12:07 AM
why do physicists have to be so counter-culture ugh
homogeneity is defined for the 2nd slot in physics but in the 1st slot for mathematicians
@Obliv a space with an inner product is simply called an inner product space
Hilbert space is an inner product space that is complete with respect to the induced metric
Yeah I read that far on wiki and realized I don't know what metrics or completeness means so I stopped
"most general" is something I wouldn't concern myself with
@Obliv no wonder the riesz theory proof in axler made no sense to me
@zetaspace I wouldn't call it a generalization since it loses linear structure for example
@Obliv a metric is any notion of distance i.e. its a function $d:X\times X\to \mathbb{R}$ on $X$ that satisfies $d(x, y) = 0$ iff $x = y$, $d(x, y) = d(y, x)$ and $d(x, z)\leq d(x, y)+d(y, z)$ for all $x, y, z\in X$
distance from $x$ to $y$ is $0$ iff $x$ and $y$ are the same
distance from $x$ to $y$ is the same as from $y$ to $x$
and triangle inequality represents the phenomena that happens on the plane for example with triangles
12:14 AM
this map $d$ resembles an inner product somehow
no it doesn't
on an inner product space we take $d(x, y) = \sqrt{\langle x-y, x-y\rangle}$
or $d(x, y) = \|x-y\|$ where $\|x\| = \sqrt{\langle x, x\rangle}$
$d$ is a metric from Minkowski's inequality
Now completeness means that every Cauchy sequence $x_n$ with respect to $d$ is convergent
that is if for every $\varepsilon > 0$ there exists $N$ such that for all $N\leq m, n$ we have $d(x_n, x_m) < \varepsilon$ (i.e. $(x_n)$ is Cauchy), then there exists $x\in X$ such that for all $\varepsilon > 0$ there exists $N$ such that for all $N\leq n$ we have $d(x, x_n)<\varepsilon$ (i.e. $(x_n)$ is convergent)
@Jakobian this is equivalent to $\sqrt{(x_1-y_1)^2+(x_2-y_2)^2}$ for vectors $x,y$?
@Jakobian can there be non-cauchy sequences that converge?
also do all cauchy sequences converge? I'd think so
12:42 AM
@Obliv no
@Obliv not always
@Obliv if your inner product space is $\mathbb{R}^2$ with dot product then yes
note however that this is not the only possible inner product space
@Obliv Consider any sequence of rationals which converges to something irrational. This does not coverage in the rationals, though it is Cauchy there.
@Obliv No. Can you prove it?
If you want an example of an inner product space where not every Cauchy sequence converges, take any infinite-dimensional inner product space and let $e_1, e_2, e_3, ...$ be a countably infinite amount of independent vectors. Their linear span is an inner product space which is not complete.
@XanderHenderson I didn't read the definition properly and thought $\frac{(-1)^{i}}{x^{i}}$ would converge but not always be decreasing, but yeah the absolute value bar changes that
@Obliv I haven't read back the entire conversation (and I am not likely to). What does $(-1)^i/x^i$ have to do with anything?
it would be an example of a convergent series where for no $N \in \mathbb{N}$ does $m,n >N \implies x_m-x_n < \epsilon$
12:55 AM
Huh?
I don't understand...
I can make the difference between $1/x^m$ and $1/x^n$ arbitrarily small by choosing large enough $m$ and $n$...
(And if you do something silly like leave off the absolute value, it is even easier, since any negative number you like is smaller than a positive $\varepsilon$.)
@XanderHenderson Oh I guess idk why I thought the abs bars did anything important then.
@Obliv They do add something important in the general theory. In this particular case, if you f*ck up and forget the absolute value, however, you just end up with a bad proof.
@XanderHenderson I don't know either, and I read the conversation

