12:35 AM
Alright finally Laplacian time
So to see the change of variables going on
$r^2 = x^2 + y^2$, take partial derivatives wrt $x$ and get $2rr_x = 2x$
So $r_x = \frac{x}{r}$, and then $r_{xx} = \frac{r - r_x x}{r^2} = \frac{r - \frac{x^2}{r}}{r^2} = \frac{r^2 - x^2}{r^3} = \frac{y^2}{r^3}$
Similarly $r_y = \frac{y}{r}$ and $r_{yy} = \frac{x^2}{r^3}$
Or lemme see exactly what I need to do here
So chain rule gives $f_x = f_r r_x+ f_{\theta}\theta_x$ and $f_y = f_r r_y + f_{\theta} \theta_y$
And $f_{xx} = (f_r)_x r_x + f_r r_{xx} + (f_{\theta})_x \theta_x + f_{\theta}\theta_{xx}$
$f_{yy} = (f_r)_y r_y + f_r r_{yy} + (f_{\theta})_y \theta_y + f_{\theta} \theta_{yy}$
So now we have to compute $(f_r)_x$ and $(f_{\theta})_x$
$(f_r)_x = f_{rr}r_x + f_{\theta r} \theta_x$
$(f_{\theta})_x = f_{\theta r} r_x + f_{\theta \theta} \theta_x$
And similar for $y$
So putting that all together
$f_{xx} = (f_{rr}r_x + f_{\theta r}\theta_x)r_x + f_r r_{xx} + (f_{\theta r} r_x + f_{\theta\theta}\theta_x)\theta_x + f_{\theta}\theta_{xx}$
$f_{yy} = (f_{rr}r_y + f_{\theta r}\theta_y)r_y + f_r r_{yy} + (f_{\theta r}r_y + f_{\theta\theta}\theta_y)\theta_y + f_{\theta}\theta_{yy}$

gross

So those are the quantities I want. I have $r_x$, $r_{xx}$, $r_y$, $r_{yy}$
So now $\tan(\theta) = \frac{y}{x}$, take partials wrt $y$
$\sec^2(\theta) \theta_y = \frac{1}{x}$, so $\theta_y = \frac{\cos^2(\theta)}{x} = \frac{x}{r^2}$
And $\theta_{yy} = \frac{-2xy}{r^4}$
Now taking partials wrt $x$ is gonna look different, we get $\sec^2(\theta) \theta_x = -\frac{y}{x^2}$
So $\theta_x = -\frac{y}{r^2}$, and $\theta_{xx} = \frac{2xy}{r^4}$
Conveniently $\theta_{xx} + \theta_{yy} = 0$
And $\theta_x^2 + \theta_y^2 = \frac{1}{r^2}$
So now putting everything together, help me God
I was called now I'm back
$f_{xx} = f_{rr}r_x^2 + 2f_{\theta r}\theta_xr_x + f_{\theta\theta}\theta_x^2 + f_rr_{xx} + f_{\theta}\theta_{xx}$
$f_{yy} = f_{rr}r_y^2 + 2f_{\theta r}\theta_yr_y + f_{\theta\theta}\theta_y^2 + f_rr_{yy} + f_{\theta}\theta_{yy}$
Now we have some convenient facts
Well first let's see
$r_x^2 + r_y^2 = 1$
$r_{xx}+r_{yy} = \frac{1}{r}$
$\theta_{xx}+\theta_{yy} = 0$
$\theta_x^2 + \theta_y^2 = \frac{1}{r^2}$
$r_x\theta_x + r_y\theta_y = 0$
So finally
$f_{xx}+f_{yy} = f_{rr} + \frac{1}{r}f_r + \frac{1}{r^2}f_{\theta\theta}$
So now in polar coordinates, the hyperbolic Laplacian is $(1-r^2)^2(f_{rr} + \frac{1}{r}f_r + \frac{1}{r^2}f_{\theta\theta})$

1:14 AM
Yes, the answer is right. I can get there in mere lines with forms :)

Lmao, maybe I wish I had done that. Though I guess since we're using a different metric it's a bit trickier for me to reason about Laplacian on forms
It should be $\star d \star d f$ right?

If you think briefly like a physicist, unit analysis checks out. Also you can see $\log r$ is a solution, as cx analysis tells you it must be.
Yes. Well, you didn't take the metric into account when you did yours. You just had a final formula. Here $\star$ builds in the metric if you want to.

