Hi everyone, what is the short cut for right arrow on mathjax.

\to

only that

what if \to\rightarrow\

@MikeMiller can u reply on my previous text on the chat.

I have no idea what you're asking. \to outputs $\to$.

That is a right arrow.

like right arrow to show mapping from A to B. what is the short hand to show right arrow on mathjax

Neman, it's \to

As Mike has been saying.

thanks, now have a look and give me explanations:math.stackexchange.com/questions/1798842/…

I agree with your second comment: if we specify $(-1,1]$ as the codomain, the function is not well-defined.

What is $g(\pi)$, for instance?

3.14

No, that is $\pi$. I mean, what do you get out when you put $\pi$ into $g$?

0.99

No.

What is the difference between the codomain and the range?

Basically, the codomain is bigger.

The range is the actual outputs of a function, taken over all possible inputs.

A function is a map $f: A \to B$, where $A$ and $B$ are sets (you might like to think of them as sets of numbers, at this stage in your career). In calculus, you mostly think of functions $f: \Bbb R \to \Bbb R$: it takes real numbers as input, and spits out real numbers as output.

The codomain is simply a set in which the range lives.

It's a useful formalism when the range is not very natural or obvious; you can still define a function without needing to understand its behavior beforehand.

Now, for $f(x)=x^2$, you see that all of its outputs live in $[0,\infty)$. That is the range of the function. But baked into the function itself is $B$, the 'codomain' of $f$: where it maps to. When I tell you "Consider the function $f: \Bbb R \to \Bbb R$, given as $f(x)=x^2$", the codomain is $\Bbb R$, even though the range is smaller.

It may seem silly to do this thing where we build the codomain into the function, but you'll have to take my word that it's both useful and important for now.

but couldn't you say that the codomain is all complex numbers?

Yep

so what exactly is the point?

But the point is

You have to specify a codomain beforehand

I think I get it

What I don't understand is when you said that the function is f(x) = $x^2$, the range is [0,$/infty).

What about f(i)

or f(a+bi)

i is not in the domain.

oh

I don't like restricting the domain to real numbers

Hopefully soon I'll be in more complex classes where I won't have to :)

Restricting to $\Bbb R$ is not always the most useful thing to do, but there are cases in which it's nice to not have to worry with all of $\Bbb C$.

For instance, the factorial function, we generally restrict the range to $\Bbb N$.

Because, after all, what is (-2)! ?

infinity

haha

:P

I've explored the gamma function quite a bit

but I see what you mean.

Surely you mean $-\infty$ then :P

it depends on which side you approach it from

I think

It's been a while since I've looked at the graph

Right, so this is why I chose a negative integer :P

Even if you go to Gamma, you still can't define it at -2

true

But I have a (I can't think of the mathematical term)

guess

would conjecture be the right word?

Depends

What comes next in the sentence :P

Anyway, I guess that infinity is actually the same as negative infinity

Congratulations, you're a geometer.

If you define them to be the same, then 1/x is continuous and everyone is happy

looking up what geometer means lol

?

What does that have to do with geometry?

So infinity and negative infinity being the same turns out to be great for geometry.

Less great for other things, like number theory for instance :P

oh.

interesting

Actually, I may retract my statement about number theory. It's certainly bad for order theory, and the intersection of number theory and order theory is the only part of number theory I've ever had a taste for ^.^

Off topic, but you and Mike Miller seem to be kind of in charge here

lol no

we're just the ones who have been talking while you're here :P

oh lol

If you hang around long enough, you'll start to understand the regulars.

There are moderators, though nobody really runs the room unless someone gets rowdy. I'm just a loudmouth.

I said that because Mike Miller was on for a really long time

Now you're gonna make me feel bad for being unproductive.

Yeah Mike does anyone harrass you about working? :P

lol

I'm supposed to be doing a physics project right now

so you're not alone :P

I'm supposed to be writing a blog post about a conference I went to

Let he who is without sin cast the first stone.

I've been trying to work out the supposedly-simplest computation in this paper. It's been about three hours since I started.

