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2:39 AM
I am going to install Ubuntu Mate 14.10. It is the distro's first release, and it might become my distro of choice!
 
2:53 AM
@Chris'ssis here:
Show that $\frac{\log(n+1)}{n}\le\frac{\log(n)}{n-1}$ for $n\ge2$:
$$
\begin{align}
\left(1+\frac1n\right)^n&\le n+1\tag{1}\\[6pt]
(n+1)^{n-1}&\le n^n\tag{2}\\[12pt]
(n-1)\log(n+1)&\le n\log(n)\tag{3}\\[6pt]
\frac{\log(n+1)}{n}&\le\frac{\log(n)}{n-1}\tag{4}
\end{align}
$$
Explanation:
$\begin{array}{l}
(1):&\text{the left side is $\lt e$}\\
(2):&\text{multiply by $n^n/(n+1)$}\\
(3):&\text{take logs}\\
(4):&\text{divide by $n(n-1)$}
\end{array}$
$$
\begin{align}
\sum_{n=2}^\infty\frac{\log(n)}{n!}
@Chris'ssis: Sorry, I had to leave for a meeting and then we went to dinner. The proof above was shower-math.
 
I find it a minor inconvenience that after clicking on (see full text), the down arrow on the right hand side scroll bar is blocked by "link my next chat message as a reply to this"
 
@IceBoy at least you can still drag the scroll bar. If it were the other way around, you could never link the messages.
 
@robjohn there is a "reply to this message" on the left hand side arrow
 
@IceBoy Then it doesn't matter, there is a backup for each one.
@IceBoy So did you see the question that this series was for?
 
3:08 AM
It's just a minor inconvenience :)
@robjohn yes, I have been following along...
 
$\sqrt{2\sqrt[3]{3\sqrt[4]{4\dots}}}\lt2$
 
Hello Professor @TedShifrin they say imitation is the highest form of flattery, I see that Hippa's duplication has taken that to a dizzying new height :D
 
@IceBoy What have they duplicated?
 
Hippa was the person who made the Ted Shifrin clone account, @robjohn
 
@robjohn almost everything
5 hours ago, by Balarka Sen
@robjohn He has topped anon's April fools prank :D
 
3:19 AM
@IceBoy That is pretty bad. I will change it.
@IceBoy There....
 
haha
it's very accurate
 
@IceBoy That is at least distinct, however similar
 
4 hours ago, by Balarka Sen
@WeaponofChoice Can't anything be done with this? I am actually offended...
 
4:10 AM
@robjohn
A friend told me that for every $x>0$, $$\frac{1}{\sqrt {2\pi}}\int_x^\infty e^{-t^2/2}dt\geqslant \frac{1}{\sqrt{2\pi}}e^{-x^2/2}/(x+1/x)$$ which can be reduced to $$\int_0^\infty e^{-t^2/2}e^{-xt}dt\geqslant (x+1/x)^{-1}$$ but I am getting numerical evidence against this.
I think that the original exercise is that if $X$ is a random variable with $X\sim N(0,1)$ then $P(X>x)\geqslant f_X(x)/(x+1/x)$.
 
@PedroTamaroff For roughly what $x$ do you find numerical evidence against this?
 
For some $\xi\simeq 1.27$ I get the inequality reverses.
I mean, the inequality is true up to that point, then it reverses.
It might also be my software is just poor.
 
It's always (numerically likely to be) true for everything I've checked.
 
I've only checked the first inequality, not the second
 
4:19 AM
I am looking at the second.
 
Same results for the second.
 
Are you really doing it or just trolling?
 
For $x \approx 1.27$ I get $LHS-RHS \approx .08688$.
 
:-)
 
4:22 AM
Note that %42 and %44 are identical out to about 70 or so digits; one could probably actually prove that it's within some error of the actual value but I've forgotten all the error estimation I ever knew.
 
Hey, I have a quick topology question. Suppose I have (X,T) and (Y,T') topological spaces. If we have $A\subseteq X$ and $B\subseteq Y$, and $X\times Y$ with the product topology, then if (a,b) is a limit point of $A\times B$ then either a is a limit point of A or b is a limit point of B.

How would we show that this statement is true? I feel like I'm misunderstanding something, because I'm obtaining the stronger result that a must be a limit point of A and b must be a limit point of B.
 
