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mme
mme
00:00
Feels a bit Pascal's Wagery except you remove the conscious cynicism
my immediate family is roman catholic, which in these days puts the same amount of stress about what people actually believe in. my wife is basically atheist. my mom goes to mass multiple times a week but probably doesn't buy into the full doctrine.
i guess it's a bit like catholicism and the different saints
when her uncle died we lit a candle. it doesn't need to be too complicated.
mme
mme
catholics famously don't read the bible as much as protestants, which in an era with decreased churchgoing probably doesn't breed religious fervor
mass was in latin up until 1965
00:03
that's a weird thing. i used to know a priest who was really into trying to get catholics into reading the bible, with no luck.
@leslietownes very nice, I've hard sword of doom recommended by a friend just a while ago; guess I should watch it
thorgott you should watch it tonight. i can recommend other samurai movies.
kaneto shindo's "kuroneko" is another good one.
and more recently takashi miike's 13 assassins.
mme
mme
13 assassins is one of the better samurai movies since kurosawa
it's the best i've seen in ages.
mme
mme
which isn't to say it feels like you're watching yojimbo
it does really well perhaps because it's not just a kurosawa script with updated visuals
00:08
toshiro mifune is in sword of doom. he's the swordsman who drives the guy crazy because he realizes he isn't the best.
mme
mme
i haven't seen it but i'll add it to my list. my wife doesn't usually go for things which feel actiony so it'll be a hard sell
it's a fine line. it can't be an action flick but it can't be boring. but so many good movies are very boring
my wife hated it until about halfway through.
mme
mme
eg stalker is beautiful but you do have to be willing to sit there and watch the grass grow for 5m at a time
i love samurai movies because they tend to emphasize artistry over violence, and also emphasize nonviolence while sometimes being very violent. the post-WW2 generation of japanese directors was understandably very cynical about state authority.
mme
mme
that reminds me that there's a great miyakazi interview somewhere in which he poo-poos abe for wanting to remove the pacifist bit from the constitution (as you might expect, this was a really quite unpopular comment in Japan). they asked him why and he says, more or less, "we are very bad at war. no further comment."
too bad Ted left as I popped in. Hi, @Ted. I'm off aagain soon.
00:13
there are so many really good violent japanese films that are, in substance, about the insanity of violence.
shohei imamura is one of my favorite directors. his film 'vengeance is mine' is unbelievably violent and explains nothing about the main character's motivation. they asked him in an interview, 'vengeance for what?' and he just didn't answer the question.
he did a weird post 9/11 short film about a man who was so traumatized by war that he thought he was a snake.
it's on vimeo. vimeo.com/111652156
mme
mme
that's nice
i've had my fun for the night. enjoy
I'll have to pass on tonight, it's already past 2am and I'm stuck on a stupid computation
@leslietownes I should try that, too, maybe that'll warm me up to Miike as I sadly wasn't a fan of Ichi the Killer
ichi the killler is extreme. i think it is his worst film.
just too much going for shock value.
he's also made a ton of good movies.
@leslietownes I watched Kitano's Violent Cop recently, that very much falls in line with this
oh, i love that one. and kitano in general.
my wife loves violent cop, weirdly.
00:35
yeah, the juxtaposition between the extreme bouts of violence and how serenely shot it is works really well
emotionally empty in the best way possible
I've decided to watch Kitano's filmography in chronological order, looking forward to Boiling Point soon
with Ichi the Killer, I didn't really mind the extremity. it just felt like a tonally disjointed experience to me. so many characters with bizarre quirks, but nothing to really latch onto. some of the individual scenes are pretty cool, but the whole experience didn't get me invested
with kitano i like hana-bi and his zatoichi remake the best. with miike i like 13 assassins and audition. although i would not recommend the latter to people who are not horror fans.
what are your favourite movies? japanese or otherwise
my wife for example loves 13 assassins but i would never dream of showing her audition.
shintuku, some of the samurai movies mentioned above. in french film i like le corbeau and army of shadows. in US/UK films, the third man, make way for tomorrow, and night of the hunter. and the original suspiria from the late 70s. it has been remade.
i watch a whole lot of horror movies that are terrible just because i like the genre.
stanley kubricks the shining is also really good. i've probably seen that movie 20 times.
i know none of them, stuff to add to the list hehe
ah except the shining
woah, all of them are pre-1980
big statement
they stopped making good movies.
kuroneko is a really good movie.
i think the only more recent movie that i like is henry: portrait of a serial killer. which is just a remorseless film about serial killing. the guy who commissioned it wanted some exploitation flick and got something else. michael rooker has gone on to star in the walking dead and is a talented actor.
but the subject matter is very unpleasant.
