00:00 - 21:0021:00 - 00:00

12:12 AM
what about other greatest hits? Papyrus? Algerian? Hobo?

Comic Sans is famously the font you use when you don't want to be taken seriously. Algerian and Hobo I'm not familiar with, but Papyrus is reserved for people who want to make movies about blue aliens

Can someone please explain to me why I'm getting these rather convenient values using the function listed in cell 18 as an argument to circular and inverse circular functions? Why are the results "nice" numbers like approx. 1 or approximately the original function even? desmos.com/calculator/zkfagfdxwx
This doesn't make too much sense to me other than the somewhat vague connection between $e^{x+iy}$ and all of these functions here.
@robjohn You will find the reciprocal approximation in cell 17 and the recursive formula to improve its accuracy at the very bottom in cells [39, 41]. I believe it will be of use to you.
I honestly have never come across something like this before, but hopefully someone else here has.
Also, I have nothing more to say concerning reciprocals now apart from recreation :)
Could this actually be related to the hyperbolic functions since $\frac{1}{x}$ is a hyperbola?

1:00 AM
Rithaniel: or menus for juice or smoothie bars

1 hour later…
2:22 AM
@Rithaniel I have used Comic Sans

@robjohn You and your darn wobbles.

Hello, quick question pls. How we can have integral from
$$\int_{j-1}^j{\frac{1}{j}}<\int_{j-1}^j{\frac{1}{x}dx}$$
Then how we can have this:
$$\sum_{j=2}^n{\int_{j-1}^j{\frac{1}{x}dx}}=\int_1^n{\frac{1}{x}dx}$$
I just got these in final calculations
In the former, we will get a bunch of integrals:
$$\sum_{j=2}^n{\int_{j-1}^j{\frac{1}{x}dx}}=\int_{2-1}^2{\frac{1}{x}dx}+\int_{3-1}^3{\frac{1}{x}dx}+\int_{4-1}^4{\frac{1}{x}dx}+\cdots =\int_1^n{\frac{1}{x}dx}?$$

@Avra the left side is incomplete
@Avra this is correct

@robjohn. How the last one though equals
$$\int_1^n{\frac{1}{x}dx}?$$

@Avra in the latter

2:34 AM
@robjohn. In the latter it's true because on left side we will have -1/j and in the right sid, 1/j - 1/(j-1) and thus larger?

@Avra former
@Avra latter

What do you think of the second please?

@Avra You don't see how this is true?

$$\sum_{j=2}^n{\int_{j-1}^j{\frac{1}{x}dx}}=\int_1^n{\frac{1}{x}dx}?$$
Logically yes
It's all integral from 1 to n :{(

$$\int_1^2\frac1x\,\mathrm{d}x+\int_2^3\frac1x\,\mathrm{d}x+\cdots+\int_{n-1}^n\frac1x\,\mathrm{d}x=\int_1^n\frac1x\,\mathrm{d}x$$

2:38 AM
I know in my head, but algenrically ambigues for me
@robjohn. I swear never take this in my cal before!
Never saw a bunch of integrals connected into one

do you see that $$[1,2]\cup[2,3]\cup\cdots\cup[n-1,n]=[1,n]$$

yes

that is all that is going on

is it legal to split integral in this way?
So we are treating it like sets as you did

Do you see that $$\int_a^bf(x)\,\mathrm{d}x+\int_b^cf(x)\,\mathrm{d}x=\int_a^cf(x)\,\mathrm{d}x$$

2:41 AM
We took this yes

The other is just proven using that and induction

Thanks, I know what you are talking about. Our isntructor just gave us rules back then
No deep explanation into stuff like these. Thanks

$$\color{#C00}{\int_a^bf(x)\,\mathrm{d}x+\int_b^cf(x)\,\mathrm{d}x}+\int_c^df(x)\,\mathrm{d}x=\color{#C00}{\int_a^cf(x)\,\mathrm{d}x}+\int_c^df(x)\,\mathrm{d}x=\int_a^df(x)\,\mathrm{d}x$$

very similar to this:
1

Thanks for your question. I will continue from inductive step. Inductive step: Assume $P(k)$, then we want to show it holds for the inductive step $P(k+1)$: \bigcup_{j=1}^{k+1} A_j \subseteq \bigcup_{j=1}^{k+1} B_j = \left(A_1 \bigcup A_2 \bigcup ... \bigcup A_k\right) \bigcup A_{k+1} \subseteq...

do you see how that goes for 3 integrals? we build on the case for 2 integrals

2:45 AM
Yes...
We keep building up from pair of integrals like induction

yes

you are amazing thanks

@Avra that depends on the local laws

@LukasHeger. Wie ghets :) Got it now. Thanks

ganz gut und selbst?

