Mathematics - 2023-08-30 (page 1 of 2)

00:00

There doesn’t seem to be a specific text for the class. Topics are: Basic logic: implications, proofs (direct, indirect, induction, contradiction, contraposition), defining sets: reals, integers, etc.
Convergence of sequences, series
Continuity of functions
Differential calculus: Define, compute derivatives, Taylor's theorem
Integral calculus: Define general Riemann integral, compute special integrals (including improper integrals)
Uniform convergence, continuity: Interchanging limits, sums, integrals

Ted Shifrin

Yeah, this is all single-variable “baby” analysis.

Cotton Headed Ninnymuggins

The 2nd semester of the class is: Topology of n-dimensional Euclidean space. Rigorous treatment of multivariable differentiation and integration, including chain rule, Taylor's Theorem, implicit function theorem, Fubini's Theorem, change of variables, Stokes' Theorem.

Ted Shifrin

That part is more rigorous multivariable. This sounds like the old Buck Adv Calculus book.

Cotton Headed Ninnymuggins

He cites several books, not any in particular. A couple are Elias Zakon - Mathematical Analysis I, and Terence Tao - Analysis I

冥王 Hades

@robjohn Don’t worry I found it. Here:

GratefulDisciple

00:11

@leslietownes that's where i first encountered rate my professors stockholm syndrome. Makes one shudder to think whether the same students might be experiencing stockholm syndrome toward their dad/mom, their church, or (horror of horror) their future spouse !

Cotton Headed Ninnymuggins

What is Stockholm syndrome?

I also don’t think it’s hard for people to make multiple Rate My Professor accounts to write multiple reviews for the same professor

The lowest RMP I’ve seen is a 2/5 which one of my physics professors had

GratefulDisciple

@CottonHeadedNinnymuggins It's a psychological coping mechanism in a power imbalance situation where one is trapped, so instead of seeing clearly how one is abused, to preserve one's sanity one develops reasons to praise the "captor". Named after a real hostage situation here. So the victim would rationalize that they benefit although in reality they do not. They can go as far as defending and protecting those who abused them, which makes it tragic.

Xander Henderson

@GratefulDisciple [footnote] No mainstream psychological or medical organization recognizes it as a diagnosable condition.

Abdo Ismail

I'm just here to tell everyone to have a good night! <3

Cotton Headed Ninnymuggins

@GratefulDisciple interesting. I’ll have a psychology course this fall for the first time.

@AbdoIsmail you too

GratefulDisciple

00:20

@CottonHeadedNinnymuggins Great. Ask your professor, he/she should know.

Ted Shifrin

I think the teacher/student relationship Is truly not an abusive one except in very rare occasions. Being intellectually demanding is not abuse.

Night, Abdo.

Cotton Headed Ninnymuggins

I agree^

GratefulDisciple

@XanderHenderson Is that so. That's interesting. I guess only a psychoanalyst in a one-to-one therapy sessions might help one able to see it. What I have seen in others is those who defend their "church" leaders who I would categorize as cult leaders.

@TedShifrin Completely agree. It's really how they relate to their students human-to-human.

Ted Shifrin

On the church I totally agree. Same with MAGA, clearly.

Xander Henderson

@GratefulDisciple I feel like I've lost something in your train of thought. All I said was that no medical institution recognizes it as a diagnosable condition.

GratefulDisciple

00:25

@TedShifrin Yes, which worries me very much !

Xander Henderson

The wikipedia article you linked seems to make that point at least three times in the opening spiel.

Ted Shifrin

I guess some profs do have almost a cultish following. I wonder if some perceived me that way .

Cotton Headed Ninnymuggins

And that the FBI only observes about 8% of hostages showing signs of the “syndrome”

冥王 Hades

Wonder if Koro watched Gear 5

GratefulDisciple

@XanderHenderson I can agree about its not being objective enough to deserve a DSM entry. But I think even though its vague, it can be useful as a name / label when helping a possible victim. One famous (fiction) case is 007 movie The World Is Not Enough.

