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00:00 - 14:0014:00 - 00:00

14:00
How long it takes to compute that? Well the same time as for $e^x$ plus the time for the rotation in the complex plane...
Well that's assuming you're not using purely integers.
$\exp(x)$ for some integer is rather trivial.
to get the mantissa, I can just multiply by some power of 10. I then end up with a large integer that has all the digits; I would just have to add the decimal point myself.
I'm only going to indulge you for a few comments until you start going off the deep end again -- what's the issue with computing $e^{ix}$?
Well, it could just be my misconceptions about complex exponentiation...
14:05
Having trouble with exercise 6
I know 2xn is Fibonacci
F_{n+1}
How would you easily compute $(\cos(1) + i\sin(1))^{x}$ efficiently to get something very accurate for real-time applications? That is, on the order of nanoseconds.
Ideally tens of nanoseconds.
$e^{ix} = \sum_{n=0}^{\infty} \dfrac{(ix)^n}{n!}$ has a very fast convergence rate
Is $x$ real? Also this is different from $e^{ix}$.
When $x$ is very small that's nearly $1+ix$
and then you can use repeated squaring
@JoeShmo $n$ should start at 0.
14:08
fixed, thanks
It is equal to $e^{ix}$, no? Since $\cos1+i\sin1=e^i$
Yeah, for $e^{ix}$, I'm using the identity $(e^{i})^{x}$ here because it's simpler.
Well simpler in terms of computing sine and cosine from it.
Could use $(e^{x})^{i}$, but then you need $x^{i}$ which seem to require the complex exponential anyways to compute.
why not just use $e^{ix}$..?
@AkivaWeinberger, yes its just $e^{ix}$
@JoeShmo well, I would, but how many terms do I need to get something like 22 decimal places of accuracy? 100? 1000?
So, ideally, I would have something that can compute $e^{ix}$ in constant time as a rational approximation for all real values of x.
And then ideally by improving the approximation of one or more transcendental constants, I could get more digits out of the circular functions as needed but in the same amount of time regardless of whether I need 1000 decimal places or only 12.
Does that make sense?
Or have I "gone off the deep end"?
More terms of the infinite series definition of $e^{ix}$ means more operations which means more time, not constant time. That's why I don't want to use it.
22 decimals is constant time
14:24
Yes, but it requires a different implementation than 1000 decimal places... I want a function that can give any accuracy or precision, including infinite precision were it possible to have a computer with infinite memory.
But in constant time.
not happening.
Why?
because $e$ is an irrational number. Any numerical algorithm will have to run longer for better precision
Yeah, but I conjecture that any irrational number can be approximated as the ratio of two integers.
Or at the very least, two reals.
To get more digits, I should think it is pretty straightforward... you just improve the approximation by changing only the constants themselves. The number of operations remains constant and therefore the irrational number we are approximating is computed in constant time.
oh, well in that case use your conjecture to compute $e^{ix}$ in constant time.
14:35
Consider for example how I implement $\exp(x)$ for integers. We know that $2^{\log_{2}(e)x} = e^{x}$ for $x \in \Bbb{Z}$, right?
@AMDG That's obvious, $\pi$ is approximated as a ratio of two integers by $3/1$, $31/10$,$314/100$ etc., and you can do the same for any irrational number. But the complexity of dividing two integers is a function of their length, not a constant
Also computing better approximations is not constant time, in my silly example computing better approximations is just as hard as computing the irrational number directly
$2^x$ is trivially implemented as bitshifts as 1 << x, and $log_2(e)$ is a transcendental constant, but I can just take out my calculator and compute $log_2(e)$ and choose some arbitrary number of digits that is sufficient to give an integer approximation of $\exp(x)$. In theory, this means I can compute $a^{b}$ in constant time since I now have $\exp(x)$.
@AlessandroCodenotti Sure, but it doesn't necessarily have to be a function like that which computes $\frac{a}{10^{b}}$. There might be some function which can generate the representation of those quantities one or more digits at a time.
Just like how computers primarily manipulate binary strings to perform real computations.
40 mins ago, by Akiva Weinberger
user image
Got the answer
(Ex 6, 3xn)
For odd n, the answer is 0
Perhaps there is a base 10 string function that can compute circular functions easily.
