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12:00 AM
@robjohn oh ! Thank you :)
@robjohn Yes, can you show that?
@PedroTamaroff just a sec...
x\int_x^\infty e^{-t^2/2}\,\mathrm{d}t
&\le\int_x^\infty e^{-t^2/2}\,t\,\mathrm{d}t\\
Integrate both sides of the preceding
\int_s^\infty e^{-x^2/2}\,\mathrm{d}x
&\ge\int_s^\infty x\int_x^\infty e^{-t^2/2}\,\mathrm{d}t\,\mathrm{d}x\\
\left(1+\frac12s^2\right)\int_s^\infty e^{-x^2/2}\,\mathrm{d}x
&\ge\frac12\int_s^\infty t^2e^{-t^2/2}\,\mathrm{d}t\\
@robjohn Nice !! :D
@r9m Yesterday was the first time I'd gotten the upper bound. The lower bound is pretty common knowledge. It was prompted by Pedro's question before my post (by a few hours)
12:17 AM
@r9m Did you see this one?
@r9m Now see this one
A: A Challenging Logarithmic Integral $\int_0^1 \frac{\log(x)\log(1-x)\log^2(1+x)}{x}dx$

Omran KoubaLet the considered integral be denoted by $I$. Our starting point is to reduce the number of logarithms of different arguments in the integrand. Thus, using the fact that $6ab^2=(a+b)^3-2a^3+(a-b)^3$ we obtain \begin{align*} 6I&=\underbrace{\int_0^1\frac{\log x}{x}\log^3(1-x^2)dx}_{x\leftarrow \s...