Well the metric is what brought in $(1-r^2)^2$ factor
I guess it's probably notable in this case that even though the Laplacian is different, it seems like the harmonic functions are the same?
Anyway let's try with forms
So the trick here is to ask what the star operator looks like
So $df = f_r dr + f_{\theta}d_{\theta}$

But you need the metric to figure out $\star$
What's the orthornormal basis for $1$-forms?

Hyperbolic angles should be the same as Euclidean ones so I guess it's just taking an orthonormal basis for 1-forms in the Euclidean case and then multiplying by $(1-r^2)^2$
Basically $(1-x^2-y^2) dx$ and $(1-x^2-y^2) dy$

1:28 AM
What's $dx\otimes dx + dy\otimes dy$ in polar?
You're right that you will just stretch from the Euclidean case.

Hmm, this might be where I need to cry uncle a bit since I haven't really ever thought about tensor (as opposed to exterior) products of forms
My naive approach to the overall problem would be to say that $\star dx = (1-x^2-y^2)dy$ and $\star dy = -(1-x^2-y^2)dx$

1:53 AM
So now let's take the Poisson kernel
$\frac{1-r^2}{1-2r\cos(\theta - \phi) + r^2}$, where $\phi$ is just some fixed guy
So let's compute
$$\frac{1-r^2}{1-2r\cos(\theta)\cos(\phi)-2r\sin(\theta)\sin(\phi) + r^2}$$
Normally it seems like this is a thing when $\phi = 0$ but Helgason wants it with a possible $\phi$ in there for what he's doing
So let's show this guy is a Laplace eigenfunction

The Euclidean metric is $dr\otimes dr + r^2 d\theta\otimes d\theta$. Work it out,
You know tensor product from algebra, just forget skew-symmetry.

2:17 AM
Alright, I will do that. Though for now I'll finish up the Poisson kernel
So let's take the derivative of that monster wrt r
Let's call that guy $P$
$$P_r = \frac{-2r+4r^2\cos(\theta-\phi)-2r^3 - (-2\cos(\theta-\phi)+ 2r)(1-r^2)}{(1-2r\cos(\theta-\phi)+r^2)^2}$$
Simplify that to get
$$P_r = \frac{-4r+(2r^2 + 2)\cos(\theta - \phi)}{(1-2r\cos(\theta-\phi) + r^2)^2}$$
$$P_{rr} = \frac{-4+4r\cos(\theta-\phi)}{(1-2r\cos(\theta-\phi)+r^2)^2} - \frac{2(-4r +(2r^2+2)\cos(\theta-\phi))(-2\cos(\theta-\phi)+2r)}{(1-2r\cos(\theta-\phi) + r^2)^3}$$
If this is wrong I'm going to cry
$$P_{\theta} = \frac{-(1-r^2)(2r\sin(\theta-\phi))}{(1-2r\cos(\theta-\phi)+r^2)^2}$$
$$P_{\theta\theta} = \frac{-(1-r^2)(2r\cos(\theta-\phi))(1-2r\cos(\theta-\phi)+r^2)^2 +2(1-2r\cos(\theta-\phi)+r^2)(2r\sin(\theta-\phi)(1-r^2)(2r\sin(\theta-\phi))}{(1-2r\cos(\theta-\phi)+r^2)^4}$$
$$P_{\theta\theta} = -\frac{(1-r^2)(2r\cos(\theta-\phi))}{(1-2r\cos(\theta-\phi)+r^2)^2} + \frac{8r^2\sin^2(\theta-\phi)(1-r^2)}{(1-2r\cos(\theta-\phi)+r^2)^3}$$
Okay this has a chance of working
I really hope I didn't drop a sign somewhere
So $\frac{1}{r}P_r + \frac{1}{r^2}P_{\theta\theta}$ first
$$\frac{1}{r}P_r + \frac{1}{r^2}P_{\theta\theta} = \frac{4r\cos(\theta-\phi)-4}{(1-2r\cos(\theta-\phi)+r^2)^2} + \frac{8\sin^2(\theta-\phi)(1-r^2)}{(1-2r\cos(\theta-\phi)+r^2)^3}$$

3:03 AM
oh my lord

1 hour later…
4:32 AM

Lmao
I'm also worried this doesn't even work but I had office hours

4:49 AM
0

Many sources assert that China appears to have eliminated COVID-19 within the country, except for the odd imported case or cluster, which is quickly isolated and eliminated. For instance, Worldometer's coronavirus tracker, based on data provided by each country's government, shows no new cases fo...