@MikeMiller Ok, I've reduced it to showing that for any $a>0$, $$\int_0^\infty e^{-at^2/2}e^{-t}dt\geqslant 1/(1+a)$$
 
@PedroTamaroff I don't know how to show that. You've got lots of better people to ask here than me :)
@MichaelDyrud Can you sketch your proof attempt?
(Also can you define limit point? The definition varies between authors)
 
@MikeMiller I don't think it is a hard task, but it is bothersome.
 
4:53 AM
Ubuntu Mate 14.10 is excellent so far. It brings back the goodness of Ubuntu before Unity shell and Gnome shell spoilt things.
 
Do quaternions, octonions $n$-ions in general have cardinality $2^{\aleph_0}$ because they're $n$ copies of $\Bbb R$ respectively?
 
@UserX Their underlying vector space is $\Bbb R^n$ (for appropriate $n$), which has the same cardinality as $\Bbb R$, yes.
 
@PedroTamaroff I know that $$ \begin{align} \int_x^\infty e^{-t^2/2}\,\mathrm{d}t &\le\frac1x\int_x^\infty e^{-t^2/2}\,t\,\mathrm{d}t\\ &=\frac1xe^{-x^2/2} \end{align} $$
 
@Ryker I have created a separate chatroom where I can tell you what I know about this.
 
@MartinSleziak keeping secrets? ;-)
 
5:03 AM
@robjohn No, just do not want to disturb here.
 
@robjohn I should create separate chatrooms to share my secrets.
 
@JasperLoy does anyone want to know your secrets?
 
And the response is probably going to be pretty long, so it would be difficult to read it here, where it would be interrupted after each message by an unrelated conversation.
 
@robjohn Yes. For example, Sarah, lol.
 
@robjohn Are you seriously calling writing something in publicly accessible and easily visible chatroom and announcing this also in the main chatroom keeping secrets?
 
5:05 AM
@MartinSleziak That's what linked comments are for... but there is no reason against a separated chatroom. I was just being annoying :-)
 
@MartinSleziak It's just a joke.
@robjohn You can't annoy me. I have attained unannoyability, lol.
 
3 mins ago, by robjohn
@MartinSleziak keeping secrets? ;-)
note the smiley
 
Even with linked comments, if I wrote here I would have here 5 messages interrupted by 10 unrelated messages. (Maybe less, since we wouldn't have discussion about my secrets, but still, I think it would be readable.)
 
@martin If you have any secrets though, feel free to share them with me.
 
Yes, I have notice the smiley face on the right. I have seen also frowned avatar on the left, so they pretty much cancel each other ;-)
Of course, my response was not meant to be serious.
 
5:08 AM
Complex number exam in 3 mins
 
@UserX Good luck. And try to quit smoking, lol.
 
Last time I scored 96/100 because one subquestion asked for the complex numbers that have min, max distance, I read that fast and calculated the min, max distance
The biggest frustration is if you get a completely correct test back, with a less than 100 score haha
@JasperLoy thanks
 
Well, it's just a test, no need to worry too much about marks.
 
5:22 AM
@JasperLoy I'll explain you why this test's mark is important after I finish
 
A question out of curiosity. Top user Andre has never asked a question O.o how come?
 
5:47 AM
@TheArtist Perhaps this discussion on meta is interesting in relation to this: Why are there active users who do not ask questions?
 
6:07 AM
@PedroTamaroff I can show that $$ x\int_x^\infty e^{-t^2/2}\,\mathrm{d}t \le e^{-x^2/2} \le\left(x+\frac1x\right)\int_x^\infty e^{-t^2/2}\,\mathrm{d}t $$
so what your friend said was true.
 
6:36 AM
@JasperLoy this exam counts on my score to get in medschool indirectly
The mark of this exam lets my teacher know that I know complex numbers, thus she ll get me a good grade in the semester
And I avoid solving problems on the blackboard because I skip trivial steps and do some tricks, and I don't wanna explain each time to everyone how I factor this or how I saw that etc
So the teacher sees how good I am only through this semester exam
And then (the mean value of two semester grades plus the grade I'll write in finals)/2 is one of the 6 grades that get the mean, and that's my final score for medschool. At least 18.7+ is required for this
 
Am I missing something: Need to prove that the $n$th derivative of the sum of two functions is the sum of the $n$th derivative of the functions.
I did this with inductions; but worried that I missed something.
 
Link?
 
