Mathematics - 2014-09-02 (page 1 of 3)

@JasperLoy yes

@JasperLoy this view is somewhat accurate for other forms of media that are consumed on the basis of interest. there is a dynamic force at work: if a piece of art is obscure, piracy can lend it the attention it needs to increase its sales, but if it's popular and mainstream, piracy can limit the money that would otherwise be made off of it (although the offset in revenue is only a small fraction of the piracy taking place; most that pirate wouldn't buy if they couldn't pirate).

this works for many books in particular, but not textbooks I don't think, because they are not generally obtained on the basis of a person being interested in the content, but rather on the basis that they need the book for a class. and it's no longer a fractional offset in revenue - it's far more proportional to the actual level of piracy, without the benefit of meaningfully changing the popularity levels of textbooks.

12:37 AM

right

software (e.g. Adobe) is a good example of where piracy can boost later sales

1:49 AM

@TedShifrin We were talking a while back about nontrivial maps between real projective spaces; I never made any progress on the complex case, so I posted it as a question.

Coincidentally, there was this interesting question from yesterday.

FractalHand

3:48 AM

0

Well, I know it is repetitive.I have read the proof from different textbooks.But sometimes I feel doubtful about it all.Every time I try to prove it for myself, I fail at some points.I'm asking those guys who are good at geometry, what would you do when you face proving such things?I'm doubtful a...

Any input will be appreciated !

@blue This is the first time you said so much to me.

only in recent memory

So yesterday I drank 4 mugs of coffee.

I use hot water to reduce the itch on my skin.

It's OK as long as it is not too hot.

are you putting the coffee on your skin or something?

No, they are unrelated statements.

The past few months I have been thinking about the perfect set of books to study for GRE and quals. I think that list has changed about 10 times. I still have not finalised it yet.

Although I know very little math, I probably know more math books than anyone else, lol.

It's confusing that advanced calculus, mathematical analysis and real analysis can all refer to the same set of topics.

I read that James Stewart built a 20 million dollar house. I wonder how much of that came from the sales of his Calculus.

And there are so many versions of his Calculus.

I am quite enjoying this monologue, lol.

4:12 AM

heh

This chat is a bit like my personal diary.

I say whatever I want to say.

hello, @blue

hello

Nobody says hello to me =(

how's things

4:15 AM

We are the 3 blues of this chat.

hello implies change, @Jasper, and you are omnipresent

so I'm actually teaching classes now

intermediate and college algebra, and helping out as a TA with abstract algebra

what classes?

Wow, it must be calculus?

nice

4:17 AM

An undergrad TA here is unheard of.

I have given up on answering lhf, there are too few these days

on paper I'm grading papers for combinatorics, because it was less paperwork for everyone to fill out...

haha

WTF

yeah, when I TA'd I was completely unofficial

I was the grader, and also of my own volition helped out in class

i don't know what I'll be TAing in the fall

probably some form of calculus, if I'm lucky discrete math

Calculus is like one of the hardest branches of math, lol.

There are too many formulas to remember

4:22 AM

...

like what

though perhaps blue's response is the wiser here

Like all those formulas for arc length, surface area of revolution, etc.

memory is aided by understanding

thus, understanding can be a form of information compression

But it seems that often what is given in a calculus text is justification and not proof.

i like that line

4:25 AM

For example, some people try to prove that the area under a curve is given by the integral without going into the axiomatic definition of area. In such case it is probably better to take that as a definition of area instead.

5:07 AM

I'm being asked to show that for $x\in\mathbb{R}$, $x$ is irrational if and only if $1/x$ is irrational.