冥王 Hades

00:33

I don’t feel like going to college today

GratefulDisciple

@CottonHeadedNinnymuggins Which seems to be common sense. Only egregious and long-term hostages would show the sign.

Xander Henderson

@GratefulDisciple (1) I'm not sure that I can agree. The goal of labeling something is to use that label to offer potential treatments and prognoses. If the label doesn't offer any help with treatment, or any idea of what to expect, then it doesn't seem like a very useful label. Which is slightly tangential---I don't know enough to be able to say whether or not "Stokholm syndrome" is a useful label. The medical community, however, seems to believe that, for the moment, it is not.

(2) When your "famous example" is a fictional character, it doesn't really help your case. There are a lot of fictional ailments which don't map to anything in reality. "Zombie-ism", for example.

GratefulDisciple

@XanderHenderson Yes, I also don't know enough. I'm just using it as a layman, which professionals may get annoyed by me :-) . Maybe like how some users automatically say "it's a bug" when reporting a problem.

冥王 Hades

Some real life diseases/conditions are even weirder than fictional ones. I only recently learned that Treeman syndrome is a real thing

GratefulDisciple

@冥王Hades Wow, you're right. Just saw some pictures. Terrifying.

冥王 Hades

00:39

There’s a reason why I didn’t link it

GratefulDisciple

@XanderHenderson OK. Guilty as charged. Anyway, I need a label when I see someone got hooked by a cult Christian teaching that no matter how I reason with him/her, still cannot recognize how that teaching is out of the mainstream.

Xander Henderson

@GratefulDisciple I mean, one does not need to be mentally ill in order to be taken in by a cult.

GratefulDisciple

@XanderHenderson Yes, and I wouldn't category someone with a "Stockholm syndrome" mentally ill either, more like having a distorted reality.

Xander Henderson

Cults tend to be much more about the charm, charisma, and control of the leader (and the structures of power they build around themselves) than the susceptibility or mental health of the followers.

冥王 Hades

MAGA?

GratefulDisciple

00:46

@XanderHenderson What label would you use for the true-believers followers of the cult leader?

Xander Henderson

@GratefulDisciple Do they require a label?

Are all followers suffering from an ailment which can be treated following some standardizable protocol?

GratefulDisciple

@XanderHenderson Not for name-calling them, but to have a handle on their "condition". And also not to claim we have a specific treatment for them. More like labelling an emotion, I guess, like said, broken-hearted, etc.

Xander Henderson

Also, what's wrong with "true-believers" as a label?

冥王 Hades

I mean, I call them Maggots. Does that help?

Xander Henderson

@冥王Hades No.

GratefulDisciple

00:51

@XanderHenderson Well.... I recognize that in popular parlance "true believer" is quite established. Even my dad used it before I understand what it means. But among academic theologians, we don't like that because we would like to use it positively in the sense of "genuine believer" instead of "cultural Christian" for example.

冥王 Hades

I refuse to call them “true-believers”

GratefulDisciple

Another reason: it devalues the word "truth". I would rather use the label: "die hard gullible follower", implying someone who cannot think for themselves.

Xander Henderson

@GratefulDisciple I mean, you're the one who used the term "true believer".

GratefulDisciple

@XanderHenderson Again, guilty as charged :-). Maybe subconsciously I have resigned to the popularity of the term in the negative sense.

Xander Henderson

@GratefulDisciple Don't you think that this rather dehumanizes the cult followers? You seem to want to completely take away their agency. This is what the cult leader does to control them, no?

Again, I would suggest that cults are much more about the leader than the followers.

GratefulDisciple

00:55

@XanderHenderson The real examples I'm thinking about, and the sad thing about it, is that the leader does NOT take away their agency. We're not talking about Waco style cult here.

Xander Henderson

And just imagine trying to deprogram someone if you are starting from "Well, you are a gullible idiot who can't think for themselves. No wonder you got cultified." Always good to blame the victims...

