
Sorry, I meant $a = -9.8\mathrm{m/s^2}$
My teacher just flipped the sign of the acceleration to resolve this issue, but what is the justification for this change?
I'm being taught that objects moving downward will take positive gravitational acceleration while objects moving upward will take negative gravitational acceleration.
But that doesn't sit right with me because an universal perspective isn't being maintained
 
what's v ?
 
The final velocity
 
So you're using energy conservation right ?
 
I think there's just a missing sign
 
${1\over 2}m v^2 + mgz$ is constant
 
4:43 PM
the gravitational potential has derivative the negative of the force
 
Neh it's a confusion about what the quantities are
The point is, the higher you are, the higher your gravitational potential energy is
 
I'm not sure what you mean by energy conservation
 
Do you know where $v^2 = u^2 + 2as$ comes from ?
Or is it just a formula you pull out of a magic hat?
 
ultimately all of physics is a magic hat
 
Fair point
 
4:48 PM
I got it from $s = \frac{u+v}2t$ and $v=u+at$
 
Are those given ?
 
I don't get where that first one is coming from
 
They're the part of the equations of motion
@Astyx Yes
 
@Thorgott It's true when t is the total time it takes for the thing to hit the ground
@Typo You've got a sign error
It's "final position - initial position"
here s is the initial position, hence you get $0-s = {u+v\over 2}t$
(which makes sense, because the RHS is <0 since v<0 and t>0)
 
Oh I see
I was never told about there being more to that formula, instead just told to bypass the problem by flipping the sign of the acceleration like that
 
5:04 PM
The fundamental thing to know are Newton's laws
And how to integrate polynomials
If x is the position, you know that x'' = a
Then x' = at +u
and $x = at^2/2 +ut + x_0$
here u=0 and $x_0=s$
 
Oh my goood, that's eye opening
I can't hold in my grin
 
You're looking for the time $t_f$ when $x =0 = at_f^2/2+s$ which gives $t_f^2=-2s/a$
Here we're interested in $t_f >0$, hence $t_f = \sqrt{-2s/a}$
Plug that back in $x'$ to find the final speed, $v=x_f' = \sqrt{-2sa}$
One interesting thing is that we could had a solution with $t_f$ negative
This just shows that the object should have been thrown with that speed at $t=t_f<0$ for it to have this height and zero velocity at $t=0$
 
5:20 PM
Thank you so much for the help
I've learned about basic calculus and am now learning about basic mechanics. To see that these two topics are so linked is really insightful
 
Glad to help
 
Salut @Astyx
 
Salut!
Quoi de neuf ?
 
5:42 PM
De neuf? Quelque chose de neuf existe vraiment? :)
 
Qu'est-ce que la nouveauté ? Vous avez 4h
L'usage de la calculatrice est interdit
 
4h? Oh merde ...
Et de l'ordinateur en plus?
 
C'est ce qui est écrit sur les sujets d'examen
En particulier les concours des écoles normales supérieurs en lettres sont connus pour avoir des sujets très abstraits.
 
En ce cas-là, à quoi sert une calculatrice?
 
Absolument à rien
Enfin c'est possible de mettre des références/citations dedans
C'est pourquoi c'est interdit de les utiliser
Mais je pense que si qqn a besoin d'une calculatrice pour ça, il n'a de toutes façons aucune chance
 
5:56 PM
Ah, oui. On pourrait mettre tout un livre dedans.
 
Aussi il y a des calculatrices connectés
Enfin l'image plus haut est un exemple de sujet aux ENS
 
"n'est pas autorisé" ≠ "est interdit" :D
 
Et forcément, il y a des candidats qui ont expliqué pourquoi l'usage de la calculatrice n'était pas autorisé
6 heures aussi
 
Ah oui, 6 h aussi.
Tu rates comme autorité :P
 
J'aurais jamais pu faire lettres
 
6:50 PM
Is this "Chambre pour le français"?
 
Not unless you want us to sleep. :)
 
camera est omnium linguarum ^^
 
Can't you give us an ablative absolute, while you're at it?
 
hoc quaesito ita feci
 
Wait. Where's the ablative?
 
7:01 PM
hōc = ablative of hoc
quaesītō = ablative of quaesītum
 
Oh ... Well, my Latin is rusty. I last studied it in 1968.
 
wow that's a long time ago
 
You think? :D
 
did you mistake ablative for another case or...?
 
I don't recognize any of these words at this point.
 