00:53
heheh
leslie, that's from 1986
"recent"
i'm old, OK?
i just love the idea of commissioning what you think is going to be a teen slasher flick and what you get is someone remorselessly killing people for almost two hours. that must have been a fun conversation.
some of the pixar movies are OK. i really liked ratatouille.
for the sake of balance, I'll recommend a recent film I like: One Cut of the Dead
is that streaming somewhere in the US, thorgott?
00:56
japanese zombie comedy
difficult genre
i'm not familiar with the director but i will look into it.
let's see how the crowd reacts
there's been more recent stuffed that i liked. 28 days later was really good.
and the david copperfield remake from a year ago.
i tend to lean toward very violent movies, and my wife tends to lean toward romantic whatever. we don't have a lot of common ground.
I think it's on Amazon Prime
it does look like it's streamable on amazon prime.
01:00
looks like good reviews
@leslietownes the middle ground is Bonnie and Clyde?
yeah, my wife did like that one.
a "recent" one I liked is short cuts
robert altman is a genius. have you seen 3 women?
3 women and McCabe and Mrs Miller are two of the best films of all time.
i saw the two of them at a double feature at the castro, pre-pandemic. i miss hanging out there.
all added to the list
i should do that more often, choose movies based on director
oh he's the guy that did MASH
01:21
@leslietownes There’s a Castro in Long Beach?
 
2 hours later…
Bob
Bob
03:13
Hi
is anybody around?
03:25
finally! the day when knowing how to count in binary is useful has come
you can make sure you're listing all elements of a power set counting by in binary
04:15
-1
Q: What is the usage of studying trigonometric equation?

S.M.TI have solved many Q regarding trigonometry in which we have to equate LHS to RHS. For example , $\frac{1}{\csc A-\cot A} - \frac{1}{\sin A} = + \frac{1}{\sin A} - \frac{1}{\csc A+\cot A}$ and many more. As I see this equation and if I try to visualize this. Let us take a triangle with sides 3,4,...

Please share your answers regarding my Q.
@S.M.T an example of the usage of sec, csc, and cot, is for integrals that are solved using trigonometric substitution
for example $\int \frac{1}{\sqrt{x^2+1}} \ dx$
you will very likely encounter an integral similar to this one, in any career that uses calculus
greetings fellow math-tickians
04:36
Given $f$ differentiable everywhere in $\mathbb R$ such that $f'(x)\le l\lt 1$ for all $x$, where $l$ us a fixed real number. I'm looking for an example of such $f$ which satisfies the aforementioned hypothesis and yet fails to have any fixed point.
I thought $f(x)=x-10e^x$ could work but almost immediately realized that we must bound $f'$ by $l$, which is $\lt 1$.
Had the hypothesis included $|f'(x)|\le l\lt 1$, then I would have had no problem in proving the iterative sequence $(x_n)$ defined as $x_n=f(x_{n-1})$ for all $n\ge 2$, given $x_1$, a Cauchy sequence.
04:55
I managed to come up with a counterexample: Let $l=\frac 12$, $f(x)=-10x^4+\frac 12 x -10$ then $f$ has no fixed point.
no this example is wrong :'(
05:23
Topology masters?
0
Q: $f^{-1}(C)$ is not a finite set, whenever $C$ is a closed subset of a topological monoid, and $C$ is saturated as a set (but not necessarily a monoid)

I'm an alien Im an eagle alienLet $X, Y$ be topological spaces on the ring of integers such that $Y$ is a topological monoid (multiplicative, on $\Bbb{Z}$) and let $C \subset Y$ be a closed set that is multiplicatively saturated: $xy \in C \implies x,y \in C$, but not necessarily closed under multiplication. In other words $...

