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12:00 AM
Now MO?
 
Nah, MO is out of my league so I don't even look there.
Not always but whenever it's within reach someone gets there first anyway. Lots of experts who are constantly on the lookout.
 
I am not convinced, but yeah.
 
12:42 AM
@TedShifrin I believe that I see the post in question, but it doesn't seem to be part of a pattern, and may be entirely unrelated to anyone's understanding of community standards. For what it is worth, I am generally pretty down on hint answers (because an answer should be an answer), but your hint, modulo the word "hint", looks like an answer to me.
(My general issue with "hints" is that they are used as a way of getting around low-quality posts---in this case, I don't think that this is an issue.)
In any case, if it continues, please raise a flag.
 
i just wanted some attention from ted, and i got it, so, mission accomplished.
now to downvote some xander posts
 
@BalarkaSen Please don't do this (or, at least, please don't say aloud that this is what you are doing). Voting in order to "counter" someone else's vote goes against the spirit of voting, which is meant to be about the content, not the user or the actions of other users. Again, for what it is worth, "I thought the answer was good and have upvoted it" is fine.
"I upvoted to counter someone else's downvote" is not so good.
 
Let $Y_n = n \inf_{1 \le i \le n} \{X_i\}$. Let $p_{Y_n}$ be the distribution function of $Y_n$. What is $\mathbb{E}(Y_n)$? $\mathbb{E}(Y_n) = \int_\infty^\infty f(y) p_{Y_n}(y) dy$?
 
@Xander wow, haha, that answer didn't go over at all
 
@leslietownes Nope.
 
12:48 AM
It's clear that $Y_n \rightarrow$ an exponential rv in distribution.
 
And no one left a single comment, so I have no idea why it is so hated.
 
oh, and $X_i \sim U[0,1]$
 
@JoeShmo \sim ?
 
there we go :-)
 
Anywhoodles, I need to get dinner. Then maybe sleep. Tomorrow is a long day.
 
12:50 AM
@XanderHenderson there's a typo in the equation immediately before the "finally"
 
the reason I ask is because $f$ is taken to be continuous and bounded here. but I don't think I need continuity, just boundedness for dominated convergence.
oops, we are looking for $\mathbb{E}(f(Y_n))$
 
@leslietownes Hrm... i am obviously too tired. I don't see it. If I decide that I care, I'll have a look in the morning.
 
there's a 2^10 that jumps from a denominator to a numerator
 
Oi... f'k'in dyscalculia. I know that it is there now, and I still can't see it. Gah...!
 
12:56 AM
Can anyone help this problem? math.stackexchange.com/q/4261805/854335
 
@XanderHenderson What you have written is what I meant
It cannot be helped that it incidentally also goes against the spirit of voting.
Hard to preserve spirit of any idealist concept in real life
 
1:18 AM
I have given up on the model of upvoting stackexchange proposes.
 
Fallen unto nihil?
 
No. I just think that internet points are overall bad.
What have you been up to, Balarka? Have you ever dabbled in information geometry?
You were going full out probability when you were taking that course, it would be natural with your previous background. :P
 
I have briefly heard about it, but I do not know much. Something about Fischer information matrix being a Riemannian metric.
I have stopped thinking about probability for some time and am back to h-principles. I am learning about moment maps and symplectic quotients.
 
Nice!
 
Well, OK, I am doing three things at once but they're all related in some way
But mainly that
 
1:24 AM
I still need to get around to h-principles one day.
 
We could read the Eliashberg-Murphy paper
Contact stuff is up your line
 
Which one? With Borman (?)
 
Yeah
 
Would be a neat paper. I watched a talk today that used it in a proof.
It's quite long, though, I think.
 