7:05 PM
hoc = this thing
quaesito = having been asked
ita = in this way
feci = I did
so the whole thing means "with this having been asked, I did it so"
 
bows to esteemed Leaky
 
@TedShifrin Romans were still around then right?
 
I don't understand why Latin is seen as like a prestige language or whatnot
it's just another language
 
@Astyx But their togas were getting worn.
Another language that no one has spoken in many centuries.
I loved learning it, though.
As a gift to my 9th grade Latin teacher (who was less than stellar), I bestowed upon him my own typed solutions manual to all the exercises in the Jenney text we used.
waits for Astyx to appreciate the weak pun
 
not a few centuries ago everyone was writing in Latin
why did everyone stop
not many*
 
7:09 PM
Well, that was still almost a few centuries ago.
Because it was overly pompous to write in a language that no one spoke?
 
I thought everyone learnt it in school
I don't know if the situation is like English nowadays
everyone learns English in school
 
I don't believe all Americans studied Latin in the 19th and 20th centuries.
Perhaps that was so in parts of Europe. Surely the Greeks didn't?
That's actually a good historical question.
 
maybe only the rich learnt latin
were Euler and Gauss rich?
 
> More than 149,000 Latin students took the 2007 National Latin Exam. In 2006, 3,333 students took the AP Latin Literature exam.
maybe in the future there will be a second renaissance
and people will place more importance on Latin again or whatnot
 
7:17 PM
I cannot seem to find a paper, can anyone help me find it? It's Y. Ihara's "Discrete subgroups of $PL(2, k_p)$"
 
@BalarkaSen this? (link to direct download)
 
Howdy, a @Balarka.
Balarka has gone p-adic.
 
This seems right, @LeakyNun. Thanks!
 
np
i searched "y ihara discrete subgroups"
 
Hi @Ted
@LeakyNun Lol this one links the one I was thinking of see reference [2]
But it's a "to appear"
 
7:21 PM
oh lmfao
 
Oh oh.
 
apparently it appeared in Proc. Symp. in Pure Math.
 
is that legal
 
Yes.
 
@BalarkaSen I take it as the problem having been solved then
 
7:22 PM
It used to be legal to reference preprints and private communication, too.
 
I proved the Riemann Hypothesis
see private communications
 
Still can't find it rofl
Why did he have to use PL(2, k_p) in the title
This shit makes it hard to find
 
@BalarkaSen the paper I linked was on March 11, 1966
maybe you need to find somewhere near
 
That's even older than 1968.
 
@inproceedings {MR0205952,
    AUTHOR = {Ihara, Yasutaka},
     TITLE = {Discrete subgroups of {${\rm PL}(2,\,k_{\wp })$}},
 BOOKTITLE = {Algebraic {G}roups and {D}iscontinuous {S}ubgroups ({P}roc.
              {S}ympos. {P}ure {M}ath., {B}oulder, {C}olo., 1965)},
     PAGES = {272--278},
 PUBLISHER = {Amer. Math. Soc., Providence, R.I.},
      YEAR = {1966},
   MRCLASS = {10.21 (14.49)},
  MRNUMBER = {0205952},
MRREVIEWER = {M. Eichler},
}
@BalarkaSen this is the bibtex ^
 
7:25 PM
Thanks!
 
How does that help?
 
P.274 on pdf
(printed as P.272)
@TedShifrin it allows me to search "algebraic groups and discontinuous subgroups"
 
Cool, I was just going to track down the DOI
 
it's only 7 pages long lol
imagine wasting so much time to find 7 pages
 
lol yeah
Now that I have the paper hopefully I'll be able to understand what I want from it
 
7:28 PM
Interesting that the whole book is on-line. I wonder if that's legal :P
Some of the deepest results have been in very short papers. They don't all have to be thousands of pages on monsters.
 
Yeah, this one is supposed to prove existence of an infinite family of Ramanujan graphs
 
But it uses theory of automorphic forms lol which I do not know shit about
 
rip
@EdwardEvans is that not covered in Cassels--Fröhlich?
 
I haven't read Cassels-Fröhlich lol, but I guess it's covered in lots of places
 
7:32 PM
it's the standard reference for this kind of stuff
 
it's in Serre for example
 
8:00 PM
does a metric allow you to determine all the isometries?
 
@EdwardEvans whoa I didn't know about this def
 
In terms of Galois cohomology?
 
yeah I don't know about Galois cohomology
I only know about central simple algebrae
 
Ah okay, me neither tbh hahaha, but I think the usual definition is the one you know about CSAs
and the isomorphism with the 2nd Galois cohomology dude is a theorem
 
I only understand like half of part 3 of your link lol
 
8:11 PM
yeah the document is kinda the wrong way around right?
 
