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10:00 AM
I wanna find height programmatic, don't wanna fix it
It may raise issue like
if someone is measuring from 2nd floor then what in that case ?
 
2nd floor
 
tan(angle) = distance / h, so distance = h * tan(angle)
 
Ohh ya :)
h and d are unknown only angle is known
 
and what m.k. says is that from the angle alone you can only determine the ratio of the distance and the height.
 
yep
if the angle stays the same and you get higher, the distance gets longer
 
10:05 AM
Yeah
 
@JonasTeuwen I think you're talking about strong equivalence
I'm talking about topological equivalence
that they do not generate the same open sets
 
@tb Finally, I started writing in your template. Thank you. I am just morphing it!
 
Does this question look right? I am not very good in english, and I had some problems formulating myself.
 
@N3buchadnezzar TeX!?!
 
@N3buchadnezzar: First thing I notice is that you should LaTex some things
 
10:07 AM
@N3buchadnezzar Some misspellings are there. Shall one of us go ahead with the edit?
 
@tb Hey
@KannappanSampath hey
 
Hi @Ben
 
hi ben
 
hey
@tb You and Jonas were discussing what I was trying to give an example of?
 
@KannappanSampath I texed up some things, but please clearify if you want to !
I really have to leave now, but is the question itself clear? Eg what I want to do?
 
10:09 AM
I'm trying to show that the space $C[0,1]$ with the sup metric and $L^2$ metric are not equivalent in the sense of topological equivalence: en.wikipedia.org/wiki/Equivalence_of_metrics
 
Some spellings: mathematical
fantastic
G eogebra (The first line!)
 
Like I said feel free to edit, I really, really, really have to go
=)
 
OK. I'll edit it then.
 
@BenjaminLim well, you certainly have $\|f\|_2 \leq \|f\|_\infty$, right?
 
oh wait
yes
because
 
10:12 AM
$\|f\|_{2}^2 = \int_{0}^1 |f|^2 \leq \|f\|_{\infty}^2 \int_{0}^1 1 = \|f\|_{\infty}^2$.
 
$\int_0^1 |f|^2 \leq \sup|f|^2$
So now
 
Do we need a geogebra tag?
 
Why not?
We have mathematica, matlab, octave, etc.
 
Sorry to interrupt the discussion.
@tb OK, then, may be I'll create one....
 
Well, if it doesn't exist already, I'd not do it.
 
10:16 AM
It does not exist. I was going to press the button but I saw (1) and I changed to this tab!
 
@BenjaminLim this tells you that every $\|\cdot\|_2$-ball contains a $\|\cdot\|_\infty$-ball of the same radius.
 
yes
ball about $0$
 
doesn't matter, the norm is translation invariant.
 
But I'm trying to show that it is not necessarily true that we have the other direction
Like a $||.||_\infty$ ball need not contain a $||.||_2$ ball.
 
so, you're trying to show that every $\|\cdot\|_2$-ball contains elements of arbitrarily large $\|\cdot\|_\infty$-norm.
 
10:19 AM
yeah
My example is
the function that on $[0,1]$ looks like $\Lambda$
The spiky thing
 
Hey, J. M.!
 
Hey, hey. I'll be here for a short while.
 
cool!
 
@JM Hello
 
10:21 AM
@BenjaminLim well, one function is not enough...
 
@JM How is the binangkal
letchum
 
@BenjaminLim ?
 
What's wrong? You just need to produce one function that is not in the other for any given $\delta$.
 
@JM Binangkal is cebuano
@JM letchun?
 
10:22 AM
@BenjaminLim i.e., a family of functions :)
 
so it has to be like $f_n$
no problem that was what I was thinking anyway
 
@BenjaminLim Oh, sorry. I don't speak that dialect...
 
@JM Balut, lechon?
 
@BenjaminLim Well, that I understand. I haven't eaten any of those in a while...
 
@JM So many cebuanos here in canberra they're trying to teach me to speak bisaya
 
10:24 AM
@BenjaminLim :D They're an enthusiastic sort.
 
@tb $f_n(x) = \begin{cases} 0,& -1 \leq x \leq 0 \\ 2nx,& 0 < x \leq (2n)^{-1}\\ 2- 2nx, & (2n)^{-1} < x \leq n^{-1} \\ 0, & n^{-1} < x \leq 1 \end{cases}$
@JM We have basketball league that is like 99.9% philippinos
 
so, now compute it's $L^2$-norm.
(and why do you define the function on $[-1,0]$?)
 
