Mathematics - 2016-02-21 (page 1 of 2)

GGG

00:17

Why does $x \equiv 3 \pmod 7 \text{ and } x \equiv 1 \pmod{11}$ imply that $x \equiv 45 \pmod{77}$?

idonutunderstand

@GGG by the Chinese remainder theorem

GGG

@idonutunderstand is there a way to get that implication without the CRT?

idonutunderstand

Why do you need to get it without the CRT?

GGG

00:37

Well, I tried to learn the CRT, but didn't understand what they meant by things like $p^{-1}\mod{q}$ unless that actually means dividing by $p$. (In this link crypto.stanford.edu/pbc/notes/numbertheory/crt.html).

idonutunderstand

00:53

Maybe a simple way to explain what $p^{-1}\mod{q}$ is would be to say it is the integer r between 0 and q such that $p*r \equiv 1 \pmod q$?

Also the set of all congruence classes of the integers for a modulus n is a commutative ring and you can define the inverse that way: en.wikipedia.org/wiki/Modular_arithmetic#Integers_modulo_n

GGG

I think I get the idea now, thank you. However I keep getting $47 \mod{77}$ instead of $45 \mod{77}$.

idonutunderstand

How are you arriving at $47 \mod{77}$?

Also my previous statement about the commutative ring is not quite right I think because one must first define the inverse before one even knows it is a commutative ring.

GGG

01:15

$11r = 1\mod{7}$ gives me $r = 2$. And $7r = 1\mod{11}$ gives me $r = 8$. Thus $11^{-1}\mod{7} = 2$ and $7^{-1}\mod{11} = 8$.

Is that correct so far?

idonutunderstand

yes

GGG

01:36

So $x \equiv 3 \times 11 \times 2 +7 \times 8 \mod{77}$ so $x \equiv 122 \mod{77}$ therefore $x = 45 \mod{77}$.

Thank you very much @idonutunderstand

idonutunderstand

ya

Akiva Weinberger

02:47

If a mathematician invented Celsius, freezing would be $0$ and boiling would be $1$.

Balarka Sen

06:06

@MikeMiller You asked for a proper diffeomorphism of $\Bbb R$ whose derivative is not uniformly bounded. I believe $f(x) = x + x^3$ works: $f'(x) = 1 + 3x^2$ which is nowhere zero, and $f' > 0$, so $f$ is strictly increasing forcing injectivity. It is not hard to see (either by staring at the graph or solving the equation) that $f$ is also surjective. So $f$ has an inverse, and that inverse is also continuously differentiable because $f' \neq 0$.

Thus, $f$ is a diffeomorphism - it is clear than $f$ is also proper. $f'$ is unbounded because $f$ becomes arbitrarily steep towards infinity.

Balarka Sen

06:36

Hi @L33ter.

L33ter

Hi @BalarkaSen how are you doing

I have economics exam after tomorrow

Balarka Sen

I am good. Wish you luck on the exam!

L33ter

Didn't study open the book or study it for a month it is easy though. It is 100 level economics just bunch of introductory stuff. I just feel you know I shouldn't be doing these stuff haha.

Art elective sucks

Balarka Sen

But economics is not arts!

L33ter

It is considered as art here

Balarka Sen

06:39

That's silly.

L33ter

I guess any social science here is considered art.

I do agree anything that doesn't follow the scientific method shouldn't be classified as science.

Balarka Sen

Maybe political science and nonsensical stuff. Economics doesn't classify as arts, though.

@L33ter But economics is just applied mathematics. It should be classified as science.

L33ter

Yeah, if you look at it from this way.

Semiclassical

economics is a weird mix of social science, applied math, and politics.

Balarka Sen

Also, definition of "scientific method" is arguable. E.g., I think biology atleast at a high school level doesn't deserve to be called science, because there's no logical thinking involved.

Hi @Semiclassical.

Semiclassical

06:44

hi

L33ter

I guess it depends, but for that course that I am taking it just bunch of non-sense stuff.

Mike Miller

@Balarka: That works. You have another.

@Semiclassical: What's a center manifold? Have you heard of that?

Balarka Sen

@MikeMiller I know. I have to leave for a couple hours (family gathering nonsense), but as soon as I get away I'll think and come up with a solution.

Semiclassical

haven't encountered it before, or at least it doesn't come to mind

Mike Miller

Ok, the book I'm reading cited a mathematical physics journal for the proof, so I figured it was worth a shot.