For even n, it's the integer closest to $$\dfrac{(1+\sqrt3)^{n+1}}{2^{n/2+1}\cdot\sqrt3}$$
14:51
The other thing I'd like to point out is that if we don't need angles, sine and cosine are trivial. The reason is obvious, I'm sure.
Also, I don't know precisely how far I can get with this, but it would seem that I can get rational approximations of circular functions somehow by just dividing the area of a circle evenly into circular sectors and ensuring one of the radii is on an axis.
I missed most of the conversation but you can do $1+ix/2^n$ squared $n$ times
for large $n$
Thank you, I will keep that in mind.
I don't know... I mean I could just settle for my cosine approximation that gets to 10^(-7) in accuracy and call it a day for my standard library, but I like the challenge, as well as the clear benefits, of finding such a definition for the circular functions as I have just described.
Everyone benefits! :D
Alternatively, we could develop screens that implement polar coordinates in an analog format instead of using square grids of pixels, in which case I wouldn't really care about circular functions.
Polar Coordinates master race. Euclidean plane go burr.
15:09
can you link us to your library on git
I guess, but... it's far from complete and I don't have a demo available yet.
Could take weeks, months, or years more.
The only interesting thing to look at right now is bit_math.h which really doesn't have much.
does anyone know of a more elementary argument to prove the following? Let $a,b,c,a',b',c' \in \mathbb{C}$ and suppose $f_1 = ax+by+c$ and $f_2 = a'x+b'x+c'$ are non-constant, then the zero locus of $f_1$ and $f_2$ are equal iff $(a,b,c) = \lambda (a',b',c')$ for some non-zero complex $\lambda$: 1. $f_1,f_2$ are irreducible, so by Hilbert's Nullstellensatz, they have the same zero loci iff they are associate
perhaps there is a way to do it by just thinking about row-rank of a certain matrix?
sorry, there is a typo in the above, $f_2 = a'x +b'y + c' $ is what it should be
okay, this is the argument I can come up with via thinking about a certain matrix, but I think it is a bit ugly: since the solution space of $\begin{matrix}a & b\\a'&b'\end{matrix}\right)$ where by that I mean that matrix times some vector $= (c,c')$ is such that its space of translations has dimension $ 1$, since $ax + bx + c = 0$ and $a'x + b'x + c' = 0$ have the same solutions
, and are non-constant, the rows in that matrix are multiples of each other and so $ax_0 + by_0 = -c = \lambda (a' x_0 + b'y_0) = -\lambda c'$
besides these two arguments is there anything simpler
15:28
@AMDG Actually I'm designing the language of the metacompiler right now which is directly related to my studies that I mentioned about composition and decomposition of algebraic expressions. Current idea that I intend to implement is defining behaviors implicitly by relating three or more states.
Huh, I did not know that pings me.... lol. I was just trying to provide a continuation.
so what happens when a non-catholic wants to use catholicus?
is it a runtime, or compilation error?
It's quite simple: you use it according to reason. I actually haven't decided on the name of the language. That was the original name.
I don't know know what you're talking about see
I don't know what you mean by runtime or compilation error.
test
- - - t e s t - - -
15:33
how did you do that?
thanks
It's ---
lol
Fun fact, "bureau" is an anagram of the first six letters of "U R BEAUtiful"
🤣
15:35
anyone know a simpler way to answer my question? it feels like it should be some really basic linear algebra
It just means that when someone who is not Catholic wants to use my software, you just follow the laws of the Church strictly with regards to that software insofar as you are able which is informally summarized as "just using reason" and tersely as "do not use my software to do evil".
 
2 hours later…
17:50
Hey chat.
Hey Lucas
who is this chat you speak of and where is he
I forgot how to prove that $\mathbb Q (\sqrt[2]2, \sqrt[3]2, \ldots)$ is an algebraic extension over $\mathbb Q$. I tried to show that each element $\alpha$ belongs to some $\mathbb Q(\sqrt[n_1]2, \ldots, \sqrt[n_k] 2)$, but I didn't succeed. Any ideas?
You're right though; any element of that field is a finite linear combination of the elements $2^{1/n}$, $n \geq 2$.
By definition of span in linear algebra.
@BalarkaSen you meant $2^{m/n}$?
17:55
Yes, thanks.
Oh man, no fun at all. I don't know how to describe that field besides the definition
"the smallest field that contains $\mathbb{Q}$ and every $n$-th root of 2"
> I don't know how to describe that field besides the definition
Story of my life.