I'm just looking at $C[0,1]$, any closed and bounded interval will do
 
Sure, but stick to $[0,1]$ is all I'm saying. Doesn't matter.
@JM one brilliant idea on meta. :)
 
@JM Hi.
 
10:29 AM
@MattN :D
 
What's so funny about saying hi? : )
 
@tb Sorry I actually defined it for $[-1,1]$
 
@MattN That's my wide grin. ;)
 
: )
 
@tb Apparently he doesn't know how idiosyncratic most of us are...
 
10:31 AM
@JM exactly...
 
@tb That someone seemingly random posts a seemingly random link as an answer?
 
Maybe he does not know what is $\lim_{\rightarrow}$
 
@MattN well, that someone Googles limit and intuition and posts a link to a calculus site.
 
@tb The area is $2\int_0^{1/n} 4n^2x^2 dx$
$4/3n$
So as $n \rightarrow \infty$
the area goes to zero
@tb So sorry man
Oh it is actually $(8/3n)$
@tb the $n$ is in the denominator
it's $n^2/n^3$
 
Then write $8/(3n)$.
 
10:35 AM
$8/(3n)$
 
thank you :)
 
@tb Sorry man because of this field theory assignment my eyes are gone now
 
Hey What is the meaning of $ that you are appending both sides of the equation ?
 
So this means that for all $\delta > 0$, we have a function $f $ in the set $B(\delta)$ which is
$B(\delta) = \{y \in C[0,1] : \int_0^1 |y|^2 < \delta\}$
 
@Dharmendra using this bookmarklet we get the formulas rendered this way (it's LaTeX)
 
10:36 AM
such that $f \notin \{g \in C[0,1] : \sup|g| < 1/2\}$, taking $\epsilon = 1/2$.
@tb That does it right?
 
More or less.
 
Done.
@tb Next few weeks I have to take my eyes of AC
Got like a field theory mid sem and analysis mid sem in the space of three days
@tb You fixed your shower?
 
And, the electric socket, as well?
 
@tb My analysis is not so good when we go to function spaces and stuff
 
I would scale the functions in such a way that their $L^2$ norm is $1$, i.e. consider $g_n = f_n / \|f_n\|_2$. Then you can see that the $\|\cdot\|_\infty$ norm of $g_n$ tends to infinity.
 
10:39 AM
@tb Hmm, okay thanks.
 
@KannappanSampath naah, not yet. I'm lazy these days.
 
@tb Wait don't you want to say for all $\delta > 0$, we have a function whose $L^2$ norm is always less than $\delta$ such that....(insert here)
 
Maybe I should say especially lazy, because I'm always lazy.
 
so why do you want to normalise?
 
@tb No you re not. You probably must be exploring new things!
 
10:41 AM
@BenjaminLim then take $\delta \cdot g_n$...
 
$\delta g_n$ for???
 
@KannappanSampath seen the LHF I've been going after recently?
 
@tb Hmm, no... What do you mean? You told me you started doing GGT. So, I had that in mind under the hypothesis that you're doing this stuff only now or something of that sort!
 
@BenjaminLim because I find it more intuitive to say that the closed $\|\cdot\|_2$-unit ball contains elements of arbitrarily large $\|\cdot\|_\infty$-norm.
So it cannot be contained in any $\|\cdot\|_\infty$ ball
 
oh ok. @tb I'm still stuck with that lemma in my head in munkres on when two topologies are finer than each other
 
10:45 AM
@KannappanSampath Oh, when I said "I started out" I meant when I started doing math on my own like 12 years ago...
 
@tb Sorry, but I assumed because you do functional analysis. Sorry for wrongly assuming!
 
@KannappanSampath I have to be away from AC a bit
 
@KannappanSampath that's a myth! :)
 
@tb Don't you work in homological algebra?
 
I am lost.
 
10:47 AM
@BenjaminLim I'm not quite sure what you're saying. We know that every $\|\cdot\|_2$-ball contains an $\|\cdot\|_\infty$-ball, so the $\|\cdot\|_\infty$-topology is finer. Now we produced an example showing that there can't be an inclusion in the other direction, so the topology is strictly finer.
 