I see several routes to solution, but I'm not sure how best to formally deal with $x=0$ in each of them

just assume they meant R\{0}

Huy

Anyone with a good taste for interior design? :D

5:21 AM

@blue Is this still fine or should I just go back and prove it by contrapositive instead: i.imgur.com/cjziS9x.png

(I realize this is one half of the proof and the other implication needs shown as well)

contraposition is slicker, to show not(P) iff not(Q) we can just show P iff Q

Yeah...was there anything wrong formally with what I had? (I had originally tried to do it like that only to try to avoid $x=0$)

5:54 AM

I'm having trouble coming up with a route to prove "If $x$ is rational and $y$ irrational, then $x+y$ is irrational."

a lot of similar problems are trivial..

assume otherwise, consider (x+y)-x

since Q is an additive subgroup of R, this is a special case of cosets being disjoint

'otherwise' is "x is rational and y is irrational and x+y is rational", then...ok I see

:P

thanks

Greetings

Huy

Good morning, @Chris'ssis.

@Huy Hi

@r9m did you finish the job on that series? :D

nerdy

7:35 AM

You guys have an idea on how to prove that a sequence $x_n$ has a subsequence converging to a real number c iff $ \forall \varepsilon>0, |x_n - c| < \varepsilon$ for infinitely many terms n

such an infinite set of n's determines a subsequence, and vice-versa

nerdy

You have an idea on how would " for infinitely many terms n" be expressed more rigorously

?

Would it go like this : $\forall \varepsilon>0 ,\forall n ,\exists k>n ( |x_k -c|<\varepsilon ) ?$

Huy

It would work, for example.

"for infinitely many n" (not "terms" n, since n isn't a term of the sequence, it's a variable index) is already rigorous. not sure if there's a standard way in symbols to say "there exists infinitely many"

nerdy

8:33 AM

I think in proving that a sequence $ x_n$ has a subsequence converging to a real number c iff $\forall \varepsilon>0, |x_n - c| < \varepsilon $ for infinitely many terms n , we are indirectly using something like the LUB axiom ( completness axiom ) or one of its consequence ( like the Nested Interval theorem )

9:08 AM

@Chris'ssis Hi !!! :D ya I did it !! :-) .. but the coefficient of $\zeta(2)\zeta(3)$ should be double of what you showed me last night .. the rest is perfect :D :-)

@r9m Sorry? Let me check that ... Everything is perfect there ...

:17456868 Yeah, this I got too and showed you.

@Chris'ssis ah !! I must have seen it wrong :( sorry .. I saw $\frac{9}{2}$ last night :(

@r9m Well, sorry for you then. :-)

..

@r9m Congratulations! :-)

9:13 AM

Thank you :D

sorry .. the thing is I made silly mistake 3 times while adding up the final results .. I couldn't see my stupidity last night :) .. but in the morning I figured it out .. :))

@Chris'ssis is the $\sum\limits_{n=1}^{\infty} \psi^{(1)}(n)^4$ tough ? :O

@r9m Try it and see. :-)

@Chris'ssis awwhh !! you must be LYAO atm :'( :(

@r9m :-))))

@Chris'ssis my proof is long and ugly .. I had to decompose it into 4 sums and compute them individually :( .. is your proof short ? :-)

@r9m I'll show it to you when I put it on paper. It's a crazy awesome proof.

9:24 AM

@Chris'ssis but you said its going to be in THE BOOK .. !! :O

@r9m Yeah, it is, but I also need to write all things in latex. I think I can show it to you without publishing that on the blog, right? ;-)

@r9m There will be $3$ versions in my book, all of them awesomely computed.

@Chris'ssis OH MY GOD !!! :D

@r9m :D

@Chris'ssis trim trim tra la la la :) okay .. agreed :) I won't put it until your books are published :))

@r9m Do as you wish. I'd prefer you not to do it, but it's your decision.

@r9m OK :D

@r9m The book is also meant to blows up minds ... :-)

9:30 AM

@Chris'ssis like Stark's Jericho ?! :D

@r9m lollllll :-)

@Chris'ssis That scene from Iron Man 1 is my wallpaper =P .. 'I present to you .. the Jericho .. '

@r9m Iron Man 1 was very nice! :-)

@Chris'ssis just nice ?!! that's a Crazy Awesome movie :D ..

