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19:02
I haven't heard of immanants. What are they about? @TobiasKildetoft
@Evinda Are you familiar with the other permanents?
ohh woops, the immanants are the general ones, the permanent is the special one that looks like the determinant
Can anyone help me? I have a little class 10th question
(all of which one easily finds out by googling immanant)
19:04
permanent of a matrix= "write down the formula for its determinant, then flip all minus signs to positive."
minus signs coming from the determinant, mind, not the matrix elements themselves
A ok, I haven't heard of permanents before
I forget what applications they have.
If sin x = anything.. say 3/4
Find x
permanents have some combinatorial interpretation iirc
The only one I know of is in quantum mechanics. When you want to write down an n-fermion wavefunction (e.g. a bunch of electrons in an atom) you write down what's known as a slater determinant
19:07
Like rooks on a chessboard with forbidden cells or similar problems maybe
this ensures that the wavefunction is antisymmetric under exchange of fermions, i.e. if i swap two indices it should flip the sign
Uhh.. can anyone spare little time to answer my question?
by contrast, a system of bosons should be symmetric under exchange.
@ArmaGeddON Do you know about sin^{-1} aka arcsine?
@TobiasKildetoft So at the topic you suggested we find the number of steps needed so that we find the permanent of a matrix?
So you can't do a determinant in the boson case. For that, you'd need a permanent instead.
19:08
@Evinda Yes, the complexity of computing the permanent is as far as I recall of interest (actually, I recall it being NP complete)
there's an entire wikipedia article on computing the permanent, in fact, which is easily found via google
@Semiclassical So in what cases do we compute the permanent?
@Semiclassical Ok, I will look for it
Semiclassical.. I read this.. x = (sin 3/4)^-1 but
Check out the main page on permanents for applications.
Doesnt seem to be working fpr me
19:10
@TobiasKildetoft Is there a lot say about this topic?
Did you mean sqrt(3)/2 and not 3/4?
@Evinda Have you looked at the wikipedia page yet?
When you say it doesn't seem to be working, do you mean that it's not coming out to the number you expect?
If so, you might check whether your calculator is working in radians or degrees.
There shouldn't be a nice answer for $\sin x=3/4$, though. It'll be some approximation.
for $\sin x=\sqrt{3}/2$, by contrast, the answer is nice (in either radians or degrees). hence Tobias's query.
19:12
Approx value is fine for me
ok. then what number are you expecting to get, and what are you getting instead?
@ArmaGeddON You write x = (sin 3/4)^-1. That's not what sin^-1(3/4) means though.
Hello chat
Balarka actually I read the formula wrong.. u r right ..the latter one is correct
But anyone tell me about it plz.. all I know is that its called inverse sine :p
How to calculate inverse sin?
19:20
Practically? Use a calculator.
In exams
Or shall I leave it in terms of inverse sine only?
Depends on what you're taking the inverse sine of. If it doesn't have a nice answer, then you'd indeed leave it like that.
You should not leave sin^-1(sqrt(3)/2) just like that, for example. Or sin^-1(1/2).
There are just a handful of nice cases, most relating to 30-60-90 or 45-45-90 triangles.
or sin^{-1}(1/sqrt(2)).
all of those have nice answers which you can see by using the unit circle.
on the other hand, some of them are just plain obscure. for instance, $\sin^{-1}(1/2*\sqrt{5-\sqrt{5}})=36$ degrees. that's something you certainly would -not- be expected to know.
it's true, but it'd be way too hard for a 10th grade exam.
19:24
M so confused.. not those standard angles.. in these sample questions I came accross 3/4, 2.6 and other weird angles
(it has to do with pentagons)
@ArmaGeddON That's unusual. Can you produce the questions?
arccos(3/4) doesn't have a nice answer as far as I remember.
and sin(x)=2.6 definitely doesn't.
Its a physics question. .
Refraction of light
Oh, then just leave it as sin^-1(3/4).
19:26
If it's a physics question and they're giving those numbers, then they probably want a numerical result.
In which case, -use your calculator-
10th grade does not allow calculator, @SemiC.
Depends on the school.
Ok, I guess.
Plus, for a physics versus math course?
Yup and its gonna be a mcq paper..if all values are numeric.. m gone
19:27
if it's multiple choice and requires numbers without calculators, that's dickish.