2 hours later…
6:52 AM
Alright finally
Verified it
So this guy is harmonic for sure
Call this guy $P(z,b)$ for Helgason reasons
And then apparently $\Delta_z(P(z,b)^{\lambda}) = 4\lambda(\lambda-1)P(z,b)^{\lambda}$

7:34 AM
$\lceil \log_b (x+1) \rceil = \lfloor \log_b (x) \rfloor +1$ right?

1 hour later…
8:52 AM
Suppose that $X$ is an homogeneous topological space. It's easy to see that it is not true that every (necessarily continuous) function between finite subspaces of $X$ extends to an automorphism of $X$, but is true that for every pair of points there is an automorphism of $X$ swapping them? Maybe I'm missing something obvious

9:17 AM
Hi
Can someone tell me, which form of Riemann's Hypothesis needs to be proved? Is the one for 0 > c > 1 enough?

9:35 AM
"Good morning"

morning

What happened that made someone star all of @Amin 's computations lol

9:51 AM
Anyone alive here?

@Tomasz what do you mean by "the form with $0 < c < 1$" ?

Any other good music to suggest? @Edward

Sybreed - Slave Design
timeless album

The one that's extended from the basic n^-s

I don't know what you mean by that lol

9:53 AM
Sorry I don't know these symbols. I'm learning now to get those million bucks.

Good luck

Thanks

Where does your question come from?

I found a solution and now want to know if it's ok

you haven't even said what the question is

10:04 AM
There's the basic form of SUM (n ^ -s) and then there's the one that has alternating signs. And there's the function equation. So I want to know if the one with alternating signs is enough for a proof, as from what I've been reading that should do.

Depending on where you read that, it's probably true
lol

I must be a funny carrot.

The zeta function is related to the alternating zeta function (also called Dirichlet's eta function) by $\eta(s) = (1 - 2^{1-s})\zeta(s)$ so a zero of the zeta function is a zero of $\frac{\eta(s)}{1 - 2^{1-s}}$ shrugs
Analytic number theory is "not yet" my thing

@AlessandroCodenotti Take Z with the topology consisting of intervals $(n, \infty)$ for $-\infty \leq n \leq \infty$. Then f: Z -> Z is continuous iff it is nondecreading, I believe
So f(n) = n+k is continuous for any k and so this is homogeneous
But f can never swap two points

but the inverse of f(n) is not continuous then
I was also playing around with orders to find a counterexample

2 hours later…
11:48 AM
@AlessandroCodenotti No, I disagree, translation is always continuous
Nondecreading meaning $x \geq y \implies f(x) \geq f(y)$
Not $f(x) \geq x$
So translation is a homeomorphism

Ah of course, sorry for the brainfart

You're good

12:04 PM
@MikeMiller I think it's obvious that the next thing I am supposed to learn is Banach representations of noncompact groups

12:44 PM
@EdwardEvans someone wanted to push away your awful "norm wildberger" joke

Eigenfunctions of the hyperbolic Laplacian I assume

1:22 PM
@user2103480 and lo, it hath returned

Good (deleted) question mike

Am I banned
Looks like someone deleted it for me, I thought I got flagged

I decided its better to try to understand Weierstrass preparation

In the message history it says that "Feeds" deleted it
Maybe it automatically filters some swear words? I have no idea

weird
fucking weird
what the fuck is this weirdness
lets see
@EdwardEvans Rivers of Nihil is good

1:31 PM
ye man

@AlessandroCodenotti No. The message was flagged.

I see, thanks for clearing up the confusion
So whoever mods deletes a message it shows as if deleted by the bot?

If the net flag count of the message reaches six or if the message is flagged by a moderator, the message is deleted by Feeds.
If a moderator manually deletes the message, it will show this moderator instead.

@EdwardEvans what the hell this is death metal jazz
that should not exist

hahaha

1:40 PM
@BalarkaSen Can I introduce you to between the buried and me

oh i have heard of colors but never listened to them

thanks will do

@Edward you might like it too if you don't know them already

2:04 PM
What are interesting examples of spaces such that any (necessarily continuous) bijection between finite subspaces extends to an automorphism of the space? Does $\Bbb R^2$ have this property? Does this property imply trivial $\pi_1$?

