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12:07 AM
needs a $-{1 \over 12}$ article.
gg ffs
with appropriate accent
it's pretty good, too.
the other eclectic feature of math wiki is the citations will sometimes be links to free articles, sometimes just citations to books with no link even if the book is freely available, paywalled content, and formerly paywalled content that has been liberated and is on some random website somewhere.
it's helpful, in a lot of other fields you don't see a lot of links to anything.
also cites to really old papers that probably aren't available anywhere except at the libraries of the people who wrote them.
12:58 AM
@copper.hat this is close.
of course the page on the zeta function also mentions it
1:49 AM
its "sensational math"
the sort of thing one tweets about
1 hour later…
2:49 AM
Ah, I didn't see that leslie found it.
that's top-tier wikipedia math in my book
3:08 AM
western spies can pronounce mississippi quite easily
The Conway base 13 function is a function created by British mathematician John H. Conway as a counterexample to the converse of the intermediate value theorem. In other words, it is a function that satisfies a particular intermediate-value property—on any interval (a, b), the function f takes every value between f(a) and f(b)—but is not continuous. == Purpose == The Conway base 13 function was created as part of a "produce" activity: in this case, the challenge was to produce a simple-to-understand function which takes on every real value in every interval, that is, it is an everywhere surjective...
why was base 13 chosen?
what would have happened had base 10 been chosen?
beats me. i think ive shared my favorite example of that, which i like because there is a 'calculus' style formula for it. umv.science.upjs.sk/analyza/texty/predmety/MANb/distancne/…
3:37 AM
@leslietownes I am confused at 2nd para. What is surjective on every open interval? I say $f: A\to B$ is surjective if range f= B.
meaning that the range of its restriction to any open interval is R
choose any (a,b) you like, that formula maps it onto R
it's not quite the conway function but similar ideas
Ahh and such function can't be continuous because R is closed in R so $f^{-1}(R)$ should also be closed which is not the case. Right?
any important inequalities for $\sqrt[n]{n}$?
lots of ways to see it. f^{-1}(R) in this case is R. maybe helpful to look at the inverse images of points. or think in terms of limits in the domain.
math.stackexchange.com/questions/28348/… evaluates the limit, which includes information about inequalities. i don't know if any particular one jumps out at me.
oh thank you!
do you have like a searching technique on mathse? how did you do that so fast
i searched and found nothing
3:48 AM
i use google. it syncs with math.se, usually gives top results, and anything recent will be on there and perhaps more easily searched.
oh, do you search writing the mathjax?
@leslietownes Let $f: (a,b)\to R$ be surjective then $f^{-1} (R)$ is (a,b) for any $a<b$, then why $R$?
in any case thanks for the tip, i'll start using google
f means too many things there.
let $f$ be a function
3:50 AM
there's an f with domain R and codomain R. then there is the formally different function obtaioned by restricting the rule of f to (a,b) only.
inverse images of R under these different functions may be generally different although they are not in this case. but i was not using a different notation for the restriction so maybe that was notationally ambiguous above.
hey guys i've been wrestling with this problem for a while now: i've been tryin to determine whether the integral of sin^2(pi*(x+1/x) from 0 to infinity diverges. I just made a post about my solution, and I was hoping someone here could take a quick look.
tbh its more about making my solution more rigorous
@leslietownes (f is surjective on all non empty open intervals):= if $a,b\in \mathbb R$ such that $a<b$ and $f: (a,b)\to \mathbb R$ is such that range (f)=R (this is as per definition suggested above). So I assume on the contrary that $f$ is continuous then $f^{-1}(R)=(a,b)=(0,1)=(0,2)$ as a, b are arbitrary so from here how do we get contradiction.
i don't fully understand (0,1) = (0,2). maybe go the other way. if f is surjective on any open interval it's also surjective on any closed interval [a,b] with a < b, but the image of a closed and bounded interval under a continuous map is bounded.
inverse images get confusing when there's more than one domain your different rules could be inverse imaging into.
maybe notationally at least.
@leslietownes great!! I got that.