hi @robjohn @studentmath @Pedro @Chris'ssis
@TedShifrin Hey, Ted!
@r9m Well, why did I show you this? Good question, indeed.
@TedShifrin Hi
@robjohn did you meet $$\int_0^{\infty} \frac{\cos(x^{\alpha})-\cos(x^{\beta})}{x} \ dx$$ version before? Maybe on some MSE post?
@Chris'ssis I don't think so...
@robjohn Using that integral I can get some amazing result. It's too late here to continue now and at 7 o'clock I need to get up.
12:26 AM
@Chris'ssis without much effort we can assume $\beta=1$ and get the whole thing
@robjohn What do you mean by getting the whole thing? You refer to using my previous result, right?
@Chris'ssis $$\frac1\beta\int_0^\infty\frac{\cos(x^{\alpha/\beta})-\cos(x)}{x}\mathrm{d}x$$
@robjohn Ah, yeah, sure, that's clear.
@r9m my point is that I don't think Kouba got inspired from sos's work and I think Kouba is a math god in the real sense, although, of course, I might be tempted to say the 2 works are a tiny bit similar. :-)
@r9m Besides that I'm mainly focused to criticize ideas, attitudes, not criticizing a person I don't know in the real life, I have no idea how you are.
out for some sleep
1:24 AM
@Chris'ssis It was not even my intention to criticize idea/work in the first place (I don't believe I am capable/qualified enough to criticize someone's idea) .. -_- my one and only intention was to share information ! ..
be your own critique' .. that's what I believe in ..
1:36 AM
@Chris'ssis Prof. Kouba mentions in the beginning of his solution to the same problem in Assymetry Journal Vol 5 (problem was proposed by Cody from I&S) that the use of the cubic identity was inspired from sos's proof in the forum .. :) so chill !! :-)
@Chris'ssis if not Lord, perhaps Lady Unreasonable ? :D :P
@r9m Is Chris's still talking about that?
@Committingtoachallenge I don't know :-) but I sure am not .. :)
1:54 AM
@robjohn we have done very little probability theory in our UG course .. this is the first time for me seeing this inequality is related to Mills Ratio (googling around leads me to references of results form probability) :-)
2:15 AM
@robjohn cool :-)
@Sawarnik didn't delete it .. simply changed the domain name O-O
@J.M. Yeah true haha.
I went out and walked for about 6 hours, lol.
@Jasper Wow, are your legs dying? My knees hurt like hell after walking for $3$ hours
Nope, used to it. I will start running in Nov.
2:23 AM
@JasperLoy Wasn't this $3$ hours ago?
@Committingtoachallenge Yes, I reached home 3 hours ago, lol.
@JasperLoy Good work :). Walking always makes me feel good
@Committingtoachallenge So you have a girlfriend?
@JasperLoy Yes, in 5 days I will have lived with her for 2 years and 3 months
@Committingtoachallenge Good, good. When are you getting married?
2:33 AM
@JasperLoy Haha, probably in $4$ years :P
@Committingtoachallenge OK. I might never find someone, lol.
@JasperLoy Don't say that, I am sure you will. Once you start uni again it will seem more likely
@Committingtoachallenge I hope so. It would be a miracle in itself if I get well by the end of next year, as targeted.
Are you taking anti-depressants? (Obviously don't have to answer if you don't want to)
@Committingtoachallenge I have stopped taking them for a month.
2:38 AM
@JasperLoy Good. I did a few courses at Uni as electives in Physiological and Cognitive Psychology at once of the top 10 university in the field, and they essentially said they were equally or less effective than spending more time in the sun :\
Going for a daily run was significantly more effective in decreasing depressive symptoms, and intermittent fasting had some mixed but encouraging results
@Committingtoachallenge Well, different people will tell you different things. But I now belong to the group that believes that psychiatric medications for depression and anxiety are largely useless.
Anti-dpressants are. Anti-anxiety does work though
Anxiety treatment is simple, since the neurotransmitters are well understood, but depression is far more complex
4 hours later…
6:20 AM
6:46 AM
That may or may not be my computers password xD.
I usually type in my password and then turn the monitor on, but apparently the chat box was selected
7:12 AM
@r9m lol, I didn't know that, but still ... that identity is a well-known identity, I don't think that I'd need to learn it from sos (like many many others things).
@r9m about the part with criticism, I don't think you got my point. Anyway.
@r9m This seems an endless discussion, so I dropped it. As a last thing to say on the matter, my proof to Au-Yeung series remains the best one in the world, and I'm sure this is just a small think amongst others that I'm going to accomplish in the next period of time. ;)
@r9m I ask apologize if needed that I got such results after only doing some serious math from the beginning of 2013. (the time is short for such results, I know ...)
hmmm, I hope to compute soon $$\sum_{n=1}^{\infty} \left(\frac{H_n}{n}\right)^3$$ this one wont' be any easier ...
Back some hours later (some work to do)
7:37 AM
What is $H_n$ above @Chris's
@Chris'ssis It looks nice & easy :)
@Committingtoachallenge It's the harmonic number. Well, if I compute it nicely I'm going to publish my proof.
I won the lottery. Not much, the smallest amount possible, but now I wonder how unlikely this event is.
So you want $\sum \limits_{n=1}^\infty \left( \frac{\sum \limits_{k=1}^n \frac1k}{n}\right)^3$
@Committingtoachallenge you have $3$ instead of $2$
@r9m Here?
@MatsGranvik How much?
7:44 AM
@Chris'ssis Sorry, yeah that is a beautiful problem :)
@Sawarnik 6 euros 90 cent
@Committingtoachallenge Yeah, it is, indeed.
Anything is nice :D
yes it is
7:47 AM
@Chris'ssis Do you have any idea what it equals?
@Committingtoachallenge Yeah, I have that result somewhere. (it's available in a paper by Flajolet)
If 5 numbers are chosen among 50 numbers and 2 additional numbers are chosen among 10 numbers, what is the probability to get 2 numbers right among 50 number and 1 number right among ten numbers?

Is the probability then one in Binomial[50, 2]*Binomial[10, 1] = 12250, or is there something to more to this probability? Seems a bit too much since I have not bought that many lottery tickets.
hello. off topic. Consider $f:\lbrace a\rbrace \to\lbrace b\rbrace$ for some real numbers $a$ and $b$. I am unsure whether this is necessarily continuous, whether its vacuously continuous or if it is defined to be non-continuous. On one hand, I feel that it is vacuous since I can "draw it without picking up my pencil," on the other hand, I can't take a limit. Any insight here?
nevermind, i wouldn't count that as continuous, even though I can draw it without picking up my pencil.
8:24 AM
@Chris'ssis: First question today was yours...
So the probability is, according to wikipedia:

Binomial[5, 2]*Binomial[50 - 5, 5 - 2]/Binomial[50, 2] =~ 115
1 in 116
I got that wrong. I forgot the additional numbers.
Binomial[5, 2]*Binomial[50 - 5, 5 - 2]/Binomial[50, 2]*
Binomial[2, 1]*Binomial[10 - 2, 2 - 1]/Binomial[10, 1]=45208/245
2 hours later…
10:07 AM
@robjohn Back. You refer to that series I posted above? :-)
I was about to miss the food for dogs, I did all things in a hurry, but I don't think I could have returned home without it. My dogs love me very much, if I'm away from home, they wait me near the gate no matter how long, they don't eat, don't drink but they only wait for me.
@Chris'ssis I get $\left(\frac1\beta-\frac1\alpha\right)\gamma$
@Chris'ssis No, I answered one on main from a week ago.
@robjohn Oh, I see now.
@robjohn That's true.
10:22 AM
Q: riammanam manifold

razi aLemma 4.6. Let ∇ be a linear connection on M . There is a unique con- nection in each tensor bundle T k l M , also denoted ∇ , such that the following conditions are satisfied. ( a ) On TM , ∇ agrees with the given connection. ( b ) On T 0 M , ∇ is given by ordinary differentiation of functions: ...

You mean "Riemannian Manifold"? Some people spell things as they hear them.
It's not my question, I just found it funny
10:53 AM
Reading a meta post's comment I saw this statement; Indeed, any use of L'Hopital's rule on $$\lim_{x\to 0} \frac{f(x)}{x}$$ where $f(0)=0$ is by definition circular
It's obviously circular for $f=\sin$
But I don't understand how it's circular for EVERY $f$? How do we know we'll need $\displaystyle \lim_{x\to 0} \frac{f(x)}{x}$ to evaluate $f'$ by the definition of the derivative?
@UserX: The derivative of $f$ at $x$ is $$\lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$$ and thus for $x=0$ $$\lim_{h \to 0} \frac{f(h)}{h}.$$
Hey @Huy
Hi, @Sawarnik.
@UserX lol
@Chris'ssis :)
11:03 AM
Lol Greece is weird. Yesterday a teacher in a highschool cut his wrists in the middle of a class because 2 students didn't leave the class when asked.
@UserX: How does that exactly reflect on the whole country Greece?
@UserX That didn't make sense :/
@Huy How does it not?
@UserX How does it?
@Sawarnik How does it not?
11:06 AM
When your program works
@UserX: Very creative.
I think a teacher should also have a strong personality, and very important, to respect kids. Many bad things comes from the lack of respect. Of course, the students must respect their teachers.
@Huy When I read the article, the first thing that came to my mind was that the image of Greece will deteriorate.
@UserX: I'm glad my brain works differently than yours.
If I were a teacher, I'd probably be tough with the problem, but very respectul at the same time. The lack of respect is the source of many unpleasant things.
11:09 AM
@Chris'ssis if you were a teacher you would the arrogant mathematician that doesn't care about anyone or anything.
@UserX How?!?
Which post are you referring to @Balarka ?
Click the little arrow on the left.
On mobile
4 mins ago, by UserX
@Sawarnik How does it not?
11:11 AM
@UserX lolll, I don't think so ... (this is a rule of mine I always follow: I respect you, you respect me, or you respect me I respect you).
Maybe it doesn't by that statement alone(although it's obvious that the teacher was allowed to teach even though he had psychological problems) but with the whole article, you can see that he was diagnosed as unfit to teach untill he was rediagnosed as cured, yet he was allowed to teach anyway.
