Mathematics - 2021-07-29 (page 1 of 2)

copper.hat

12:07 AM

needs a $-{1 \over 12}$ article.

leslie townes

they got one. en.wikipedia.org/wiki/1_%2B_2_%2B_3_%2B_4_%2B_%E2%8B%AF

copper.hat

gg ffs

with appropriate accent

leslie townes

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.

robjohn

12:58 AM

@copper.hat this is close.

of course the page on the zeta function also mentions it

copper.hat

1:49 AM

its "sensational math"

the sort of thing one tweets about

robjohn

2:49 AM

Ah, I didn't see that leslie found it.

leslie townes

that's top-tier wikipedia math in my book

russian bot

3:08 AM

western spies can pronounce mississippi quite easily

Koro

Conway base 13 function

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?

leslie townes

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/…

Koro

looking

Koro

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.

leslie townes

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

Koro

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?

shintuku

any important inequalities for $\sqrt[n]{n}$?

leslie townes

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.

shintuku

oh thank you!

do you have like a searching technique on mathse? how did you do that so fast

i searched and found nothing

leslie townes

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.

shintuku

oh, do you search writing the mathjax?

Koro

@leslietownes Let $f: (a,b)\to R$ be surjective then $f^{-1} (R)$ is (a,b) for any $a<b$, then why $R$?

shintuku

in any case thanks for the tip, i'll start using google

leslie townes

f means too many things there.

Koro

let $f$ be a function

leslie townes

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.

David Choi

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.
https://math.stackexchange.com/questions/4211726/prove-int-infty-0-sin2-pix-frac1xdx-diverges

tbh its more about making my solution more rigorous

Koro

@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.

leslie townes

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.

Koro

@leslietownes great!! I got that.

Thanks a lot @Leslie :)

leslie townes

counterexamples and weird examples in topology and real analysis are sometimes fun.

Koro

4:01 AM

I didn't think of closed and bounded $[a,b]$

leslie townes

weird constructions, one-off ideas.

Koro

:)

copper.hat

4:20 AM

the mse site was offline for a while.

i did another pse, not even a convex one this time.

shintuku

what's a pse?

leslie townes

a misspelled pee or psq.

maybe both at the same time. copper contains mutitudes.

shintuku

ah, of course, pee or psq

robjohn

4:37 AM

@shintuku how about this?

shintuku

very neat!

copper.hat

4:50 AM

i did mean psq. i am afraid to think of what i might come up with for pse.

Semiclassical

identifying 8-dimensional polytopes is not easy

copper.hat

i haven't seen many

Semiclassical

esp. when the source says it's regular and it's not :(

copper.hat

most that i know are named bob

then again, if it was named mary i guess we could call it a polytrope.

robjohn

What about Polly Taupe?

is she okay?

LearningCHelpMe

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?

robjohn

nope

LearningCHelpMe

Semiclassical

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)

LearningCHelpMe

@Semiclassical Thanks. still not seeing it, but will think about it some more

Semiclassical

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

LearningCHelpMe

5:07 AM

Ahh I see

Thanks

Semiclassical

yup

if you want to check your result, note that $\tan(\pi/8)=\sqrt{2}-1$ and $\tan(3\pi/8)=\sqrt{2}+1$

shintuku

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 $ab \geq ab$?

robjohn

Have you noticed that $\frac\pi8=\frac\pi2-\frac{3\pi}8$?

Semiclassical

that too

robjohn

what does that mean about $\tan\left(\frac\pi8\right)\tan\left(\frac{3\pi}8\right)$?

Semiclassical

5:17 AM

nice

David Choi

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

leslie townes

could also write it as $\int_{-\delta}^{\delta} f(x) \, dx$ if this is a one variable thing.

David Choi

yep its a one variable thing

ahhh i see

leslie townes

they're using an inequality to define the interval instead of writing it in the other way.

David Choi

didn't know that was a thing lmao

leslie townes

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.

robjohn

intervals are often written as inequalities

copper.hat

@shintuku was that a math joke?

shintuku

i am a math joke

leslie townes

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.

shintuku

but that statement was not a math joke

copper.hat

5:22 AM

$\int_{ x \in [a,b]} dx f(x)$.

robjohn

@copper.hat my eyes hurt! I can't unsee that!

shintuku

it was a question of authentic mathematical innocence

copper.hat

what would be the point in showing $ab \ge ab$?