Simple indeed.

eech

I'm running out of time for my physics, so I gotta go

see you later.

jeje you too

@EricStucky It kinda sounds like Jesus just wants to be the first to throw the stones there

XD

is anyone here expert in linear algebra

?

"expert"

if you have a question; ask it :)

expert = "I can pretend to have some authority in that field."

:)

Is there any measure for how close a matrix is in being positive semidefinite?

I mean, you can always count how many eigenvalues are negative.

Maybe even add up the negative eigenvalues if you want something a touch more sophisticated.

Any particular reason you want this?

semi definite, not definite

pos semidef <--> all eigs nonnegative.

Now you have me doubting myself

you mean all eig values greter than 0 means positive definite?

that I know

So, I'm a complete noob to the Archemedian property and I'm tutoring someone in Analysis

Hmm, so wiki only states this for hermetian matrices.

I need to figure out how for each $t > 0$, there is an $n \in \mathbb{N}$ such that $$\dfrac{1}{n^2} < t$$

(My statement, Raj: yours is true.)

To me, it's not very obvious

What's your version of Arch, Clarinestist?

@EricStucky For $x, y \in \mathbb{R}$, $x > 0$, there is an $n \in \mathbb{N}$ such that $nx > y$.

So I'm thinking

Okay, for $t > 0$, we have $nt > 1$

That's my gut instinct

but I have no idea where to go from there

You can work with that; maybe call it $N$ since it's definitely not the $n$ you want.

@Eric I happen to lok at your home page (profile) now and I see jump functions. What is your interest in these

:P someone was asking me about them at length

@EricStucky Okay, where would you suggest going from here?

ah ok

Well, if $N$ is a square number, you agree we're done?

@EricStucky Yeah, that's pretty obvious.

Now if $N$ isn't square

If we make $N$ smaller, does the inequality still hold?

@EricStucky : thanks for the lead, you seem right

@EricStucky Lost me there

So, Raj, here is a stupid thing you could do:

?

Hmm, actually, this only works for real matrices.

@Clarinetist Suppose $N' > N$. Is $1/(N')^2 < 1/t$?

The function $f:\Bbb R^n\to \Bbb R$ given by $x^TMx$ is somewhere negative, somewhere nonnegative, on the unit sphere.

Take $\min(f,0)$ and integrate this over $S^n$.

whoops, typo that :P

Haha Mike that is a lot more slick than where I was going :P

@EricStucky @MikeMiller Let's see. We're given $t > 1/N$. If $N^{\prime} > N$, then obviously $t > 1/N > 1/N^{\prime}$ but I don't see how we can incorporate the squared term

@EricStucky @MikeMiller Oh duh

Sorry

Let $N$ be your favorite number. Can you pick a square bigger than $N$?

since $(N^{\prime})^2 > N^{\prime} > N$

we get $t > 1/N > 1/N^{\prime} > 1/(N^{\prime})^2$

How's that look?

Looks good, although

No reason not to take $N'=N+1$, just to be a tad more explicit.

And actually, if you jigger the equalities correctly, you can take $N'=N$.

or just $N' = N^2$ and be done with it :p

@MikeMiller Yeah, that's what I was thinking

So a write up

By Archimedean, we have for $n \in \mathbb{N}$, $nt > 1$. If $n$ is a square number, the claim follows. If $n$ isn't square, let $N = n^2$. (How do we know this exists, by the way? Is there some well-ordering involved?)

No, it's just the fact that $\Bbb R$ is a field.

@EricStucky Oh duh

Thank you

Even less than that, really; it's the fact that multiplication is well-defined $\Bbb R^2\to\Bbb R$.

@EricStucky So basically, if $n$ isn't square, let $N = n^2$. Then it follows that $N^2 > N > n$, which, for $t > 0$, implies that $N^2 t > nt > 1$. Thus the claim follows. How's this?

@Rajesh: You see I basically only have one idea; I just keep smacking it until it works :P

But the complex case presents some genuine difficulties, it seems.

You have too many squares, Clarinetist.

You only need $N>n$.