@UserX: And that can only happen in Greece?
@Huy that HAPPENED in Greece which is already infamous as disorganised.
@UserX: How is Greece infamous as disorganised?
A professor humbled in our country
11:18 AM
I could not help but notice from the Map that this room does not have an African, Middle Eastern or Russian presence.
@Huy We are in a period of crisis. Of course it is disorganised. @Chris'ssis where are you from? Universities here are waaay different
@UserX Romania
@UserX: To me, almost every country seems disorganised.
@Huy Do you think Germany is disorganised?
@Chris'ssis: I'm sorry but I always imagine Romania to be Transylvania; with bats, werewolves, Count Dracula and Van Helsing
11:22 AM
@UserX: Yes.
@Chris'ssis: Teenagers dissin' a prof. is common in every hood. It wasn't as bad as what I've imagined Romaina to be like :D
@Nick: I've never seen any of that in any "hood".
@Nick Our country is $\Large \text{full of very very nice girls}$. :D
@Nick dissin' ?
@Huy: I didn't watch the entire video. I just saw the badass over there stand up and threaten the prof who just stood there calmly like a brick wall. Strong man, that teacher.
@Chris'ssis: I'm only interested in $\int$ Integirls $\int$
@G.T.R: dissin'is a verb which corresponds to the act of intentionally insulting or harassing a person through words, actions and or other media (like jedi mind tricks)
@Huy: I see kids like that everyday. They are socially deviant; in a negative way. They like to "stand up to the man" and assume a role of dominance; king of their domain. What is sad is that their attitudes and academical performance may not have any correlation with each other causing the possible scenario of a jerk with good grades and a geek with horrible grades.
Tis' true. One's desire to win can be far more beneficial to success than one's attitude.
On a less depressing note, all throughout history, abusive men and women have been penalized by society. See the case of Hitler and many others.
11:49 AM
How do I do a Horner's scheme with LaTeX?
12:08 PM
Hi @Nick @User @G.T.R @Huy @Chris'ssis and @Ice
:D hi pal @Sawarni
@Sawarnik Hi
@UserX: Use arrays. Go ask the tex room or refer the latex guide we all use.
@Nick Orly? :P
@Sawarnik: Yes, I can be interested in Integrals too. As long as they're easy to do.
12:11 PM
@Nick somtimes? and othertimes? :P
@Sawarnik: Other times I'm interested in derivatives.
Poor joke.
... What joke?
@Nick Well, we should ask @sab about it :D
He will fill Africa.
One man to fill Africa... O_O
12:17 PM
How can I use the Newton-Leibniz formula to evaluate $\int_{-x}^x e^t dt, x\in\mathbb{R}$ as doesn't it only work of the limits of integration are constants?
@user112495: They are constant in this case.
@Huy Oh, so do I just say i'm fixing an x, and then evaluate it?
@user112495: Sure.
@Huy: Isn't that integrals answer = $e^{x} - \frac{1}{e^x}$ ?
@Nick: Sure.
12:20 PM
@Huy: Sure
@Nick: Sure?
@IceBoy: Sure :|
This is so annoying.
@Studentmath: Sure?
Less, moreso this thing I am trying to prove.
I don't mind being sure for sure
12:32 PM
Assure to me a stoppage to this unsurely sure madness!
12:34 PM
\ _O/
What @Stud and me said are just a symbolic representation of the conversation :