Homeomorphism group of any manifold acts $n$-transitively on the manifold, for any $n$.

That's different from what I'm asking

Confused.
How is that different

surely $\mathbb{R}^2$ does

No wait, sorry

2:07 PM
I can't argue it without handwaving, but surely it does

@BalarkaSen You can take any n-point set to any other, he's asking to permute them with specified permutation
But the proof of n-transitivity already proves the latter

Oh. Yeah

I don't remember, in the definition of say $2$-transitive can you swap two points?

No, but you can do this easily by modifying the proof.

If $a \in [0,1)$, does $\sum_{k=1}^\infty \frac{a^k}{k}$ converge?

2:09 PM
Ok but now I'm confused

Eh, $n$-transitive does mean that, you can just take a permutation of the original collection of points.
Whatever

Pick 4 points on $S^1\subseteq\Bbb R^2\setminus\{0\}$ in clockwise order $a,b,c,d$, how can the permutation swapping $b,d$ and fixing $a,c$ be extended to the whole space?

$a, c$ are antipodal, right? Imagine a reflection of the circle.
Fixing $S^0 = \{a, c\}$

and if they're not directly antipodal, you just do some squishing

2:13 PM
Ah of course I see
Ok I'm going back to complete sequences of covers, I can't work with spaces you can actually see without getting confused apparently

But wait, there's no self-homeo of $\Bbb R$ realizing $(13)(24)$, right?
Order-preservation

@AlessandroCodenotti mood

It has to be either increasing or decreasing

@BalarkaSen right

So what is this $k$-transitivity I am talking about? I don't know. I mean $\text{Homeo}(M)$ acts transitively on the configuration space of $k$-points, $(M^k \setminus \{\text{various diagonals}\})/S_k$.

2:17 PM
That was my example too for homogeneous does not imply this property
Maybe all $n$ up to the dim of the manifold?
@BalarkaSen Hmm what does that mean?

One should be able to write some proof for $\Bbb R^2$

You might be talking about $k$-homogeneous actions (transitive on subsets of size $k$)

is it actually true for $\mathbb{R}^2$, I'm not sure anymore

I need to leave for a while, sorry

@Thorogtt Yes, I think so.

2:26 PM
I'm convinced you can take any $n$-element subset of $\mathbb{R}^2$ to any other $n$-element subset via homeomorphism, but I'm starting to be worried about permutations

Let $1, 2, \cdots, n$ be points on the real axis in $\Bbb R^2$. Any permutation is a product of transpositions, so it suffices to extend transpositions without affecting the other points.
Now given two points you can draw a topological disk around it which doesn't hit any of the other points
On a disk $D^2$ say $p, q$ are two points on the same diameter
Push the diameter so that $p$ and $q$ reverse order
I'll draw a picture
You can take 1st picture to 2nd picture by an isotopy, not just a homeomorphism.
Doing this for each transposition extends your permutation
I think this is what they call a braid diagram
@Alessandro This proves it for $\Bbb R^2$ hence for $\Bbb R^n$ for all $n \geq 2$, hence for all manifolds of dimension $\geq 2$ because you can always take finitely many points to be inside a chart by a homeomorphism (by $k$-homogeneity if you wish, thanks @Tobias)
(The braid diagram connection is as follows; the $n$-stranded braid group $B_n$ is MCG of $D^2$ with $n$ punctures; draw the essential curves between the punctures and see the image of these curves under the homeomorphism corresponding to the braid element - this configuration is reminiscent of what I am drawing and determines the braid uniquely)

2:52 PM
can you build a manifold based on information on the boundary?

3:05 PM
I have an $n\times n$ matrix where the diagonals are equal to one and the off diagonals all have magnitude smaller than one. Is this matrix invertible? What is the quickest way to prove it?