@EricStucky Lol I just realized that

Thank you

And, just to be even more slick

You don't have to consider the n square case separately.

Presumably this analysis class is your student's intro to proofs class: please talk to them about this process as well. There is the act of discovery, and then there is the act of writing, and then there is the act of cleanup. Ideally you'd do it all, but in a pinch you can get partial credit :)

@EricStucky Yeah, it is. I've been trying to emphasize that with them. Unfortunately, I imagine taking an analysis class during a summer session, quite frankly, isn't enough time to really drill that into someone

Oh yeah, summer session :/

You're probably right

Taking an analysis course during the summer seems like a preposterous idea IMO

^ that

Why in the world would it be offered during the summer? IDK

my matrices are real. Is ratio os sum of positive eigen values to sum of negative eigen values would sound good? @EricStucky

Hmm I'm not sure

What's the point of PSD, anyway? I mean, my idea assumes the positive values don't matter at all

But maybe the relative values are important.

Hmm actually no

Are your matrices also real-diagonalizable?

I mean, you might still have complex eigenvalues even if they're diagonalizable.

This was a problem with my old idea, too :/

Wait, what is this about positive semi-definite matrices?

Raj wants to take a real matrix, and ask how "close" it is to being PSD

Associating some number to that measurement.

yep

Hmm... how interesting

Yeah Raj, at the end of the day the primary reason I care about PSD matrices is because their bilinear forms are inner products.

Quick Google search: sciencedirect.com/science/article/pii/0024379588902236

I have matrices (real) coming from a physical measurement, in hypothesis they are PSD, but in reality all things happen, so i want to check for PSDness to avoid bad data

Oooh then Clarintest's idea is best.

Well, their stolen idea :P

can someone explain this codomain g: R to (-1,]. well the codomain(-1,], does this means -1 is not included in the codomain?

Yes, Surdz.

Raj: Unless you have some sort of absolute guarantee about the possible range of values for the matrix, in which case you'll want to normalize that measurement.

@Clarinetist : thanks a lot. Do you happen to find on google or you already have interest in this topic. I am asking as I would like more help in case it is

@RajeshDachiraju Nah, it was Google

ok

what did you look for?

@RajeshDachiraju From a stats perspective, I can understand why this would be important, though. (I'm working on a M.S. statistics.) Covariance matrices should always be PSD, for example

so if I find a real number equal to -1 in the range this means the function is not accurate?

ok

Yes, Surdz.

@RajeshDachiraju google.com/…

'Night everyone! Thanks for the help @EricStucky and @MikeMiller

npnp g'night :)

eech I should be doing that too :/

But this post still isn't done XD

Thnaks @Clarinetist : your reference was a game changing event for my future experiments. Hope I find something useful.

Done! g'night chat :)

A quick question, how can I use the saddle-point method to integrate $\int_{-\infty }^{\infty } \exp(-x^2) \exp \left(i \left(\frac{2000 x^3}{3}+50 x\right)\right) \, dx$ ?

Hi @TedShifrin. How's Chicago?

If the area covered by a vector is $ A\left(\vec{u}\right) = u_x \hat{\mathrm{e}}_x \wedge u_y \hat{\mathrm{e}}_y \wedge \ldots $ then does it make sense to notate the surface normal as $ A\left(\frac{\partial}{\partial \vec{v}}\right) \vec{u} = \frac{\partial \vec{u}}{\partial v_x} \wedge \frac{\partial \vec{u}}{\partial v_y} \wedge \ldots $?

Hi @all

Hi @EricStucky

I found a wide collection of algorithms for finding a nearest correlation matrix to a given symmetric matrix

but I need an additional constraint that the nesrest corr matrix should not be nearly singular!

Can a positive definite matrix be singular?

I got it, i just need to see that all eigen values are positive for non singularity

There is a wealth of information and useful algorithms here

nickhigham.wordpress.com/2013/02/13/…

math.stackexchange.com/questions/1798418/… help is appreciated :)

Does anyone want to add some more proofs to this question, just for fun? :P

Wow... I posted a wrong answer, but did not notice why it was wrong for less than a minute. It got 3 downvotes in the 56 seconds it was up. No comments.