2 hours later…
3:02 AM
based on sumsets

5 hours later…
7:42 AM
peep

1 hour later…
8:53 AM
poop
9:13 AM
@SineoftheTime I tried to do the binky exercise which could have been useful to me too, but for $x$ I could look for a particular solution of the form: $Ax + B$ or would just $Ax$ also be fine?
if $f$ is a polynomial of degree $d$ then I always try with a polynomial of degree $d$, if $f$ is $e^{\alpha t}$ then I try setting $Ae^{\alpha t}$, if $f$ is $\sin(\alpha t)$ or $\cos(\alpha t)$ then I try setting $A\sin(\alpha t) + B\cos(\alpha t)$
9:30 AM
in general search for a polynomial of the same degree, $Ax+B$, but here you can see that $y_p=x$
For example if I had $g(x) = -5x^2 - 5x$ I will have to associate the particular solution $Ax^2 + Bx + C$? Or $Ax^2 + Bx$
@SineoftheTime Yes, that's what I did, but it wasn't very clear to me why I had to add B too
$Ax^2+Bx+C$
Why ?
@Pizza because a particular solution is a polynomial and you don't know a priori that $B=0$
Are we using this? if $f$ is a polynomial of degree $d$ then I always try with a polynomial of degree $d$
9:32 AM
yes
Ah okok clear!
but for example in that binky exercise could I also have used the parameter variation method?
Although I think it becomes much more complicated due to the integrals to solve
yes you can use it
but you have exponentials and polynomials, you already have a particular solution
I'm going to read what it says here, thanks so much for the help anyway!
9:53 AM
hi
see this movie called Mirage
it's a SciFi+romance+thriller+psychology movie
a storm connects a room at two different points of time, and now the future can affect the past which consequently affects the future

3 hours later…
12:43 PM
> Problem: Let $X_1$ and $X_2$ be independent, $U(0,1)$-distributed random variables, and let $Y$ denote the point that is closest to an endpoint (i.e. $0$ or $1$). Determine the distribution of $Y$.
The event that the closest point lies in the interval $(0,y]$ for $0<y<1/2$ is simply (1) that $X_1\leq y$ and $X_2<1-y$ or (2) that $X_2\leq y$ and $X_1<1-y$, i.e. \begin{align}P(Y\leq y)&=P(X_1\leq y,X_2<1-y)+P(X_2\leq y,X_1<1-y) \\ &=y(1-y)+y(1-y)\\ &=2y-2y^2,\end{align}for $0<y<1/2$.
Differentiating I obtain the density $2-4y$ for $0<y<1/2$, which is indeed the answer.
But what is the event that the closest point lies in the interval $(0,y]$ for $1/2<y<1$?
1:31 PM
We have, $$P(Y\leq y)=1-P(Y>y),$$but what is $\{Y>y\}$ in terms of $X_1,X_2$ for $1/2<y<1$?
I solved it I think. Thanks!
There's a lot of symmetry which I did not see.