Semiclassical

06:48

ah. well, glancing at the Wiki page, it seems straightforward enough

i've definitely encountered that kind of thing in physics, just not under that name

especially in the context of RG flow (and I almost think i have seen that term there)

Mike Miller

I'll look at that... The definition I saw in the book was pretty unmotivated so I should looj at that

Semiclassical

what was the claim?

the entry at Scholarpedia for center manifold also looks good

Mike Miller

Center manifolds exist for a certain gradient flow (on an infinite-dimensional space)

I just want to understand why I care

Semiclassical

ahh.

wikipedia has the following statement in the intro paragraph: "Center manifolds play an important role in: bifurcation theory because interesting behavior takes place on the center manifold; and multiscale mathematics because the long time dynamics often are attracted to a relatively simple center manifold."

Mike Miller

I don't know what that means haha

Semiclassical

06:55

also, I mentioned RG flow? well, that's pretty much the go-to example of a flow in an infinite-dimensional space in physics

you start with a physical system with infinitely many parameters available, but with only a few of them being finite. in doing RG flow, you change which particular system you're interested in, and that amounts to some trajectory in that infinite-dimensional parameter space.

and depending on the dynamics involved in that, you learn a lot about the relevant physics

(i don't actually do RG flow stuff myself, so that's a half-remembered description)

Mike Miller

Yeah, I think I should try to ynderstand s finite dimensional example

I have a good intuition/understanding for my setting so I'll try to transfer the finite dimensional understanding

Semiclassical

yeah, that's sensible. i talk about that infinite-dimensional example, but in terms of how i intuit things i really only have the finite case in mind

Mike Miller

I think that's true for everyone

Semiclassical

yeah.

Mike Miller

I learned something cool earlier

Let me tell you when I'm free in a little bit

Semiclassical

07:06

i think what the center manifold concept practically says is: If I consider a flow in some parameter space, the number of 'relevant' coordinates can be a lot smaller than the dimensionality of the system

Mike Miller

Seems like that, yeah

Semiclassical

@MikeMiller you'll have to tell me tomorrow, it's 1am here and i should sleep

enthdegree

I have a simple kind of crank-ish question: what are the consequences of a negative answer to the global regularity conjecture for Naiver Stokes?

Mike Miller

@Semiclassical: Sure thing

Semiclassical

later

enthdegree

07:17

nevermind

Adel Saleh

09:53

hi there :), anyone online?

skull petrol

hi

enthdegree

hi

Evinda

10:59

Hi @robjohn
Do you maybe have an ideaa about my question: http://math.stackexchange.com/questions/1662005/can-we-just-set-it ?

Brooks

11:42

Hi

I have some problems in understanding singular complex, and I asked a question on the site, could anyone help me?

there is a answer on it, but I still have problem on it

user243301

13:45

I am trying learn some easy facts about the **Discrete Fourier Transform** (like as you can read in Wikipedia, concerning the definition, the inverse transform, also I see some easy example with online tools, Plancherel and Parseval, and I remember what is a phase and an amplitude for a cosine function). On the other hand I have a lot of doubts, and I've read in a book that refers an experiement (**Marek Wolf**,*1/f noise in the distribution of prime numbers*, I believe that the author put in our hand a copy in researchgate.net), and I believe that I have no abilities to understand and repr…

Mambo

15:02

@BalarkaSen The function $f:x + x^3$ has invertible derivative at every point and also $(f')^{-1}$ is uniformly bounded. Did you get an example where such a condition is not satisfied instead function's unboundedness on each unbounded set is satisfied?

Balarka Sen

15:35

Whoops, alright. You want $\Bbb R \to \Bbb R$?

Mambo

$\Bbb R^n$ for any $n$

Moshe

15:47

Anyone around?

I'm trying to work out a differentiation problem by parts.

Mike Miller

@Balarka: Just work one out, it shouldn't take too long.

Balarka Sen

So we want a proper $C^1$ function $f : \Bbb R \to \Bbb R$ such that $f'$ is invertible everywhere and $1/f'$ is not uniformly bounded. Hmm.

Evinda

Hey @robjohn
Do we use the fact that $\int_{\mathbb{R}^n} \delta(x-y) dy=1$ ?

Mike Miller

16:04

Make it smooth.

Mambo

@BalarkaSen Sorry I went for dinner. Yes exactly!

Mike Miller

16:36

@Balarka: I have something else for you if you're bored of that.