Thanks a lot @Leslie :)
counterexamples and weird examples in topology and real analysis are sometimes fun.
4:01 AM
I didn't think of closed and bounded $[a,b]$
weird constructions, one-off ideas.
4:20 AM
the mse site was offline for a while.
i did another pse, not even a convex one this time.
what's a pse?
a misspelled pee or psq.
maybe both at the same time. copper contains mutitudes.
ah, of course, pee or psq
4:37 AM
@shintuku how about this?
very neat!
4:50 AM
i did mean psq. i am afraid to think of what i might come up with for pse.
identifying 8-dimensional polytopes is not easy
i haven't seen many
esp. when the source says it's regular and it's not :(
most that i know are named bob
then again, if it was named mary i guess we could call it a polytrope.
What about Polly Taupe?
is she okay?
4:55 AM
Any hints on how to go about part (c)? The roots $z$ are the following $\tan \theta$ where $\theta = \frac{\pi}{8}, \frac{3\pi}{8}, \frac{5\pi}{8} and \frac{7\pi}{8}$
is the photo showing?
presumably one of the other roots is of the form $z_2=i \tan(3\pi/8)$
or something like that
oh. you did that already
@LearningCHelpMe Hint: $z_4 = i \tan(7\pi/8) = -i\tan(\pi/8)=z_1$ and similarly for $z_3=i\tan(5\pi/8)$
(should be $z_4=-z_1$, oops)
@Semiclassical Thanks. still not seeing it, but will think about it some more
the other hint is to write the polynomial as $(z-z_1)(z-z_2)(z-z_3)(z-z_4)$ and expand it out
5:07 AM
Ahh I see
if you want to check your result, note that $\tan(\pi/8)=\sqrt{2}-1$ and $\tan(3\pi/8)=\sqrt{2}+1$
in the proof of $|a| + |b| \geq |a+b|$, is there a reason we can't just take the square of both sides $\sqrt{a^2}+ \sqrt{b^2} \geq \sqrt{(a+b)^2}$ to arrive at $ab \geq ab$?
Have you noticed that $\frac\pi8=\frac\pi2-\frac{3\pi}8$?
that too
what does that mean about $\tan\left(\frac\pi8\right)\tan\left(\frac{3\pi}8\right)$?
5:17 AM
guys what does this notation mean $int_{|x|<\delta} f(x) dx$?
i saw this in an answer but i didn't get what the inequality meant when used as the limits of integration
could also write it as $\int_{-\delta}^{\delta} f(x) \, dx$ if this is a one variable thing.
yep its a one variable thing
ahhh i see
they're using an inequality to define the interval instead of writing it in the other way.
didn't know that was a thing lmao
5:21 AM
only a pervert would write $\int_{a \leq x \leq b} f(x) \, dx$ but bet someone out there is doing that too.
intervals are often written as inequalities
@shintuku was that a math joke?
i am a math joke
whenever something like this comes up i always look at what i did in my dissertation and it's usually the opposite of what i think it would be.
but that statement was not a math joke
5:22 AM
$\int_{ x \in [a,b]} dx f(x)$.
@copper.hat my eyes hurt! I can't unsee that!
it was a question of authentic mathematical innocence
what would be the point in showing $ab \ge ab$?
also another thing: how does one rigorously show that $\sin^2(\pi(x+\frac{1}{x}))$ can be approximated as $\sin^2(\pi*x)$ as $x \to \infty$? Is it enough to just state that $x + \frac{1}{x} \approx x$ as $x \to \infty$?
@shintuku take a good look at the left side and its square
5:23 AM
@robjohn i was going to do $\int_{ x \in [a,b]} m(dx) f(x)$ but decided not to :-)
well, from $|a| + |b| \geq |a+b|$ we get "if and only if $ab \geq ab$", so the former would be true by virtue of the latter
@robjohn ok will meditate on it a bit
i don't seem to have used explicit interval related bounds at all. maybe once or twice a letter like $U$ defined elsewhere and then i write $\int_U$.
@shintuku Square $|a|+|b|$
copper you should be committed.