@tb I was just saying that I was stuck with that lemma in munkres about a condition for a topology to be finer than another
 
Why not pay someone to do that socket for you? Saves time and the possibility of an accident.
 
@BenjaminLim I don't know... I don't know what the things I'm doing is actually called and I don't care much. I would say I do neither and both.
 
@tb That's tautological :D :D :D
How can you do both and not do both???
 
@MattN because no one would do it for me because of that crappy law and plugging six pieces of cable is not that dangerous, IMO.
(and why spend 100 bucks when I can get away with less than 5?)
 
10:51 AM
@tb As if they'd think of the law when you phone them up. And as if they'd bring a measure tape with them : D
 
@tb In australia labour is very expensive so most of the time people fix their own stuff. For example at my uncle's house I have to help him dig the garden and stuff.
 
@tb Obvious: because it's a boring thing to do.
 
I don't find it that boring. I find it more inconvenient to wait one morning for some guy to come to do something I could do myself in less than 10 minutes.
 
@MattN In australia you have no choice. Labour is very expensive so you have to do almost everything by yourself
all but finitely many things by yourself I mean
 
@BenjaminLim that was more or less my point. I don't do either of them seriously enough to say I am working in those fields but what I'm doing has to do with them...
 
10:54 AM
Less than 10 minutes? Heh. To me it sounds as if you have to remove the mirrored cupboard thingie (that alone takes 10 minutes) and then fiddle with cables.
@BenjaminLim Sucks : )
 
@tb I see. You're liking working in between both. That's nice. I have had visions of doing some stuff in like topology.
I always see people writing like $H^1(a,b)$ on the board....
 
0
Q: plotting phase of a signal adding delay

MazzyI'm trying to plot the phase of this signal $s(f)=A^2T^2sinc^2(Tf)e^{-(j\pi Tf)}$ my problems came when I add the delay.I understand how to plot it without delay but adding the delay I didn't understand how plot.some help?

 
@MattN no, I have to remove the lamp shade: that takes maybe half a minute. Then I have to undo two screws and take something off, one more minute. Then I have to take a cable with three wires and plug those into six prepared thingies (which the guy left there, of course). Then I have to re-install the thingie, fix two screws and put the shade back on. Ten minutes are amply sufficient for that. :)
 
And also please don't parachute in like that @Mazzy
 
10:58 AM
@KannappanSampath Hope That Helps, and maybe his initials or some boy-scout salutation.
 
@MattN Try doing paving under the hot sun. Twenty passes over the tiles even the NZ earthquake won't shake em'
 
@KannappanSampath excuse me
 
@BenjaminLim No thanks, not so keen on skin cancer. That's definitely worth paying for.
 
who undervoted my question?
 
10:59 AM
@KannappanSampath smh.com.au/world/…
read that
absolutely amazing
 
@Mazzy I didn't, but the question is not very clear to me. What is the delay?
 
Neither did I.
 
the delay is the exponential
 
@Mazzy: stuff like that, you should be noting in your question...
 
I'm italian so maybe I made a mistake to express myself
it's obviously that an exponential function is a delay in the frequency world
 
11:06 AM
@Mazzy Not to most of us; only very few here do signal processing. That's why you need to be more forthcoming.
 
@Ben Absolutely remarkable!
 
@KannappanSampath I was just like wow man
 
Very apt title, I have nothing much to say. Spellbound!
 
@KannappanSampath He still loves his mother
he left home when he was 5, crazy man
 
Yes, I never can imagine a day, when I do not call my mother on skype and talking to her, I am 17! I never had the guts, I believe! : )
 
11:13 AM
@KannappanSampath You speak in tamil to your mum?
 
@BenjaminLim Yes. : )
bye I should be off now!
 
Byee!
It was not actually that difficult. But "Requires a little bit of thinking" is not true.
 
@MattN More likely, you and the one who said that define "little" differently. ;)
 
@JM I'm sure we define "little" a little differently.
 
@JM Yes. But I used to think that he uses my definitions when he talks to me.
 
11:22 AM
@MattN well, because you chose to use covering spaces, which are an unnecessary diversion. You can just use the functional equation of the exponential function.
 
@MattN As you can see t.b. is amply demonstrating my point. ;)
 
I can see that.
But it's ok. Next time he causes suffering I'll just stick him back into the trunk.
 