"Up High!"

"Too high!"

"Down low!"

"Too slow, dude"
A very meaningful conversation
@BalarkaSen: What is said was a symbolic representation of a Dalek farting.
and suddenly I felt that one can be easily done ... some ideas came to mind ...
12:37 PM
@Nick What did you think of galois theory, then?
Cool stuff, no?
Deceptively easy?
@Chris'ssis So your country is full of very very nice girls like you? LOL.
@JasperLoy Yeah, my country is full of nice girls. ;)
@Chris'ssis Mine is full of silly girls, lol.
I need a brainstorm so badly
12:49 PM
@Studentmath what do you have in mind?
@Balarka afraid you can't help...
It's in probability-combinatorics
oh, probability stuff. ugh.
@Chris'ssis I haven't tried it. Do you have a simple solution? It often happens that when a difficult solution is presented, a simple one is hiding.
@robjohn Maybe that applies to FLT too, lol.
@robjohn I'm working on it, I have some nice idea.
@JasperLoy you're tough with the girls in your country ... :-)
12:54 PM
@Chris'ssis They don't like me either, lol.
@JasperLoy Well, if you can solve $x^{37} + y^{37} = z^{37}$ simply... you're welcome.
@BalarkaSen 37 is a very special number. It is the number of strokes in my Chinese name.
@JasperLoy There could be, but it has undergone a lot of scrutiny, so it is less likely.
@JasperLoy $\Bbb Z[\sqrt{37}]$ is a Dedekind domain, but yet not a UFD
Ubuntu Mate is very very good.
1:00 PM
@JasperLoy The girls do not work like the math theorems.
@Chris'ssis So, what is your estimate for your book's publication date?
@JasperLoy The end of the next year (the best estimation).
@Chris'ssis That is also when my mental problems will be completely solved, hopefully.
@JasperLoy lol, but all things you do happen sometime during the next week, month, year ...
@Chris'ssis Yes. I have not started running yet, but I will start next month, which is Nov, lol.
1:08 PM
Hello there.
Have you fixed the problem with your balls?
throws tables at @Pedro
Disks. Disks.
So what's up?
@PedroTamaroff Nothing much. Haven't thought about anything serious. Do you have some topology problems?
Hmm. Yes.
1:11 PM
Suppose $X$ is a metric space with the property that every infinite subset of $X$ has a llimit point in $X$, and take an open cover $\mathcal O$ of $X$. Then there exists $\varepsilon >0$ such that for any $x\in X$, $B(x,\varepsilon)\subseteq O$ for some $O\in\mathcal O$.
This number is called a Lebesgue number for $\mathcal O$.
And I trust that open cover of some set $A$ is a collection of open subsets $\mathcal{A} = \{B_i\}$ satisfying $A = \cup B_i$, right?
@BalarkaSen Yes.
Proceed by contradiction.
Cool. I suspected that from the terminology.
If no $\varepsilon >0$ exists, what can you deduce?
1:15 PM
OK, I am going to think about it.
@Pedro No hints!
It is a very nice exercise.
@PedroTamaroff What book did you use for metric space?
@MarcGato I didn't use a specific book.
@Marc Try Simmons.
@PedroTamaroff Ok.
@BalarkaSen Thanks
1:25 PM
@MarcGato I read Rudin, Apostol, Mendelson and Kolmogorov among others.
@PedroTamaroff Thank you I will take a look :-)
@Pedro I have to go at the moment, so I'll ping it to you when I figure it out.
Just so you know, I'd appreciate if you can send me some nice exercises on topology time to time.
@BalarkaSen Let E a normed vector space, $f$ a continuous linear form on $E$, $x_0\in E$ such that $f(x_0)\ne 0$. Show that $\Vert f\Vert=\frac{\vert f(x_0)\vert}{d(x_o,\ker f)}$ (operator norm here). I hope it's nice :-)
1:32 PM
@MarcGato Cool beans. In fact you can do something first: $f$ is continuous iff $\ker f$ is closed.
@PedroTamaroff It's a cool exercise but here we can do this exercise without this fact. I have to go. Good bye
1:50 PM
@Nick Generally, otherway round.
2:09 PM
How is everyone on this fine night?(12:09AM Australia)
Fine thanks, how are you?
Very good thank you Ice Man
I have spent all day travelling about doing business related things and now I am very tired
I didn't do any maths :(
It's 17:12 in Greece. Just finished my saturday exams, now I have 4 days free of classes because we have a national holiday this thursday and a celebration this monday
Are you a highschool student or university student @UserX
Last year of highschool
2:17 PM
And you want to go into medical school right?
I'll use these 4 days to cram 60 pages of introduction in set theory
What author is that @UserX
19:47 in India, the British have just left :D
Naive Set Theory, Paul R. Halmos
Is it any good? It's the shortest I could find, don't have the time for a long and detailed book
one of the best
2:21 PM
@UserX: Put that into your set of must read books.
Truly Ice boy? I haven't heard of it myself
you will :-)
Well I have now. SO I should have said "I hadn't heard of it"
Then my 4 days won't go in vain
2:41 PM
I wish I had time off to do some more of my challenge, but I am so busy for the next three weeks. After that I will start smashing through the challenge
What challenge, @Committing?
What's the challenge?
Oh just read it in your profile. If you think you can finish ALL exercises of 3 Rudin books(don't know about the rest but I know Rudin) I would say you're delusional at the least.
3:01 PM
@Nick Do they leave everyday?
@Nick Have you read it? Should I order it?
@Huy :D
@UserX Oh. True.
@Sawarnik: Sup?
@Huy Triangles.
@Sawarnik: "The British" is what I call my kitten. It's not my kitten, just a stray that wanders around my building compound. It's gone back to it's nest :D Also, 1947 is the year in which India got its independence, I thought it would be funny to say at the time.
@Sawarnik: No, Halmos Set Theory is one of those serendipity books. Read it when it comes to you.
@Nick Then at least your grammar was wrong.
But how is the kitten so timely?
@Nick Serendipity :/ So have you read it?
...comes in 10 min...
when it comes to these classic textbooks: "You only get out of it what you put into it"
3:10 PM
@Sawarnik: Maybe a englishman may have gotten onto a plane at the same time. Ohk, I won't cover it up, my grammar was wrong.
@Sawarnik: A bit, it's in my school library.
@Sawarnik: Also, what @IceBoy just said! He's right! Absolutely. Follow that wise man's words.
Why are most posts in MO community wiki?
Can someone tell me whether the following method is correct. Let $D:\mathbb{R}[X] \rightarrow \mathbb{R}[X]$ be the differential operator $D(f(X)) = f'(X)$ for any $t\in\mathbb{R}$. Prove that $e^{tD}(f(X)) = f(X+t)$.