$a^2 + 2\sqrt{a^2}\sqrt{b^2} + b^2$
5:27 AM
i am committed to getting committed
@shintuku that's better
but it is equal to $a^2 + 2ab + b^2$, so we get $a^2 + 2ab + b^2 \geq a^2 + 2ab + b^2$
do we lose the properties of the absolute value in doing this?
@shintuku what are you trying to prove?
$|a| + |b| \geq |a+b|$ i have the author's proof and i know i'm wrong, i'm just trying to make sense of why the logic prevents resolving $\sqrt{a^2}\sqrt{b^2}$ into $ab$
at some point we should lose an $\iff$ somewhere
$|a| + |b| \geq |a+b| $
$\iff \left( \sqrt{a^2} + \sqrt{b^2} \right)^2 \geq \left( \sqrt{(a+b)^2} \right)^2$
$\iff a^2 + 2\sqrt{a^2}\sqrt{b^2}+b^2\geq a^2 + 2ab+b^2$
$\iff a^2+2ab+b^2 \geq a^2 + 2ab + b^2$
whoa, that broke the side of my screen
5:37 AM
my bad
what if $a<0, b>0$?
there has to be a false $\iff$ inference here somewhere right?
@copper.hat i'll check
yes, you are forgetting $|ab|$ not $ab$.
@shintuku $\sqrt{a^2}\sqrt{b^2}=|ab|$
$\sqrt{a^2} \sqrt{b^2} = |ab|$.
5:41 AM
@copper.hat I had just going to have said that!
i was still picking up the bits from the right hand side of my screen :-)
ah, of course! $\sqrt{a^2}\sqrt{b^2} \iff ab$ is false for, as you said, $a<0, b>0$
thanks a lot
@shintuku So the final result of all that is that $|ab|\ge ab$
i would say $\sqrt{a^2} \sqrt{b^2} = ab $ is false, as the right & left hand sides are not true or false statements
thank you very much, highly appreciated
6:05 AM
not sure if this kind of question is allowed here but i need help understanding an answer on this site: math.stackexchange.com/a/3577929/923312
ok so in mark's answer: he writes:
\int_{-1}^1\,f(x)\,\frac{t}{t^2+x^2}\,dx&=\int_{-1}^1\,\left(f(x)-f(0)\right)\,\frac{t}{t^2+x^2}\,dx+f(0)\int_{-1}^1 \frac{t}{t^2+x^2}\,dx\\\\
&=2\arctan(1/t)f(0)+\int_{|x|<\delta} \,\left(f(x)-f(0)\right)\,\frac{t}{t^2+x^2}\,dx\\\\
&+\int_{\delta<|x|<1} \,\left(f(x)-f(0)\right)\,\frac{t}{t^2+x^2}\,dx\tag1
and says that the third term on the right obviously goes to 0 as $t \to 0+$
but my smooth brain ass doesn't understand why that's obvious
i am afraid that i am unable to answer because of the disturbing visual
its probably better to see his answer from the link i provided
im sorry about that
jk, the smooth brain ass
blyat i hate it
but it would be very nice if someone recommended a nice mathematical logic book for me
another thing my smooth brain doesn't understand is this line:
$$|\int_{|x|<\delta} \,\left(f(x)-f(0)\right)\,\frac{t}{t^2+x^2}\,dx| \le \pi \varepsilon$$
i understand how mark derives it, but i don't rlly understand his point. is it that since $\epsilon$ is an arbitrary number greater than $0$, we can essentially make the absolute value of the second term go to 0 by making ϵ as small as we like?
6:14 AM
@DavidChoi note that $|f(x)-f(0)| $ is bounded on $[-1,1]$ by continuity. Note that ${ 1 \over t^2+x^2} \le {1 \over \delta^2}$.
then you have some constant bound multiplied by $t$ which is going to zero as they say.
for the second one $|f(x)-f(0)| \le \epsilon$.
and the rest of the integral is $2 \arctan { \delta \over t} $ and this is bounded by ${ \pi}$.
hopefully this will crinkle your brain a little.