It's union of the fields $\Bbb Q(2^{1/2}, \cdots, 2^{1/n})$, $n \geq 2$
Oh my God, that was dumb.
Thanks @BalarkaSen. Now it's obvious :|
Nah, this one is genuinely confusing the first time. Glad to help.
17:59
Where in the world did the notation tg(x) become a thing for $\tan(x)$?
@BalarkaSen I'm kinda sad. I mean, I studied Galois theory. I'm using this one as an example of algebraic (but not finite) extension in a conference.
Forgot 90% of what I've studied, like whole part that relates normal extensions and normal subgroups in Galois groups
Good time to revisit, maybe, yeah. It's very easy to forget Galois theory. I religiously forget and relearn it every 2 or 3 years.
And it was the first actual math I learnt as well.
Galois theory is hard and nontrivial, even if it looks elementary
I only learn Galois theory when I have a duel to the death with someone the next day
Hahah
No need to revisit again
rip our poor boy
18:03
I have never learned Galois theory and probably will never study it formally.
Can't forget what you don't know. Checkmate.
AMDG is a flat earther and a creationist maybe
perfect attitude
Imagine using inductive reasoning
But will always be remembered not knowing when reading word Galois?
If F and G are two sheaves over a top space X, what's the comparison between the fiber of Hom(F,P) at $p\in X$ and Hom($F_p$, $G_p$) (with $F_p$ and $G_p$ being the fibers at p of $F$ and $G$) ?
I'm having trouble seeing what $Hom(F,G)_p$ is
Yeah its not going to have an easy description in general
18:08
What can I say? I'm a doer and a thinker. I only want to know as much as I need to know, nothing more, nothing less.
How do you know wether you need to know something or not ?
Or do you not need to know that?
You obviously have to know what you need to know in order to know what you need to know.
I guess you could ascend this chain of reasoning and ignore everything
Stalks behave well with tensor products. Hom(F, G) = F^* o G canonically when F is free
So I expect if F is quasicoherent everything goes through
I've been told there's an injection from one to the other but I'm not sure which
18:10
@AMDG Was just kidding I understand nothing about Galois. I am also more practical load my memory only with essential routines compiling takes longer though!
Hom(F, G)_p = Hom(F_p, G_p) if F is has a 2-step free resolution I'm convinced
I don't know anything in greater generality
If I look at $F_p\from F(U)\to^{\Phi(U)}G(U)\to G_p \to Y$
How do I do a reverse arrow ?
What are you trying to do?
There's a natural map Hom(F, G)_p -> Hom(F_p, G_p)
Why ?
@OOOVincentOOO You must be using C++. The compilation for us C programmers who don't use C++ see a difference on the order of milliseconds. ;)
18:12
That's my question
That's obvious
You just restrict over some open set U of p, localize the map at p
Then take a limit over U
You know what you need to know based on what path God has placed you on. I am a programmer, so I need to know all about computers and software. God usually speaks with ideas, not words.
temple os
Why is this a homomorphism ?
I get that reaction a lot lol
18:16
@Astyx thats just a small check innit
am i seriously missing something
You're probably right it's obvious
I'm just not familiar enough to really understand what I need to check
If I take $\tilde f$ in $F_p$, I take a representative $f\in F(U)$, take it to G(U) with $\Phi(U)(f)$ and take the limit at p $\tilde{\Phi(U)(f)}$
(the last tilde is over the whole $\Phi(U)(f)$)
@BalarkaSen TempleOS is a cool concept, but it obviously isn't practical for anything beyond a hobby to mess with. I'm working on a metacompiler that works with "NAND trees" that represent patterns, and the language will compile to NAND trees.
So this gives me a map $F_p\to G_p$ (and I know how to check it's well defined)
You can represent anything and everything as a decision tree. That's the principle this whole thing works on. You give it an input pattern, a conversion pattern, and it spits out an output pattern. Simple yet powerful.
Why is it a homomorphism ?
18:27
@AMDG, I think your questions would be better answered in the computer science forums
I don't have any questions about CS, I was just explaining what I'm working on.
Now, what does $\Phi$ usually mean in something like this: $\tgh(\Phi)$?
you might find answers to the questions you are asking in the CS circles
Yeah, you're right.