2:48 AM
@LukasHeger. Liebe für alle
@LukasHeger. Wünschte, ob robjohn Deutsch spricht

only a little

@robjohn. You see us :0

3:28 AM
er ist ein amerikanischer Spion

hi

@leslietownes. niemand sieht uns

hi fin

@fin. Hallo
@leslietownes. How is your toddler today?
Revived from flu hopefully ?

still not 100%, but much better than the previous day

3:38 AM
it has been over 7 days I guess?
Flue takes from 1 week to 2 weeks

it's been maybe 5 or 6, yeah

great flue is 1/x as x->\infity :/
hopefully all good soon. the worst thing when the toddler is 2 or 1 year old
Where you know nothing as to what he/she is complaining about :(
At least at older ages, they can express something

yeah. she's good about describing something about what's wrong, most of the time. she hasn't learned how to blow her nose, which will be a big step.

You can understand the signal and signal to noise ration is lowwwwwwww
That's alone is struggle to teach

we put her to bed almost an hour ago and she's still playing with one of her toys in a completely dark room
the cat is keeping her company

3:43 AM
You should cam your kid so that she sees how much you struggle so that she would appreciate that :/
the cat is evil
:/
Cats don't go to sleep at night?
Never paid attention :0

we should do that. our baby monitor does take video, and can record to a micro SD card
the cat's asleep at her feet on the bed

gosh
I love cats at house not dogs
cats are very intelligent

i like some of my friends' dogs, but only to visit. i would only want a cat at home. cats are a little better about taking care of themselves.

@leslietownes. We are 99.99 matching on examples, animals, ... :/
dogs are messy agreeee
mathematician enemy is being messy

they can be fun, but they are more work. and i'm lazy, like a cat.

3:51 AM
mathematician want his/her wife/huband be a rectangle with no mistakes
sharp
:/

eh, nobody's perfect.

@leslietownes. From mathematician eyes, I can predict your behaviour as x-> infinity
Don't play with me
Even if there is something missing I will interpolate that
No escape from mathematican

i was driving around with expired registration on my license plate. i'd paid to update the registration, but forgotten to put it on/in my car. i fixed that today.
when i mentioned this to my wife, i learned that her registration expired the month before mine, and that she hadn't even received an updated registration (although she paid for it, probably stolen in the mail). this troubles me more than it troubles her.

Horrible
I was told to insure my license plate :/

and I don't have a driving license :'(

4:02 AM
i am generally a more anarchistic person than my wife, but i do keep track of the bureaucratic stuff. things like tickets for random driving infractions, fines for overdue library books, etc. bother me. they don't bother everyone.
i am holding the universe together with my regard for rules and regulations.

@Koro. From mathematian eye, possibibility is uniform, so I can get a chance for a license plate. I don't have 0 chance for anything

driving is annoying. i went without a license for as long as i could manage.

@leslietownes. Please don't mention that. Both of my brothers wife are mathematician...I can tell you you are angle
:/
@leslietownes. Crazy drivers horrify me, so I would prefer to keep everything up to date
Specially at night around drunk people

this is a good opportunity for ambiguity without punctuation. two brothers and two wives, or one brother and two wives?

If you are early owl, then is much better
@leslietownes. 2 brothers and 2 wives :'
one for each. one for each. one for each...once upon a time

4:10 AM
prof Ted: one question: why wasn't it felt that bounded variation chapter should also be incorporated in Spivak's Calculus?

That isn’t even in most analysis courses. There is more in Spivak than I could cover in 120 classes.

my guess is that the Bounded Variation Society couldn't afford the hefty promotional fees.