@r9m hehe, indeed :-)

9:37 AM

@Chris'ssis hehe instead of saying @Huy Hi .. you could have just said Huy =P (although I have an ambiguity about the pronunciation of Huy .. is it pronounced with a stress on 'u' or like 'Hai' ?!)

2

@r9m :-))))))))

10:48 AM

@Chris'ssis $\sum\limits_{n=1}^{\infty} \psi^{(1)}(n)^4$ seems very difficult .. it requires evaluation of double harmonic sums of the kind $\sum \frac{H_n}{n^s}H_n^{(r)}$ and $\sum \frac{H_n^{(p)}}{n^q}$ for $p,q,r,s$ varying between $3$ to $6$ .. I don't have elementary solutions for such sums :( $\sum\limits_{n=1}^{\infty} \psi^{(1)}(n)^3$ I could manage because all $p,q,r,s$ were $1,2,3$ or $4$ (I have elementary solutions for them) :| ..

@r9m hehe, that's the idea, to be very difficult and find brilliant solutions. :-)

@Chris'ssis okay :-) ,, you must have a lot crazy amount of patience .. evaluating the 3 to 6 Euler single and double sums .. _/_ :O !!

@r9m maybe sos can help a bit?

@Chris'ssis y u say sos ?! :( that guy isn't talking to me at all :((

@r9m OK :-)

10:56 AM

for some reason he's ignoring me :( ...

@r9m Some sins?

@r9m Sorry for that. :-(

@Chris'ssis not one that I can identify at least :( .. I wonder if its something that I might have done/said that might have ticked him off .. :| Heaven knows what :( ..

@r9m I suppose you can survive this situation ... (I mean it's not a disaster)

@Chris'ssis survival ?! =P My avatar is a Ninja .. that's at least what I'm supposed to be good at :P

@r9m :D

11:02 AM

I felt stoopid for some reason .. calling myself a Ninja .. :P I can't do squats properly :P LOL

:-))))))

11:29 AM

@r9m: One can be a Ninja without being a ninja. That is the beauty of it's Zen.

@Nick how so ? :o

11:43 AM

@Khallil wat

@r9m Nuh-huh. There can be only one Book in this world.

12:23 PM

Bah! Last night I spent a long time proving that the volume of a regular $n$-simplex is $$\left(\frac s{\sqrt2}\right)^n\frac{\sqrt{n+1}}{n!}$$ only to find out that was not the question asked.

They wanted the volume of an expanded $n$-simplex

I've never heard of that before.

Ouch!

Manasi

@BalarkaSen : How was your interview?

12:57 PM

Very bad these days to me, although I'm very creative, I cannot focus on things. I force myself to continue the work on my project.

I think I'll take more breaks, and longer ones.

Good idea :-)

@Manasi Excellent.

@robjohn What is an expanded simplex?

Never heard of it.

s i m p l e x

:D

?

joke

1:07 PM

I don't get it.

s__________i____________m___________plex

Here's a cool continued fraction someone showed me today : $$5 = \sqrt{11 + 2\sqrt{22+3\sqrt{33+\cdots}}}$$

@skullpatrol I still don't get it

expand

aha

old (a + b)^n joke

yep

1:16 PM

um, where's the tex syntax doc for describing a limit? i just failed the first question in my mock exam :(

found it

is it considered "not cricket" to post a link to a troubling question here, in order to get help quicker than otherwise?

never mind, posted it anyway. felt it was too long for here

@topper It's an algebra mistake : you can't multiply one side by (1 - x) and leave the other fixed.

The correct statement after step 2 is $$\lim_{x \to 1} \; ( 1 - x) f(x) = \frac{3}{1 + x + x^2} - 1$$

1:32 PM

@BalarkaSen Firstly, thanks for looking. Secondly, I feel stupid. What I intended to do was to multiply the denominators and the numerators by $(1-x)$ (effectively one - so as not to change the value of the limit). Let me try that again before I say any more here

I'd suggest simplifying $\cfrac{3}{1-x^3} - \cfrac1{1-x}$.