6:55 AM
@Martin They follow Erdős
 
7:23 AM
@ParthKohli How have you been?
 
How is my display picture? :D
 
7:38 AM
@Martin Sleziak thanks
 
 
1 hour later…
8:41 AM
Hi @UserX
How did it go?
@BalarkaSen Fine now :)
@BalarkaSen offended .. lol .. ok, not now :D
 
@Sawarnik apart from an arithmetic mistake it was nice. Plus it had an interesting question
@Sawarnik if $a_1,a_2,a_3 \in C$ such that they belong in the unitary circle in the complex plane and $|a_0-a_1$=2, |a_1-a_2|=1$ prove that $|a_2-a_0|=\sqrt 3$
 
@UserX Your LaTeX is messed up
 
8:58 AM
@Hippalectryon I read the codes only, I can't render the LaTeX. I guess the mess-up is at the end so ; $|a_0-a_1|=2, |a_1-a_2|=1$ prove that $|a_2-a_0=\sqrt 3$
 
I know. I'm on mobile, my brother took my pc. If you can render LaTeX on this site in chrome for android, please tell me the way.
 
It works on mobile too
 
My url bar is too small for the script and I can't add the bookmark.
 
You can't add a bookmark ? :c
What kind of phone is that
 
9:06 AM
Samsung s4 :/
I can add a bookmark, I can't run it without visiting a blank page though
 
That sucks
 
9:25 AM
@UserX now what'd be a "unitary circle"? do you mean $a_1, a_2, a_3$ are on the "unit" circle, i.e., $|a_1| = |a_2| = |a_3| = 1$?
@Hippalectryon!
 
@BalarkaSen!
 
You're not a game anymore!
 
@BalarkaSen true
 
@Hippalectryon I am working on a wonderful problem on point-set topology.
 
Which is ?
 
9:29 AM
Trying to classify closed sets in a metric spaces (with or without extra structures) which can be separated by open spheres.
Turns out I need a notion of convexity in general. Prof points out to geodesic metric spaces -- would get loads to know about that stuff next time we meet.
Excited actually.
 
-___-
 
@Hippa Er..?
 
Sounds boring to me
 
You and your matrix groups. thunks @Hippa
 
9:37 AM
I find matrices pretty boring without any motivation like being a K-algebra over a given field.
I even think of vector spaces as modules over fields :p
 
@BalarkaSen yes that's what I mean
 
@UserX Polar coordinates.
 
@BalarkaSen I have two ways of solving it. One uses euclidian geometry, the other Sine Law.
The euclidian geometry one uses the fact that an angle inscribed in a semicircle is alwats a right angle.
 
Well, let $a_i = \cos(\theta_i) + i\sin(\theta_i)$
 
9:49 AM
The sine law is pretty straightforward. What's the polar coordinates one?
 
Then apply the addition laws.
@UserX Every complex number $(x, y)$ can essentially be parametrized as $x = r\cos(\theta)$ and $y = r\sin(\theta)$. That is called polar coordinates.
 
10:07 AM
@BalarkaSen I know polar coordinates. Couldn't use them on the test though as they're not being taught.
 
Oh, OK.
 
 
1 hour later…
11:16 AM
Greetings
@robjohn Good job!
 
11:56 AM
@Chris'ssis Hello
 
@Hippalectryon Hi
 
@Hippalectryon Hi. Ted's quite angry :P
 
I saw that
 
:P
@Hippa Besides, did you know he is on FB :O
 
I don't have FB
It's plain useless to me
 
12:08 PM
@Hippalectryon Why? :O
 
why not ?
 
Why is Ted angry?
 
No idea ;P
 
@UserX Are you on FB?
@UserX Because profs are ;)
Because professors are ;D
 
Don't get it. Yes I have fb
Although I use it only for chatting as it is quite famous between my peers, I never post etc
I uploaded most of my pics because girls asked me to too. I'm not active in fb
 
12:15 PM
Got disconnected :<
@UserX Me too :)
@UserX Can we be friends there?
@Hippalectryon What wud you do now? :P
 
@Sawarnik Taco. taco must be the answer.
 
@Hippalectryon :/
 
If we accept the negation of the continuum hypothesis, wouldn't that influence cardinal arithmetic?
 
12:38 PM
Oo
 
What can we say about the convergence of $$\int_0^{\infty} \frac{\sin(x)}{x+\log(x)} \ dx$$?
 