1 hour later…
2:56 PM
hey
I have. amuck question
can reference angles be used to evaluate inverse functions that have x values that are not with the domain, I am aware that you can do so with coterminal angles but is it too for refrence angles too?
so like sin^-1(200 degrees)= sin^-1 (20 degrees)?
well first of all inverse functions are usually not functions of the angle
arcsin and arccos have domains on bounded intervals, you can consider them as multi-functions but that's not because you extend the domain but the range
@FedericoRuck I am very confused by your question. What do you mean by a reference angle? Why are you trying to evaluate things like $\operatorname{arcsin}(200^\circ)$?
of course you can define whatever you want some extension of arcsin to be, but no one does that as far as I know
no I mean like look up the definition of reference angles and hwo to calculate them
@FedericoRuck I don't understand what you are asking.
3:03 PM
like for example if I have sin^-1(380) we ca do 380-360=20 degrees to find the coterminal angle that fits the domain of sin^-1 and we can then evlate sin^-1(20)=sin^-1(380) can we do this type fo manipulation with reference angles too? Thanks
@FedericoRuck I've looked it up but it doesn't seem like your question is any clearer
No.
$380^\circ$ is not in the domain of $\operatorname{arcsin}$.
yea it is not that's the point
so we transform it to 20
so than we can evaluate it
The domain of $\operatorname{arcsin}$ is $[-1,1]$. This domain is generally regarded as unitless. It isn't measured in degrees or radians or whatever.
5 mins ago, by Jakobian
well first of all inverse functions are usually not functions of the angle
3:04 PM
I have a screenshot of my textbook but can't send it
The sine or cosine of a number is a ratio, without units.
yeas ok but simplify arcsin^-1(sin5Ο/3)
(Or, rather, in the most elementary setting, these objects are ratios---at a higher level... maybe not; but they are still unitless).
you see that even if it is not in th domain we use this property to simplify
yes that is true but I am omginign different concepts?
its $\arcsin(\sin(x))$
this is a different function
3:06 PM
@FedericoRuck That's a completely different object!
yea ok
but still my question reamins
By definition, $\operatorname{arcsin}(x)$ is a number $y$ in $[-\pi/2,\pi/2]$ such that $\sin(x) = y$.
we have to transform the value of sin 5Ο/3 before plugging it in sin inbverse
yes, because you can't take an inverse of $\sin$
$\arcsin$ is not inverse of $\sin$, but inverse of $\sin$ restricted to particular domain
so $\arcsin(\sin(x)) \neq x$ in general
So $\operatorname{arcsin}( \sin( 5\pi/3) )$ is number $y$ (representing an angle) between $-\pi/2$ and $\pi/2$ such that $\sin(y) = \sin(5\pi/3)$.
3:08 PM
the actual graph of $\arcsin(\sin(x))$ looks like a zig-zag curve
So, what $y$ has these properties (i.e. $y \in [-\pi/2, \pi/2]$ and $\sin(y) = \sin(5\pi/3)$)?
anyway... time to make myself some coffee, I hope your confusion is alleviated
you had two different helpful views so something must have hit, right
sorry guys can't relly see the format you re writing in hwo to do that?
LATEX in chat: tinyurl.com/cfqcvpc
3:53 PM
complex valued metric vs. complex metric?
@zetaspace No.
4:52 PM
Does anyone here have a copy of Atiyah and Macdonald? I have a question about the tensor product of two modules.
@Joe Nope.
(Though the correct logician's answer is "I don't know".)
I do but I haven't read that much of it
@Joe yes. I have online copies of almost every book :P
if its just a question about tensor product of modules then I believe I can help
@Jakobian: I am stuck on Corollary 2.13. I don't know what the sentence "Let $M_0$ be the submodule of $M$ generated by the $x_i$ and all the elements of $M$ which occur as first coordinates in these generators of $D$, and define $N_0$ similarly." Do you want me to write down the surrounding text, or can you find it?
Specifically, I don't know the "first coordinate" of a generator of $D$ is.
Maybe I am misreading, but the generators of D consists of certain formal linear combinations, like $(x +x',y) - (x,y) - (x',y)$ for some $x\in M,y\in N$, and I don't know what the first coordinate of such a combination is supposed to mean.
5:09 PM
I think they might be referring not to any sort of the first coordinate of $(x+x', y)-(x, y)-(x', y)$, but to $x+x', x, x'$
then these will be elements also in the corresponding construction for the free $A$-module with $M_0\times N_0$ as generators
@Joe
@Jakobian: Okay, thanks for your help. I will try to follow through on this idea
How might I construct an $(M_t,g_t)$ with $g_t=f_t(y)dx^2+f_t(x)dy^2$ s.t. the intersection loci of the 1-parameter family of surfaces, generate $M_t$ itself?
5:26 PM
I was trying to study this differential form: $\omega(x,y) = \frac{[xy-(1-xy)\log(1-xy)]}{1-xy}\text{dx} + \frac{x^2}{1-xy}\text{dy}$, regarding the closure of the differential form I read that the partial derivatives must be calculated and see if they match. Is there any other more method to see this thing or do I only have this way?