Balarka Sen

@MikeMiller Yeah I tried a few piecewise things, threw $e^x$ and $\arctan(x)$ at it but I can't find an example :S

I do think I can make a piecewise thing work though. But not sure if there's already something easy out there.

Most of my examples are proper restricted to the positive x-axis, but not to the negative one (and vice versa).

Mike Miller

You're probably thinking too hard. All we're asking for is a function whose derivative gets arbitrarily close to zero - $x^{1/3}$ does that. Modify it near zero to fix the obvious problem.

Balarka Sen

I don't see how to modify $x^{1/3}$ near zero to make $f'$ nonzero.

And yet make $f'$ go arbitrary close to $0$ at $0$.

Mike Miller

What? You can't possibly make it go arbitrarily close to 0 near 0. That's a compact set.

The only way you can make the derivarive get very small is to go to infinity.

All you're doing near 0 is to smooth it out.

Balarka Sen

Ah, I misunderstood you. $f'$ gets arbitrarily close to $0$ at $\infty$, true.

Mike Miller

16:44

You can either modify it by hand or use some amount of invocation of the Whitney extension theorem. In any case, that's something to do later.

Question. What direction does $\nabla f$ point?

Balarka Sen

I should have found this out by myself :( I thought about $x^{1/3}$, but discarded it immediately as $f'$ is not defined at $0$.

Mike Miller

?? $f'$ isn't zero, it's not defined

Balarka Sen

I am sorry, that's what I was trying to say.

(was thinking $x^3$)

@MikeMiller What is $f$ here?

Mike Miller

Smooth function on $\Bbb R^n$.

Balarka Sen

Arbitrary smooth function?

Mike Miller

16:48

Yes.

Balarka Sen

$\nabla f(a)$ points at direction at which $f$ has the greatest rate of change at $a$. Vector $v$ s.t. $\|D_vf(a)\|$ is maximum.

Owatch

Hey

I have a question, it will be easy for you all.

Ted Shifrin

hi @Owatch

Owatch

No. I can't do that..

Ted Shifrin

Well, we got far that time.

Mike Miller

16:50

@BalarkaSen: OK. So, here's a more-or-less obvious application of the gradient. Suppose I'm standing on a point in $\Bbb R^n$, and I want to head towards a (local) minimum of $f$. Where do I go?

Owatch

I wanted to say $1 < n$, if $n \in \mathbb{N}^{+}$

Ted Shifrin

Nice trap question, @MikeM.

Owatch

But that is not true..

Balarka Sen

Move forward along $-\nabla f(a)$, right?

Mike Miller

@Ted: I suppose I've trapped myself, because I don't see what the trap is.

Ted Shifrin

16:51

I think I already gave @Balarka the task to think about this phenomenon, and he solved it ... but that was months ago.

Mike Miller

@Balarka: Infinitesimally, yup. Obviously I should not head in that direction forever.

Balarka Sen

@MikeMiller Nope.

Ted Shifrin

Oh, no, I gave him a slightly different variant.

Mike Miller

So, let's see, suppose I have a curve $\gamma(t)$, that "infinitesimally" is going in the right direction. How do I want to state this? Well, I want $\gamma'(t) = -\nabla(f)(\gamma(t))$.

Ted Shifrin

@MikeM, @Balarka: Can you guarantee me that you'll end up at a local minimum?

Mike Miller

16:52

@Ted: Shhh!!

Or, I guess don't sh, maybe I should think about how I want to phrase this.

Ted Shifrin

Balarka actually did give me an example of a very closely related phenomenon.

Balarka Sen

I think I gave you an example where you take arbitrarily long time to end up at a local minimum/maximum.

Ted Shifrin

Right, @Balarka, that you did.

Balarka Sen

@MikeMiller Right, true.

Mike Miller

I mean, when you're doing this, you won't literally ever reach the critical point.

You should be able to prove that.

Ted Shifrin

16:55

"The" critical point? There's by assumption a unique such?

Mike Miller

A critical point.

Owatch

Alright. I think I'm in the clear here. Sorry I didn't really articulate my question. Hello back Ted.

Mike Miller

(I would call such a $\gamma$ a "gradient flow line"; normal people would probably call it "a path of steepest descent")

Ted Shifrin

LOL, glad to be of service, @Owatch. That's the best way :)

Mike Miller

We call the zero set of $\nabla f$ the "Critical set" of $f$.

Balarka Sen

16:55

@MikeMiller I am not sure what is the question you are asking.

Mike Miller

@Balarka: It's unimportant, so whatever.