@Euler2 can you look at my post? Someone downvoted...
For realz. Makes no sense
it's a valid question I had
@I'manalienImaneaglealien um.. the new policies are a bit too strict
@Euler2 I know right
they want motivation, context and/or work
05:37
you being Euler, they must be so futuristic to you :)
otherwise your question can end up in CURED
The motivation is twin primes, but no one wants to hear that so I left it out
The work? I'm asking for how to approach it essentially, it's such a high level question that of course I did a lot of work on it
they want it
I'll try to put in an attempt. Thanks @Euler2 you're my math hero
nay
ok
a good math hero should be... a bit 'experienced'
I will be an expert when pigs fly
05:48
You're speaking obtusely
What's your angle
hehe
You're stoned, like me :)
yes and no
The duality of the universe
^_^
duality is my primary characteristic
or... secondary
06:06
@Euler 2
Say you have a topological monoid $Y$
and a continuous map $X \to Y$
call it $f$
and this certain closed subset $C \subset Y$ while infinite, we suppose that $f^{-1}(C)$ is finite!
Also, the set $C$ is such that $xy \in C \implies x,y \in C$ (opposite of multiplicative closure, called "saturated")
Is such a thing possible?
I'm assuming it's not, so I look for a proof by contradiction
But I need some additional conditions most likely.
If you can prove that there's a contradiction, then you've proven twin primes with my setup which I haven't posted yet, but the topologies are:
$X = (\Bbb{Z}, \tau_0), Y = (\Bbb{Z}, \tau_1)$ where $\tau_0 = $ arbitrary unions of submonoids (multiplicative) are open, and $\tau_1 = $ arbitrary unions of ideals are open.
The map $f: X \to Y$ is $f(n) = n^2 -1$ and I can easily prove that it is continuous
Another continuous map out of $Y$ is $\Omega : Y \to (\Bbb{Z}\cup \infty, \tau_2) = Z$ where $\tau_2 = \{ [a, \infty] : a \in \Bbb{Z}\}$ are all the infinite intervals. $\Omega(q_1 \cdots q_r) = r$ is from number theory, $q_i$ are all prime.
beware
this is what leslie says about me
15 hours ago, by leslie townes
you're the non-expert who offers the view of the common man on the proceedings.
Since $\Omega$ is continuous as well, we get that $(\Omega \circ f)^{-1}((-\infty,2])$ is infinite is the same thing as the twin primes.
But $\Omega^{-1}((-\infty, 2]) = C$ is our closed set
and it is saturated
ask someone worthy or the version of me in 2 years
I'm trying to do a topological proof of infinitude of twin primes
06:15
That's all one needs to know (the above) to make it work
However, I don't know if it will work
Furstenberg's infinitude of primes is different since it uses cosets of ideals as open sets
in fact doing so gives you a topological ring
*a basis for open sets
So I've reduced the question down to the first few lines of the above paragraph regarding a saturated closed set
idk what has happened to me maybe i am ignorant but whenever I see something which can 'prove' a conjecture i disagree
You're probably right
doesn't hurt to try though
 
2 hours later…
08:30
It hurts when you think about same proof for 3 hours straight and you start to doubt that you are not made for mathematics. You feel like you are worthless and start to be suicidal and after you give up then suddenly you know answer and you will start to realize that is so simple. This repeats all the time. All of the things a person goes through in life cause suffering and they cannot do anything about it. Instead, they have to accept that it is there.
Most suffering is caused by a tendency to crave or desire things.
People should not be too focused on wanting many different things as the enjoyment won’t last.
People must try to stop craving as much as they can in order to work to end suffering.
Before I am banned I will teach you how to end suffering :u:
Right action ,speech, livelihood, mindfulness, effort, concentration, understanding and intention.