Question is
jesus man, language :P
pffft.
on the note of language, have you seen today's SMBC?
Hmm, I wonder if "dickish" or "Jesus" is the more offensive from a Christian point of view
At what angle must a ray of light incident on the glass so that it is refracted at an angle of 30 deg. Given refractive index of glass = 1.5
19:29
@Semiclassical Nope, let me google it
Okay. What's the relevant physics?
(I know what it is, but I want you to say it :) )
Refraction of light
There's a specific thing you want to use here (a law, a rule or whatever you want to call it)
Sure. What principle is applicable here to get the angle?
@BalarkaSen The fact that I recognize the stuff in the background as valid vector calculus...not sure how I feel about that :P
sin I / sin r = 1.5
19:31
right, Snell's law
Sin r is 30 deg
Yup snells law
careful. r is 30 degrees, sin r is not.
Yh. My bad
what is sin(30 deg)?
In any case his answer is correct
19:33
Sin r is 1/2
Cmon m not a nerd
You're right, the answer would be sin^{-1}(3/4)
Actually
@Semiclassical I guess I don't find this very funny :/
When they say that it's refracted at an angle of 30 degrees---angle with respect to what?
If it's to the vertical, you're right.
If it's to the horizontal, you're not.
19:34
Air / vacuum
Not what I asked.
When doing snell's law, the angle in there is relative to the vertical.
However, you could also specify that direction in terms of its angle to the horizontal plane.
So if you're 10 degrees from vertical, it's 80 degrees from horizontal.
Without that clarification, the question is ambiguous.
Its horizontal
They didnt teach is vertical yet
either we're talking on different wavelengths, or they've taught the subject in a very strange way.
19:38
Looks like we went off topic..
Eh, it's not off-topic
it goes to whether or not r in Snell's law is really 30 degrees, or whether it should be in fact 60 degrees
In these sample papers they wrote I = inverse sin of 3/4
Ah, okay then.
And then directly answer 48 in next line
...without a calculator, that's just
eugh.
19:40
Impossible?
not impossible, but not fair.
Hmm ik..
I could probably show that sin(48 degrees) is approximately 0.75 given enough gtime.
(or a calculator)
However, was that a multiple choice problem?
what were the other choices?
19:42
32, 48, 63, 72
hmm.
i guess you could work it out from those on the following basis.
I just rounded these angles for convenience
Difference in options is big so approximation can work
But I need to sleep now
Good call.
Thanks for help semi.. will catch u later
19:44
1:20am here
Btw whats ur nationality?
USA
it's 1:44 pm here, so yeah
K..cool
we're not exactly in sync :P
Bbl hope I catch u tomorrow :)
how do you trigger a captcha at MSE?
19:51
don't you need to fill a captcha to make an account?
mmh, no i mean if you post something, it happened once to me, but the captcha made no sense
it was "i'm no robot" lol
any sophisticated spambot would recognize that i suppose
@TedShifrin Heya
Hi @Balarka
(@Semiclassic: Isn't Snell's Law always done with the angle from the vertical — orthogonal to the interface?)
He was just wondering if that's the angle the question was talking of and not the angle from the horizontal
I've never seen a math or physics book do it otherly.
19:59
You misunderstand me. You're correct that the Snell's law is done with angles from the vertical; but the question might talk about the other angle.
morning
In which case you have to plug in 90 - that angle before applying the law.
G'night, @MikeM
@BalarkaSen Well, since I didn't see the original question, I shouldn't have said anything. But I understand complements of angles :P
Hello chat.
Fargle!
20:00
How goes it, @Ted?
It goes well. I still have another letter of rec to write, though :( ... How're you doing?
I forgot that it was called complement. Yeah, I was just clarifying.
How's Rudin faring?
@Balarka: It's not complimentary.
@TedShifrin everything ok?
I dunno, Null. Should it not be?
20:03
Hi @Ted
That's at least better than the violet pun Akiva made that day
@Ted: I'm doing well. Rudin is going a little more slowly now. The thing I'm really having trouble with is picking exercises that are non-trivial, but not too, er, advanced.
Hi @Alessandro, @Fargle
Hi @Balarka!
Heya @Balarka, what's up?
20:03
@Fargle: Well, you should try to do all levels. You are welcome to ask me for recommendations/comments.