For a quadratic equation, we have f(x) = ax^2 + bx + c. Where a is not equal to 0
Why is there no condition for b not equal to 0?
hmmm ok ok..... wait @copper.hat why does ${ 1 \over t^2+x^2} \le {1 \over \delta^2}$?
we're talking about when $\delta<|x|<1$ right?
as u can see i'm quite smooth brain right now lmaooo
the integral is over $\delta < |x| < 1$.
so $\delta^2 \le x^2 \le x^2 +t^2$. now invert.
6:32 AM
also for the second one, since |second term| < $\pi \epsilon$, we can make $\epsilon$ as small as we like since its arbitrary, and this implies that the second term goes to 0 right?
well, this is how you show it goes to zero.
bed time for me. good luck! good night.
alrighty thanks for all your help! @copper.hat
2 hours later…
8:18 AM
Hi Guys! I want to calculate the probability of geometric brownian motion to be within and interval [a, b] at time t. To do so I integrate the pdf as specified here: en.wikipedia.org/wiki/Geometric_Brownian_motion within a and b. Does it make sense? My results seem to be way to small
2 hours later…
10:03 AM
oh, math life is way easier when you understand that $|x-x_0|<\delta$ is meant to translate $x$ is going towards $x_0$
limit composition becomes obvious that way
2 hours later…
11:36 AM
any faster way to determine the sequence given by $a_n = n(C+1-n)$ with $n \leq C$ is such that we always have $a_n \geq C$ other than to factor $n(C+1-n) - C$ into $(n-1)(C-n)$?
11:55 AM
@shintuku. What course is this? Linear algebra/algebra ?
it's real analysis or maybe even calculus
@shintuku. Never saw it in Calculus
maybe a second or third class in calculus, you eventually have to do series for taylor polynomials, and you need sequences for that
or, of course, any class in real analysis
hello! I noticed math.stackexchange.com/questions/514388/… is the go-to thread questions about the limit of the nth root of n factorial are directed to, but it seems the thread isn't specifically about this limit
just thought maybe it would be of interest to someone. cheers!
12:15 PM
2 messages moved from CURED
well that's awkward
1:01 PM
A single number is technically a sum with just one term yeah?
1:25 PM
Consider the continuous function $f:E \to \mathbb{R}$, with accumulation point $x_0$, the sequence $\{f(e_n)\}$ composed of the sequence $\{e_n\} \subset E$ approaching $x_0$ and the sequential limit $\lim \limits_{n \to \infty} f(e_n) = L$. I'm trying to prove this last limit implies $\lim \limits_{x \to x_0} f(x) = L$. I'm wondering: do I have to prove this holds for the x that are in between the $e_n$ terms?
@shintuku you are correct. The sequential limit does not imply the complete limit.
yes, my bad
hm, alright. I'm trying to understand in what sense the author says there is a sequential limit definition identical to the epsilon-delta one
"We can write $\lim \limits_{x \to x_0} f(x) = L$ if for every sequence $\{e_n\}$ of points of $E$ with $e_n \neq x_0$ and $e_n \to x_0$ as $n \to \infty$, we have $\lim \limits_{n \to \infty} f(e_n) = L$
seems like "$x_0$ is an accumulation point of E" is doing some work here
You can define the sequential limit with the same form. $\lim\limits_{n\to\infty}f(n)=L\iff\forall\epsilon\gt0,\exists N\gt0:n\ge N\implies|f(n)-L|\le\epsilon$
but, what happens to the x in between the natural numbers? the author is saying it is equivalent with the epsilon-delta definition
I've proven the $\lim \limits_{x \to x_0} f(x) = L \implies \lim \limits_{n \to \infty}f(e_n) = L$, but I'm having a hard time doing the opposite way
Let $f(x)=\sin(\pi/x)$, then $\lim\limits_{n\to\infty}f(1/n)=0$
but $\lim\limits_{x\to0}f(x)\ne0$
1:36 PM
the key is "for every sequence"
of course
every sequence implies every x
thanks a lot!