OK I am ready to get banned for 30 min for being off topic :-)
09:24
damn
redemption before one's death
almost feels biblical
good morning
09:52
$S=\bigg\{\begin{pmatrix}
e^a & e^b \\ e^c & e^d
\end{pmatrix}: a,b,c,d \in \Bbb Z : e^{ad-bc}=e\bigg\}$
$\begin{pmatrix} e^2 & e^1 \\ e^5 & e^3 \end{pmatrix} \begin{pmatrix} e^1 & e^1 \\ e^5 & e^6 \end{pmatrix}$
$=\begin{pmatrix} e^7 & e^8 \\ e^{20} & e^{23} \end{pmatrix}$
the first row times the first column is done as: $e^{2(1)+5(1)}$
so $S$ under this operation forms a group right?
10:55
This is the same thing as $S=\bigg\{\begin{pmatrix} a & b \\ c & d \end{pmatrix}: a,b,c,d \in \Bbb Z : ad-bc=1\bigg\}$ with the usual multiplication as group operation
11:10
@muad long time no see my friend
11:45
@BBischof what is google wave?
 
1 hour later…
13:14
@EvilJohnRennie if you're suicidal after struggling with a problem for 3 hours, it's not the mathematics causing your suicidality
 
2 hours later…
14:49
eesh, yeah. i recommend therapy or religious guidance of your choosing. unless you're using that as a figure of speech.
in math i was constantly annoyed by whenever i had a lemma or subproblem it was never easy. it wasn't like the stuff in textbooks. it was frustrating but did not cause me to doubt myself.
15:26
@EvilJohnRennie Question: what is off-topic in chat?
anything to do with mathematics.
@leslietownes I am hoping it is hyperbole
@leslietownes pretty much, from what I can tell ;-)
yes, it is concerning. but people do use that metaphorically.
15:59
I am always off-topic.
day care is off this week so it's complete anarchy at home.
Sounds no different from your usual.
she has this toy truck and she's trying to see how far she can push it. she has had some good 20-30 foot runs but mostly she crashes it into the wall and is mildly damaging the paint.
we've resigned ourselves to a lot of mild property damage and won't bother fixing it until she is old enough to do better.
She’s going to be literally hell on wheels.
she's a lot. she was taping her drawings to the wall and saying "i'm just making this secure." i don't know where she gets that.
she turns 3 in october.
she's going to go to the zoo today. according to her, they have three different owls and an ocelot and a kit fox.
i'm not sure about the kit fox. they do have a barn owl, a great horned owl, and some burrowing owls.
she's obsessed with owls for some reason.
16:14
Has she read Winnie the Pooh? Owl is the wisest by far.
oh, that's a great suggestion. she mostly looks at pictures of birds. we have a book with like 50 owls in it, and another book with 100 different birds in it. she loves ID'ing them.
it's often difficult for her to follow a narrative but she can identify things.
She's mature enough to get the Winnie the Pooh stories.
At least partly.
None of this Disney crap.
some of her great aunts have given us disney stuff and we donate it to charity.
You only get eeyore when you're older
it's in keeping with the philosophy of her day care, which is, no branded characters.
16:17
I've always been Eeyore.
"Nice day today, isn't it Eeyore?" ... Sullenly, "If it doesn't rain."
yeah, i'm eeyore too.
Oh, not again!
it keeps happening.
16:19
who doesn't love Eeyore?
her great aunts also give her princess stuff and things that are strongly gendered toward being pretty. we donate that to charity too although maybe i should just throw it out. it's weird to be like "not for me, but maybe for someone else's kid."
I don't see anything wrong with the latter. It's a shame to throw out stuff that someone might really use.
that's true. if a kid can't have toys unless they're free, it's different. we can pick and choose.
@Thorgott Eeyore doesn't love Eeyore :(
16:21
We won't talk about how Owl "found" Eeyore's tail and turned it into a doorbell.
my grandmother had a ryhme. the wise old owl lived in an oak, the more he saw the less he spoke. the less he spoke the more he heard, we should all be like that wise old bird. i thought it was a nice rhyme but later realized she was just telling me to shut up.
We need your grandmother in here all the time.
she would be 111 if she were alive. she was a character.
she also said i should be a lawyer, again, hitting on the theme that i talked too much.
she's buried in a beautiful cemetery in massachusetts that has people from the 1600s buried in there. i think it's not open to more people anymore. i visited when i was in law school, and a groundskeeper said 'we're closed today' and i said 'my grandmother is in here,' and he said 'really?' he couldn't believe it.
it's mostly like weird pilgrims and stuff in there
there was this weird genre of doing extremely creepy skulls and stuff on tombstones, in the early days. nothing like what you'd want to see on a tombstone
Where is the cemetery?
franklin, MA.