At least not if you allow negative entries. Haven't thought too much about if they are all non-negative

It's not true
You need stronger conditions, look up being diagonally dominant

It is true for diagonally dominant matrices
I got sniped :/

I'm going to read what you wrote earlier now
Makes sense

3:22 PM
I just noticed something. Perhaps I'm wrong, but if you have an isosceles triangle with two equal sides of length a, and one side b, and a is equal in length to the angle, can't you compute sine and cosine from that triangle's lengths sine angle doesn't care about the length of the sides?

counterexample with all positive entries

since*

nvm, small error, sec

This would appear to be true since all circles are similar, and for all isosceles triangles, two points will always be on the circumference of a circle, so then you just create an arbitrary isosceles triangle with lengths of desired angle, and then compute the intersection of a side length with one of the sides on either the x or y axis to get sine/cosine, right?

@AlessandroCodenotti This depends on the dimension being larger than 1
All it has to do with is the fact that points can only be cut points in dim 1
Every connected manifold is homogeneous in the following strong form: for $p, q \in M$ there is a compactly supported homeomorphism $f: M \to M$ with $f(p) = q$
So you iteratively construct a homeo which takes $p$ to $q$, and then do the same story for points not in $M \setminus \{p\}$
Your new homeomorphisms extend by the compactly supported assumption
The idea only can go wrong in 1D when the deletion process can disconnect

4:03 PM
It would seem what I said is true. If the angle is equal to the two side lengths which is a, then the smaller side b must be proportional and therefore easily computed from the two side lengths (and the angle bisector of said triangle is either the x or y axis). Then, we use b to find a line passing through the unit circle. Then sine/cosine should be the intersection of that line and the unit circle. Please correct me if I'm wrong.
Specifically, b/2 is delta y of said line, and delta x is the length of the angle bisector which is computed from length a.
So does this mean I don't need to use approximations anymore to calculate sine/cosine? I can just calculate it directly now for all values of x using this method? It's so incredibly simple. I find it hard to believe. I just have to test this. I hope someone else will verify as well.

4:47 PM
Hope this is the right place for this: I have a right isosceles triangle where the base is half the length of the height (h = x units) and I'm trying to find the smallest integer x where its possible to pack 100 circles (r = 5 units) into this triangle with no overlap. Doing some googling/WAing hasn't really helped, none of the pages that might have helped go up to n = 100 (trying to pack n circles into isosceles triangles). Wondering if anyone had any hints/help?

1 hour later…
5:51 PM
I should hope this here that I rendered should be somewhat readable and legible. imgur.com/gallery/S0a8sPK
(Obviously not to scale)

Never thought about looking at nested brackets as words constructed with () as magma multiplication

The point is that you want all possible strings and all possible ways of regarding them as a product

make sense

7:00 PM
@AMDG it appears that both the radius of the circle and the angle between $AB$ and $AC$ are $2\alpha$. Is that true?

No, refresh the page and see my addendums and notes.
I'm not very good at writing things down in formal math speak, so please bear with me. I'm a programmer, not a mathematician. :)

and so what is the question?

Well I posted it because I wanted others to verify that this actually yields cos(alpha) and sin(alpha). I'm not certain of it yet.
Because it would seem that simply setting the lengths of the sides equal to 2alpha should mean that the remaining side b must also be proportional to their angles as well as a direct correspondence if I have read correctly about the nature of isosceles triangles.

howdy @robjohn

@TedShifrin Hey, Ted! How are things?

7:12 PM
If I ignore the country and the world, fine. You?

@TedShifrin that sounds like something to try... Doing okay.
@AMDG whenever you have a right triangle with one of the non-right angles equal to $\alpha$, the opposite side is $\sin(\alpha)$ times the hypotenuse, and the adjacent side is $\cos(\alpha)$ times the hypotenuse.

Oh yeah, that would simplify things... I was thinking of it geometrically, but my reasoning is that any isosceles triangle has two points on a circle, and all circles are similar, so if you could use only the length of a to compute the length of b via a formula or constant proportion not requiring sine or cosine functions, then the intersection of the resulting lines with the unit circle would yield sine and cosine.
So I could have something that is completely wrong, but it definitely gets me closer to an answer if it is wrong anyways.

Does $\int (1-x^2)^{\alpha} dx$ have a closed form, where $\alpha > -1$? I am trying to show that $\int_{0}^{1} (1-x^2)dx$ is finite.

Thing is, though, @robjohn , that would be for any right triangle which is in the unit circle, correct? So these isosceles triangles are not required to be on the unit circle, however, they all intersect with it.
Because the length 2alpha could be greater than 1 or less than 1
err any right triangle with a hypotenuse at the origin and the circle* to be specific

Whoops, I am trying to show that $\int_{0}^{1} (1-x^2)^{\alpha} dx$ is finite.