If you have it for every sequence, prove the full limit by contradiction.
will do, thanks!
Hi guys, someone knows about the Spectral Matrix with 2 parameters, instead than 1?
I am working on this by 5 days and I am k.o. :S
: If P(x) = x^2 + k^2. Then , Find P(P(x))
I got P(x^2 + k^2) but what should be the next step ?
I have to eliminate the P also
1:52 PM
k is a constant @SrijanM.T
@SrijanM.T what is P(y)?
@shintuku ?
P(x) = x^2 + k^2
P(y) = ?
y^2 + k^2
Let y = 2x. P(y) = ?
I’m thinking. 1 min
@shintuku P(2x) = (y/2)^2 + k^2 ?
I think
Also , it can be written as P(2x) = (2x)^2 + k^2 @shintuku
2:00 PM
your first equality is incorrect. Take P(2x) = (y/2)^2 + k^2, and substitute y = 2x on the right hand side. What do you get?
@shintuku It is wrong. I get x^2 again
exactly, so it is incorrect that P(2x) = (y/2)^2 + k^2
we know P(y) = y^2 + k^2. Let y = 2x. Then, y^2 + k^2 = ?
@shintuku P(2x) =( 2x)^2 + k^2
so, if y = 2x, P(2x) = P(y) = y^2 + k^2 = (2x)^2 + k^2, right?
@shintuku Yes
2:04 PM
Let y=x^2. P(y) = ?
P(x^2) = (x^2)^2. I think
hm. we are missing k!
P(x^2 + k^2) = (x^2 + k^2)^2 + k^2 @shintuku I think my Q ans is this
@shintuku Hooray. Thanks a lot
2:09 PM
@shintuku What do you do in life ?
Someone knows about barycentric interpolation with a function evaluated on two parameters, i.e. f(x,y) ?
2:43 PM
Can someone give me a hint on how can I prove that the function f is increasing if f is differentiable on [a,b] and f' is not identically zero on any subinterval of [a,b]? Small hint or a theorem that i should study will be a great help. Thanks!
3:19 PM
@robjohn I will give you one variable and one variable only, $z=hu$.
3:29 PM
unique, i think you need more hypotheses. consider f(x) = -x on [1,2]
other than that you are probably looking at darboux's theorem.
there is a difference between what my reputation page and the reputation menu shows. a little bug somewhere.
3:56 PM
you have to deduct the millions of illegal votes.
Hi everyone
Hello John
I have a matrix m = {{t,t+1},{t-1,t}}. I wanted to find an expression for m^n
in terms of t and n
How can we do this in mathematica?
4:14 PM
have you tried just asking for the nth power of that? i don't know mma very well.
for all but two values of t, 1 and -1, it is diagonalizable. it wouldn't surprise me if it had a diagonalization command you could use in that case. and then it's just two other cases.
actually, I dont have mathematica lol, I dont know anything about it
but my friend has one
If I can get the proper command he can run it for me
you might be able to trick wolfram alpha into doing it even if it is not a stock command.
@leslietownes wolfram wasnt able to handle a variable power
@leslietownes oh wow
I messed up maybe
i wasn't expecting it to work either.
the stated formula doesn't make sense when t is 1 or -1 (when the matrix isn't diagonalizable), so maybe it is doing a diagonalization procedure to get that.
or maybe that formula simplifies into something that does work but i kinda doubt it.
I'm not advanced in topology - Does this answer prove that there is no space \Bbb R^{3/2}?
4:29 PM
yeah it's just ignoring those cases.
for those cases if you just write out a few powers you see the pattern. and WA can handle them if you type in those specific matrices and not the thing with t in it.
@vitamind it proves it says it proves. writing R^{3/2} is meaningless unless you specify what it should mean
@Thorgott Did I understand it right that there are no spaces X, which satisfie X\times X=\mathbb R^n if n is not a multiple of 2?