Cool! What was the talk on
It's 81 pages. Not the shortest paper in the world, I guess, but could be worse
 
1:28 AM
To be honest, it ended and I wondered exactly what it was on. It was a bit everywhere. They used it while approximating (con)foliations with contact structures. That's the most notable idea I grasped.
But sure, if it was something you were meaning to read, I wouldn't mind trying.
But if it is a compromise, then I am sure we could find something better. :P
 
Aha. Yeah I would like to read it, but I am open to suggestions about anything else you'd like
 
If you want to read it, then let's do it. :)
 
Great. You told me last time about some place you could put the whole thing up and annotate/scribble from both ends, or am I misremembering?
 
We could use hypothes.is to annotate it. You do need to register for an account, though (if you are against that sort of thing). We can do the annotations publicly or privately in a group (whichever you prefer).
 
I can sign up. Let me do that.
Two things, by the way: (1) I might be going through the paper at snail's pace so that might cause some frustrations on your end; partly because I already am doing too many things and (2) I know ~ 0 contact geometry beyond word of mouth so you might have to explain me from time to time.
 
1:34 AM
It's okay, I am probably just as slow, if not slower.
 
Good for me!
 
After signing up, we can leave annotations with LaTeX on the arxiv pdf easily. You just go to the arxiv pdf, and either load hypothes.is through the browser addon, or you can load the page up through hypothes.is's link converter thing on their site (I usually do the former).
arxiv.org/pdf/1404.6157.pdf Just double-checking that this is the one you had in mind?
 
That's the one. Just signed up on hypothes.is.
 
@BalarkaSen No no no! "I am voting to counter another's vote" is against the spirit of voting (that is what you said). "I am voting because I think the post is good (whether or not someone else voted in some way)" is entirely appropriate.
Also, why am I still here? I need to not be here.
 
I am voting because the post is good and it does not deserve a downvote.
 
1:39 AM
@BalarkaSen Great, I have left a little annotation on the abstract for you to see how it will look.
 
@anakhro How do I access your annotation?
Oh I see
 
See it now?
 
Yeah
How does this work man? You created a public annotation?
 
Yes.
 
Ah yeah it says that here
 
1:44 AM
Great, mathmode delimiters are \( and \)
 
Excellent stuff
I should really use this more!
 
Like proper LaTeX should be. :P
 
Lol
 
Yes, I should use this more, too. But I find it hard to find people to read a paper with.
You can make private annotations and everything, so if you don't want something to show up yet or at all, then you can do that so it shows only for you.
 
Ah I see so lets see I'd like to create a private annotation group with you
I guess I know your username now so I can just make a group or something
Ah OK it gives a link I can share, so I'll send it to you on discord
Wait did you remove yourself from discord? I don't see you on my friends list anymore
 
1:50 AM
What was your name on discord again, I cannot locate our conversation.
 
Ibsen#4948
 
I did tidy up my friends list, but I don't know why I would remove you, unless I didn't recognize the username anymore.
 
i went from bsen to Ibsen
:p
 
Oh that could be why. Says you are not accepting friend requests right now.
 
oh damn one second
Try now
 
1:55 AM
Great! All set.
 
Cool
 
TIL wolfram recognizes a meme constant, the Wadsworth constant.
 
What makes it a meme?
 
Oh it's literally a meme. Heh.
 
2:00 AM
Wolfram dropping hints about my search for an optimal quotient algorithm :P
It just so happens that the constant listed there is a coefficient that gives an approximation of 1/3 to 1 ulp for double precision float.
(Using the series I posted earlier)
That unfortunately is not so for other values we wish to approximate, however, I only need enough precision for integer arithmetic and arbitrary precision arithmetic afterwards.
You surprisingly do not need much precision for integer arithmetic over a wide domain, even $[0, 2^{64}]$.
I still have yet to just try integrating and using the natural logarithm instead.
I want to maximize the precision as much as I can using this first and then I'll integrate it.
If anyone gets any ideas, please let me know. :D
 
@AMDG what exactly are you working on?
 
A fast quotient algorithm. I'm looking for one.
This is about as close as I've gotten thus far because of the fact that it is parallelizable.
 
Like integer division?
 