$e^x = 1 + x + \frac{x^2}{2!} + \cdots + \frac{x^k}{k!} + \cdots$

$e^{tD} = 1 + tD + \frac{t^2D^2}{2!} + \cdots + \frac{t^kD^k}{k!} + \cdots$

$e^{tD}f(X) = 1 + tDf(X) + \frac{t^2D^2f(X)}{2!} + \cdots + \frac{t^kD^kf(X)}{k!} + \cdots$

$= 1 + tf'(X) + \frac{t^2f''(X)}{2!} + \cdots + \frac{t^kf^{(k)}(X)}{k!} + \cdots$
Differential operator*
I'm not sure whether I'm making any assumptions that I'm not allowed to make (assuming that this method is otherwise valid, which it might not be).
For example, should I write out f(x) as a taylor series as well, or is this sufficient?
When you multiply by $f(X)$ on the third step, you can't have $1+\mathcal{D}f(X) \dots$
Why didn't the 1 become $f(X)$
3:23 PM
@UserX ah yeah, sorry, I meant $f(X)+tDf(x)$
@UserX with that continued to the line below as well.
An irrevelant hint, use mathcal for the differential operator so it doesn't render as italic font and be similar to a variable
@user112495 Why do the power series(assuming infinitely differentiable and taylor series convergence) $ 1 + tf'(X) + \frac{t^2f''(X)}{2!} + \cdots + \frac{t^kf^{(k)}(X)}{k!} + \cdots$ equal $f(X+t)$?
@UserX Isn't this kind of just by definition?
@Sawarnik: How so, triangles?
@UserX Sorry, I think I see where i've gone wrong. I think I need to also express f(x), f'(x) etc. as power series.
@Sawarnik: Did you hear Taylor Swift's most recent album yet? It's awesome! :3
3:37 PM
@user112495 I don't see how the last equality holds. Try expressing $f(X+t)$ as power series, that will help you understand when you can have that last equality
@Huy: Taylor Swift should release an album called "Series" and the first track should "Expansion"
@usee112459 it's enough to show that the taylor series of $f(x+t)$ are the same as that last equality. They indeed are, so it's easy. Remember that $x$ and $t$ are dummy variables so you can interchange their places in the taylor series(given that you change do all the mappings $x \mapsto t, t \mapsto x$ and not leave any term unchanged$
@Nick: Not that one.
Back again :/ This connection's so unreliable.
3:49 PM
@UserX Oh, so can I just say $e^{t \mathcal{D}}f(x) = f(t) + xf'(t) + \frac{f''(t)}{2!}x^2 + \cdots + \frac{f^{(k)}{(t)}}{k!}x^k + \cdots$
Evaluate the taylor series of $f(X+t)$ as it may not be considered trivial. Then you can say the series and the error term are identical, thus equal(can't think of a proof but neither of a counterexample to this statement(
Can someone explain the birthday probability math that Bill Nye is trying to explain: youtu.be/TUcZTcSY7Zo?t=14m18s
@UserX If I use those mappings in the taylor series, then would that not represent $e^{X \mathcal{D}}f(t)$ instead?
You can avoid the mappings if you evaluate the taylor series of $f(X+t)$ around $t=0$ now that I think of it
So yea, don't do the mappings
@UserX Ah yeah, that would make sense :p. Thanks!

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