1) Suppose the critical set of $f$ is compact. Show that a gradient flow line accumulates somewhere on the critical set.

2) If it's not compact, find an example where it doesn't. Feel free to work on the upper half-plane or something if that's more convenient.

Ted Shifrin

heya DogAteMy :)

Akiva Weinberger

Hi

Owatch

Okay, one last thing then.

Mike Miller

Now, note in 1) that I never said you actually have a limit. You might "Spiral in" towards the critical set. (You should at least be able to draw an example where this happens, even if you can't write down the explicit function.)

Akiva Weinberger

16:57

¿Qué pasa?

Owatch

If $ 0 \leq i < i+1 \leq n$ and $n \in \mathbb{N}^{+}$. Can I say: $0 \leq i < n$ ?

Ted Shifrin

Mike is playing our favorite game: Stump the Balarka.

Sure, @Owatch.

Mike Miller

3) Suppose $f$ is real analytic. Use the Łojasiewicz inequality to prove that $\lim_{t \to \infty} \gamma(t)$ exists and is a point on the critical set. (That is to say, you don't spiral in: you have an actual unique limit.)

Owatch

Yes!

Balarka Sen

That's (2), not (3), by the way. Also, cool.

Ted Shifrin

16:58

No, there's a 2) up there.

Mike Miller

I think 2 is the least interesting question of the bunch, since in every case I care about, the critical set is compact.

Balarka Sen

Oops, yup.

Mike Miller

In many applications of gradient flow lines you'll see people assume that the critical set is finite. The Lojasiewicz inequality says you don't have to make an assumption like that.

Ted Shifrin

I've actually never encountered that before, @MikeM.

Mike Miller

Encountered the problem, or the assumption on the critical set?

Ted Shifrin

17:00

What came up in my research stuff was Whitney conditions on stratifications. Not quite related, but ...

the inequality, I meant, Mike.

Akiva Weinberger

@Owatch You need $i$ to be an integer, also, if you want them to be equivalent. Consider $i=1.5$ and $n=2$. The first set of inequalities is false but the second is true.

Ted Shifrin

Of course, yes, Owatch is dealing only with integers here.

Owatch

i is an integer.

Ted Shifrin

We had $i\in\Bbb N^+$ before.

Owatch

$i \in \mathbb{Z}$, and $n \in \mathbb{N}^{+}$

No?

n was.

Ted Shifrin

17:01

Oh, oops. The DogAteMy's complaint is valid :P

Akiva Weinberger

OK

Mike Miller

I learned it yesterday in an infinite-dimensional context. The authors didn't want to assume (or perturb so that) the critical set of the Chern-Simons functional was nondegenerate, eg finite. So they used an infinite-dimensional generalization of Lojasiewicz, due to Leon Simon, to prove that finite energy instantons have a unique limit on the critical set, no matter how nasty it is.

Owatch

Should be ok then.

Ted Shifrin

Leon Simon's done some good stuff.

Mike Miller

Yes, and it got me all excited by this inequality.

So I'm talking about it as much as I can so that I don't forget it.

Ted Shifrin

17:03

Yeah, that'll work. :D

Balarka Sen

When you say zero set of $\nabla f$, you mean the zero locus of the function $\nabla f : \Bbb R^n \to \Bbb R^n$ given by $a \mapsto \nabla f(a)$, correct? ($f$ here is a function $\Bbb R^n \to \Bbb R$).

Ted Shifrin

To interrupt, sure.

Balarka Sen

ok. So it's just the collection of critical points of $f$.

Ted Shifrin

Yuppers.

Mike Miller

BTW, for this problem you should assume the domain of $\gamma$ is $[0,\infty)$. I don't care about the backwards time direction.

Balarka Sen

17:08

Right.

Mike Miller

@Ted: Feel free to interrupt, I'm busy cleaning up from the rowdy poker boys.

Ted Shifrin

Make sure you sweep up all the cigarette butts.

I'm off for a walk to the bank. BBIAB.

Mike Miller

They're not that rowdy.

Balarka Sen

Sorry, I was away. I am onto problem (1).

Akiva Weinberger

17:24

@MikeMiller Yet

Balarka Sen

17:40

@MikeMiller Should it be immediate to me why the gradient flow line accumulates arround the critical set? Or do I have to do nontrivial amount of thinking to figure out?

Mike Miller

@Balarka: Everything here requires a nontrivial amount of thinking, because what's the point of trivial problems? But I think I need more assumptions.