Yes
And if you can do integer divides, you can use that to do fixed-point and floating-point divides as well.
 
What O() do you need for it to be "fast"?
 
2:09 AM
The theoretical optimum for O(M(n)) is O(n log n), so that is what I need. However, it doesn't matter the complexity if the formula is recursive. What I need is something for which I can work on terms independently of each other in order for it to be fast.
It also can't require that I actually have to multiply by something other than powers of two.
If I need that, then it's game over because HW mul is still 3c, and I can't do four of them at a time like I can bit shifts (which is a multiply by a power of two).
 
Wild. I have never been good at optimization of algorithms, so I am useless to you. But bonne chance.
 
Merci
@robjohn You wouldn't happen to have anything to say about this, would you?
 
2:26 AM
Huh, apparently setting to n=1 fixes the precision issues... desmos.com/calculator/8dfiziwfbx
It's so accurate in fact that I only need one term and just need to adjust a coefficient...
 
'Internet sites and other applications for questions and answers'
'Internet sites and other applications for asynchronous collaboration and sharing knowledge'
 
2:46 AM
So, I have this function $c^Tx+\lambda^T(Ax-b)$ that I'm trying to minimize with respect to $x$ as a function of $\lambda$. $A\in\mathbb{R}^{m\times n}$, $b,\lambda\in\mathbb{R}^m$, and $x,c\in\mathbb{R}^n$. $0\leq x,\lambda$ are variable vectors and all other vectors are constants.
However, I can't find any reason this function has to have a finite infimum. In fact, I'm pretty sure that it's usually unbounded.
Wait, I think I don't even need to take that step
 
3:20 AM
Hello
Guys tell me list of approximation function like taylor series, fourier series ...etc
 
too many to count, really. you've hit on some big ones. the choice of what approximation to use often depends on the application.
 
do you learn these on numerical analysis?
 
that's probably where you'd first encounter a lot of them in a series of courses at a school, yeah. or some more general course involving numerical methods
 
3:49 AM
3
Q: Is "reverse homology" $\ker g \subset \text{im} f$ possible?

SmokenSieEinBitteChebaHitBits$\forall$ has $\exists$ as a "left adjoint" in Topos Theory. Let $M \xrightarrow{f} M' \xrightarrow{g} M''$ be sequence of $A$-modules and $A$-module homomorphisms. Now suppose we don't have $gf = 0$ or in other words no ability to compute traditional Homology at $M'$, but instead we have sort...

I put a bounty on that
The highest: 500 points
Perhaps @Euler2 could you answer it?
@BalarkaSen are you an homological guru?
Let's just say HA for short
Usually $d^2 = 0$ is the condition for a complex to have a homology defined.
That's $\ker d \supset \text{im} d$
However, you can have $\ker d = \operatorname{im} d$ and that's called exact at $\operatorname{dom} d$.
I'm saying take the first inequality and do $\ker d \subset \operatorname{im} d$ instead. It seems to work. See write up
that's all of the idea, you just have to learn the details now
Oh man, I finally got an upvote on that answer
lol, waited forever on that one :)
 
4:08 AM
LaTeX ( LAH-tekh or LAY-tekh, often stylized as LaTeX) is a software system for document preparation. When writing, the writer uses plain text as opposed to the formatted text found in "What You See Is What You Get" word processors like Microsoft Word, LibreOffice Writer and Apple Pages. The writer uses markup tagging conventions to define the general structure of a document (such as article, book, and letter), to stylise text throughout a document (such as bold and italics), and to add citations and cross-references. A TeX distribution such as TeX Live or MiKTeX is used to produce an output file...
Author is leslie
 
Lamport. This is Leslie Townsend. They're unique identifiers don't equate
You're proved wrong. QED
I always call it LahTeK, never LayTech. That's reserved for sex toys.
 
5:05 AM
It is not proved until you use proof by voting. Thus your proof is unacceptable.
 