I think you have to use the definition @Pizza
@SineoftheTime ah ok so: $\frac{\partial}{\partial y} \alpha(x,y) = \frac{\partial}{\partial x} \beta(x,y)$
5:44 PM
if $\alpha$ is the coefficient of $dx$, then yes
you have to verify if it's true
you don't have to let them equal
@Jakobian: I asked a question on the main site about proving the proposition just by using the universal property of tensor products, rather than the concrete construction, if you are interested.
@SineoftheTime Don't I have to verify that the two partial derivatives coincide to say that the form is closed?
@SineoftheTime π thanks
$\alpha_y(x,y)\overset {?}{=}\beta_x(x,y)$
5:56 PM
@Joe thanks for notifying me. I'm not particularly interested, just wanted to help you. I'd be more interested in seeing different ways to prove something that's related to topology
@Jakobian: No worries at all. Thanks for your help!
6:11 PM
Is geometry duly consequential of one's own rational opinions?
What?
@SineoftheTime I found that the form is closed, now to study whether the form is exact I read that Poincaré's theorem could be applied. So if $A$ is a simply connected set
Xander it means, is geometry invented or discovered?
@zetaspace language is supposed to convey meaning. Some people try to find art in trying to make beautiful expressions, but ultimately, if your goal is to convey meaning then your question should be as clear as possible
Clarity of speech is a sign of intelligence
3
@Pizza simply connected or star domain
what is $A$ in your case?
6:23 PM
βAny intelligent fool can make things bigger, and more complex. It takes a touch of genius β and a lot of courage to move in the opposite direction.β
~ Schumacher, renowned economist.
@SineoftheTime $A : \{(x,y) \in \Bbb R^2 | 1 - xy > 0\}$
ok, so the domain of definition
is it simply connected?
@SineoftheTime I don't know the meaning of "simply connected set" very well but the domain in question doesn't have holes so yes?
@zetaspace Is mathematics invented or discovered? Ask a philosopher.
@Jakobian I have an extremely low IQ, maybe 95 at best and it's known that IQ is not very malleable so there's not much I can do. So I'm fine being a fool, as that is how I was made
6:29 PM
@Pizza yes
it's simply connected
Let me put it differently. Its unreasonable to complicate things unnecessarily if all you want is an answer to your question
@SineoftheTime Could you by any chance give me a clearer explanation of simply connected set?
@zetaspace 95 is not low IQ, but okay. Also its only one measure of intelligence and so on
consider a generic closed curve in $A$, if you can shrink the curve into a point without going out of $A$, then $A$ is simply connected
> So I'm fine being a fool, as that is how I was made
I find that naive
6:32 PM
or as you said, if $A$ does not have holes then it's simply connected
but this does not work in $\mathbb R^3$
@SineoftheTime So for example $A : \{(x,y) \in \Bbb R^2 | 1 < \sqrt{x^2 + y^2} < 2 \}$ it is not
@SineoftheTime Ah ok!
@Pizza right, it's not
@zetaspace When someone points out that you are behaving in a non-constructive manner, you have a couple of options. For example, you could take the input on board and choose to alter your behaviour in order to be more constructive. Alternatively, you can choose to do what you are doing now, and double down, saying things like "Whelp, nothing I can about it! Just the way I was made!"
I would suggest that if you are going to continue to be non-constructive, you might not be very happy here.
@Pizza consider the circle with radius $1.5$, this is a closed curve but you can't "compress" it into a point staying in $A$
Yes. Its a prime example of rationalization
6:37 PM
@XanderHenderson Okay I will stop being non-constructive - I apologize
@SineoftheTime I have read that, however, if the set $A$ is not simply connected, it can be divided into simply connected components
and therefore in each component the differential form is exact
if it's closed
Yes
However, I think I understand, I'll see as I continue with the exercises
A thousand thanks!
np
do you use a book for ex or just the one of previous exams ?
@SineoftheTime The old book does not have these kinds of exercises
For now I either take them online or from the theory book which has the examples
But I have the file with the exercises that I still have to consult
I should learn these topics then in the end I do all the exercises I can and see if I know how to do them otherwise it means I'm not ready enough yet
6:54 PM
you can find exercises in Giusti's book
and also Marcellini Sbordone
but you have to be careful and select the correct exercises
Yes
@SineoftheTime Yes, I have the sbordone exercises
Now for this new topic I prefer to start by seeing some exercises with solutions and try to do it
So I can see if I'm doing the steps well
7:11 PM
that's a good idea
@SineoftheTime π
8:11 PM
Interesting question
8:30 PM
Take a one parameter class of twisted Kähler metrics $G_t$ with $A,B$ convergent on some vertical strips in $\Bbb C$ where