And by "I think" I mean "I know".

Balarka Sen

So compactness of the critical set is not sufficient?

Mike Miller

Maybe assume in addition that the curve has finite energy, $\int \|\gamma'(t)\|^2 < \infty$.

Balarka Sen

Ah, OK.

Mike Miller

@Balarka: No, let $f$ be like $1/(1+x^2)$. Obviously you just flow off to infinity.

Yes, ok, this is the desired assumption.

Sir Cumference

17:44

Sexy prime

In mathematics, sexy primes are prime numbers that differ from each other by six. For example, the numbers 5 and 11 are both sexy primes, because they differ by 6. If p + 2 or p + 4 (where p is the lower prime) is also prime, then the sexy prime is part of a prime triplet. The term "sexy prime" stems from the Latin word for six: sex. == n# notation == As used in this article, n# stands for the product 2 · 3 · 5 · 7 · … of all the primes ≤ n. == Types of groupings == === Sexy prime pairs === The sexy primes (sequences A023201 and A046117 in OEIS) below 500 are: (5,11), (7,13), (11,1...

Huh, didn't think this would be a thing

Balarka Sen

(I was just asking whether I should be able to solve it after sufficient amount of starting, like e.g., your irrotational vs. conservative vector field problems or whether it was really very nontrivial like e.g., the exotic sphere problems. If later, I could ponder on it rather than stare)

@MikeMiller Hmm.

Oh yeah, true.

Mike Miller

You mean "if the latter". You should be able to solve all these, but I don't know what the difference between the two cases above is. But no big deal.

Note, BTW, that finite length implies finite energy - but I am not willing to assume finite length.

Balarka Sen

I agree.

Mike Miller

4) Figure out if finite energy is equivalent to accumulating on the critical set. (I think it is but don't have a proof off the top of my head.)

Ted Shifrin

<-- observes Mike keeps distracting @Balarka from all the problems he's already supposed to do :D

Mike Miller

17:53

5) Show a finite length gradient flow line indeed does automatically have a unique limit.

Ted Shifrin

Can't get your 225B students to do these, Mike? :)

Mike Miller

@Ted: He distracts me from all the problems I'm supposed to do, so it's fair.

@Ted: Haven't tried. But I doubt any of them would do any problem I assigned.

Ted Shifrin

Last time I taught Riemannian geometry, less than half the class made serious efforts on the homework.

But the ones that did learned a lot, at least ...

Owatch

I hate proofs.

Ted Shifrin

If you're an engineer, that's ok, @Owatch. If you're a math major, then it isn't.

Owatch

17:56

Every block I think I've gotten away from courses requiring proofs. Not a single time yet.

I'm not a math major.

Ted Shifrin

Of course, a certain amount of proofs really does help you problem-solve better.

Owatch

Usually I don't know what I'm doing when I start.

So thats no good.

I actually know what I need to do now in the one I'm working on. But it's been long enough that I've been working at it to get here.

All damn afternoon..

Ted Shifrin

As with methods of integration and lots of applied things, practice is key in learning mathematics ... proofs, too.

Owatch

I don't have to do those kind of proofs actually. These proofs are for program correctness and stuff. It's like another course I took which just required I prove things using logical symbols like conjunctions, disjunctions, negations, and letters. Also existential and for-all quantifiers.

Ted Shifrin

Well, yes, theoretical computer science is really mathematics.

Owatch

18:02

Yeah yeah. Belongs on this spectrum: xkcd.com/435

Probably before the physics guy.

Akiva Weinberger

@Owatch The QR code on your profile is wrong. It links to unisung.com, but that doesn't exist anymore. I'm guessing you want it to link to unisung.blogspot.com.

Owatch

Yeah I didn't renew the domain.

I'll have to take that down.

You can ignore anything linked via my network profile. It'll be updated to just a hello or something soon. Change will take some time to propagate. If you're trying to get the App I have linked on the blog, that's down too. My iOS license expired and they took my App down.

Akiva Weinberger

:(

Honestly, I was just curious what the QR code was.

Owatch

Yeah, I thought it was a cool way to link things.

Akiva Weinberger

I don't actually know how to scan a QR code on my screen. I just opened your profile on my computer and scanned it with my phone.

Owatch

18:18

Yeah, that's how you do it. I don't know about websites that do it for you, but there may be some you can just 'drag-and-drop' QR codes into.