@AlexClark yeah. I've never seen voting go so fast, but that so many could see the mistake so fast.

I've never gotten upvotes so fast!!

@AlexClark Got the "proper" accept of my paper now.

@AlexClark Communications in Algebra

not the most prestigious, but the paper is also not that amazing

@AlexClark Not sure how good a metric it is, especially for more specialized journals

for math journals as far as I recall anything above 1 is decent while anything above 2 is near the top

but again, there is a big difference between general math journals and ones that specialize in a subfield

yeah, I think it is a B on the old Australian ranking

@AlexClark I am not subscribed to any journals, but I can access most through the university here

@AlexClark I have an rss feed for group theory and representation from arXiv, which gets me most of what I need

of course, if I lack access from the university, there is always sci-hub

Are you familiar with the Australian ranking of journals, or was that already phased out when you started?

@AlexClark I only read a few of them, but I keep an eye on what gets uploaded in those two subjects (I used to have a feed for many more, but those were never interesting to me)

It is a list of math journals with a "grade" for each, A*, A, B, C, but I think they stopped using it a few years ago (and it was pretty much outdated as soon as it was made, at least for some journals)

austms.org.au/Rankings/AustMS_final_ranked.html

But it is still a decent place to start to get an idea of how good a journal is, at least as long as the journal is old enough (which most are)

It is a B (the list is ordered alphabetically, so it is only the grade that matters)

which is fine but nothing special

(which also describes the paper)

right

Well, there is also ungraded (you must never go there)

not necessarily

For one thing it might just be newer than the list. But for old enough journals it means you should not publish there

Thanks! =]

@Inarito Those two pages give two different things, no? The first is the average value of a function, the second is the average rate of change.

The average rate of change is the average value of the derivative, I think.

Nice, I managed to find an example of how a certain thing fails to hold for "small" primes, but which is known to hold for large enough primes (relative to the group one is working with). So now I have plenty of work to do, trying to figure out precisely how to correct this failure (which is for an analogue of BGG reciprocity for algebraic groups)

@AlexClark No, it is an analogue of BGG reciprocity

(BGG reciprocity is indeed in Category O)

this setup is for an algebraic group in positive characteristic

Does anyone else have problems with this link?

@robjohn Doesn't load for me

yes

now it did :D

@TobiasKildetoft okay, so it is not just me.

Quick question, I got -2 rep from user being removed, how is such a number possible? None of questions I had accepted answers from had their answers removed (accepting is 2 rep)

Suppose $A = \{ x \text{ mod } 2 : x \in \mathbb{Z} \}$ then is $2A = \{ (2x) \text{ mod }2: x \in \mathbb{Z}\}$ or $\{ (x\text{ mod }2)(2): x \in \mathbb{Z}\}$

what is $(x\ \rm{mod}\ 2)(2)$ supposed to mean?

@s.harp Take an integer modulo 2 and multiply it by 2

And by mod $2$, I assume you mean to return either $0$ or $1$?

so the set $A$ is just the set $\{0,1\}$.

Yes to both

would you have $2\cdot(1\ \rm{mod} \ 2)$ be equal $0$ or $2$?

in which case it would never be the first option, as that would be the set $\{0\}$.

@TobiasKildetoft Ahh

@s.harp that can't be 0

why not, multiplication $\mathbb Z_2$ has $2 x = 0$, if you are taking modulo $2$ the most natural way to view the result is not as an element of $\mathbb Z$ but as an element of $\mathbb Z_2$

$\exists k \in \mathbb{Z}: 2 \cdot (1 \text{ mod } 2) = 2 \cdot (1+2k)$ and this cannot be $0$, unless I'm misunderstanding the notation

@JesterTran I think the issue is that you are treating mod as an operation that returns an integer. This almost never ends well

@s.harp are you saying $2\cdot (1 \text{ mod } 2) = (2\cdot 1) \text{ mod } 2$?

@AlexClark you see your responses page?

@TobiasKildetoft but doesn't it?