1:45 PM

@BalarkaSen Got it now. I hate how slapdash I am. But I keep telling myself every balls-up is a learning experience. And that's not empty rhetoric.

2:07 PM

@BalarkaSen you need to look at the question. I reread the question and still I am not sure.

okay, let's take $$\lim_{x \to 1} f(x) = \frac{\sqrt{x^2-1}}{x-1}$$. what would be your first step (only) when solving this?

my answer was $\infty$, and that's what the answer in my solution book is (although they got there differently). but wolfram says "two sided limit does not exist". wolframalpha.com/input/…

@topper You're right about $\infty$, that should suffice to prove that the limit doesn't exist. What Wolfram tells is that not only the left and right limit converge, but they aren't equal either.

i.e., if you approach $x \to 1$ from the left then the limit tends to $-i\infty$

You can easily see that by yourself : take $x$ to be something very close to $1$ yet $x < 1$, e.g., $x = 0.999$ (say). Then $x^2 - 1 \leq 0$ but then square root of some negative number is imaginary.

2:24 PM

@BalarkaSen Got it. In good news, after learning about limits weeks ago, learning a whole lot of other material in between, forgetting about limits, and now exercising them for the first time, I have gotten my last two questions right! Next...

If you understand it, great. It's not about remembering a lot of stuff.

Keep the geometric interpretation with you.

just need to exercise it much more. tonight will be hours of limit problems. confident that i can get there

Oh wow. $$\lim_{x \to 1} f(x) = \frac{x-1}{\sqrt{x-1}}$$. I got zero. Wolfram gets zero. Given solution is $2$

I simply multiplied top and bottom by $\sqrt{x-1}$

Obviously that's incorrect. You are right about it.

The book must have a typo.

Haha! Oh wow. I'm buzzing! It's such a small victory but still...

Great work. Keep that up =)

2:29 PM

I don't know how OT we get here but I'm reminded of Alan Shearer the footballer talking on Match of the Day one week, saying that nothing breeds confidence like winning...

nullgeppetto

2:40 PM

Hi people, if $\int_{\Bbb{R}^n} mathbf{x}f(mathbf{x}) \mathrm{d}mathbf{x}$ is equal to the mean vector of a multivariate Gaussian distribution, then what is $\int_{\Bbb{R}^n} \|mathbf{x}\|^2f(mathbf{x}) \mathrm{d}mathbf{x}$? Thanks a lot!

3:05 PM

@topper: Math is like a marathon, except it's the hare who takes a moment to think that wins the race.

@Ted: No one needs sympathy when they have a good book.

@topper: Here's one to think about. (Don't use Wolfram): $$\lim_{x \to 0} x^x$$

How do I arrange equalities?

\begin{arrange}... or something? :)

@rehband: what do you mean by arranging and what do you mean by equalities.

@Nick Im talking about the MSE latex editor

I would like to write a chain of equalities below each other

nicely arranged

@rehband: You can do it with // in the case of inline. For multiline, Idk.

@Nick Thx

3:48 PM

@robjohn I just discovered other magnificent series ... and now I'm thinking of a crazy generalization.

@robjohn At any rate, some things, although very nice to me, I won't add to my book since they might be too hard.

Or there is another option ... I add them to my book and put there very brilliant proofs that make them easy ...

Stumped by $$\lim_{x \to \infty}\frac{1+3+5+...+(2x-1)}{x+3} - x$$. So far, have thought of getting a lowest common denominator which I think looks like $$\lim_{x \to \infty}\frac{1+3+5+...+(2x-1)-x^2-3x}{x+3} $$. But then it gets indeterminate. Grateful for any first steps to solving this, as I don't think mine were very useful

:17463002 I really think you are a genius! Your book will be well received!

I think the numerator becomes a problem, with $2x - x^2$ etc, when x tends to infinity

@Chris'ssis I think you should add all questions and all proofs to the book, make it as big as you want.