Hi @Alessandro
@TedShifrin Thanks. I still need to get through chapter 3--I've been trying to make it a habit to read math books more slowly, and to digest proofs as they come.
Better yet, @Fargle. Read a proof, "understand" it, close the book, and try to write it yourself.
That's great advice. I'm a bit embarrassed I hadn't thought of that yet, haha.
Sometimes you might even try to write your own proof before you read the one in the book. Rudin is way too slick for my taste most of the time.
That's why I like Spivak better for most of this stuff.
@TedShifrin yes
20:07
I know, @robjohn :P And hi :)
@TedShifrin heya
Quick question for the room because I'm too lazy to think right now: What are the curves in $\Bbb R^2-\{0\}$ that make a constant angle with lines through the origin (or with circles centered at the origin)?
sigh, I guess j need to take this
need? it looks interesting, though.
20:12
it hits a few of my interests, tho ofc is not directly applicable to research
I thought you were learning automorphic forms for one of your friends.
haha I quit that class
Uh huh ...
too much work, I don't like him that much
LOL, I figured.
20:15
probably what will happen with this class too, since our new faculty member is giving a khovanov homology class in the same quarter
@Fargle Not much yet.
Busy times ahead, @Balarka?
nah, exams are over. i can chillax.
Awesome.
No one's answering my question about curves making a constant angle with lines through the origin? :(
20:22
Too hard.
Kidding, of course.
It's coming from an MSE question asking in essence for the geodesics in the punctured plane with metric $1/r^2$ times the usual one.
Just submitted a rec at UCLA for one of my former students, @MikeM.
@MikeMiller Oddly, I haven't yet figured the right way to look at this to solve it.
assume you are arclength parametrized. you want $(x'(t), y'(t)) \cdot (x(t), y(t))/\sqrt{x^2 +y^2}$ to be constant, yeah?
@Balarka: Angle doesn't need the conformal fudge factor.
Oh, but let's see.
You're doing the dot product, so I need the correct lengths in there.
OK, I agree.
20:33
Squaring $\alpha' \cdot \alpha/\|\alpha\|$ and differentiating would give you a differential equation but solving that would be a pain. I don't see anything slick coming out of this.
Switch to polar coordinates, of course.
Ah, yeah, that's better.
You get $r'=c$, but you have to impose the arclength restriction, which is $x'^2+y'^2=1/r^2$, I think.
Right, agreed.
No, actually, I don't agree any more.
The unit radial vector field is just $(x,y)$ in this metric.
I'll work this out after lunch.
20:37
Alright. I'm chickening out though :P
I should do spectral sequences
If you say so.
Bubye.
bon appetit
@Ted: Who? Maybe I'll see them at open house.
20:54
My vote right now is that in polar coordinates it's any curve of the form $r(\theta)=m(\theta-\theta_0)$
A spiral does satisfy that, yeah, I think.
in fact, I think it'd have $m=\tan \phi$ where $\phi$ is the constant angle to the circles.
hello
Up to rotation there's clearly only one such curve for each angle so if yours satisfy the condition...
(so long as $\tan \phi$ finite, of course. when $\phi=\pi/2$ you've got radial lines i.e. $r=$const)
The only thing I need to convince myself of is that it's $r'(\theta)$ that's constant rather than $\frac{1}{r}r'(\theta)$
20:59
Your solution is exactly what Ted figured though, $r' = c$
Yeah.
Actually, though, I think it really might be $\frac{r'(\theta)}{r(\theta)}=c$.
Shrug. I guess; too lazy to do the calculus right now
Since if I think in terms of differentials i'd best be working in terms of the changes in arc length and radius, i.e. $rd\theta$ and $dr$.
I'm not too lazy to do plotting, thankfully.
which tag has most of the easy questions?
^ vote to close: opinion-based.
21:03
geometric-topology
okay, here's an r'=c spiral
That's with $c=\tan(\pi/4)$.
Doesn't look terribly constant-angle
Your condition is probably th right one
so (log r)' = c, so r = e^{ct}
right.
and it looks more correct as well:
these would still be spirals tho
that's with c=tan(pi/10)
21:05
yeah, better
main problem is that it grows rapidly due to the exponential dependence
that's what i get if i go down to c=tan(pi/60)
As a final argument, from Mathworld one has: "The logarithmic spiral is also known as the growth spiral, equiangular spiral, and spira mirabilis." [emphasis added]
So I think that settles it.
proof by mathworld and mathematica
What's that?