Ok thanks
4:51 PM
vaguely reminds me of math.dartmouth.edu/~doyle/docs/three/three.pdf although only vaguely. you can divide sets by three.
i don't often remember papers i've read in set theory but that one is beautifully written.
funny footnote on p. 1 about the exposition.
somewhat rude not to include the figures. i wonder if that was just a tex compilation error.
I'm at work
is anyone else at work - or is it just me?
sort of
5:07 PM
can a non-periodic function cross the real axis infinitely many times but never equal zero?
What is usually meant by "a group where the word problem is efficiently solvable"?. Thanks a lot in advance.
@geocalc33 Sure.
@Hasini No idea. Where are you seeing this phrase?
the imaginary and real parts of the function just never intersect at the same place on the real axis?
@geocalc33 You have given no restrictions on the function. Is it continuous? holomorphic? What?
You just asked about "a function".
5:13 PM
@copper.hat I mentioned that earlier today to the other mods. I will post to meta, now that I know it's not just me :-)
It is also not entirely clear to me what you mean by "crossing the real axis". I have a very clear idea about what it means for a function from R to R to cross an axis (or any curve), but it is not clear to me what is means for a complex valued function to cross an axis.
holomorphic @XanderHenderson
@XanderHenderson I saw it in this article researchgate.net/profile/Vitaly-Romankov/publication/…
@robjohn Thanks! No need to just on my account.
In page numbered 3(Chapter 2, 2nd paragraph last line), it says "In particular, the word problem is efficiently solvable for G" @XanderHenderson Thanks.
5:16 PM
@XanderHenderson thank you
you've given me a lot to think about
to say that the word problem is solvable means that there is a deterministic algorithm solving it
what "efficiently" means is up to the beholder, that's not a universally agreed upon thing
'efficiently solvable' gets scare quotes in that set theory paper i linked to up above, for that reason.
oh wait he uses 'effectively computable.'
@Thorgott On page 3 of the linked paper, it seems to assert that "efficient" means linear or quadratic.
i think there's a spectrum of all of that and you really have to look at who's saying it or what problem they are trying to solve
But then they use "efficiently" again to describe a normal form.
So... shrugs.
Way outside of my wheelhouse, so I am going to go hide now.
5:21 PM
so really, its ineffectively computable
it seems to just be an overview, so that's probably why they're being very imprecise
though I have not and will not read the rest of the paper
Hmmm, okay, word problem then means like computing the product of elements like $h_1 h_2^2 h_3$ and obtaining the group element represented, where $h_1,h_2,h_3 \in G$
Something like that?
deciding whether a word is the identity, i think. 'obtaining the group element represented' is a bit of a broader and more ambiguous problem without something resembling an enumeration or canonical form of the elements.
but exactly that idea.
yes, it's deciding whether a word represents the identity
it would be a nice to have paper metrics that show (a) how many people had actually read a given paragraph, (b) understood it in a meaningful way and (c) thought it made sense.
5:24 PM
@leslietownes Yes I also have the problem that what problem should be solved, when they say word problem
a usual strategy to do as such is having a computable "normal form" for elements and reducing a word to its normal form will determine whether it's trivial or not
the author seems to assume the existence of such a normal form
im trying to numerically investigate a complex function
is anyone here a really good complex plotter?
Hmm, yes I've heard the usual word problem to be determining whether it represents identity
i have a few conspiracies and scams i am hatching but that is not what you mean.
But in that paper I can't see any connection with identity..
5:26 PM
@leslietownes lol
only do quaternions, sorry. complex is too simple.
maybe i will put something on indeed
complex plotter needed. must be able to plot at least 30 complex functions. seasonal opportunity
the word problem is by definition about the identity
Yes, thanks a lot @Thorgott
But how is it used in this article is not clear..
@geocalc33 you need to get real
5:34 PM
They get a element $w(a_1, \cdots, a_k)$ and find a marginal tuple $\bar{c}=(c_1, \cdots, c_k)$ such that $w(c_1 a_1, \cdots, c_k a_k)=w(a_1, \cdots, a_k)$
5:44 PM
@copper.hat Here is the meta post.
@robjohn Thanks!
@copper.hat Does that look similar to what is happening to you?