Brennan.Tobias

18:41

wow

this thread has become incredibly low quality math.stackexchange.com/questions/1665899/…

user166748

Guys please give your views to this question.

3

Q: Differenciating a function with respect to another function confusion

I am having problem solving the following question- Differenciate $\tan^{(-1)}{(\sqrt{1-x^2}/x)}$ with respect to $\cos^{(-1)}{(2x\sqrt {1-x^2})}$, where $x$ is not $0$. My attempt - I took the tan function as $a$ and cos one as $b$. Now we need ${(da/dx)/(db/dx)} $ So here if I substitute $...

I do not understand if my book is correct or wrong

GGG

By the Chinese remainder theorem the solutions to $a^2 = 1 \mod{15}$ are given by $a^2 = 1 \mod{5}$ and $a^2 = 1 \mod{3}$. So we need to solve $a = \pm 1 \mod{5}$ and $a = \pm 1 \mod{3}$. Let $p$ and $q$ be comprime. By the Chinese remainder theorem the solution to the system of equations $x = a \mod{p}$ and $x = b \mod {q}$ is $x = aqq_1+bpp_1 \mod{pq}$ where $p_1 = p^{-1}\mod{q}$ and $q_1 = q^{-1} \mod{p}$ (where $p^{-1}\mod{q}$ means the integer $r$ such that $pr = 1 \mod{q}$).

Using this we have $a = (\pm 3 \cdot 5^{-1}\mod{3}\pm 5 \cdot 3^{-1} \mod{5})\mod{15}$ and since $5^{-1}\mod{3} = 2$ and $3^{-1} \mod{5} = 2$ this gives $\pm 6 \pm 10 \mod{15}$ which gives $a = 1,4,11,14$.

How do I write the above proof in a clear manner? I'm hopeless at writing proofs! :[

Brennan.Tobias

@ggg what are you working on?

could this be of any use to you math.stackexchange.com/questions/1661650/…

sorry ignore that

i see you're doing a different problem

GGG

18:59

@Brennan.Tobias I'm trying to find all the integer solutions to $a^2 = 1 \mod{15}$ which I believe I've done, but is the way I've written it okay?

Brennan.Tobias

one easy way would be to try a=1,2,3,4,...,15

@GGG is that any use to you

GGG

@TobiasKildetoft Yes, actually -- all the solutions can be found that way. But I was was using the CRT so that I know how to solve problems like this when the modulo is bigger.

Julian Rachman

Anyone? math.stackexchange.com/questions/1664962/…

Brennan.Tobias

well another way to do it is this

you want to solve a^2-1 = 0 mod 15

that's the same as 15 | (a-1)(a+1)

so clearly a=1 and a=-1 work

but you might also have 3|a-1 & 5|a+1 (or the reverse)

sharaf zaman

limit of product= product of limits can it be applied to a function which is in multiplication with 1/x^2

like

Brennan.Tobias

19:10

that will find you the other two 'extra' solutions

sharaf zaman

limit (x tends to 0) x/sin(x)*1/x^2 = limit( x tends to 0) x/sin(x)*lim(x tends to 0) 1/x^2

is it correct?

Brennan.Tobias

@sharafzaman, you can do that at points a function is continous at

s.harp

quick question, is the group of deck transformations on a covering space always isomorphic to a subgroup of the fundamental group?

Brennan.Tobias

if f and g are cont. at y then lim x->y (f(x)g(x)) = (lim x->y f(x))(lim x->y g(x))

GGG

Thanks, @TobiasKildetoft. That's a nice way to do it.

sharaf zaman

19:14

can it be a function whose limit exists but not continuous

Brennan.Tobias

no

sharaf zaman

okay! but limit x->0 {1/x^2} exists because LHL=RHL

Brennan.Tobias

no it doesn't thats division by zero

sharaf zaman

then even lim x->0 sin(x)/x should not exist

Brennan.Tobias

isn't that 1?

because near 0, sin(x) is like x

so its just x/x = 1

Julian Rachman

19:19

If someone could check this out and verify it, that would be awesome: math.stackexchange.com/questions/1664962/…

sharaf zaman

but sin(0)=0 so 0/0 if taken in other way

Brennan.Tobias

im not being rigorous just showing intuition

sharaf zaman

yeah! i understood sorry!