16:31
dinner the other night had mashed potatoes, boiled potatoes, french fries & potatoes anna as some of the sides. pure heaven.
nothing stereotypical about any of this.
or the fact that i would have loved that.
conan obrien had a story once where his family was fighting and someone threw a potato at someone else and it flew outside the window. and he thought, how could you be any obviously more of a stereotype.
OMG, copper, that's too much.
to be fair, it is no longer typical. mostly a nod from my sister in law to me :-)
16:35
ah, about 15 miles south of where I went to high school, @leslie.
so you're going to be 10 pounds heavier when you return, @copper.
my whole wife's side of the family was around franklin and later framingham.
Framingham is next door to Natick, where I was.
my aunt had a lot of friends in natick.
Probably after my time.
she's also out of her mind now, so who knows who those friends might have been.
it's a very beautiful part of massachusetts.
16:39
Oh, maybe she is about my age, then.
she's 78.
Oh, wow, ten years older.
There's a big age gap there — she must have been a much older sister.
i dunno, my mom is 76. we have babies very late in my family.
Oh, indeed.
I was a late arrival, too. My mom was 30 when I was born, 38 when my sister was.
my daughter was born when i was 38 and my wife was 36. it's much more common now than it would have been in those days.
they put a flag on all of the maternity documents, "AMA" meaning Advanced Maternal Age. i think they could avoid that. there's already an age field.
16:53
$(z_0 )^{2n+1} =1 $
Bob
Bob
I know somebody who is applying to grad school in Comp Sci
He has a learning disability
He thinks that will help him get in
that does not seem right to me
So , $z_0^1=z_0^3= z_0^5=....=1$ what else am I missing..?
Where are you getting that nonsense?
Bob
Bob
17:09
@TedShifrin Is your question for me?
No. For @Rover.
Don't forget you're working with complex numbers!
These are crazy questions, though. Someone has nothing better to do than to make up these artificial questions.
$z_0= cos(2\pi /2n+1 )+isin(2\pi /2n+1) = e^{i2\pi /2n+1}$ so, $z_0 ^{2n+1} = e^{ i 2 \pi} =1$
Yes, I agree that $z_0^{2n+1}=1$, of course.
And if we put n=0,n=1... we get those result don't we..?
$n$ is fixed.
17:18
Ohhh ...is it ? Okay.
You should draw some pictures (say for $n=1,2$) of points on the unit circle.
You need some understanding of roots of unity to play with this problem. It is sneaky. I understand what's going on, but I don't see the last stuff.
i'm also struggling with it.
Here's the first thing to understand. $z$ is pure imaginary.
$z$ is pure imaginary...?
Yes.
Hint: What is $1+z_0+z_0^2+\dots+z_0^{2n}$?
17:22
Zero ?
Right.
So now see that $z+\bar z$ is that sum.
But I don't see what's going on with $2z\pm 1$.
As I said, someone has too much time on his hands making up these questions.
@TedShifrin Really? But, I got the answer to question I think...
Maybe you found a trick I'm not seeing yet.
Yeah , got it.
@TedShifrin From this only.
I'm trying to compute it directly for $n=1$. I get $0$.
17:29
I too got 0
How does knowing $z_0$ is an $(2n+1)$th root of unity help you with $(2z-1)^{2n+1} + (2z+1)^{2n+1}$?
$1+z_0+z_0^2+\dots+z_0^n +z_0^n(z_0+z_0^2+...z_0^{n})$=0
Oh, that's too obvious. Good for you.
1+(z-1/2)+$z_0^n$(z-1/2)
Ya
OK, I still don't see it.
17:37
1+(z-1/2)+$z_0^n$(z-1/2)=0
So, $(z+1/2)=-z_0^n(z-1/2)$
Your dollar signs are all messed up.
Oh, cool. Now I see it. Good job. So why are you asking us if you've already solved it?
No, I solved it after you gave that hint:
Ohhh. Great! Well done!