@JasperLoy Thank you for encouragement. :-)

3:59 PM

I have to agree with Jasper! It will turn out fantastic and I can't wait to learn from the book :)

$$\int_{0}^{1/2} \log^2(x) \left \{\frac{1}{x}\right \} \ dx=3- \log(2)-\frac{1}{2}\log^2(2) - \frac{1}{3} \log^3(2)- 2\gamma-2 \gamma_1-\gamma_2$$

I'm planning to compute the deformed Au-Yeung series too (it's a modified version).

@Chris'ssis Wow. What are $\gamma_1, \gamma_2$ ?

@rehband Stieltjes constants

@Chris'ssis I would like to ask you what your favourite analysis book is.

4:15 PM

@JasperLoy To be more specific, one of them is: An introduction to the theory of infinite series (1908)

(by Bromwich)

@Chris'ssis Wow, that book is ancient. Of course, I never heard of it.

@JasperLoy Some ancient books are really crazy awesome.

@JasperLoy archive.org/details/introductiontoth00bromuoft

I know about Knopp's Theory and Applications of Infinite Series.

Hi, I was wondering if all elements of $\mathbb{R}$ are limit points of $\mathbb{R}$? It appears to me that it is true, because for any point $r$, we can always find other points that satisfy $|x-r| < \epsilon$ (due to the fact that $\mathbb{Q}$ and $\mathbb{I}$ are dense in $\mathbb{R}$). But I'm not sure if this counts as a valid proof, and I've tried to search it on Google, but I couldn't find anything.

@Hunter It is true.

4:20 PM

@JasperLoy thanks. Is my reasoning/proof correct?

@Hunter You should take $0<|x-r|<\epsilon$ instead. And what does $x$ mean?

@JasperLoy I was about to mention it. :-)

@Hunter Also, $\mathbb I$ is not standard notation.

@Hunter Your proof is overly complicated. Just consider the sequence of points $(r+\frac 1n)$.

@JasperLoy of course, thanks. "And what does $x$ mean?" I'm not sure what you mean, $x \in \mathbb{R}$ and $x \neq r$. By $\mathbb{I}$ I mean the set of irrational numbers, what is the typical notation?

I'm using notation of Understanding Analysis by Stephen Abbott

@Hunter Well, it isn't too widely used. =)

@Hunter So your $x$ is a real number distinct from $r$? Then that is correct.

4:25 PM

@Chris'ssis Ping me when you post cool stuff, that way I can see it when I come back :)

@JasperLoy ok, thanks for letting me know. How would you denote it. (Thanks again for your help.)

@Hippalectryon OK :D

@Chris'ssis Do you have some cool stuff with roots of roots of roots .... ?

@Hunter I would use $\mathbb {R} - \mathbb {Q}$ instead.

@Hippalectryon Yeap ...

4:26 PM

Like $\sqrt{n+\sqrt{n+\sqrt{n+\sqrt{n+\sqrt{n+\sqrt{n+\sqrt{n+...}}}}}}}$

But Harder, better, cooler, :D

@topper: Your question looks weird. What's the given answer?

@Hippalectryon: the approximation of $\pi$ with trigonometric functions probably involves that.

@Nick The above converges to $\sqrt{n}$ i think

@Hippalectryon: I have no idea how you got that. If it's intuitive, tell me. If it's complicated, I'll learn later.

@Hippalectryon Aren't you tired of laying traps here? :-)

@JasperLoy: Use $\sim$ instead of $-$ while denoting set difference. It gives you style.

4:33 PM

@Nick I prefer -, lol.

Okay, I had a great patch, now I'm flagging. Why can't I just substitute 2 in $$\lim_{x \to 2}\frac{x^2+5x-14}{x^2-5x+6}$$? Why do I have to factorize it? It's not indeterminate when I substitute - I get $-\frac{10}{10}$

@topper You sub wrongly. It is 0/0

@JasperLoy Oh the humanity. I'd delete it but it can serve as a reminder. Sorry

I'm sort of laughing

@topper How did you get 10/10 ?