The notation is encountered here on page 2: arxiv.org/ftp/arxiv/papers/1102/1102.1563.pdf
@Astyx xymatrix would fit well
 
2 hours later…
20:14
Can someone prove this? "Any planar graph will always have vertices with degree <6"
@Sophie Can you relate the number of edges to vertices in a graph?
but what if the graph is infinite?
does it still hold?
No
Consider the graph whose vertices are the integer points
There are two horizontal edges and vertical edges emanating from each point as well as an up-right diagonal edge
Then every vertex has degree 6
That's a funny way of saying it's a triangular lattice lol
"Triangular lattice" strikes me as very vague
But if I communicated a picture that's great
20:27
yeah I got it
I'm actually not sure if you can get unbounded min-degree for infinite planar graphs
It sounds really hard to get over 6
do it on H^2
geodesic tessellations with as many guys coming at a vertex as you want
Makes sense
@Sophie Only if you want to preserve the geometry of the ambient in some sense
planar graph is a very weak notion for infinite graphs
20:52
Anyone has any clue on how to solve $(y^3+ty^2+y) dt = (t^3+yt^2+t)dy$ ?
Hello everyone, I am stuck on an exercise in Group Theory and am looking to find how this thing is called:
Let N be the normalizer of a finite group G, let K be an arbitrary subgroup of G.
So then we have K*={Nk : k in K}.
Is there a name for K*?
I know I need to find an integrating factor to make it exact, but any attempt I've made has been hopeless.
@Everstudent well it's isomorphic to NK/N which is also isomorphic to K/(N cap K)
I tried multiplying everything by factor $u(t)$ and $u(y)$ but nothing I could solve appeared. Multiplying everything by $u(t,y)$ leads to a partial differential equation too hard to solve.
@Everstudent it's also the image of K under the canonical map G -> G/N
20:56
@LeakyNun
Wait, I made a mistake in my definition:
N is the normalizer of H which is a subgroup of a finite group G. K is an arbitrary subgroup of G.
Then we have K*={Nk : k in K}.
everything I said applies
So far I've proven that there is a bijection between {Nk : k in K} and {aHa^{-1} : a in K}
But I do not see why |K*| divides |K|
The chapter with the exercises has not defined quotient groups yet, so I am thinking there should be way to do it without them.
Basically I do not see why the number cosets of N of the form Nk should divide |K|
I mean, N is not a subgroup of K
I can probably use cosets of the form (N cap K)k and then it will work, but I wonder if there's a simpler way
11 mins ago, by Leaky Nun
@Everstudent it's also the image of K under the canonical map G -> G/N
order of image divides order
@LeakyNun true, so I get your reasoning, but it's in the chapter on Homorphisms, so I shouldn't need to use Quotient groups to solve it. So I am wondering if the author had something else in mind
I mean either that or he put the exercise in the wrong chapter, lol.
21:19
is (-infinity , -1/2] u [1/2, infinity) open?
no
21:50
in Constructive Feedback, 3 mins ago, by Shaun
The following answer of mine was downvoted. I'm not sure why.
-1
A: Do elements $x^2$ and $y$ commute in group $G = \langle x,y \mid x^4, y^{10}, xyx^{-1}y^{-3} \rangle?$

ShaunNo. gap> F:=FreeGroup(2); <free group on the generators [ f1, f2 ]> gap> rels:=[(F.1)^4, (F.2)^(10), (F.1)*(F.2)*(F.1)^(-1)*(F.2)^(-3)]; [ f1^4, f2^10, f1*f2*f1^-1*f2^-3 ] gap> G:=F/rels; <fp group on the generators [ f1, f2 ]> gap> Display(G); generators = [ f1, f2 ] relators = [ f1^4, f2^10, f1...

There's a high possibility it's downvoted because someone has seen it's not a piece of "TeXed" mathematics and hasn't bothered to check that it's gap code
Thank you, @EdwardEvans.
I mean, that's not a definitive reason lol
I hope something like that is the case. What could I have done better?
@EdwardEvans . . .
Idk, something like "Here is some ***--^:):)>> $\huge{\textbf{gap code}}$ <<(:(:^--***that refutes your claim:"
2
22:57
Wow, those are some big small letters.
23:55
If $f:[0,1]\rightarrow \mathbb{R}$ is such borel measurable and that $f>0$ and $\int_{0}^{1}f^n dx= a$ for each n, for some $a\geq 0$, what can we conclude about the function f?
we're supposed to relate this to the notion of almost surely functions
but i'm not sure how
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