David Choi

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 $x \to \infty$?

robjohn

@shintuku take a good look at the left side and its square

copper.hat

5:23 AM

@robjohn i was going to do $\int_{ x \in [a,b]} m(dx) f(x)$ but decided not to :-)

shintuku

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

leslie townes

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 $\int_U$.

robjohn

@shintuku Square $|a|+|b|$

leslie townes

copper you should be committed.

shintuku

$a^2 + 2\sqrt{a^2}\sqrt{b^2} + b^2$

copper.hat

5:27 AM

i am committed to getting committed

robjohn

@shintuku that's better

shintuku

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?

copper.hat

@shintuku what are you trying to prove?

shintuku

$|a| + |b| \geq |a+b|$ 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| $
$\iff \left( \sqrt{a^2} + \sqrt{b^2} \right)^2 \geq \left( \sqrt{(a+b)^2} \right)^2$
$\iff a^2 + 2\sqrt{a^2}\sqrt{b^2}+b^2\geq a^2 + 2ab+b^2$
$\iff a^2+2ab+b^2 \geq a^2 + 2ab + b^2$

copper.hat

whoa, that broke the side of my screen

shintuku

5:37 AM

my bad

copper.hat

what if $a<0, b>0$?

shintuku

there has to be a false $\iff$ inference here somewhere right?

@copper.hat i'll check

copper.hat

yes, you are forgetting $|ab|$ not $ab$.

robjohn

@shintuku $\sqrt{a^2}\sqrt{b^2}=|ab|$

copper.hat

$\sqrt{a^2} \sqrt{b^2} = |ab|$.

bingo

robjohn

5:41 AM

@copper.hat I had just going to have said that!

copper.hat

i was still picking up the bits from the right hand side of my screen :-)

shintuku

ah, of course! $\sqrt{a^2}\sqrt{b^2} \iff ab$ is false for, as you said, $a<0, b>0$

thanks a lot

robjohn

@shintuku So the final result of all that is that $|ab|\ge ab$

copper.hat

i would say $\sqrt{a^2} \sqrt{b^2} = ab $ is false, as the right & left hand sides are not true or false statements

shintuku

thank you very much, highly appreciated

David Choi

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:

\begin{align}
\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\\\\
&=\int_{-1}^1\,\left(f(x)-f(0)\right)\,\frac{t}{t^2+x^2}\,dx+2\arctan(1/t)f(0)\\\\
&=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
\end{align}

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

copper.hat

i am afraid that i am unable to answer because of the disturbing visual

David Choi

its probably better to see his answer from the link i provided

im sorry about that

copper.hat

jk, the smooth brain ass

russian bot

blyat i hate it
but it would be very nice if someone recommended a nice mathematical logic book for me

David Choi

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?

copper.hat

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.

Srijan M.T

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?

David Choi

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

copper.hat

the integral is over $\delta < |x| < 1$.

so $\delta^2 \le x^2 \le x^2 +t^2$. now invert.

David Choi

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?

copper.hat

well, this is how you show it goes to zero.

bed time for me. good luck! good night.

David Choi

alrighty thanks for all your help! @copper.hat

wklm

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

shintuku

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

shintuku

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 $(n-1)(C-n)$?

Avra

11:55 AM

@shintuku. What course is this? Linear algebra/algebra ?

shintuku

it's real analysis or maybe even calculus

Avra

@shintuku. Never saw it in Calculus

shintuku

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!

Saad

12:15 PM

← 2 messages moved from CURED

shintuku

well that's awkward

Kiro

1:01 PM

A single number is technically a sum with just one term yeah?

shintuku

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?

robjohn

@shintuku you are correct. The sequential limit does not imply the complete limit.

shintuku

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

robjohn

You can define the sequential limit with the same form. $\lim\limits_{n\to\infty}f(n)=L\iff\forall\epsilon\gt0,\exists N\gt0:n\ge N\implies|f(n)-L|\le\epsilon$

shintuku

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

robjohn

Let $f(x)=\sin(\pi/x)$, then $\lim\limits_{n\to\infty}f(1/n)=0$

but $\lim\limits_{x\to0}f(x)\ne0$

Thorgott

1:36 PM

the key is "for every sequence"

shintuku

ahhhh

of course

every sequence implies every x

thanks a lot!

robjohn

If you have it for every sequence, prove the full limit by contradiction.

shintuku

will do, thanks!