It belongs more with measure theory.
@leslie the munchkin got the flu from day care?

yeah, isn't it basically the lebesgue decomposition of a measure?
i guess some people might be interested in it for generalizations of the FTC, independent of any general concern with measure theory. those people should buy their own books.
ted, we don't know if it's the flu. might just be a long-running sinus infection. whatever it is, definitely from day care.
i got it too, as a bonus. had a splitting sinus headache that did not go away with OTC meds for about two days. made it difficult to sleep. it is gone now.

I’ve had sinus guck and drip for a week. It’s that time of the year.

4:22 AM
we know it's from day care because they haven't sent her home, so i'm guessing all of the kids have it. the only hard-and-fast rule with those folks is you can't run a temperature and be at day care. munchkin had a temp last weekend but it was gone by monday.
daughter is finally asleep. cat is curled in a tight little ball at her feet.

4:46 AM
why is $p(x) e^{-x^2}$ bounded for any polynomial $p$?

hey, so what's a good place to self learn algebraic topology very well? hatcher seems to have very less details for a beginner, so that isn't working for me

@epsilon-emperor Greenberg is good

twink, suffices to check for any fixed power of x. implied by knowing that x^n/(e^(x^2)) goes to 0 as x goes to infinity. could prove that e.g. by induction on n using l'hopital's rule.

ok thank you

@Twink thank you!

4:59 AM
twink: another way to prove that x^n/e^(x^2) goes to 0 would be to first prove that x < e^x (by whatever method you like) and note that it implies x^n e^(-x^2) < e^(nx - x^2) which again goes to 0 because nx - x^2 goes to -infinity no matter what n is.
lotsa other methods and tricks exist, no doubt, based on whatever tools feel simple or are readily at hand.

5:23 AM
some professors are such a dick

ohhh i can get a measure of the economic power of an agent by taking the inner product of his economic activity with that of the total economic activity on a $\Bbb R^l$ space of $l$ commodities. isn't that the niftiest of things

I don't understand how people like that get qualified to be an instructor

inner products are mental

the mathematics is finally meshing with the economics and everything is getting super interesting
it already was but more spice

5:34 AM
in my generation and before, irish folks were very observant of matters bureaucratic because, as the saying goes, you can't fight city hall. job applications would be denied because your letters strayed outside the box, etc. i have receipts back to the 80's, etc. someday i will have to let go.
spice was the name of a popular analog circuit simulator.

spice also sounds like a slang for some sort of illicit drug
but uh, i meant interesting
don't receipts have some sort of legal time limit validity?
hm nvm i'd expect not

I'm so stupid, I don't know why I study math
I don't even know how to solve this stupid exercise

the exercise seems difficult.

no
in one way it's obvious

yes, from limit to showing sup is finite, seems obvious.

5:49 AM
yes

conversely, if we put k=0, then f and all its derivatives are bounded.
hmm, i think contradiction should prove it the other way also.

:/

6:14 AM
So f and all its derivatives are bounded. Hence, for every $l$ there exists M such that $|f^l(x)|\lt M$. Consider $g(x)=|x^k||f^l(x)|$. If $g(x)$ does not have limit $0$ as $x\to \infty$ , then g(x) is larger than a non zero value d for infinite distinct values of $x$ (let's say they make a set S). Then by choosing larger and larger $x$ in S, we see that $g(x)$ is unbounded as $x\to \infty$ (x are in S). This contradicts the given hypothesis. @Twink
I think this should work, but I may be wrong also.

6:33 AM
@Koro Where did you use $|f^l (x)|<M$?
And what if there's a constant $c>d$ such that $g(x)<c$ on $S$?

if sup |x^2 g(x)| is some finite C, then |x^2 g(x))/x| <= C/|x| allows you to evaluate lim |x g(x)| = 0. nothing too special about the powers 2 and 1 here. a finite sup assumed on a higher power should gives you a 0 limit on any lower power.

Fantastic @Leslie
@Twink: please note that I have used a set S (collection of such x for which g(x) is >d)
Some more details should be provided in my "alleged" proof, I think.