冥王 Hades

@XanderHenderson You just called me out without knowing it…

GratefulDisciple

@XanderHenderson OK, I see now. Yes, cult IS about the leaders. And yes, "gullible" probably wouldn't be the right term to use. But internally, I don't see them as idiot, but as victim that does not WANT to think for themselves, because they have trusted the leader too much to even CONSIDER that the leader might be wrong.

So I want to help the victims, but I find many of them don't want to be "helped". They rather want to recruit me instead !

冥王 Hades

There is a very notable and specific community I have in mind that follows that example perfectly.

GratefulDisciple

What I see is cult leader operates on the followers' desire, which motivate them to do certain things, including donating money, energy, buy their books, and spread the teaching to others. One is trapped in the "movement" it's hard to get them out. When one criticizes the leader, they automatically defend the leader. I would categorize them as savvy marketer: they make the followers feel good about themselves, give them meanings, etc.

@冥王Hades I can probably guess what you have in mind.

John Zimmerman

01:20

how on earth do you notate "counting the number of curves of the family $a_t(x)$ which are greater than the number of curves of the family $b_t(x)$ up to a given magnitude?"

where $t$ indexes the set of functions.

You have to specify a path along which you count it seems. So the best I can come up with is this:

Consider a function: $f: \Bbb N \to \Bbb N$ defined as follows:

$$f(n)=\bigg\lbrace\#\lbrace a_t(x) \rbrace >\#\lbrace b_t(x) \rbrace: \# \pitchfork \mathrm{id}\bigg \rbrace.$$

where $ \# \pitchfork \mathrm{id}$ means the counting is done along the identity path

BlR

Question about eigenvectors: if I have two matrices of similarities (following 3000 objects) and I want to identify which eigenvector of matrix 1 corresponds to which in matrix 2. How do I go about it? (assuming have a spectral decomposition of 400 vectors for each matrix)
I thought of matching the vectors based on their quantiles but at that scale (3000 objects, 400 dimensions) that takes way too long.
What other more direct approach can I use?

GratefulDisciple

@XanderHenderson Reading the criticism section of the Wikipedia article, yes, I think I'm rather careless in using "Stockholm Syndrome". And yes, it's about the leaders, as theology professor Roger Olson identifies 4 cultic features in his new book announced here. Thanks for your feedback.

Thomas Finley

01:59

0

Q: Problem in understanding the reason behind the continuity of the irrational points in Thomae's Function

We consider $h:A\to\Bbb R$ where $A=(0,\infty).$ In Thomae's Function we consider: $$ \begin{align} h(x) = \begin{cases} 0 & \text{if $x$ is irrational}\\ \frac{1}{n} & \text{if $x = \frac{m}{n}$ where $\gcd(m,n) = 1$} \end{cases} \end{align} $$ It is also to be noted that $x\in A.$ I was studyi...

I need some help with this...

@leslietownes Can you please take a look at this? :?)

copper.hat

if you post a question, why don't you wait a while to get an answer instead of immediately posting same here?

Thomas Finley

@copper.hat The problem's that in here, I have friends like you all!

Thomas Finley

02:13

I think in my post, the proof is not saying that there exists at least one rational number in canonical form whose denominator is less than $n_0,$ but they are saying instead, that even if there is some rationals whose denominator is less than $n_0$ in canonical form, then the number of such rationals are finite. But I don't know whether this is the thing intended ?

But I am having a strong feeling that this is the case, precisely

copper.hat

02:30

once again i am a bit lost. in any bounded interval there are a finite number of values of the form ${p \over q}$ for a fixed $q$.

Thomas Finley

@copper.hat Check my edited post, when you have time. I have edited it, adding more and more clarity in my words, in my thoughts, in my presentation to the best of ny ability! ;)

@copper.hat that's what I said :))) in :chat.stackexchange.com/transcript/message/64294062#64294062

I just made the thing more rigorously clear...

(Hopefully)

copper.hat

any two rationals with denominator $n_0$ are separated by at least ${1 \over n_0}$, so any interval of length $2$ can have at most $2 n_0+1$ rationals with denominator $n_0$.