3

Q: $\lim_{x\to 0} (2^{\tan x} - 2^{\sin x})/(x^2 \sin x)$ without l'Hopital's rule; how is my procedure wrong?

please explain why my procedure is wrong i am not able to find out?? I know the property limit of product is product of limits (provided limit exists and i think in this case limit exists for both the functions). The actual answer for the given question is $\frac{1}{2}\log(2)$. My course book h...

calculus limits limits-without-lhopital

check out my that question why it is wron

*wrong

Julian Rachman

0

Q: Order-Preserving Adjunction?

Let there be two categories PrO (category of preordered sets) and ProM (category of preordered monoids). If we know that there exists an adjunction $F\dashv G$ for functors $F:\textbf{PrO}\to\textbf{ProM}$ and $G:\textbf{ProM}\to\textbf{PrO}$, how can we show that this adjunction is order-preserv...

abstract-algebra category-theory order-theory

(There we go)

Brennan.Tobias

I did look at that @JulianRachman but I couldn't say anything about it

Mike Miller

19:26

@s.harp: No; the double cover of a circle has deck transformation group $\Bbb Z/2$.

sharaf zaman

Anyone please answer my question it will be greateness

Julian Rachman

@Brennan.Tobias that's alright. At least you took a look. :)

sharaf zaman

no reply? i know i am loud

Akiva Weinberger

@sharafzaman It seems there are 7 answers there

sharaf zaman

@AkivaWeinberger What?

Eric Stucky

19:32

Your question has seven answers

sharaf zaman

hey! how?

s.harp

@MikeMiller I think my memory may be a bit messed up, is the statement that covers correspond to quotients of the the fundamental group true?

Eric Stucky

and @k170's answer starts with a good two-sentence explanation of what went wrong.

sharaf zaman

@EricStucky how?

Eric Stucky

Have you read the first two sentences of his answer?

Akiva Weinberger

19:34

What do you mean "how"? Seven people answered your question.

Eric Stucky

Also, user254665 points out another issue

Akiva Weinberger

You're talking about this question, right?

3

Q: $\lim_{x\to 0} (2^{\tan x} - 2^{\sin x})/(x^2 \sin x)$ without l'Hopital's rule; how is my procedure wrong?

please explain why my procedure is wrong i am not able to find out?? I know the property limit of product is product of limits (provided limit exists and i think in this case limit exists for both the functions). The actual answer for the given question is $\frac{1}{2}\log(2)$. My course book h...

calculus limits limits-without-lhopital

sharaf zaman

@AkivaWeinberger i think it has no answers because my question is why i am wrong (it is not plzz solve)

yes

Akiva Weinberger

What do you mean it has no answers? Click on the link, scroll down…

Eric Stucky

akiva

Akiva Weinberger

19:37

Yes?

Eric Stucky

notice that most of the answers are

just saying hoiw to do it correctly

which does not answer the question he asked

it's a fair complaint

sharaf zaman

@EricStucky THANKS

Eric Stucky

But as I mentioned, sharaf, there are two that do.

Akiva Weinberger

True. Not all of them, though

Basically, you replaced $\lim f(x)g(x)$ by $(\lim f(x))\cdot(\lim g(x))$, which is not always possible

sharaf zaman

really

Akiva Weinberger

19:39

Consider this "false proof":

sharaf zaman

k

Evinda

Is anyone familiar with coding theory?

Akiva Weinberger

\begin{align}\lim_{x\to\infty}1&=\lim_{x\to\infty}x\cdot\frac1x\\&= \lim_{x\to\infty}x\cdot\lim_{ x\to\infty}\frac1x\\&=\lim_{ x\to\infty}x\cdot0\\&=\lim_{ x\to\infty}0\\&=0\end{align}

@sharafzaman

sharaf zaman

yes!

Akiva Weinberger

Do you see what the mistake is there?

Evinda

19:42

Let $C$ be a linear code with minimum distance $2k$. I want to show that there is a coset of $C$ that contains at least two vectors of weight $k$.

Solving the system $Hx=0$, where $H$ is the parity matrix, we can find the coset of $C$, right?

But in this case we don't have a generator matrix of the dual code. How can we get information about the parity matrix?

Or how else can we get information about the coset of $C$?

Mike Miller

@s.harp: Covers correspond to shbgroups but the correspondence isn't given by the deck transformation group. The deck transformation group is a quotient (IIRC, it's the quotient of $\pi_1$ by the normal subgroup generated by the subgroup your cover corresponds to)

This is written down somewhere in hatcher ch1

sharaf zaman

@AkivaWeinberger ya i saw

Akiva Weinberger

@sharafzaman It's basically the same flaw as in your question

s.harp

Thanks, I had tried to use google to find the statements but I could not, I'll look into hatcher.