Yeah but there might a $c>d$
@leslietownes wow
how do you do that?
:(
how do you come up with answers so easily?
I want to be like that :(

experience.

@Twink $g(x)$ is greater than $d$ for x in S. Hmm, I think it should also be proven that $f^{l}(x)$ is away from zero (I mean my argument works if $|f^{l}(x)|\gt m$ for some $m\gt 0$.

6:46 AM
@leslietownes had you ever seen this problem before?
@Koro but if $x$ is large it doesn't mean $g(x)$ is large

i don't think so, but i've seen a lot of stuff like it.

it's just greater than $d$
but not unbounded
$d$ is fixed

@Twink: as accepted above already, if $|f^{l}(x)|\gt m\gt 0$ for all x, then $g(x)$ can be made arbitrarily large.

hmm but why would $|f^l(x)|>m$? Actually $|f^l(x)| <M$

i had that in mind but my argument was missing "how to find such m or whether such m exists or not". That's why my argument needs more work. :)

6:52 AM
Ok thanks for helping anyway :-)

the lim finite implies sup finite direction is easier because if you assume lim finite for f, you immediately get sup finite for f without considering other functions. the other way does require you to hop around a bit in the schwarz space. e.g. if you only make those assumptions for 0 <= k, l <= some fixed N, the lim finite implies sup finite for all k, l implication still works, but the other one would break down.

i think you should think more about this @Twink. I hadn't seen this question either ever before. But I hope the argument above can be fixed.
hmm, i think there is no fix to the argument, you may ignore that please.

@leslietownes do you think we can use the same argument if we change $|x|^k$ by $|p(x)|^k$ where $p$ is any polynomial
Because in another book they define the Schwartz space like that
with any polynomial $p$

feels like it ought to work, both in the sense of those two conditions still being equivalent to one another, and also in the sense that it shouldn't make a difference in the definition of the underlying space.

I hate the Schwartz space

7:01 AM
what makes this all fairly uncomplicated is that these things are being assumed for a wide range of parameters (all powers/polynomials of x, all k, all l). it gets tricky if you begin placing limits on those things.

@TedShifrin that is not polite :(

3 hours later…
9:43 AM
@Twink if $\sup\limits_{x\in\mathbb{R}}|x|^{j+1}|f^{(k)}(x)|\lt\infty$, then can you show that $\lim\limits_{|x|\to\infty}|x|^j|f^{(k)}(x)|=0$?
The first condition says that there is an $M$ so that $|x|^{j+1}|f^{(k)}(x)|\le M$; therefore, $|x|^j|f^{(k)}(x)|\le M/|x|$

2 hours later…
11:59 AM
How is the problem called? You want to cross a river with a boat, but it has a current. So you have to travel at a specific angle to cross it the fastest way possible.

12:21 PM
2

We all know that a circle is exactly defined by three distinct non-collinear points. But I need a way to solve the following problem: Given three points, calculate a circle with all three points on its border if it exists, else calculate a circle with minimum radius which has two points on its b...

2 hours later…
2:49 PM
A question for the analysis people around (which I'm thinking about making a proper question on the site): Is there such a thing as a maximal nonmeasurable set under the Lebesgue measure? I'm thinking the answer is "no," because my attempts at applying Zorn's lemma have encountered the issue that the union of nonmeasurable sets might not necessarily be nonmeasurable

3:24 PM
$[ \lim_{r\to 0} 1/r \fint f(x)-f(0) ]=0$

are you gonna clean that up? I'm not sure what youre asking

$[ \lim_{r\to 0} 1/r \fint f(x)-f(0) ] =0$ for a function f differentiable at zero
i cant

what's \fint supposed to represent?
an integral?
\int?
f \int?

i want to prove the average integral over a ball divided by R the radius of f(x)-f(0)
the limit is zero

in $\mathbb{R}$?

3:31 PM
$\ \lim_{r\to 0}1/rB(r,0) \int ( f(x)-f(0) \ =0$

there we go. except now get rid of the superfluous [ ]

where f is in R^n --> R
also its not the classic integral
its the average

ok, can you prove it when $f: \mathbb{R} \rightarrow \mathbb{R}$?
what course is that, multivariable calculus?

pdes

ok

3:33 PM
yes im tryna do it for one dimension

anyway, what is this fact in $\mathbb{R}^1$?

i can divide and multiplicate by x inside the integral to make the derivative appear

ok, and?

at zero

can you write this statement in $\mathbb{R}^1$ with open intervals?