11:47 AM
@JM may I ask you to take care of this?
 
@tb The first delete vote was mine. :)
 
Hi Skullpatrol.
 
Hi Matt.
 
Hi Gigili. Still bored?
 
@Gigili Hi
 
11:54 AM
@JM I've sort of gotten used to them coming from Asaf :)
 
@tb Apparently I was away for far too long... :)
 
@JM yeah, definitely... here's one Asaf would be grateful for...
 
@tb Answer's kaput now.
 
'Ello Matt. Uhum, sadly. Still hate weekends.
'Ello @Skullpatrol.
 
@JM Thanks, Asaf will be happy
 
12:00 PM
@Gigili Do you not have any exciting homework to do? Or any exciting books to read?
 
@tb Did you look at that link to the khan academy definition of abstract-ness?
 
no
 
Holy cow, where have the past few hours gone to?
@JM Hi!
 
@JonasTeuwen Hi!
 
Do you know Koekoek? He wrote a book on special functions 8-)!
 
12:13 PM
@JonasTeuwen Wolfram Koekoek? Not personally, but I've read his stuff on hypergeometrics.
 
Roeloef :-).
Yes, he's a colleague of mine. Was just wondering if you knew him, he wrote some books about that stuff.
I tried to feed him my integral but he spit it back, a bit like Mathematica does...
8-)
 
@JonasTeuwen Oof right, I confused him with Koepf. :)
@JonasTeuwen The Bessel-like integrals?
 
Yes.
 
@JonasTeuwen SE.
Ello Jonas : )
 
@MattN SE?
 
12:17 PM
Stackexchange.
 
Yes, what's up with that?
 
Well click the back arrow.
 
I really can't help but feel that it has to be (multivariate) hypergeometric. (By this, I include Meijer's and Fox's functions...)
 
@MattN Oh, no, I just remember it being 10:00 and now it is 12:30.
@JM No problem!
 
@JonasTeuwen Sounds like a case of abduction.
 
12:19 PM
I get a sequence of hypergeometrics...
 
whistles X files tune
 
@MattN No homework (is there any interesting homework at all?), maybe some book but weekends are supposed to be more exciting and interesting.
Maybe going out shopping or something is more interesting but I'm no in the mood. =|
 
@Gigili Do you play an instrument? There's always the possibility to sit in the nearby park and play.
 
@MattN If need be, you can lay out a hat of some sort while playing... :D (kidding, of course)
 
Yes, that's unusual here like that.
Giraffen schlafen im Durchschnitt nur 1,9 Stunden am Tag.
 
12:23 PM
@JM I'd guess that depends on your skill : ) I've not tried. I used to practice in a park to avoid disturbing my house mates but I'm also shy so I did that at 6 am when there weren't that many people in the park.
 
'Ello @JM, you're finally back I presume?
 
@Gigili Well to use a hat you have to have a permission afaik.
 
I first have to buy a nice hat.
 
Shopping it is then.
 
Shop till we drop.
 
12:27 PM
@Teddy I'm sorry for refusing to listen and think.
 
Tim
Thanks @Matt
 
Np : )
 
Tim
Thanks @tb
@tb: (1) I wonder where I can find that version of Riez representation theorem Didier used?
(2) In Didier's reply, is the weak convergence in $ L^p, p \in [1, \infty]$ wrt $L^q, 1/p+1/q=1$, both weak convergence and weak* convergence? Because $L^p$ and $L^q$ are continuous dual to each other?
@MattN :-)
 
Play me in Draw My Thing: omgpop.com/i/drawmything/1gdevx_12ydr1
 
@Gigili Doh, too late.
I give up. Looks like it's crashed. Also, I've got some stuff to do. I'll play you later if you're still there.
 
12:47 PM
@Tim (1) you just have to observe that a probability measure gives a continuous functional of norm one on $C_b(X)$ and that two distinct measures give rise to two distinct functionals (an exercise you should do!). (2) $L^p$ is reflexive for $1 \lt p \lt \infty$. For $p = 1,\infty$ this is wrong. Try to come up with counterexamples.
 
@Gigili Not quite yet; I still have a number of loose ends. (For now I'm exploiting a slight lull.)
 
@Gigili I will play.
 
This question is probably a candidate for having the wrongest possible tags.
 

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