Evinda

@AkivaWeinberger Hey!!! Do you maybe have an idea?

sharaf zaman

19:44

but this violates product of limit rule

@AkivaWeinberger product of limit = limit of products

Akiva Weinberger

That rule only applies if both limits exist, @sharafzaman

$\lim_{x\to\infty}x$ doesn't exist.

sharaf zaman

@AkivaWeinberger but LHL for that is equal to RHL

Akiva Weinberger

?

John Jack

Hi @DanielFischer

Akiva Weinberger

The full theorem is, "If $\lim f$ and $\lim g$ both exist (and are finite), then $\lim fg=(\lim f)(\lim g)$"

sharaf zaman

19:47

@AkivaWeinberger i mean for limit to exist there is only 1 restriction left hand limit should be equal to right hand limit and it is true here the why? it doesn't exist

Akiva Weinberger

In my false proof, I wrote $\lim x\frac1x=\lim x\cdot\lim\frac1x$, but $\lim x$ doesn't exist

It's not finite

Oh, I see what you mean.

With $\lim\limits_{x\to\infty}$, there is no right hand limit anyway, so that criterion doesn't make sense.

But if it were $\lim\limits_{x\to c}$…

…then the limit would exist iff the left-hand-limit existed, the right-hand-limit existed, and the left-hand-limit and right-hand-limit are equal.

In your question…

we had $\lim_{x\to0}\frac1{x^2}$ at one point.

But neither the left-hand- nor the right-hand-limit exists.

They're both infinite.

So $\lim_{x\to0}\frac1{x^2}$ doesn't exist.

sharaf zaman

@AkivaWeinberger oh do you mean if a limit has no other solution then tending to infinity then the limit will not exist

am i correct

Akiva Weinberger

Yeah

sharaf zaman

@AkivaWeinberger AWESOME!! you are expert

what's your age?

Akiva Weinberger

Somewhere between 0 and 100

sharaf zaman

19:57

hahaaaaaa!!

but i am a beginner with calculus

k bye! everyone

John Jack

20:54

How would you define a 'planar curve'?

Eric Stucky

Any particular context?

Akiva Weinberger

@JohnJack A continuous function $f:[0,1]\to\Bbb R^2$, I guess

Eric Stucky

that better be continuous, Akiva :P

Akiva Weinberger

Fixed

Eric Stucky

jeje

John Jack

21:00

There is a question which asks if $z = 9 - x^2-2y^2$ is a planar curve.

Akiva Weinberger

Oh

Then it would be something that lies on a plane and is one-dimensional

John Jack

@AkivaWeinberger Not necessarily a coordinate plane, just any plane?

Akiva Weinberger

@JohnJack I think so. But you probably learned how to figure out what shape that is.

There's a special name for that shape.

It's the shape as $x^2+2y^2=-z-9$

Paradox 101

Are there any theorems that define the relationship between the orders of distinct subgroups?

Eric Stucky

21:20

You hoping for something like a necessary condition on a multiset, for it to be the orders for subgroups for some $G$?

(I don't know anything like this, btw, just wondering)

Paradox 101

@EricStucky not exactly I need to prove that a group is cyclic if it has two subgroups with distinct orders. I've used lagrange's theorem but not sure where to go from there

John Jack

Okay cool thanks

Mike Miller

@Paradox101 Uh, that's definitely not true. You should check the statement again.

Of the problem, I mean.

Paradox 101

@MikeMiller the question statement is that a group G is finite with order n. It has two distinct subgroups with distinct orders. show that G is cyclic

Andrew Thompson

Where did you find this problem?

Mike Miller

21:30

Precisely two distinct subgroups? Meaning that it has two subgroups that are neither $G$ nor the trivial subgroup, and they both have different orders?

Because I can believe that. But the phrasing "It has two distinct subgroups" makes me think first that there is some pair of subgroups with blah

Andrew Thompson

No, then its still false, Mike.

Mike Miller

ex?

Andrew Thompson

Sorry

Might be true

Thought of something definitely cyclic.

Paradox 101

yes precisely two. So if then they have distinct orders does that imply that their orders are relatively prime?

Mike Miller

@Paradox101: No. Take $\Bbb Z/8$. Precisely two nontrivial proper subgroups.

Different orders.

Paradox 101

21:35

Ok. This is tricky. In this case is proving through contradiction going to be simpler?

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$Mathematics$ Mathematics