3:36 PM
$\lim_{r\to 0} \frac{1}{2r^2} \int_{-r}^{r} f(x)-f(0)dx$
if im not wrong

can you send me the original statement, I can't read like this. I forget the exact statement myself. It should be the mean value theorem for integrals, anyway

its only differentiable at zero
the statement is for an open interval

can you send me the statement verbatim

prove that if f is differentiable at zero
and continuous at the ball
f:R^n-->R

why can't you just send this to me verbtaim
if f is diff at 0 and continuous on the ball then the equality above holds?
just screenshot the entire statement

3:52 PM
its in greek
yes exactly as you said

oh
ok

Is there a method that has a name which might be derived from Newton's method which finds peaks in functions instead of roots? (i.e. $f'(x) = 0$) You could use a similar method by computing the perpendicular of the derivative.

I'm a little busy right now, I'll work it out and get back to you soon @Jam

cool no problem . I cant believe it i cant solve it haha

I'm thinking it could be a means of analyzing fourier transforms.

3:58 PM
@robjohn what fraction of points on an $n \times n$ grid sit at a distance $< \log (n)$ of one another?
I need to upper bound this (maybe its an overkill)

4:13 PM
@JoeShmo All points sit within $1$ of another point. What is it you really want to ask?

Prof. Rob, one day we discussed {$\sqrt n$} is dense in [0,1]
and the other day, we also discussed {$nt$} is dense in [0,1], where $t$ is irrational.

I apologize for wedging myself into the conversation, but I need a little advice from a specialist in vector-matrix calculus.

I think that since {$\sqrt 3 n$} is a subsequence of {$\sqrt n$} and the former is known to be dense, denseness of the latter follows. :)
here {.} is fractional part function.

@dtn by "known" do you mean "fixed"? Or are they arbitrary inputs?

hi everyone

4:30 PM
@robjohn, well but some points sit at a distance $> \log(n)$ of one another. hence, the question is what fraction of points sit within, and without $\log(n)$ out of the total number of possible pairwise distances I guess? Idk what I'm looking for here.. the way I just asked could be just $\sim \frac{\log(n)}{n}$ maybe, just by looking at a single row and hoping hard it might stay that way after considering all points

what is the difference between two points on a plane connected by 4 curves and two points in $\Bbb R^3$ connected by 4 curves, s.t. no curves are co-planar

the original question is: let $(X_n)_{n \ge 1}$ be rvs on the same probably space with $\mathbb{E}(X_i) = \mu$. Suppose $Cov(X_i, X_j) \le f(|i - j|)$ where $(f(n))_{n \ge 0} \rightarrow 0$ as $n \rightarrow \infty$. Prove $n^{-1} \sum_{i=1}^n X_i \rightarrow \mu$ in $L^2$.
so you get $\mathbb{E}(n^{-1} \sum_{i=1}^n X_i - \mu)^2 = Var(n^{-1} \sum_{i=1}^n X_i - \mu) = Var(n^{-1} \sum_{i=1}^n X_i) = \frac{1}{n^2} \Big[ \sum_{i=1}^n Var(X_i) + 2 \sum_{1 \le i < j \le n} Cov(X_i, X_j) \Big ]$

@anakhro Yes, just arbitrary vector and matrix

and so for that last covariance term I need to know that there is only a "small" amount of indices $(i, j)$ (looks like no more than $O(n \log(n))$) that are "close" to one another (hence I'm thinking at most of distance $O(\log(n))$ of one another). and the rest can go to to $0$ since you can pick $n$ large enough so that $f(\log(n)) < \epsilon$

Which is lim ln(n!)/n as x->+infinity?
ln(n)/n + ln(n-1)/n + ... + ln(1)/n = 0
But it's wrong I think
Graphic calculators stop at 169, 4.15 for some reason