Another cute thing from there: "The logarithmic spiral was first studied by Descartes in 1638 and Jakob Bernoulli. Bernoulli was so fascinated by the spiral that he had one engraved on his tombstone (although the engraver did not draw it true to form) together with the words "eadem mutata resurgo" ("I shall arise the same though changed")."
Wow
21:09
Just calculate the damn angles :p
I'm satisfied enough from the ODE, if I'm honest.
Now, harder question: What if we're on a sphere? (The Mathworld page has the answer)
loxodrome, iirc
:D
21:12
spoilsport
Amusingly, it's very simple on a Mercator projection. it's just straight lines.
i just remembered seeing it before
gotcha.
(I wonder what it is on a pseudosphere)
should still look swirly, but cluttering near the middle instead of cluttering towards infinity
yeah.
I'm partly wondering what happens to the spacing of the lines, since parallel lines diverge on it (negative curvature)
@Balarka: The curves are logarithmic spirals: $r = ce^{k\theta}$. It's more useful to eliminate the parameter and actually see the curves.
Oh, it seems @Semiclassic already said this above.
21:24
Does anyone know how to obtain $x$ when $x^x-x = 1 - \log_g(y)$?
@MikeM: If he gets accepted, I'll certainly make sure you know each other's names :)
@Selfless: Impossible to do so explicitly.
Hi @TedShifrin
@Ted: A fair enough condition.
Hi @Zach.
If we take $f(x) = x^x - x$, then can anything be analytically known about the inverse $f^{-1}(x)$ ?
21:28
In terms of Taylor series expansion, sure.
In terms of elementary functions, nadda.
@TedShifrin working my way though section 2 exercises
@Selfless: It's not a one-to-one function unless you limit yourself to $x>1$ or $x<1$.
Enjoying some of 'em, @Zach?
yeah :)
@TedShifrin Can this notion be translated to a finite field?
still pondering exercise 11
21:29
If you want to write up a few interesting ones carefully, I'll look at 'em, @Zach.
i think maybe desargues is involved, but im not sure.
@TedShifrin i have some of my solutions typed up on LaTeX, if you'd like to see them,
If you're asking about finite fields, you're in a much different context than the one we assumed.
@Zach: I'm skeptical re Desargues and #11.
@Semiclassical I realized, though I didn't want to influence the response. I often find insight from other areas of mathematics leading me to unexpected solutions to the overall problem.
I'm not sure I know what exponentiation means in a finite field.
21:31
> re Desargues
what?
#11, @Zach.
i don't know what you mean though
what is "re"?
well, 3*3*3=3^3 still makes sense
@TedShifrin It could be restricted to integer values (mod $2^n$ in my case)
It actually doesn't necessarily need to be a field at all
@Zach: re = regarding
21:35
oh, sorry :P
@Semiclassic: But what do you do with a finite field that is not $\Bbb Z_p$? I'm not sure what this means for other elements of $\Bbb F_{p^k}$.
Hi, @Ali. You finally got some sleep? :)
@TedShifrin Sorry, the choice of ring is rather arbitrary here. Anything finite is sufficient.
It just seems simpler to work in a field, but maybe it admits a nicer solution somewhere else!
I have no idea what it means, @Selfless.
I can't sleep because of the blister on my face
:( @Balarka
21:40
at least the doctor said it's not serious
I'm sure it's stress-related.
BTW, I've had cold sores and shingles ... stress caused the latter, for sure.
@TedShifrin Well my strategy is mind-numbingly basic, I'd end up defining it as an algorithm: To compute $x^n$, assume $n$ is always a multiple of the $1$ element, and then take $x\cdot x\cdots$ as many times as you have to subtract $1$ from $n$ to achieve $0$. (Or any algorithm that produces identical result)
yeah, very possible. Especially considering the history of psychosomatic diseases I have
@Selfless: That's fine if your ring is $\Bbb Z_m$, but won't make sense more generally.
I'm just surprised you're still alive.