Well CSA is certainly the more elementary approach
 
right
 
But I guess Galois cohom is the more fundamental approach
Also the guy who wrote probably thinks he's so cool for doing it that way
 
lmao William Stein's stuff is cool
 
8:39 PM
Galois cohomology is evil
why tf is the étale site the right notion
 
 
2 hours later…
10:49 PM
The man I've been waiting to speak to has arrived......Good day Mr. Shifrin..
are you available to provide some help right now@TedShifrin?
 
11:05 PM
What's up?
 
11:19 PM
SO I got a couple questions.....

1) I was trying to figure out the last part of the previous question we were discussing, but I'm hitting a brick wall. This was #10 - Sec 1.5 - the cross product one.

So I've arrived at a place using your suggestion and what I got. So we have the area of the paralellogram defined as $||\mathbf{a} \times \mathbf{x}|| = ||\mathbf{c}||$
Now assuming the existence of a vector $\mathbf{x_{0}}$ being orthogonal to our vector $\mathbf{a}$, we can express the area of the parallelogram as $||\mathbf{a}|| ||\mathbf{x_{0}}||$
rearranging with our original cross product we have $\frac{||\mathbf{c}||}{||\mathbf{a}||} = ||\mathbf{x_{0}}||$
also I can express the height of this paralellogram as $\frac{||\mathbf{c}||}{||\mathbf{a}||} = ||\mathbf{x}^{\perp}||$, where $\mathbf{x}^{\perp} = \mathbf{x} - \frac{b}{||\mathbf{a}||^{2}}\mathbf{a}$
so from this I end up with an equality of $||\mathbf{x_{0}}|| = ||\mathbf{x}^{\perp}||$
So from here I said observe: $||\mathbf{x_{0}}||^{2} = ||\mathbf{x}^{\perp}||^{2} \rightarrow \mathbf{x_{0}} \cdot \mathbf{x_{0}} = \mathbf{x} \cdot \mathbf{x} - \frac{b^{2}}{||\mathbf{a}||^{2}} + \frac{b^{2}}{||\mathbf{a}||^{4}} (\mathbf{a} \cdot \mathbf{a})$
 
I have said this a dozen times. You want the line through $x_0$ parallel to $a$.
You're just writing way too much crap,
 
a line through $\mathbf{x_{0}}$ parallel to $\mathbf{a}$ will be of the form $\mathbf{x_{0}} + t\mathbf{a}$...if I'm remembering correctly
 
11:34 PM
Yes. And the area is the same. We discussed this a month ago. All parallelograms have same base and equal heights.
 
So I have this line through $\mathbf{x_{0}}$, I don't see how it is helping me....
 
Those are all the $x$ solving the cross product equation.
Then ask which ones satisfy the dot product equation.
 
Ok.....but then if that is the solution. THere are a few things that are bothering me which would explain why I couldn't get this....

1) I was expecting to derive an explicit solution of some form: i.e $\mathbf{x} = \mathbf{x}(a,b,c)$, in some algebraic way.

2) assuming the existence of $\mathbf{x_{}}$, shouldn't I be able to express it in some explicit way? as in with some coordinates? That's what I was expecting to happen as well
was typing this while you typed your last statement
 
You can of course write formulas, but I was looking for geometric understanding — unique solution because .... You can certainly find a formula for $x_0$ using cross products and then solve for $t$ explicitly.
You can write every vector in that plane as a linear comb of $a$ and $c\times a$, for example.
Then use triple product formula and a bunch of algebra. Someone told you that solution ages ago.
 
before I process that last part, going to the satisfaction of the dot product.

it should be: $\mathbf{c} \cdot (\mathbf{x_{0}} + t \mathbf{a}) = 0$
 
11:50 PM
No, every vector in that plane is orthogonal to $c$. I'm talking about the dot product with $b$ in it.
 
from that I get the scalar: $t = \frac{b - (\mathbf{a} \cdot \mathbf{x_{0}})}{||\mathbf{a}||}$ after simplifying the denominator
 
What is $a\cdot x_0$? This isn’t quite right, is it? I'm doing it in my head.
 
it would be some scalar value no?. But that depends on what $\mathbf{x_{0}}$ is.
 
Review where it came from.
 

« first day (3877 days earlier)      last day (1131 days later) »