$$G_t =A_t(s)drd\bar r+B_t(r)dsd\bar s$$

Has a classification of $G_t$ in regions where $A,B$ agree already been achieved in the literature?
Under what explicit conditions can one provide such a classification?
Is it natural/interesting to focus specifically on the regions where $A,B$ overlap?
8:56 PM
> Problem: Let $X_1{,}X_2{,}\ldots {,}X_8$ be independent $\text{Exp}(1)$-distributed random variables with order statistic $(X_{(1)},X_{(2)},\ldots,X_{(8)})$. Find $$E\left(X_{\left(7\right)}\mid X_{\left(5\right)}=10\right).$$
The way I'd probably solve this exercise is to find the joint pdf of $X_{\left(7\right)}$ and $X_{\left(5\right)}$, which would be a tedious task (since I've got $8$ variables). Is there a simpler way?
I am almost certain there isn't. Meeh.
9:50 PM

2024 Mathematics Stack Exchange Moder…

1 hour later…
10:57 PM
Does anyone feel like giving me $150 to buy a cheap pen? @XanderHenderson "Cheap" pen huh @ε₯ηHades It's on sale! @XanderHenderson How "cheap" is it when it is not on sale?$180.
11:03 PM
It better be filled with liquid gold for that price
@ε₯ηHades It's actually pretty cheap, as pens go.
There is a $10,000 pen that I really like. But I'll never have that. @XanderHenderson Is there any noticeable difference between this and a$10,000 pen, besides aesthetics?
@ε₯ηHades Over $500, I have no experience. Under that price, there is a big difference between a steel nib and a gold nib, and the engineering of more expensive pens, in general, gives better ink flow. But over$100, modulo a gold nib, I don't think that there is that much difference.
I figured. That said I'm hardly in a position to call it out when my GPU alone costs over $2000 We all have our expensive hobbies. At the moment, mine are pens, whisky, and luxury cardboard rectangles. 11:10 PM @XanderHenderson I'd figured food would be your most expensive "hobby" given how you seem to know just about any obscure dish I can think of @ε₯ηHades Food is cheap if you cook it yourself. Did you "finish" your fractal welding sculpture? @user20458579510081670432 What? I think that the most i have ever expressed here is a desire to take a welding class so that I can learn the skills to do the thing. I've not done any such thing yet... @XanderHenderson Probably. Takes a lot time, and skill too - that I don't really have. Because of that I paid around 15,000 Yen for a Pizza, a little over$100, though the taste made it well worth it.
@ε₯ηHades That seems like a lot for a pizza.
11:19 PM
It was, indeed.
Oh, I thought you were working on building it, sorry.
I need to stop slacking in piano classes and actually attend them
@ε₯ηHades How about practicing? Still doing those anime piano pieces? (Welcome back, BTW. Haven't heard from you for a while)
Practice makes perfect only if you practice past the point of perfection.
@user20458579510081670432 Haven't heard that one before, but this article seems to agree with you, backed by cognitive science !
2
11:39 PM
Lombardi used to say, practice doesn't make perfect. It is a perfect practice that makes perfect...
...so get out there and win one for the Gipper!!!
@GratefulDisciple that article explains what is known about "automaticity" very well.
@user20458579510081670432 Oh, so that quote is from Vince Lombardi.
@user20458579510081670432 Re: "automaticity", All I know from my learning piano / organ is that I know I'm done practicing a piece when muscle can move automatically for all the notes comfortably at the desired speed, so that what I'm conscious of during performance is the artistic aspect of it. And that I can play without mistake not as a one-off, but more or less consistently, thus the "beyond perfect" part.
11:57 PM
As they say, "In the zone."