Thomas Finley

@copper.hat yes, but I said the same thing using a different argument in OP under the section of My thoughts under the paragraph "Doubt 1" . Isn't that a valid reasoning?

Though I think, your explanation is much more direct

copper.hat

i don't know. i'm not sure what is bothering you. If you fix $n_0$ then there is an interval around the irrational $b$ that contains no ${p \over q}$ where $|q| \le n_0$.

if every interval around $b$ had an element of the form ${p \over q}$ (and $|q| \le n_0$) then $b$ would have to be rational.

Thomas Finley

02:51

@copper.hat Nope, now nothing's bothering me, whatsoever.

But, my point is, that the reason for claiming that $(b-1,b+1)$ has a finite number of the mentioned rationals is because, a rational number is of the form $p/q$ where, $p,q\in\Bbb N$ with $(p,q)=1$ and, $q$ can take any values from $1,2,...,n_0-1$ and $p$ can be any natural number in $(b-1,b+1)$ (, if at all any natural is there in $(b-1,b+1)$). If there are $z$ natural nos. in the interval, $(b-1,b+1)$ then, the total number of the mentioned rational numbers is at most $z(n_0-1).$

I just need a confirmation, where these lines of reasoning is correct or not, @copper.hat ?

And this whole thing, comes under the umbrella of "Doubt 1" of mine

copper.hat

why do you care about computing a specific number of points? it has nothing to do with the proof?

Thomas Finley

@copper.hat To rigorously assert that the no. of the rationals whose denominator is less than $n_0$ in the interval, $(b-1,b+1)$ in canonical representation is finite

copper.hat

all that matters is that the number of points of the form ${ p \over q}$ with blah blah blah in the interval $b \pm 1 $ is finite. Since $b$ is not equal to any of these points there is an interval around $b$ that does not intersect with any of the aforementioned points.

that is all that is needed for the proof.

Thomas Finley

@copper.hat proof of "the no. of the rationals whose denominator is less than $n_0$ in the interval, $(b-1,b+1)$ in canonical representation is finite" ?

If so, then I just chose this line of reasoning:

4 mins ago, by Thomas Finley

But, my point is, that the reason for claiming that $(b-1,b+1)$ has a finite number of the mentioned rationals is because, a rational number is of the form $p/q$ where, $p,q\in\Bbb N$ with $(p,q)=1$ and, $q$ can take any values from $1,2,...,n_0-1$ and $p$ can be any natural number in $(b-1,b+1)$ (, if at all any natural is there in $(b-1,b+1)$). If there are $z$ natural nos. in the interval, $(b-1,b+1)$ then, the total number of the mentioned rational numbers is at most $z(n_0-1).$

Xander Henderson

@ThomasFinley I can't believe that this hasn't been asked and answered many times already in the site. Have you searched for other answers?

copper.hat

02:56

i would consider that obvious, just as the number of integers in any bounded interval is finite

you are making things more complicated than they need to be and i do not want to chase you down that rabbit hole

Xander Henderson

@copper.hat You know my feelings on "obvious".

copper.hat

obviously

Xander Henderson

And yet, I agree.

copper.hat

i guess if i am calling something obvious then it must be

Thomas Finley

@XanderHenderson Ha ha, yeah, I searched and I found some answers. They were focussed on Doubt 2. But I have 3 specific doubts. Out of them,Doubt 1,2 are cleared. But Doubt 3 is what remains.

copper.hat

02:59

i gave a formula for the max number of rationals with denominator $n_0$ in any interval of length a short while ago

Xander Henderson

I mean, if YOU can see it, it MUST be obvious.

copper.hat

we agree there

Xander Henderson

@copper.hat I'm lazy. "By the Archimedean property ..."

copper.hat

@ThomasFinley do we agree that the union of a finite number of finite sets is again finite?

Thomas Finley

@copper.hat yes, sure!

Xander Henderson

03:01

@ThomasFinley be very concrete: how many fractions are there with denominator 2 in [0,1]?

copper.hat

so, for each natural $q \le n_0$ the collection $\{ { k \over q} | {k \over q} \in [b-1,b+1] \}$ is finite.