@SmokenSieEinBitteChebaHitBits be aware of allergies: use nitrile.
 
@BannedUser ah, you're right. I'm fooled once again :>
@robjohn what would that be fore?
@robjohn did you check out my reverse homology post? Someone voted it up, then someone voted down. Don't they know that it's legit / cool... lol
 
@SmokenSieEinBitteChebaHitBits many people are allergic to latex. For this reason, many rubber gloves are made from nitrile rather than latex.
 
I just use LayTechs, my friend
You have to use LaTeX in any modern math app nowadays, because students are adapted to MSE requirements already
And know how to enter in MathJax or KaTeX
So my unicode approach was kind of bad (the desktop app)
QuiverDatabase supports LaTeX since Quiver rendering is KaTeX
Guy's a genius of life's philosophies
All of his roles / movies are classics
Some say Lah tek, some say Lay tek, some say po tay to, some say po tah toe, tomayto, toe mato, let's call the whole thing off!
 
5:42 AM
Hi
Can functions have more than 1 dependent variable? For example do functions of several variables have 1 dependent variable and the rest are independent?
 
 
2 hours later…
7:21 AM
Hi, There is always a tangent at a point parallel to secant adjacent to it. So geometrically it is correct that the limit of the slope of secant is the slope of tangent and so the approximation is converted into instantaneous dy/dx ?
 
8:12 AM
Is there an international standards organization recognized by the mathematical community as the most accepted standard for notation and definitions? Like, ISO? or NIST?
 
8:49 AM
@Axoren i dont think so
 
9:06 AM
Heya Leaky
 
10:00 AM
math.stackexchange.com/q/4261805/854335 Is there somebody can help this combinatorics question?
 
@Axoren it's the mathematical community. They decide the notation every time they write
 
10:43 AM
hey there chat
i'm studying measure theory and i recalled that theorem from real analysis: "$f:\mathbb{R}\to \mathbb{R}$ is continuous at $x$ iff $x_n \to x \implies f(x_n) \to f(x)$ for every sequence"
there's the topological, general version to think about continuity in terms of nets. at first it seems that if we apply the general net construction to think about the continuity of $f$ at $x$, we should get the same result. so you can simplify the argument of continuity to some kind of local basis, because $\mathbb{R}$ is a good topological space. what's the property of $\mathbb{R}$ that lets us do that? what property is sufficient to state continuity by seq's in a general topological space?
this also applies to every metric space. first countability i think?
 
In topology and related fields of mathematics, a sequential space is a topological space that satisfies a very weak axiom of countability. In any topological space ( X , τ ) , {\displaystyle (X,\tau ),} every open subset S {\displaystyle S} has the following property: if a sequence x ∙ = ( x i...
this is (by definition) the type of space for which the topology is determined by sequences
as you note, first-countable spaces are sequential, there's a couple more results in the article
 
 
2 hours later…
12:54 PM
@cOnnectOrTR12 Not quite. As I said in the Physics chat, the function has to be well-behaved.
in The h Bar, Sep 24 at 10:11, by PM 2Ring
@cOnnectOrTR12 If f is a nice smooth, continuous function, then the slope of the tangent at (x, f(x)) is the limit of the slope of the secant from (x-h, f(x-h)) to (x+h, f(x+h)). And there will always be some c in the interval x-h <= c <= x+h where the tangent at c is parallel to that secant.
in The h Bar, Sep 24 at 10:20, by PM 2Ring
So as h approaches 0, the slopes of the secants joining x-h to x+h approach the slope of the tangent at x. In other words, when we know m, the slope of the tangent at x, then we know that the slope of any small secant in the neighbourhood of x is approximately equal to m.
in The h Bar, Sep 24 at 10:27, by PM 2Ring
This is very useful. It means that if f and its 1st derivative are continuous in the neighbourhood of x, then small changes to x result in approximately linear changes to f(x). More formally, $$f(x+\Delta x) \approx f(x) + \frac{dy}{dx}\cdot\Delta x$$
 
1:32 PM
Suppose $V,W$ are subspaces of $\mathbb{R}^n$ with the property that $W \subseteq V$.
Prove that if $\textrm{dim} \ V = \textrm{dim} \ W$, then $V = W$.