John_maddon

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

Srijan M.T

: 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

shintuku

1:52 PM

k is a constant @SrijanM.T

@SrijanM.T what is P(y)?

Srijan M.T

@shintuku ?

shintuku

P(x) = x^2 + k^2

P(y) = ?

Srijan M.T

y^2 + k^2

shintuku

Let y = 2x. P(y) = ?

Srijan M.T

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

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?

Srijan M.T

@shintuku It is wrong. I get x^2 again

shintuku

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 = ?

Srijan M.T

@shintuku P(2x) =( 2x)^2 + k^2

shintuku

exactly!

so, if y = 2x, P(2x) = P(y) = y^2 + k^2 = (2x)^2 + k^2, right?

Srijan M.T

@shintuku Yes

shintuku

2:04 PM

Let y=x^2. P(y) = ?

Srijan M.T

P(x^2) = (x^2)^2. I think

shintuku

hm. we are missing k!

Srijan M.T

P(x^2 + k^2) = (x^2 + k^2)^2 + k^2 @shintuku I think my Q ans is this

shintuku

exactly!

Srijan M.T

@shintuku Hooray. Thanks a lot

shintuku

2:09 PM

np!

Srijan M.T

@shintuku What do you do in life ?

John_maddon

Someone knows about barycentric interpolation with a function evaluated on two parameters, i.e. f(x,y) ?

Unique

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!

schn

3:19 PM

@robjohn I will give you one variable and one variable only, $z=hu$.

leslie townes

3:29 PM

unique, i think you need more hypotheses. consider f(x) = -x on [1,2]

copper.hat

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.

leslie townes

3:56 PM

you have to deduct the millions of illegal votes.

satan 29

Hi everyone

Bob

Hello John

satan 29

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?

leslie townes

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.

satan 29

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

leslie townes

you might be able to trick wolfram alpha into doing it even if it is not a stock command.

or just ask it. wolframalpha.com/input/…

satan 29

@leslietownes wolfram wasnt able to handle a variable power

@leslietownes oh wow

I messed up maybe

leslie townes

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.

vitamin d

I'm not advanced in topology - Does this answer prove that there is no space \Bbb R^{3/2}?

leslie townes

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.

Thorgott

@vitamind it proves it says it proves. writing R^{3/2} is meaningless unless you specify what it should mean

vitamin d

@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?

Thorgott

yes

vitamin d

Ok thanks

*satisfy

leslie townes

4:51 PM

vaguely reminds me of math.dartmouth.edu/~doyle/docs/three/three.pdf although only vaguely. you can divide sets by three.

copper.hat

fluff

leslie townes

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.

geocalc33

I'm at work

is anyone else at work - or is it just me?

copper.hat

sort of

Hasini

5:07 PM

Hi

geocalc33

can a non-periodic function cross the real axis infinitely many times but never equal zero?

Hasini

What is usually meant by "a group where the word problem is efficiently solvable"?. Thanks a lot in advance.

Xander Henderson

@geocalc33 Sure.

@Hasini No idea. Where are you seeing this phrase?

geocalc33

the imaginary and real parts of the function just never intersect at the same place on the real axis?

Xander Henderson

@geocalc33 You have given no restrictions on the function. Is it continuous? holomorphic? What?

You just asked about "a function".

robjohn

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 :-)

Xander Henderson

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.

geocalc33

holomorphic @XanderHenderson

Hasini

@XanderHenderson I saw it in this article researchgate.net/profile/Vitaly-Romankov/publication/…

copper.hat

@robjohn Thanks! No need to just on my account.

Hasini

In page numbered 3(Chapter 2, 2nd paragraph last line), it says "In particular, the word problem is efficiently solvable for G" @XanderHenderson Thanks.

geocalc33

5:16 PM

@XanderHenderson thank you

you've given me a lot to think about

Thorgott

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

leslie townes

'efficiently solvable' gets scare quotes in that set theory paper i linked to up above, for that reason.

oh wait he uses 'effectively computable.'