Xander Henderson

Denominator 3? 4? 5? ... n?

Thomas Finley

That's actually seems to be a verbatim of a special case of the statement: "Countable union of Countable sets is Countable "

Xander Henderson

@ThomasFinley no.

Nonono.

Thomas Finley

@XanderHenderson Hehe, no I intended that for copper

copper.hat

03:03

finiteness is important here

Xander Henderson

@ThomasFinley I know.

Thomas Finley

@XanderHenderson 3 ?

Xander Henderson

But a finite union of finite sets being finite is NOT a special care of the result regarding countable sets.

Thomas Finley

0,1/2 and 1

@XanderHenderson but finite sets are countable?

copper.hat

sweet jesus

2

Xander Henderson

03:04

@ThomasFinley the question was rhetorical.

@ThomasFinley Maybe. But not all countable sets are finite.

Thomas Finley

@XanderHenderson Yes, that's why I asserted that it was a special case. But I agree with you for sure!

@XanderHenderson The answer is 1, if you are asking for rationals in Canonical representation

copper.hat

@ThomasFinley Do you agree that for any $n_0$ there is an interval about $b$ that contains no element of the form ${p \over q}$ with $q \le n_0$ and $q$ natural?

Xander Henderson

@ThomasFinley I don't know what you mean by canonical representation, and I said nothing about canonical representations.

I don't care if we over count, either.

Finite is finite.

Thomas Finley

@XanderHenderson Oh, sorry Sir. I got you wrong then .

@XanderHenderson yes

@copper.hat yes

copper.hat

so then Doubt 3 goes away, right?

Xander Henderson

03:12

@copper.hat obviously. :P

copper.hat

:-)

i have a proof of the twin primes conjecture but it won't fit in the chat box.

Thomas Finley

@copper.hat Sorry, but how? Forgive me, if I sound stupid, but I don't seem to have an option.

Xander Henderson

@copper.hat You and Fermat, brother.

copper.hat

one more time. in the interval $[b-1,b+1]$ there are a finite number of rationals with denominator $0 < q \le n_0$.

Thomas Finley

@copper.hat yes, and?

copper.hat

03:19

and none of them equal $b$. are we on board so far?

Xander Henderson

Ugh ... I'm full of fish, and chips, and wine, and allergy meds, and snot. I'm going to bed.

copper.hat

ohh, fish & chips...

sorry to hear about the snot

Xander Henderson

@copper.hat dad's whub Ah maybe for dinner.

copper.hat

@ThomasFinley are do you agree with my last two assertions?

Xander Henderson

Bed. Go away now.

Thomas Finley

03:21

@copper.hat $b\in [b-1,b+1]$ and $b=b/1$ and since, $1\leq n_0$ so, one of them seems to be equal to $b$

Xander Henderson

Byebye.

copper.hat

good night @XanderHenderson

Thomas Finley

@XanderHenderson Bye, take care!

copper.hat

@ThomasFinley $b$ is irrational, that is all you need to know, i do not know what the $b=b/1$ is about.

Thomas Finley

@copper.hat oh yes, sorry, I agree now

I agree with ur 2 assertions hereby

copper.hat

03:23

so there must be a small interval around $b$ that does not intersect with the rationals of teh form p/q and $0<q \le n_0$. right?

Thomas Finley

@copper.hat yes agreed

copper.hat

So, in this interval, there are only rationals of the form p/q with $q > n_0$ or irrationals, right?

Thomas Finley

@copper.hat yeah

copper.hat

then $|h(x)| \le {1 \over n_0+1}$ for $x$ in this interval.

if $x$ is irrational it is $0$, so obvious there, and otherwise the denominator must be $> n_0$.

Thomas Finley

@copper.hat Can't we say, $|h(x)|\leq \frac{1}{n_0}$ instead of that ?

@copper.hat yes

So, $|h(x)| \le {\frac{1}{n_0}}$

Thanks a ton,@copper.hat !

I got it!