7:26 PM
@user193319 for positive $\alpha$, that is less than $1$. Ah, you are looking at $\alpha\gt-1$
try substituting $x\mapsto1-x$; that is. $1-x^2\mapsto2x-x^2$

Okay. I'll give that a go. Thanks!
So, in terms of u-sub, I am letting $u = 1-x$?

@AMDG That is the "standard" triangle. All other right triangles are similar to such a triangle.
@user193319 yes

Oh yeah

@robjohn So that gives me $-\int_{1}^{0} (2u-u^2)^{\alpha} du$, but how do I deal with this? I am taking a course operator theory and spaces of analytic functions but I've forgotten all of my calculus...yikes!

@user193319 That is $\int_0^1u^\alpha(2-u)^\alpha\,\mathrm{d}u$
$(2-u)^\alpha\le1$ for $\alpha<0$ and $0\le u\le 1$
So it boils down to $\int_0^1u^\alpha\,\mathrm{d}u\lt\infty$
@AMDG That is one of the main ideas of trigonometry; reducing right triangles to one of these standard right triangles.

7:42 PM
But does trigonometry not work with all triangles?

@AMDG After you've broken them down to combinations of right triangles.

I thought so.

Notice that you've broken the isosceles triangle into two right triangles.

Yes

The Law of Sines and the Law of Cosines are proven by looking at right triangles.
All of trigonometry stands on the legs of right triangles.

7:49 PM
@robjohn I don't see how $(2-u)^{\alpha} \le 1$ holds. And what if $\alpha > 0$?

@user193319 If $\alpha\gt0$, then $\int_0^1\left(1-x^2\right)^\alpha\,\mathrm{d}x\le1$
@user193319 $1\le2-u\le2$ for $0\le u\le1$, so for $\alpha\le0$, we have $2^\alpha\le(2-u)^\alpha\le1$

1 hour later…
8:54 PM
@user193319 Does this make sense now? So as long as $\alpha\gt-1$, $\int_0^1u^\alpha\,\mathrm{d}u=\frac1{1+\alpha}\lt\infty$.
$\int_0^1\left(1-x^2\right)^\alpha\,\mathrm{d}x\le\max\left(\frac1{1+\alpha},1\right)$

9:17 PM
Anyone here use Mendeley?

9:57 PM

shorter than Spivak

wot
lol
Do you have a recommendation for some lecture notes/a book with something like "intro to topological groups" in the title or contents page?

What do you need to know about topological groups

10:02 PM
it's just a group with some topologicality

Compact groups have a Haar measure. T1 groups are Hausdorff. The space of connected components also forms a group

lol I mean I know the definition but I wanna learn a bit more for next semester

Chill dude there's really not a good reason to go reading crap about random topological groups

Ah fair, I really need to know some things about locally compact groups

Sounds like you also know what you need to know
Tell me

10:03 PM
except that they are locally compact topological groups
Well I'll be using locally profinite groups for most of next semester for a seminar on the Langlands correspondence

what does locally profinite mean

open neighbourhoods of 1 contain a compact open subgroup of the group

Great sounds like we're getting wildly specific
I googled introduction to locally profinite groups and got this

hahaha well, I'll be writing about cuspidal representations of GL_2(F) for a local field F and GL_2(F) is a locally profinite group (which I'll probably read at some point in the book)

Great there's 100 pages you can read
It's already way more general than you should spend your time on probably

10:09 PM
Okay, I'm only just starting to "get specific" and I'm used to getting an overview like in a first course rofl
finding good literature is a skill to be learned I guess

Sometimes clicking on the cited by... button

A'ight thanks :D

10:51 PM
@robjohn haha! faint applause

@TedShifrin Thanks. It worked for Newton...

11:08 PM
Hm, so I'm not getting the length of b just from alpha and the method I'm currently using here, so this attempt is false. Now, I can even pull a rather inaccurate arccos(x) approximation out of it, however... what if instead I use a kite with the angle bisector equal to the other two sides towards the origin? I think I'll look into that next!
Still, I just need to be able to find length b or just one of the points and then it'll be there.
In a slightly different but related topic, I've known for quite a while now that you can get exact values of sine and cosine arbitrarily via a recursive reflection process. You just fold an isosceles triangle over and over on one side which is the same as reflecting one unit vector by another.
Of course, to start the process, you need an initial sine or cosine value with the desired accuracy.