Suppose $\textrm{dim} \ V = \textrm{dim} \ W$. Since $W \subseteq V$, we have $x \in W \implies x \in V$, and therefore, the basis of $W$ we call $B_W$ is a subset of the basis of $V$, which we call $B_V$. If there is a vector in $B_V$ that is not in $B_W$, then $\textrm{dim} \ W < \textrm{dim} \ V$, which is a contradiction.
is this rigorous enough or is there a hole somewhere?
 
there is no such thing as "the" basis. I supposed you mean the right thing, but I suggest you fix your wording
 
oh, gotta add standard basis which we have since they are subsets of $\mathbb{R}^n$
thanks, important point
 
subspaces of $\mathbb{R}^n$ don't have a "standard" basis in any meaningful sense
 
1:49 PM
hm, aren't they necessarily spanned by some standard basis vectors of $\mathbb{R}^n$?
i may be very wrong, but if we have a subspace of $\mathbb{R}^n$, isn't it necessarily $\mathbb{R}^1$, or $\mathbb{R}^2$, or $\mathbb{R}^3$, etc?
oh: no, not necessarily, since we can make $\mathbb{R}^1$ from the ith component and $\mathbb{R}^1$ from the jth component
or e.g. two planes in $\mathbb{R}^3$ might not intersect
 
come on, you can draw a line in the plane that's not a coordinate axis
 
yep yep got it hehe
thanks
 
2:05 PM
combinatorics is really driving me crazy
 
2:19 PM
Is there any general parametrization formula for epicycle?
 
2:46 PM
love, as in the astronomy thing? just a vector sum of multiple parametrizations of circles of different radiii
 
 
1 hour later…
3:55 PM
Why the covariance or corelation for data points that form circle is zero ?
 
4
Q: Show that $\limsup \frac1{x_n}\cdot \limsup x_n\geq 1$

user6163Given a sequence $x_{n}$ and initial data that $0< a\leq x_{n}\leq b< \infty $, for $a,b\in \mathbb{R}$. I need to show that: $$\limsup \frac{1}{x_{n}}\cdot \limsup x_{n}\geq 1.$$ I think the the simplest way to do that is to show that $$\liminf \frac{1}{x_{n}}=\frac{1}{\limsup x_{n}}.$$ I s...

In the answer, I think that it should be $n\to \infty$ in line just after the first expression.
 
4:23 PM
koro, yes. i'd edit but don't want to bump a 10-year-old post for a one character edit :)
 
(I am once again reminded of my long-standing desire for a "minor edit" feature that can fix a post without bumping it)
 
that would be a good feature. it seems like something particularly helpful in math, where a few typos can cause the logic/meaning to go haywire. i could see in general why one might not want an SE-wide option of that nature.
 
yeah
it'd be nice if it existed for codegolf too - we've changed policies many times and there are a lot of posts that should be reformatted, but to avoid bumping, we just live with it and have old posts that don't follow the standard or slipped by without someone editing it early on
also when i went back and retagged a ton of our meta posts so they had the meta "status" tags, it would've been nice; instead i just picked a time in the dead of night, bumped like 90 posts, and then manually bumped the top few posts back up
 
codegolf being perhaps the most acute example of a typo making a difference. haha
 
well a typo is like half the answer xD
usually the code doesn't have typos it's just formatting. honestly not a big deal. i imagine it's much more significant for math though where a small typo changes everything
 
4:32 PM
you can often infer enough from context to detect and autocorrect typos, but not always
and people who are already confused about something and asking question are maybe less likely to trust their internal type-checking
i'm surprised math works as well as it does
 
$$\int_0^{x_t} \dfrac{dx}{(a-x)(b-x)}=kt$$ Can this be simplified? $a,b,k,t$ are constants.
 