@Chris'ssis traps ?

@Nick It's intuitive enough

4:36 PM

@JasperLoy It was simple. I made $(5)(2) = 20$

@topper lol

@Nick so we define $U_0=\sqrt{n}$ and $U_{n+1}=\sqrt{n+U_n}$

Q: $\frac{x}{10!} = \frac{1}{8!} + \frac{1}{9!}$

@topper: I've done worse mistakes. I mean seriously retarded mistakes. It's quite amazing how long a person can go without realizing something is a mistake.\

8

I have a pretty simple straightforward question. Q) Find the value of $x$ in the following: $$\frac{x}{10!} = \frac{1}{8!} + \frac{1}{9!}$$ Instinctively, I do the quickest thing I know how to do. $$\times10!\implies x = \frac{10!}{8!} + \frac{10!}{9!} = 9 \times 10 + 9 = 99$$ That seemed p...

intuition factorial fractions

@Nick Then by a quick recursion we show that $\sqrt{n}\leq U_n\leq n$ (the RHS)

Hmmm

Is it circular logic to use L'hôpital to determine $\log (1+x)/x$ ?

4:39 PM

@Nick Then we use $U_{n+1}=\sqrt{n+U_n}$ and $U_n\leq n$ twice and it's done

@Hippalectryon: I shall dub thy method Recursive Converging Sandwich, Sir Hippo.

I mean if we define the logarithm as the area under $1/t$ from $0$ to $x$, then the derivative follows from FTC.

@N3buchadnezzar: You can lie to yourself about it but I wouldn't upvote it as an answer, if you know what I mean.

Vibhav Pant

So,there's this integration formula: $\int\frac{1}{\sqrt{x^2+a^2}}=\ln|x+\sqrt{x^2+a^2}|$

Now, this can be derived using hyperbolic trig substitution

So, should I mug up this formula, or learn hyperbolic substitution?

Learn hyperbolic substitution. It's really useful, @VibhavPant.

Vibhav Pant

4:45 PM

yeah, guess Ill do that

A trigonometric substitution will also crack that integral.

$x = a\tan \theta$

That's the way I've been taught. The trig cracking.

The hyperbolic method is pretty much the same thing, @Nick. ^_^

@robjohn I think I know now what the best way is, the most elegant one to compute $$\sum_{n=1}^{\infty} (-1)^{n} \frac{H_n}{n}$$

@Nick My hips dont lie

4:47 PM

Shakira, Shakira!

That's another great song.

^_^

The last time I computed an integral was in high school

@JasperLoy So it is no longer an integral part of your life?

2

@JasperLoy: Me too. I never seem to try anywhere else.

@N3buchadnezzar Nope, I hate antiderivatives.

instantrimshot.com

4:49 PM

@JasperLoy: Integrals are Definite Integrals. Antiderivitives are indefinite integrals. Which one do you hate?

@Nick All of them.

@robjohn All is very elegantly done by using the Cauchy product.

I thought integrals referred to both definite and indefinite ... integrals.

Hence, integrals.

From, integrals.

@Khallil: It's mathspeak. People here draw fine lines on vocabulary.

Fine lines at the edge of the tide.

The world is ever-changing, and we often change with the times.

Today's enemy is tomorrow's ally.

4:53 PM

... ok, Nixon. I think we got it. But let's reconsider uniting with Bin-Laden.

I was quoting Metal Gear Solid!

@Chris'ssis Not my way?

I was quoting myself after dissolving some solid metal car part in conc. HCL.

Dangerous stuff!

$\text{H}^{+} + \text{Cl}^{-}$

It's a strong acid as it dissociates very easily, right?

@robjohn Well, your way is very nice. I mean the Cauchy product offers a solution without pen and paper, very easily. Just consider this one $$a_n = b_n = \frac{(-1)^{n}}{n}$$

4:56 PM

Hi everyone

Would I be right in saying that the sequence 1/n has a convergent subsequence?