Xander Henderson

@Thorgott On page 3 of the linked paper, it seems to assert that "efficient" means linear or quadratic.

leslie townes

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

Xander Henderson

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.

copper.hat

5:21 PM

so really, its ineffectively computable

Thorgott

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

Hasini

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?

leslie townes

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.

Thorgott

yes, it's deciding whether a word represents the identity

copper.hat

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.

Hasini

5:24 PM

@leslietownes Yes I also have the problem that what problem should be solved, when they say word problem

Thorgott

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

geocalc33

im trying to numerically investigate a complex function

is anyone here a really good complex plotter?

Hasini

Hmm, yes I've heard the usual word problem to be determining whether it represents identity

leslie townes

i have a few conspiracies and scams i am hatching but that is not what you mean.

Hasini

But in that paper I can't see any connection with identity..

geocalc33

5:26 PM

@leslietownes lol

copper.hat

only do quaternions, sorry. complex is too simple.

geocalc33

maybe i will put something on indeed

complex plotter needed. must be able to plot at least 30 complex functions. seasonal opportunity

Thorgott

the word problem is by definition about the identity

see en.wikipedia.org/wiki/Word_problem_for_groups

Hasini

Yes, thanks a lot @Thorgott

But how is it used in this article is not clear..

copper.hat

@geocalc33 you need to get real

Hasini

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 $w(c_1 a_1, \cdots, c_k a_k)=w(a_1, \cdots, a_k)$

robjohn

5:44 PM

@copper.hat Here is the meta post.

copper.hat

@robjohn Thanks!

robjohn

@copper.hat Does that look similar to what is happening to you?

copper.hat

@robjohn yes, but on a smaller scale :-)

i mean mine is the smaller scale

robjohn

6:00 PM

@copper.hat I have a retina screen, and the screen shots come out pretty huge.

copper.hat

they are reasonably sized on my tv screen & on my laptop.

big, but it makes the point

leslie townes

now the hacker knows both your IP and your OS'es, copper. look out.

copper.hat

:-) my ip is not that hard to find :-)

leslie townes

well, one of them anyway.

copper hat's cryptocurrency funded darkweb consultancy and its global network, not so much

robjohn

@leslietownes I don't think that is copper's IP. 4515 is a bit out of the 0-255 range.

Xander Henderson

6:04 PM

@geocalc33 I use this one in my classes: people.ucsc.edu/~wbolden/complex/…

robjohn

that's got to be the Chrome version

leslie townes

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.

Xander Henderson

@leslietownes You are playing with fire, here. Just don't say his name three times.

leslie townes

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.

Xander Henderson

@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.

leslie townes

6:08 PM

enjoy. what a nice image. i have yet to do that with my daughter because she can't sit still.

o.9

Must be nice being a pro hacker

copper.hat

i think the bot guy shows up as another user at the moment.

o.9

very nice

copper.hat

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

leslie townes

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.

copper.hat

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.

geocalc33

@XanderHenderson thanks, but the link isn't working for me ;/

copper.hat

i am still juvenile, just a lot older.

i misspoll things so frequently that my autocorrect has accepted my maspillings

o.9

The new mod icons are very nice

I hope whoever was behind it got payed a lot of casheroney

copper.hat

i suspect they are behind it themselves and are paying themselves the big bucks

o.9

Deserved :)

Hopefully they pay themselves some more to bring back the old design

hyper-neutrino

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.

copper.hat

oh, i must have misunderstood, where are the new mod badges to be seen?

o.9

maybe they need to pay themselves extra to bring them to the main sites

on Meta

copper.hat

i see, thx

o.9

you are welcome

ExercisingMathematician

8:53 PM

Black - Wonderful life - Remasterizado - Donate

►

robjohn

@copper.hat I want to get over my impatiance, NOW!

copper.hat

9:31 PM

:-)

@ExercisingMathematician ???

leslie townes

an Exorcising Mathematician might be able to help with that

pritchard

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!

leslie townes

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.

geocalc33

@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

leslie townes

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.

geocalc33

@leslietownes I have an idea

Transcript for

$Mathematics$ Mathematics