is x_t an unknown to be solved for?
if a and b are different you can evaluate the integral in terms of the logarithm function and the algebra might get messy. if a = b the integral is 1/(a - x_t) - 1/a
when a isn't b, the partial fraction decomposition of 1/((a-x)(b-x)) is 1/(b-a) [1/(a-x) - 1/(b-x)]
 
@leslietownes Thanks :-) It's from chemistry so I guess they don't want that.
 
i mean you can remove the integrals, or even put it in the form x_t = [function of a, b, k, and t], but if you want to solve for something else, maybe neither of those counts as 'simplified'
 
4:52 PM
so here is a lame question
i'm trying to find the easiest way to show that the equation $\cos\theta=\tan\theta\sqrt{\sin^2\theta+a}$ has solution $\cot\theta=\sqrt{1+a}$ for $a>0$ and $\theta\in(0,\pi/2)$
just to make life easy for my students
but it seems inordinately tedious no matter what i do
best i can see at the moment is to solve for $a=\cos^2\theta\cot^2\theta-\sin^2\theta$
and show that this simplifies to $\cot^2\theta-1$
hmm. i guess $\cos^2\theta\cot^2\theta-\sin^2\theta=(1-\sin^2\theta)\cot^2\theta-\sin^2\theta = \cot^2\theta-\cos^2\theta-\sin^2\theta=\cot^2\theta-1$ isn't the worst
 
i bet it isn't even the worst thing i see today.
 
@hyper-neutrino Spammers would love that. But I guess you'd just have to restrict it to people with the edit privilege. FWIW, one of the recent changes to the review queues now allow an edit reviewer to mark a question edit as trivial so that the 1st edit to a closed question no longer automatically pushes the question to the reopen queue. So changes to the edit software are possible.
OTOH, the general consensus is that it's better for edits to bump posts in order to make it easier to catch mistaken and malicious edits.
 
I have a matrix $A$ and I want a new form of matrix $A^*$ such that (1) the pivot columns of $A$ are on the left of $A^*$, (2) the free variable columns of $A$ that are in between pivot columns of $A$ are in the middle of $A^*$, and (3) other free variable columns of $A$ are on the left of $A^*$
is there a way to define column operations that will allow me to treat $A^*$ exactly as if it was $A$ with regards to matrix-vector multiplication, s.t. $A^*x = Ax$?
 
@PM2Ring well you'd need the rep for editing and also probably make it an even higher rep threshold to silently edit than normal
like, just as a random example, maybe like 10k to edit silently and like 5k on your own posts or something like that? idk
@PM2Ring that's true
recent edit review queue? 🤔
 
5:10 PM
@hyper-neutrino I guess that'd work. Maybe just push all silent edits to the existing edit review queue. But maybe it needs its own queue, since there's a perennial problem of people robo-approving bad suggested edits.
 
oh, I actually need to define a new sort of matrix-vector multiplication and can do whatever I want with my column operations
 
shin, the kinds of operations you have in mind aren't fully clear to me, or your purpose, but you can't generally expect Ax = Bx if all you know is that B is the result of applying elementary column operations to A.
other 'column operations,' i guess the sky is the limit.
 
I have a set of vectors $B$ that span solutions to $A\vec x = 0$ such that, given a vector in $B$, there is a single component corresponding to a free variable set to $1$, and other components corresponding to free variables are set to $0$, while components corresponding to pivot variables are whatever.
I'm trying to prove that $B$ is a basis for the null space of $A$
and I need to write the most general form of $A$ I can imagine
 
this sounds like a transposed version of the argument that the nonzero rows of a matrix in row echelon form are linearly independent.
i might be missing something.
 
Hi all
 
5:17 PM
the spanning is the hard part
 
I'm trying to understand the solution to this problem: Whoops, wrong image
 
i'm confused, the above begins with "i have a set of vectors B that span solutions of Ax = 0." are you not assuming that span(B) is the nullspace of A?
 