It most likely converges
 
To what?
 
Well, we're sure it converges lol
Since $\int_1^{\infty} \frac{\sin(x)}{x+\log(x)} \ dx\le \int_0^{\infty} \frac{\sin(x)}{x} \ dx=\pi/2$
 
@Hippalectryon Can we be sure of that? I wouldn't be at all ...
 
12:42 PM
Doesn't what I said above work ? @Chris'ssis
 
@Chris'ssis you are integrating over a singularity at $x=e^{-W(1)}$
 
@UserX Yeah, that point is important.
 
@Chris'ssis Owait no it does not converge
 
@Hippalectryon :-)
 
Olol
Nice one
 
12:45 PM
:D
 
@Chris'ssis can we use Cauchy's principal value?
 
@UserX Well, we can split that integral into 2 integrals and treat them separately ...
(one converges, the other one diverges)
 
@Nick Hey
 
$\displaystyle\int_0^\infty\frac{2 \tan{\left (\frac{x}{2} \right )}}{\left(x + \log{\left (x \right )}\right) \left(\tan^{2}{\left (\frac{x}{2} \right )} + 1\right)}$ olol such simplification :3
 
Or we can better split into 3 integrals ... (or I try to find a better way)
 
12:50 PM
@Chris'ssis Or wait can't we just
@Chris'ssis Look at the integral between $0$ and $1$
 
@Hippalectryon Yeap.
 
@Chris'ssis that was way too obvious. Can we manipulate the divergent one in a way we can assign a value to it? That integral would be $\int_0^{e^{-W(1)}} blabla dx$
Forgot the integrand midway of typing
 
And use $\dfrac{\sin{\left (x \right )}}{x + \log{\left (x \right )}}\sim_0\frac{x}{\log{\left (x \right )}} - \frac{x^{2}}{\log^{2}{\left (x \right )}}$ @Chris'ssis
 
@Hippalectryon Did we really need that?
 
Probably not
 
12:54 PM
@Hippalectryon You're right, no :D
 
All we needed was the Taco screwdriver
 
@Hippalectryon LOL :-))))
 
@Sawarnik: नमस्ते
 
@Nick Kaisa tha Deepavali, kuch kiye?
 
Got a question concerning the Burali-Forti special Paradox. It starts with Let $\Omega$ be the set of all ordinals. Since $\Omega$ carries all the properties of an ordinal number it is an ordinal number itself.
How do we define sets being numbers, let alone ordinal numbers and why does $\Omega$ have the properties of ordinal numbers? It's elements do but a set has no reason to behave according to it's elements.
 
12:57 PM
@Sawarnik: We got those fire sticks. I forgot what they were called. Other than that, there weren't any other fireworks. So, yes, Diwali was good.
 
@Nick Nothing else :O :O
You mean, churchuri?
 
Oh, extra question irrevelant to the paradox. What's the cardinality of $\Omega$?
 
1:11 PM
@Sawarnik: Again, I have no idea what it's called. And yes, its good for yourself and the environment to use as little fireworks as possible.
 
@Nick I know, I too scared of them so yes save environment :D
 
@Sawarnik:^ what are they called. In my language, it's called a poothiri. Literally translates to flower fire.
 
@Nick Yeah, churchuri.
 
I gotta write a 600 word essay in the next hour :/
 
@UserX Try an "essay" on $$\int_0^{\infty} \frac{\sin(x)}{x+\log(x)} \ dx$$
:-)
 
1:16 PM
@UserX Every set of ordinals numbers is well-ordered.
 
@UserX Why?
@Chris'ssis lol
 
:-)
 
@MartinSleizak I don't understand which question you answered. Is the least element of $\Omega$ $\aleph_0$by definition?
@Chris'ssis I can't possibly write more than 50 words about your integral. All I see is that it diverges and the cause of its divergence
 
@UserX The last element of $\Omega$ is 0.
 
Ohhhh it doesn't say infinite ordinals
 
1:22 PM
I do not think that you need to show that $\Omega$ itself is an ordinal to get the paradox. It suffices that it is well-ordered (and thus isomorophic to some ordinal).
 
@MartinSleziak my problem is that I don't understand how it could be an ordinal
 
But if you already know that union of any set of ordinals is an ordinal, then you can use this to show that $\Omega$ is again an ordinal.
Well, it can't be ordinal. (You are trying to get a contradiction.)
But if you have any set of ordinals which is downward closed, then it is also an ordinal.
(I hope I did not miss something here.)
 