Let me write an answer to my own question, and after that I'll share it with u loud and clear ha ha!

copper.hat

03:30

@ThomasFinley great!

Thomas Finley

03:46

0

A: Problem in understanding the reason behind the continuity of the irrational points in Thomae's Function

The proof given was : If $b$ is an irrational number and $\epsilon > 0,$ then (by the Archimedean Property) there is a natural number $n_0$ such that $1/n_0 < \epsilon.$ There are only a finite number of rationals with denominator less than $n_0$ in the interval $(b-1,b+1).$ (Why?) Hence $\delta >...

@copper.hat I have already written this answer. Without you, I couldn't have written it this explicitly. Thanks again! You may check it out and suggest some improvements if needed.

Any suggestions for improving this answer by the community folks are welcome

Also, @copper.hat Is my answer correct? I tried my best tho :D

Thomas Finley

04:42

Hello there! I need just a little confirmation 'gain. The point is:

I wanna prove this "There are only a finite number of rationals with denominator less than $n_0$ in the interval $(b-1,b+1).$"

it is asserted that, that if such rational numbers exist they must be finite.

Proof : We know that there are, finite number of natural numbers less than $n_0$ i.e $n_0-1$ natural numbers are there. Similarly, there exist a finite number of natural numbers in the interval $(b-1,b+1)$ say, $z$. Thus, the total number of rationals with dwnominator less than $n_0$ in the interval $(b-1,b+1)$ is **at most** $z(n_0-1),$ which is finite. Hence, they made such a statement.

Petəíŕd The Linux Wizard

Welcome back @amWhy :)

Thomas Finley

Is my proof, valid?

@robjohn and @leslietownes Can you guys check my proof , please ? :?)

Thomas Finley

06:04

A user said, the proof above is incorrect.

But I don't know why he said that.

He asked me to consider the interval (1,3) and said, if $n_0=5$ then my approximation given the result $4$.

But I replied by saying, that we must inclide the endpoints as the endpoints are natural numbers due to which the approximation gives exactly 12.

Was I wrong?

one potato two potato

06:41

FLP is a good book, although it's not very kind. It contains many important and useful concepts and statements.

reading FLP is like looking at a work of art. I need to look long (or very long?) enough to see what they're saying.

Koro

07:27

what is the meaning of image of $\pi_2(X)$ in $\pi_2(X,A), A\subset X$?

I mean image under what map?

pi_2 is 2nd homotopy group.

I think we can regard pi_2 X as subset of the other.

an element [g] of $\pi_2(X,A)$ is represented by: $g:(I^2, I\times 0, J)\to (X,A, a_0)$, where $J= \{0,1\}\times I\cup I\times 1$. And that [f] in $pi_2(X)$ is represented by $f:(I^2, \partial I^2)\to (X,a_0)$.

Jakobian

09:10

@Koro not the other way around?

$\pi_2(X, A)$ is constructed as equivalence on maps $D^2\to X$ which map $S^1$ to $A$. Any map $S^2\to X$ can be treated as a map $D^2\to X$ which maps $S^1$ to the base point of $X$

ah okay I see

we need the basepoint in $A$ but as long as that happens there is a canonical map, which should induce a map between equivalence classes

well, that probably depends a little on how you define those, using spheres or otherwise

not a big difference

user 726941

11:00

Joel David Hamkins: Philosophy of mathematics and truth

►

Questions asked by a "Cambridge PhD drop-out."

🤯

sunny

11:14

@robjohn my doubt is still how we go from $\sin(1/2)\le\limsup\limits_{n\to\infty}|a_n|\le1$ to $\limsup\limits_{n\to\infty}|a_n|^{1/n}=1$. You and Jakobian mentioned the inequality $\sin(1/2)\leq \sup_{k\geq n} |a_k|$, but I do not see how this inequality is derived. Moreover, if this inequality does hold, how do we get that $n$th root in there so we can take the limit as $n\to\infty$.