4:47 PM
HOW DO i post pictures here

upload button next to the typing box

i dont see it
also hi semiclassical its me Meow mIX
i used to frequent here like 5 years ago lol

or, well, ok. maybe we can fix $k \in \mathbb{N}$. and let $n \rightarrow \infty$. then the number of pairs increases as $n \rightarrow \infty$, but the proprtion still goes to $0$.
I'm going to squint, pretend that this works, and hope for the best.🤞

5:09 PM
nope, there are $O(n^2)$ such pairs anyway with a fixed $k$

5:29 PM
mm, i've seen this somewhere before.
mm, no, the thing i was thinking of wasn't it. hypotheses way too different and with operators and not random variables. it's some idea like that, though, joe.

no I'm retarded
it's trivial I way overcomplicated it
fix $k$, then there are $O(k(n-k) + k^2)$ such pairs
which is linear in $n$, which is what we want

5:55 PM
just by counting.

6:14 PM
lim ln(n!)/n as x->+infinity

it's ln(n!)/n

@Curio Here's a hint: Think of $\ln(n!)$ as an upper/lower estimate for $\int_1^n\ln x\,dx$.
Your method is not valid because there are $n$ terms and $n$ is varying. This seems to be a theme of mistakes on here lately.

it's a powerful mistake theme. i wonder if it is evidence of a malevolent deity devoted to manifesting this theme on earth.

6:30 PM
lmfao

If so, that malevolent one is likely you!
Hearing no rebuttal, I take the point to be accepted by the court.

i have the right to remain silent

d/dx (x^2) = d/dx (x + x + x + ... + x (x times)) = (d/dx x + d/dx x + d/dx x + ... + d/dx x) (x times) = (1 + 1 + 1 + ... 1) (x times) = x

I love adding a number to itself $\pi$ times.

6:44 PM
asking someone to add a number to itself pi times is a polite way of asking them to go sit in the corner and count to a billion
which is a polite way of saying a few more unpleasant things

There you go playing the malevolent one.

in undergrad there was a stubborn russian dude who insisted that the halting problem was false or something
or something else that was idiotic

we are supposed to get a little bit of rain today. i parked outside for the free car wash.

or no, that a certain construction doesn't exist or something

i hope there's enough rain that it doesn't just make my car dirtier.
every school seems to have somebody like that every few years.

6:47 PM
and prof tried to explain to him that it was impossible and the dude was hogging up the entire lecture for attention
so everyone's talking about it after class, and prof is frustrated, so I proposed that he encourages him next time to go find an example
but really, really encourage him.

now that's malevolent.

@leslietownes I might move mine later if it pours. We are limited to 2 hours of street parking.

which in my view is the appropriate way of dealing with a troll

Sometimes class trolls have mental issues in my experience.

encourage your students to achieve the impossible, as they say

6:52 PM
just doing that, all by itself, is a form of mild illness. they are lucky if the only harm is that they annoy people. some of those people waste a whole lot of personal time and become very, very unhappy if they feel that nobody is listening to them.

We have our share of trolls in this chat.

ahem. thankfully, everybody listens to them.

Oh?

you never hear about campaigns to ban actual math textbooks from the classroom. i exclude K-12 stuff with fluff inside from this, because i'm sure that's been done. someone should pick a standard undergrad book and wage a scorched earth social media campaign against it.

@TedShifrin I think anyone can be a troll, given the mood strikes them

7:02 PM
yeah, anyone can troll. i think the more intense case that joe was describing is something beyond that. the word 'crank' is sometimes used. i can choose to be irritating and at least amuse myself for a minute. many of those folks aren't deliberately choosing to do that. they can't shut it off and they aren't having fun.
or maybe they are. let's ask them! can i get an email address, or better yet, a telephone number? i want to discuss the validity of cantor's diagonal argument.

right exactly. he was an interesting character that guy
mental issues certainly describes him correctly
leslie can you please cc me

what are the 4 1 dimensional irreducible complex representations of the dihedral group $D_n$ where $n$ is even?
I know sign permutation, and trivial permutation
what else?