Isn't my $\Omega$ a set of ordinals then? Why does this become a contradiction?
 
Yes, it is a set of ordinals.
It is well-ordered by the relation $\in$.
 
But if I have any set of ordinals which is downward closed, then it is also an ordinal. Thus $\Omega$ is an ordinal. Did I just prove a paradox or what?
 
1:27 PM
So there is an ordinal $\alpha$ which is the order type of $(\Omega,\in)$.
Then $\alpha$ is one of elements of $\Omega$ (as it is an ordinal).
 
@Sawarnik: According to Wikipedia: Churchuri (Persian: چورچوري‎, also Romanized as Chūrchūrī) is a village in Qushkhaneh-ye Pain Rural District, Qushkhaneh District, Shirvan County, North Khorasan Province, Iran. At the 2006 census, its population was 417, in 74 families.
 
And we get that $\Omega$ is isomorphic to its proper initial segment $\Omega_\alpha$.
Maybe I am overcomplicating things and there is and easier way to get to the paradox.
 
Well I'm with you so fae
Far*
 
@Nick According to Wikipedia: Nick [nit͡sk] is a village in the administrative district of Gmina Lidzbark, within Działdowo County, Warmian-Masurian Voivodeship, in northern Poland.[1] It lies approximately 9 kilometres (6 mi) south-east of Lidzbark, 17 km (11 mi) west of Działdowo, and 74 km (46 mi) south-west of the regional capital Olsztyn.
 
But if you follow the account given on Wikipedia, once we prove that $\Omega$ is an ordinal, the rest seems to be clear.
They've taken successor of $\Omega$ to get a contradiction.
 
1:32 PM
Is there any way to avoid using the axiom of choice?
 
I don't see where we have used AC there.
 
Well ordering
 
Well ordering of what?
If something is ordinal, then it is well-ordered. No need of choice there.
It is a part of the definition. (If we work with the von Neumann's definition.)
 
$\Omega$ is well ordered under the relation $\in$
 
Because it is an ordinal. (And this is a part of the definition of an ordinal.)
 
1:37 PM
Just noticed that. Okay then.
But
It doesn't come from the definition of ordinals, but from the definition of ordinals comparison
 
@RandomVariable have you seen this one before? It's about testing its convergence.
$$\int_0^{\infty} \frac{\sin(x)}{x+\log(x)} \ dx$$
 
Wikipedia says this as definition of an ordinal: A set S is an ordinal if and only if S is strictly well-ordered with respect to set membership and every element of S is also a subset of S.
 
@Sawarnik: touché
 
Coulé ?
 
I am so lost in ordered sets definitions
 
1:45 PM
@Hippalectryon: I was done with @Sawarnik but if you insist, En garde!
 
@hippa You should not have "impersonated" Ted.
 
@Nick Should I remind you that I am french ? :P
 
What is going on? :O
 
@JasperLoy Why would he care :c he never met his 'double'
 
What is coule, touche and en garde? @Hippa
 
1:46 PM
Anyway, let no one try to impersonate me, please.
It has happened before, lol.
 
@Hippalectryon: So, what if you have a french grip, eh? I have an italian grip, monsieur!
 
@MartinSleziak * An ordinal number is defined as the order type of a well ordered set.* Without the ordinal comparison definition, it does not follow directly from this definition that a set of ordinals are well ordered.
 
@Nick Grip ??
 
And now I have 40 minutes to write a 600 word essay. I'm out
 
@UserX I thought the you were working with von Neumann's definition.
 
1:49 PM
@UserX I always had difficulty writing long essays. Most long essays I have read are full of shit.
 
Good luck with your essay!
 
@Hippalectryon: I refer to thy hilt, knave.
 
What people can say in ten words I usually say in five.
 
@MartinSleziak It's the definition used by Dauben, Moore and Suppes. I don't know the von Neumann definition. Thanks.
 
@Nick gtg running, i'll be back to finish you off >:c
ADVENTURE TIME
 
1:52 PM
C'mon grab your friends!
We'll go to very distant lands :D
 
My ... what ?
gtggtgtgtggtggtgtg seeya
gone
 
@Hippalectryon: Your hilt, dude. Don't you fence?
It's mathematical.
 

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