Anyway, I have this approach, but it'd be awesome to understand yours as well.

rogerroger

11:28

is there an inequality for $(x+y)^{1/2}$ similar to $(x+y)^2 \geq x^2+y^2$ where $x,y\geq 0$

Jakobian

@sunny you see how $\sup_{k\geq n} |a_k|$ is a decreasing sequence?

sunny

yes

Jakobian

And it converges to $\limsup_{n\to\infty} |a_n|$

sunny

indeed

Jakobian

and we know this is greater than $\sin(1/2)$

sunny

11:34

the limit of $b_n=\sup_{k\geq n} |a_k|$ is greater than $\sin(1/2)$, yes

Jakobian

about the $n$th root, anyway

The function $x\mapsto x^{1/n}$ is increasing, from $[0, \infty)$ onto $[0, \infty)$

If $A\subseteq [0, \infty)$, can you see how this implies $A^{1/n} = \{x^{1/n} : x \in A\}$ is such that $\sup A^{1/n} = (\sup A)^{1/n}$

you can switch between the $n$th root and the suprema

sunny

I will try to look at this again thoroughly, indeed, this was my doubt: $\sup A^{1/n} = (\sup A)^{1/n}$.

Jakobian

$(\sup A)^{1/n}$ is definitely an upper bound for $A^{1/n}$

is it the least upper bound?

If $x^{1/n}\leq y$ for all $x\in A$, then definitely $x\leq y^n$ for all $x\in A$. So $\sup A\leq y^n$, that is $(\sup A)^{1/n}\leq y$

so indeed, the two numbers (possibly infinite) are equal

@sunny what did I use in above argument

sunny

the monotonicity of $x^n$?

Jakobian

we used that $x\mapsto x^{1/n}$ is a monotone bijection

so this will hold in a more general setting

sunny

11:47

@Jakobian how come this is true though?

Jakobian

I spared enough details, I think

figure it out

sunny

ok

Jakobian

Well I'll spell the argument out anyway just in case.
Let $x\in A$, then $x^{1/n}\leq (\sup A)^{1/n}$ since $x\leq \sup A$

since this holds for all $x\in A$, $(\sup A)^{1/n}$ is an upper bound for $A^{1/n}$

robjohn

@sunny if $|\sin(k)|\le\sin(1/2)$, then $|\sin(k+1)|\ge\sin(1/2)$; are we good on that? if so, then, is it clear that $\sup\limits_{k\ge n}|\sin(k)|\ge\sin(1/2)$ for all $n$? So we have $\limsup\limits_{n\to\infty}|\sin(n)|=\lim\limits_{n\to\infty}\sup\limits_{k\ge n}|\sin(k)|\ge\sin(1/2)$.

sunny

@robjohn clear

@Jakobian thank you 👌

robjohn

11:57

Okay, so what is the question about $\limsup\limits_{n\to\infty}|\sin(n)|^{1/n}=1$

$\limsup\limits_{n\to\infty}|\sin(n)|^{1/n}=\lim\limits_{n\to\infty}\sup\limits_{k\ge n}|\sin(k)|^{1/k}\ge\lim\limits_{n\to\infty}\sin(1/2)^{1/n}=1$

user 726941

Are you ready for the blue moon this month? @robjohn

robjohn

@user726941 It's essentially here. It will be past full when it rises tonight.

sunny

@robjohn it has clarified now a lot, thank you, I just didn't understand at first how $\sup\limits_{k\ge n}|\sin(k)|^{1/k}\ge \sin(1/2)^{1/n}$ follows from $\sup\limits_{k\ge n}|\sin(k)|\ge\sin(1/2)$, but this follows I guess simply from the monotonicity of $f(x)=x^{1/n}$.

robjohn

Look. $|\sin(k)|\ge\sin(1/2)$ at least every other $k$. so $|\sin(k)|^{1/k}\ge\sin(1/2)^{1/k}$ at least every other $k$. So $\sup\limits_{k\ge n}|\sin(k)|^{1/k}\ge\sin(1/2)^{1/(n-1)}$

Take the limit as $n\to\infty$

Nicolás A.