With problems like this, it's very hard to see the right trick. I get a C on my effort.
 
2 hours later…
19:37
@robjohn , when you wrote your answer, were you aware of the fact that $hu=t-x$, i.e. the variable $hu$ does not depend on $h$ at all? This may prompt the suspicion that the limit in equation 3 should be with respect to $t$, however, this would probably make little sense since the Taylor expansion is centered at $t$. The other independent variable, $n$, does not seem to be a function of $hu$.
19:52
@schn. You mean the limit should be $t \to something$ please?
$t\to x$ probably.
@schn. Please read this, "When we see o((hu)m), we need to determine the independent variable(s) from context. In the question, the integration is with respect to u, so the independent variable might appear to be u. However, the article is looking at letting h→0 in the estimate, while u is integrated over R, so"
@robjohn said that we first need to determine the independent variable as this is the one that will go in the limit $t ~or~ h~ or~ u \to something$. He inferred that it's $h$ from the answer as I see it.
The answer was privided in the question as it seems
20:15
@Avra , you're maybe right. By $o\!\left((hu)^m\right)$ I meant what @robjohn specified in the clarification, i.e $f(hu)\in o\!\left(x^m\right)$. However, is this function also an element in the set specified in equation 3?
20:26
@Avra , which answer are you referring to?
20:39
@schn $hu$ is not a variable. It is a product of variables. If you can make a substitution $v=hu$ without violating any other bindings or scoping, then perhaps that might be useful. However, that needs to be done based on the context of the problem.
@schn $o\!\left((hu)^m\right)$ is a function of $h$ and $u$ so that $\lim\limits_{hu\to0}\frac{f(h,u)}{(hu)^m}=0$
it might be that the only appearances of $h$ and $u$ in $f$ are as a product, but that is not clear from simply specifying $o\!\left((hu)^m\right)$
If there are other variables involved, the limit given may be uniform in those other variables or it may not. That is dependent on context.
ALL that is guaranteed is $\lim\limits_{hu\to0}\frac{f(h,u)}{(hu)^m}=0$
For example, $h^3u^4\in o\!\left((hu)^2\right)$ and $h^3u^4\in o\!\left(u^3\right)$, but not uniformly in $h$ if $h$ gets big, but context may indicate that $h$ is bounded and so it may not matter. (here we are talking about little-o near $0$)
 
1 hour later…
22:03
Concerning my "nearly there", "almost an algorithm" for reciprocals: as much as what I've likely rediscovered is a true identity for integers of a limited range, it is also just as far away until the sum in the denominator is fully eradicated through algorithmic decomposition into fractions (not necessarily Egyptian fraction expansion) or some other means.
I expect that I should be able to rearrange the terms such that the complexity remains O(M(n)) and that its implementation will have the same cost as its multiplicative counterpart. I apologize if I have mislead anyone.
22:45
Oh, wait a sec... there's a laurent series that seems to converge quickly for $\frac{1}{a + b}$. How did I not see this sooner? This trivializes computations to parallel shifts and adds which could easily beat HW were it not for the fact that I don't have enough ILP... bruh, so you're telling me I've been wasting my time this whole time? Oops.
Where $a=c\operatorname{floor}\left(\frac{x}{c}\right)$,$b=\operatorname{mod}\left(x,c\right)$, and $c = 4$ chosen so that a 2-bit lookup table of qwords fills half a cache line. Values of x less than $c$ seem to be undefined, but otherwise, it works fine.
I've got a nice soft question for the community. Any number theory researcher could answer: math.stackexchange.com/questions/4226153/…
i think you're right that most of the low hanging fruit has been harvested. it is tough when working in a field as old as number theory where everyone has tried everything before.
what is the newest field of mathematics?
23:00
one of the commenters touches on category theory, there has been a lot of development in that area. i'm not sure how useful it is, divorced from applications, but a lot of people seem to be getting grants out of it.
adjacent to my field, quantum computing was this huge thing.
symplectic topology is really new I think
it was basically the rule that any paper with quantum in the title would be published, usually by a journal with quantum in the title. i'm a cynic. most of it was late 1800s recycled linear algebra.
i think there are aspects of algebraic geometry that are significantly being developed now, but this may just be an impression or vibe from the people i know who still practice mathematics.
functional analysis is functionally dead as a discipline. incremental results and tiny improvements only.