21:43
Greetings Everyone :)
@TedShifrin I can give good confidence that such a definition will be useful on any ring where the rest of my crazy idea applies :P
Well, I've said all I have to say, @Selfless. You're on your own.
(actually I might be wrong, but I can just use Zp if it comes down to it lol, so it definitely can apply)
thanks for engaging!
LOL, sure.
@MikeM Working on it.
21:48
I would like to know how can you solve calculus test faster ? Should I get stronger at algebra ? Should I learn some silly algebra tricks ? What should I do to improve my current solving time, because even if when I try to speed up, I fall in mistakes most of the time ....
Very interesting anecdote: For odd values of $x$, the function $x^x \;(\text{mod} \,2^n)$ appears to cyclically generate a random-looking permutation of the odd values.
22:09
Bounty !! ( i advice Reading comments ) math.stackexchange.com/questions/1553888/…
If $X$ is a set, is there a name for a subset $F\subset P(X)$ such that the complements of the elements of $F$ form a filter? proofwiki.org/wiki/Definition:Filter_on_Set
something like "inverse filter" or "anti-filter"?
is there a better way to calculate the continued fraction of $2^{1/3}$? Which hopefully doesn't involve $2^{1/3}$ to stupidly high precision because I hit the limit with that method
22:49
Should i accept the answer here ?? Seems too short and sketchy to me + divergent series ?? But i do not want to be mean or ignorant !! Maybe it is ok ??
1
Q: Telescoping exercise with iterations?

mickIm assuming the identities below can be proven with telescoping techniques and / or differentiation. For $x>0$ let $f(x) = \sqrt{x + {1\over2}} ,\; f(x,0) = x,\; f(x,n) = f( f(x,n-1) )$. Now we get $$ - f(x)^4 + f(x,2)^4 - f(x,3)^4 + f(x,4)^4 - ... = 1/8 - x^2 $$ Where the LHS uses Cesàro s...

@mick if it got you further, yes, if not it's imo no answer
you can also only upvote, but keep it the question unanswered ;)
@Null it is confusing to me so basically no, but Maybe my skills are to bad :/ but i assume other ppl are not getting it either.
Or Maybe you should answer ;)
Hi @TedShifrin
@TedShifrin yeah, I get it now. I was thinking of the exponent as still being in Z, when of course it would also have to be in the field
[x]^x is well defined but [x]^[x] needn't be
23:04
can someone describe me how the roots of $a^3+4b^3+4c^3-6abc$ look like in 3d space? I mean something like the roots of $Ax^2+Bxy+Cy^2+Dx+Ey+f=0$ are a conic and such
it is related to the problem of writing $\frac{1}{a+b2^{1/3}+c2^{2/3}}$ as $x+y2^{1/3}+z2^{2/3}$
23:20
is not playing a game an option in gametheory?
lo, someone upvoted a truthtable of mine haha
I did :)
@mick scary haha
Dont be a pussy
being a pussy doesn't fit well with being a vampire :D
Oh god
Anyway are there " proofs " of the prime twin conjecture based on divergent series ??
23:29
Hardy is the guy with ramunjan right?
@Null yes
Yes but was Hz gay ?
I wonder 😜
eh
haha
my conjecture is: you need much less evidence for a conjecture, than it is given by today computational standards. It makes searching for provable statements easier, but brain.exe is still neccessary :s
23:50
Hello!! Let $f : \mathbb{R}\rightarrow \mathbb{R}$ be a function with $ f(x + y) = f(x) + f(y)$ for all $x, y \in \mathbb{R}$.
I want to show that $f$ is continuous if and only if $f$ is continuous in $0$.

The direction $\Rightarrow$ is trivial.

For the direction $\Leftarrow$ : Suppose that $f$ is continuous at $0$ then $\lim_{x\rightarrow 0)f(x)=f(0)$ and from the property above we have that $f(0)=0$.
How could we continue? I got stuck right now...
@MaryStar but $bf(x)=f(bx)$ is not given?
@Null Do we need that property to show the continuity?
no, it's just that linear functions are nice ;)
@MaryStar did you try out some "counterexamples"?
No... What kind of counterexamples do you mean? @Null
well, you got a set of statements that $f$ has to fullfill, can you build something which satisfies those, but is not continuous everywhere?
(it helps to think of counterexamples ;) )
23:58
In this case it doesn't.

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