@AMDG you will need to know $l$ and $\alpha$. You can't tell $b$ from $\alpha$ alone.

Yeah, so the idea now is to have some norm equal to $\alpha$ which contains $l$ in its norm, and find a way to get to $l$ from $2\alpha$.
Now I have four right triangles to work with...

@AMDG knowing $\alpha$ gives you ratios of sides for a family of similar triangles. You need some side to know which member of the family you have, then you can get all side lengths.

That is the idea with setting the side lengths of the isosceles triangle equal to $2\alpha$, although my earlier idea that I didn't try but dismissed is actually just using the x axis as the side of the isosceles triangle directly as length $\alpha$.
so I suppose we can say that the length of the top right-triangle is just $\alpha$.

I don't see why you are setting the side length based on $\alpha$. can't you just set some side to $1$ or to something in the picture you're trying to match?

11:24 PM
Well the point is to find "exact" sine and cosine.
Functions of sine and cosine that is
I already have found an approximation using exp, and the error function is nice and predictable, but I don't know how to analyze it to remove the error to make the error be zero.
It's 10^-7 accuracy

So you are trying to compute sine and cosine?

0

A resource request. Do you have any good examples of color used in a mathematical research paper? One example I've seen is here, in reference to tensor categories: https://arxiv.org/pdf/1612.02762.pdf Other examples are of knot tricolorability and graph colorability (I'm sure there are loads). To...

Yes, but without an approximation as it were. I want something arbitrary precision where I can compute as many digits of any circular function without the precision being contingent on the number of terms of a polynomial or the like.
I did also find a relation that I can't really make any more progress on: $\cosh(i \arcosh(\cos(x))) = \cos(\arcosh(\cos(x)))$
I can get the exact function, but I can't get rid of arcosh(cos(x)) to make it cos(x)
arcosh(cos(x)) is a trivial, imaginary triangle wave.

How does this help you to compute $\cos(x)$?

How does what specifically?

11:32 PM
any of this. How about knowing $\operatorname{arccosh}(\cos(x))$?
I guess I just don't understand what you are trying to do.

well I mean... the idea is to work from something that we know to something else that we know which uses the circuluar function, then use some combination of trigonometric identities to obtain a circular function from there.
e.g. cos(arcsin(x)) = sqrt(1-x^2)

Do you have an example of what you are trying to do?

Do you mean a specific application, or how I'm going about trying to find circular functions?

just an example so that others can start to understand what you are trying to do

If you're wondering what specific application I need for this: there are a lot of uses for such a function, one of which is a kind of universal hash function, if it's trivial enough, then for use in real-time applications. I'm doing this for the challenge as well as the benefits, but mostly the challenge.
I would describe what I'm trying to do as looking for a closed form, arbitrary precision, real-valued circular function, though, in terms of a general goal.

11:42 PM
So you want to compute circular functions without series or iterations?

Yes, exactly.
And it has to take some measure of angle as its input of course, otherwise, it's no better than computing circle-line intersection.
Again, if I knew how to make a recursive relation into a function, and then use calculus to find the function when the length of the initial isosceles triangle is infinitesimally small, then I'd just do that.
Because that's an iterative method of finding exact values of sine and cosine, but with the restriction that they are multiples of the initial angle.
Here's my "clean" workspace that shows where I'm currently at in terms of an approximation: desmos.com/calculator/cqlhiywinf

@TedShifrin Opinions on this?
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A resource request. Do you have any good examples of color used in a mathematical research paper? One example I've seen is here, in reference to tensor categories: https://arxiv.org/pdf/1612.02762.pdf Other examples are of knot tricolorability and graph colorability (I'm sure there are loads). To...

Alternatively, on this:
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@AMDG The only methods I know require more iterations or terms of a series to compute more accurate approximations. Sorry.

Well that's what I'd expect. :)
But I could still use some help, though, honestly. The number of techniques of analysis available to me are limited and artificially extended by using wolfram alpha.
So far, I've only been able to really get to "one level of accuracy" meaning that I can reduce one of my approximations to a triangle wave approximation, then use a real triangle wave to remove the error, but then I can't get back up by another level which would be "level two".