Usual notation of what $(x \ \rm{mod}\ 2)$ would be $\{ k \in \mathbb Z \mid \exists n \in \mathbb Z: 2n+k=x \}$, ie the set of equivalence classes

@JesterTran It can, but the behavior becomes bad

@JesterTran then either $(x \ \rm{mod}\ 2)$ is the set of all even integers or the set of all odd integers, $2\cdot $ this will always be a subset of all even integers. You can choose different notation and identify the evens with $0$ and the odds with $1$, and then understand $2\cdot$ to just be the mulitplication of that number with $2$, but thats not what I would do.

Although I think now I have a misunderstanding of my own, $2\cdot\{ \text{evens}\} \neq \{ \text{evens}\}$ but rather $\{\text{dvisible by 4}\}$

What is a rotoreflection by 180 degrees?

ok

I see

Oooh thx

I f $T:E\rightarrow \mathbb{R}$ and $\ker T=T^{-1}(\{0\})$ if we know that $\ker T$ is closed can we deduce that $T$ is continuous ?

@DanielFischer Hi, if I have a linear functional from a normed space $E$ to $\Bbb{R}$, does the image of any open set is open ? If the linear function is not continuous then I can prove it, if $E$ is finite dimensional as well, but what aboute the infinite dimensional case ? Thanks

@JeSuis The functional must not be $0$ of course. Try to prove that every surjective linear map $E \to F$, where $E, F$ are topological vector spaces (over $\mathbb{R}$ or $\mathbb{C}$) with $F$ finite-dimensional and Hausdorff, is an open map.

@JeSuis I am skeptical when the map is not continuous.

@MikeMiller why ? I use the fact that the image of a convex set (here the open ball $B$) is a convex of $\Bbb{R}$, then an interval and not bounded because it's not continuous, and then by linearity $\phi(B)=\Bbb{R}$.

Because I didn't see how to prove it :) Sorry for interrupting and thanks for clarifying.

welcome :)

can someone help me

@DanielFischer I know the open map theorem but I need a Banach space, isn't it?

hum perhaps we only need the is banach space for $F$

have you an idea

@JeSuis That's not the open mapping theorem. This is essentially the fact that a finite-dimensional space allows only one Hausdorff vector space topology. For the open mapping theorem, the continuity of the map is part of the hypothesis. Here, we make no continuity assumptions. We make no assumptions on the topology on $E$ (beyond it being a vector space topology). It needn't be Hausdorff, it needn't be metrisable, locally convex or anything.

someone have an idea

@DanielFischer Ok, I don't really have 'advanced" knowledge in topology, I need to think.

@JeSuis Hint: consider a subspace $G \subset E$ such that the restriction to $G$ is a linear isomorphism.

@DanielFischer

@Vrouvrou I assume $T$ is linear? Then since the codomain is finite-dimensional, it follows that $T$ is continuous.

@JeSuis I don't think I've ever seen Daniel say something wrong.

@MikeMiller I'll get him one of these days ;P.

Does anyone have an idea?

anyone have any idea how to do this: $\sigma = (1~3~5)(2~4)$ , $\sigma ^{2} = (1~3~5)(2~4)(1~3~5)(2~4) = $? I tried it and got $(5~1~3)(4~2)$ but the answer is $(1~5~3)$..

in $S_n$, the exponent just means $\sigma \circ \sigma$, right?

@robjohn Do you maybe have an idea?

nvm I get it.

I've got a formula for polar derivatives that says dr/d$\theta$(sin $\theta$) + rcos($\theta$) all over dr/d$\theta$(cos $\theta$) - rsin $\theta$ is used to find derivatives. I don't get it at all. Everything I see all over the internet involves converting to Cartesian coordinates, which are not even present in this equation. I spent half an hour trying to figure it out. Any advice?

It seems that I can just derive x and y and take the usual dy/dx anyway, circumventing the polar derivative. Seems easier to me!

@JeSuis I'm much more skeptical of that than I was the previous vector space claim.

oh my. astronomy.stackexchange.com/questions/14999/does-sun-speaks-om

hi alex