7:17 PM
@Leslie:
My understanding here is correct, right?
3 hours ago, by Koro
I think that since {$\sqrt 3 n$} is a subsequence of {$\sqrt n$} and the former is known to be dense, denseness of the latter follows. :)

no
nevermind that you have a typo that just makes the statement not even mean what you intended it to mean
the rationals are a dense subset of the reals

The former is known to be dense in [0,1] as {nt} is dense in [0,1], where t is a fixed irrational.

or wth is the context
but subsets of dense sets are not necessarily dense
the natural numbers are a subset of the rationals
oh excuse me,
no youre right

Joe: I was trying to think of an easy way to prove the claim that the set {{$\sqrt n\} : n\in \mathbb N$} is dense in [0,1].

7:23 PM
where {.} means fractional part function.

yeah i dont understand the notation completely @leslietownes and parsing it may take a while
i defined $f_1: D_n\rightarrow \mathbb{C}^{*}$ to be $f_1(r)=-1$ and $f_1(s)=1$ where $r$ and $s$ are the generators of $D_n$
would that be another 1 dimensional irreducible representation?

I know that {nt} is dense in [0,1] , where t is a fixed irrational. Notation abuse here purely for brevity. {nt} may be treated as a set of all {nt} such that n is any natural number. Since {$\sqrt 3n$} is a subset of {$\sqrt n$}, and the former is dense in [0,1] so is the superset.

Leslie: I know how to prove dense-ness of {$\sqrt n$}. I have discussed that earlier with prof. robjohn.
I was wondering if what I said above is correct…

well then which direction are you going

7:31 PM
Why I want to know about correctness of the same is because then knowing $\{nt\}$ dense in [0,1] gives denseness of {sqrt n} as a corollary, we might say.

i'm not sure what you're asking.

just because $\{\sqrt{3n}\}_{n \in \mathbb{N}} \subseteq \{\sqrt{n}\}_{n \in \mathbb{N}}$ doesn't mean the prior is dense in $[0, 1]$
the other way around does work
and yes, I, too, don't know what you are asking.

Given: $\{nt\}$ is dense in [0,1], where t is a fixed irrational. This can be proven to be true. Now question is to show: $\{\sqrt n\}$ is dense in [0,1] using the given statement.

n running over positive integers everywhere?

I say yes because let S={{$n\sqrt 3$}: n is a natural number}, T={{$\sqrt n$}: n is a natural number}. Now S is a subset of T (because for every $\sqrt 3 n$ in S, we note that $\sqrt 3 n=\sqrt{3n^2}$ and RHS is in T). S is dense in [0,1] by the given statement.
Yes @Leslie, n runs over natural numbers and {.} represents fractiional part function.

7:40 PM
ok, yes, i agree. note that it is sometimes easier for people to follow and read these types of arguments if you don't use the same dummy variable in both set definitions

Since S is dense in [0,1] so is T.
This completes the proof.
I think the proof is correct. Right? @Leslie

how many times do you want me to say i agree

Haha, you already said. I didn’t see. I’m on phone.
can you say that $\sqrt 3$ times? :P
@leslietownes ahh, I had used n earlier and that was causing confusion. I see why my question was not understood earlier.

it might be a tiny bit easier to follow if you wrote a chain of set equalities like {m sqrt(3): m in N} = {sqrt(3m^2): m in N} subset {sqrt(n): n in N}. your point is that the extreme left hand side is dense in [0,1] by some theorem, so the extreme right hand side is too. but i get a little dizzy if i see n sqrt(3) and sqrt(n)

You’re right. Noted.

7:46 PM
with fractional parts everywhere (which i realize now are missing above) it is kinda visually icky no matter what, without n playing two roles
that thing about the fractional parts of nt being dense in [0,1] when t is irrational is a cool and useful result

yeah, I noted fractional parts were missing in your last comment but I understood the spirit of the comment. :)
And this one too: m+nt, where m, n run over all integers.

and the quantitative results on the distributions you get. cool stuff even if people do use it in number theory.

8:20 PM
@Leslie is now our official malevolent one.

i'm under the influence of my cat

A ghoulish influence it is. Appropriate for the season.

this is the week when her powers are at their strongest.
free car wash has begun.

Ah, it hasn't begun here yet.

not enough to make a dent in any water situation, but it'll make my car look slightly less bad

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