@JohnJack Yes, if you are talking about the set of reals.

@robjohn some simple calculations lead you immediately to the desired result.

@JasperLoy Yes if you consider the sequence x_n = 1/n then this sequence does not converge but is cauchy?

@JohnJack I need to understand what you are saying here. Are you still talking about convergence in the set of real numbers or just convergence in the set of numbers determined by the sequence? In the former case, the sequence converges to 0. In the latter case, the sequence does not converge.

The set determined by the sequence.

5:02 PM

@JohnJack Do you understand now?

@Chris'ssis If you read my answer carefully, I think you'll see that it works out to a more painstaking treatment of the same idea.

I was just being more pedantic

@Jasper yeah,...say you asked "Give ab example of a real sequence {x_n} for which inf x_n exists but which has no convergent subsequence" then would you always consider that the convergence in with respect to the set of real numbers?

@robjohn Yeah, I noted that. :-)

@JohnJack That sequence converges and is Cauchy

Michael Corleone

Let $p_0\leqslant p\leqslant p_1$. If $f\in L^p(X)$, then we can easily see that $f\chi_{|f|\geqslant 1}\in L^{p_0}(X)$:

5:07 PM

@JohnJack It depends on the context. I would need to see the entire book etc to know. But if you are talking about the real numbers, then (1/n) converges in the real numbers to 0 and it is Cauchy of course.

I didn't know Cauchy became a sequence, @JasperLoy. :0

@JasperLoy But would the default interpretation would be with respect to the set of real numbers? Wouldn't it?

@Khallil What do you think?

@JohnJack That would seem the most natural thing to do.

Michael Corleone

Let $p_0\leqslant p\leqslant p_1$. If $f\in L^p(X)$, then we can easily see that $f\chi_{\{|f|\geqslant 1\}}\in L^{p_0}(X)$. I'm having a hard time seeing why if $(f_n)$ is a sequence in $L^p(X)$ which converges to $f$, then $(f_n\chi_{\{|f|\geqslant 1\}})$ converges to $f\chi_{\{|f|\geqslant 1\}}$ in $L^{p_0}$. Can anyone help me?

I'm sorry, @JohnJack. I've not even the slightest idea.

5:10 PM

@JasperLoy Do you know of a result that states that an increasing, unbounded sequence has no convergent subsequence? Does this seem valid?

@Khallil yes

@JohnJack So (1/n) cannot be your example because it converges and so all its subsequences converge to 0.

When I said that you are 'something else', I was making reference to the idiom that you're great, @BalarkaSen!

@JasperLoy Yes understood, that was something else, I won't use that.

@JohnJack I need to think about it, you should for yourself too!

5:14 PM

@JasperLoy I am, I'm just spreading the...thought.

@JohnJack I am too lazy to think now, it's your homework, lol.

@JasperLoy :)

What do you think of the Brilliant website, @JasperLoy?

@Khallil I have no opinions.

None whatsoever?

:-(

5:19 PM

@Khallil I never bothered to visit it, somehow.

Ah, I see. I'm on it right now. It's pretty useful for getting up to speed with concepts you might be rusty on.

@Khallil It's... brilliant

Well, as per you.

The self-proclamation is slightly cringeworthy, but it's still a pretty good website, @BalarkaSen.

5:39 PM

@MikeMiller mystical greetings

@Ted If you recall our question, Jason DeVito answered it in the positive: there are indeed nontrivial maps $\Bbb{CP}^n \rightarrow \Bbb{CP}^m$ for $n>m$ for any choice of $m$; and it is likely that what $n$ we can choose depends wildly on the homotopy groups of spheres, so not much can be said. One interesting question remains: does $\Bbb{CP}^2 \rightarrow \Bbb{CP}^1$ admit any interesting maps?

Mystical greetings, @Balarka

Could you look at a 'simple' limit for me, @BalarkaSen?

OK