@robjohn yes, but on a smaller scale :-)
i mean mine is the smaller scale
6:00 PM
@copper.hat I have a retina screen, and the screen shots come out pretty huge.
they are reasonably sized on my tv screen & on my laptop.
big, but it makes the point
now the hacker knows both your IP and your OS'es, copper. look out.
:-) my ip is not that hard to find :-)
well, one of them anyway.
copper hat's cryptocurrency funded darkweb consultancy and its global network, not so much
@leslietownes I don't think that is copper's IP. 4515 is a bit out of the 0-255 range.
6:04 PM
@geocalc33 I use this one in my classes: people.ucsc.edu/~wbolden/complex/…
that's got to be the Chrome version
oh @robjohn that was a callback to a somewhat juvenile fellow who used to come in and say that he could figure out our IP addresses.
where's he been, by the way? nobody answer that question.
@leslietownes You are playing with fire, here. Just don't say his name three times.
i can figure out your math.SE profile information, xander. he's not the only dangerous guy out there.
i can also hack into the mainframe and figure out users' reputations. but not why they are listed differently in different panes.
@leslietownes Yeah, but I don't seek to anonymize myself. :D
@leslietownes OH NOES!
Right... I'm out. I promised my daughter a picnic today, so I need to put lunch together.
6:08 PM
enjoy. what a nice image. i have yet to do that with my daughter because she can't sit still.
Must be nice being a pro hacker
i think the bot guy shows up as another user at the moment.
very nice
i hate autoresponder emails telling you they will get back to you in 7,200 hours or whatever.
i'm soo impatient i get annoyed at myself for being impatient
i like them, they make me feel missed and wanted.
it's like looking forward to christmas, in a way. you wouldn't want christmas in march
absence makes the heart grow fonder.
6:17 PM
at some point in my younger juvenile days i had an autoresponder that would respond to the autoresponder. however the only person suffering as a result was myself so that ended quickly.
@XanderHenderson thanks, but the link isn't working for me ;/
i am still juvenile, just a lot older.
i misspoll things so frequently that my autocorrect has accepted my maspillings
The new mod icons are very nice
I hope whoever was behind it got payed a lot of casheroney
i suspect they are behind it themselves and are paying themselves the big bucks
Deserved :)
Hopefully they pay themselves some more to bring back the old design
6:23 PM
I am not a huge fan of the new mod badges but it's not the worst thing. Also at least they made it only on meta sites.
oh, i must have misunderstood, where are the new mod badges to be seen?
maybe they need to pay themselves extra to bring them to the main sites
on Meta
i see, thx
you are welcome
2 hours later…
@copper.hat I want to get over my impatiance, NOW!
9:31 PM
@ExercisingMathematician ???
an Exorcising Mathematician might be able to help with that
Hello, is it correct that $\mathbb{R}^n$ is not homeomorphic to a singleton for $n\geq 1$ because the cardinality of the spaces are different so there cannot exist a bijection or homeomorphism between them? I want to check if this reasoning is correct? Thanks!
yes. that works.
a homeomorphism must among other things be a bijection, and in your case, there can't be any.
there are bijections between R^n and R^m for any n, m >= 1 where the issue then gets more explicitly topological.
@ExercisingMathematician hi
are you might you be...that other person who's name starts with and/or possibly works on category theory with computer apps?
cringey diamond?
shine hence you crazy diamond?
is professional sound quality achieved by a computer by being able to process complex sounds?
so the sound is a pressure wave and the computer has to be able to take in the wave and manipulate it?
not trying to spam the chat. I don't have anymore questions lol
9:57 PM
i don't work in this area, but i think traditionally you would interpose a microphone or other sensor between the two that converts pressure wave into continuous electrical signal, then some kind of hardware samples that and turns it into digital information. the quality of the setup depending on the characteristics of everything.
i have a friend who insists on tape and would argue for reasons unknown to me that a computer should not be and maybe cannot be involved in achieving 'professional' sound quality.
he's lots of fun at parties.
@leslietownes I have an idea
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