23:17
What is functional analysis?
@leslietownes prime numbers are dead too?
As a research topic in and of themselves which includes Twin Prime Conjecture, etc.
We can't say that it's a dead area because of Zhang & Tao's results
But I bet it has slowed in progress
All the ancients with their organic diets and good drugs have long passed and they took all the easy mathematical fruits with them :|
@geocalc33 what's symplectic topology?
Is it too much to ask that it means "symple"? :)
@AMD functional = function, analysis = broken, to break. Either you break or your the math breaks. Either way, something's breaking.
@I'manalienImaneaglealien hi
Hey, mon.
What are you studying?
23:33
@I'manalienImaneaglealien I can't really say that that answers my question. Obviously it is related to functions. My question is what specifically about them is the object of functional analysis.
Also, how does one properly compute the error of a function given an infinite series approximation?
Their continuity, differentiability, the spaces the form when you take collections of all of them, and then the functions on those spaces that have related meaning such as inner product, etc.
*they form
Then the functions between the spaces, and then the space of those functions, to infinity
Ah, ok, I see. Thank you.
Going back to my question about error... I have this approximation for computing reciprocals now, and I'm not quite sure if the error is actually decreasing as x increases, or if I'm fooling myself. desmos.com/calculator/f5guhbxd3y
@I'manalienImaneaglealien
Do you know how Wolfram alpha gives a contour plot when you plot a 2d function
Sure, what about that?
Also @robjohn this may be sufficient for what you're interested in with regards to my research for fast divides. I'd say my search is over and this is the best I've found by far of anything I could scour across the (searchable) internet (via duckduckgo). I'm not interested in finding new algorithms per se; just the fastest implementations of algorithms of which new algorithms are one way of doing that.
I'm sure you can see how insanely fast this laurent series approximation is. Even absurdly large exponents can be cheaply computed with this.
23:40
just want to find the equations @I'manalienImaneaglealien for the contour curves
@AMDG which Laurent Series?
@AMDG It is hard to see where the Laurent Series is there. What is the independent variable that is usually given by $z$ in a Laurent Series? Are you talking about the geometric series that converge for $a\gt b$ and the other that converges for $a\lt b$?
@I'manalienImaneaglealien basically level sets
Hm, I don't know much about Laurent Series. I'm just calling it what Wolfram calls it.
The one listed under "Series expansion at a=\infty"
23:44
@AMDG i don't even see Laurent mentioned on that page.
Oh, well then one sec
@AMDG Ah. That is simply a geometric series
Ah, ok, thanks.
So what do you think? Good approximation, no?
@AMDG Ah, I see it. It's in a shadow font image. Non searchable
Yeah, I hate it when websites do that.
23:47
@AMDG It depends on what $a$ and $b$ are. If $b$ is close to $a$, the convergence can be slow.
See the desmos workspace. This only requires simple shifts and adds with the adds of course being O(lg2(n)) and shifts being O(n).
@I'manalienImaneaglealien what are you studying?
Statistics for college and algebraic topology in spare time, also looking at Lang's Algebra, but don't plan on reading everything in it. Just sections as I need them
If 2 doesn't work as a value, there is of course always the original 16 and also 4 that can be used with LUTs for imperfect powers of two.
23:51
21, however, seems to be the limit as it looks like the implementation of Desmos for my browser here is using double precision floats, not long doubles.
So if that's the case about convergence, then the convergence rapidly increases as x increases towards ±infty and away from $c$ (as listed in the workspace) which in this case is 2.
That would explain why this poorly approximates 1/3 but not 1/21 and so on. I don't quite get why other particular values have higher precision, though (appearing as zero to doubles).
That would also explain why choosing a modulus less than 1 would dramatically increase precision (and also allow values less than or equal to the value of $c$).
So uh... is it... good, rob?
Waits anxiously in anticipation
So you are evaluating a division using a geometric series? That seems questionable. Have you tried timing your algorithms? I would be surprised if this was faster than a machine divide.
Try computing $\frac{15}{17}$ using that series

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