I got $u \cdot v = |u||v|\cos{\theta} = 1 * 1 * \cos(60)$ = $\frac{1}{2}$
Likewise, $w \cdot v = |w||v|\cos{\theta} = 1 * 1 * \cos(60)$ = $\frac{1}{2}$
However, the book states:
 
wasn't the second bit asking for u dot w? [and not v dot w?] the angle is different.
 
hm leslie you're right, I can't begin by assuming that it spans the solutions
 
5:21 PM
@leslietownes Sorry, typo. Nevermind, I see my mistake.
I measured the angle from different points
 
a sort of cute way to see this is to note that w = v - u. so u dot w = u dot v - u dot u = u dot v - 1 can be computed in terms of u dot v
 
Type in MathJaX, dammit.
@rb3652 You have to put the vectors tail to tail!
 
w not < u dot u w = wot v w bot langle.
 
smacks leslie and leaves in a huff
 
sorry, brain is a little frazzled this morning. the cat needed some booster shots and attacked with tooth and claw while i was trying to get her into her carrier.
i always have an excuse.
 
5:25 PM
@TedShifrin Yes, yes, such a subtle thing.
 
Suppose $x = \langle x_1, \cdots, x_n \rangle$ is a solution to $A\vec x = 0$. I want to prove this vector is a linear combination of a set of linearly independent vectors $v_1, \cdots, v_n$ defined in the following way: $v_j$ has it's jth component correspond to a free variable and has as that component a 1. if any other components of $v_j$ correspond to free variables, they are equal to 0. components corresponding to a pivot variable are whatever
i think that's the proper statement
 
5:39 PM
and this is super easy if we assume the pivots in $A$ form a nice diagonal, but not that easy if they don't
 
column permutations correspond to relabeling the variables, if that helps. if you can see how to get it with a relabeled version of the variables, you can get it in general. relabeling doesn't seem to affect the property you want.
 
$\cos\theta\ne0$, otherwise the RHS is $\infty$, so we can safely multiply by $\cos\theta$.
$$\begin{align}
\cos\theta &= \tan\theta\sqrt{\sin^2\theta+a}\\
\cos^2\theta &= \sin\theta\sqrt{\sin^2\theta+a}\\
(1-\sin^2\theta)^2 &= \sin^2\theta(\sin^2\theta+a)\\
1-2\sin^2\theta+\sin^4\theta &= \sin^4\theta + a\sin^2\theta\\
1 &= (a+2)\sin^2\theta\\
\sin^2\theta &= \frac1{2+a}\\
\cos^2\theta &= \frac{1+a}{2+a}\\
\tan^2\theta &= \frac1{1+a}
\end{align}$$
Now verify that squaring didn't introduce spurious solutions.
 
oh! so a column-switching operation that also relabels the column
 
yeah, that's pretty clean
 
thanks for the tip
 
5:50 PM
@Semiclassical It was easier than I thought it'd be. My experience with ellipses has made me overly suspicious. ;)
 
Hello, I have some questions about comaximal monoids under multiplication in a commutative ring. As I understand it rings are often viewed as the ring of nice functions on some space. Monoids often arise as monoids of nowhere-vanishing functions on some subset of this space - then localization at this monoid represents “zooming in” on this subset. We say the monoids $S_1,\dots,S_n$ are comaximal when for any choice of $s_i\in S_i$ for each $i$, $\langle s_1,\dots,s_n\rangle=\langle1\rangle$.

Is it accurate to view every monoid as a monoid of nowhere-vanishing functions on some subset (for
 
@leslietownes ok thanks :)
 
Okay I guess my first question wasn’t very well thought out, probably this is not the case for the one element monoid containing just $1$
Btw I am assuming all monoids contain $1$, should have said that
Correction: what is the geometric analogue to comaximal monoids (not ideals)?
 

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