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8:00 PM
How's it going?
 
4 am here. I should go have some chocolate.
Hi @sarah lol.
 
@sarah
 
@BalarkaSen was about to do that
 
rude.
 
8:01 PM
hhhahaha
 
@sarah Today I watched "If I Stay", best movie ever.
 
@WillHunting chocolate will keep you awake
like coffee
 
I will sleep at 6 am I think.
 
@WillHunting oh? I will go see.
 
@IceBοy
Got pinged ?
 
8:02 PM
no ping
 
@Hippalectryon lol
 
@IceBoy Magic :D
 
@Chris'ssis $x^{1/x}$ from 0 to e. I had some imagination on how to solve the integral. But a step in my method diverges
 
@Hippalectryon you are a mathemagician
 
it is very very hard
Matt parker is coming to my school
 
8:04 PM
who?
 
wtf
 
what just happened?
burst in fits of laughter
hahhahahaa
 
stop that
 
that looks like a fractal
 
8:05 PM
stahp
 
oh noes @sarah is spamming again
throws a table at @Hippa
 
fu @BalarkaSen it's art
<3
 
clicks ->>> ignore user yes.
ahhh much better
 
i am going to do that too
 
ừ̴̵̶̷̸̡̢̧̨̖̗̘̙̜̝̞̟̠̣̤̥̦̩̪̫̬̭̮̯̰̱̲̳̹̺̻̼͇͈͉͍͎́̂̃̄̅̆̇̈̉̊̋̌̍̎̏̐̑̒̓̔̽̾̿̀́͂̓̈́͆͊͋͌̕̚ͅ‌​͏͓͔͕͖͙͚͐͑͒͗͛ͣ͘͜͟͢͝͞͠͡
 
8:07 PM
@Hippalectryon tell them I got rid of them
 
@BalarkaSen @Alizter She got rid of dem anyway
 
and we are getting rid of her.
 
she's nice
 
Is everyone ignoring me?
 
8:08 PM
but I'm ice
 
@WillHunting isn't Kanye west amazing.
nawt
 
@Hippa how in the hell are you doing that?
 
I saw the video of him telling a disabled person to stand up.
 
@BalarkaSen unicode
 
@BalarkaSen Magic magic :D --------- writing at the end to avoid killing other lines ----- ด้้้้้็็็็็้้้้้็็็็็้้้้้็็็็็้้้้้็็็็็้้้้้็็็็็้้้้้็็็็็้้้้้็็็็็้้้้้дด็็‌​็็็้้้้้็็็็้้้้้็็็็็้้้้้็็็็็้้้้้็็็็็้้้้้ด้้้้้็็็็็้้้้้็็็็็้้้้้็็็็็้้้‌​้้็็็็็้้้้้็็็็็้้้้้็็็็็้้้้้็็็็็้้้้้дด็็็็็้้้้้็็็็้้้้้็็็็็้้้้้็็็็็้้้‌​้้็็็็็้้้้้ด้้้้้็็็็็้้้้้็็็็็้้้้้็็็็็้้้้้็็็็็้้้้้็็็็็้้้้้็็็็็้้้้้็็็‌​็็้้้้้дด็็็็็้้้้้็็็็้้้้้็็็็็้้้้้็็็็็้้้้้็็็็็้้้้้
@sarah I saw that too !
 
8:10 PM
Was testing out.
 
Oh NO
 
He also said that Taylor Swift didn't deserve an award during her ceromony
 
Never you mind.
 
@sarah do you even math?
2
 
-_-
 
8:11 PM
Do you even shrek do anything math related on the chatroom ?
 
@Hippa
 
sure give me a problem
 
1+1 ?
-__-
 
2
 
@sarah Fancy some combinatorics?
 
8:12 PM
@sarah Oh i know
 
@sarah My question from earlier
 
@sarah
 
One by one, people.
 
8:13 PM
and who in the hell starred that?
 
I was about to say that
 
you are littering the chat room
people come here looking for math
 
It's all @Hippa's fault.
 
@Alizter 25
 
8:15 PM
not bad
@Hippalectryon your question has nothing to do with the gamma function?
 
@Alizter Uh i don't know
 
untag gamma function
 
@sarah Explicitly determine the nonempty set $S$ of nonzero natural numbers such that if $x \in S$ then $\lfloor \sqrt{x} \rfloor \in S$ and $4x \in S$.
show your reasoning.
 
it left
 
8:18 PM
and don't google.
 
Show that $\sin(10°)$ is irrational
:D
 
@Hippalectryon Isn't that easy?
 
@BalarkaSen Do it
 
@Hippalectryon who uses degrees? That's so last high school!
 
@Alizter If you want, i can add some $\frac{1}{2\pi}$ -__-
 
8:19 PM
@Hippa Niven's theorem is not hard to prove.
or just trisect 30 degrees
 
i'm actually shocked that sarah solved that problem
 
that'd do it
 
i though it was really hard :(
 
@Alizter WHat's the reasoning supposed to be ?
too lazy to think
 
$\sin(3\theta) = 4\sin^3(\theta) - 3\sin(\theta)$, right?
 
8:20 PM
I think so
 
@Hippalectryon exactly. I only knew the answer
 
then sub $\theta = 10^\circ$
$8x^3 - 6x - 1$ has no rational roots, by RRT.
 
RRT being ?
 
Rational Root Theorem.
 
Hm then I have a better problem :D
 
8:21 PM
The answer is 25, @Alizter?
 
@BalarkaSen What do you prefer ? Inequalities ? geometry ? trig functions ?
 
algebra and number theory mostly.
 
@Studentmath No it's $\huge 42$ ;)
@BalarkaSen What's larger, $\log_2(3)$ or $\log_3(5)$ ?
 
@Studentmath yes. How?
 
Oh I don't know. I just asked.
 
8:26 PM
oh
 
@Hippalectryon the former. $\log_2(3)$
 
Will try to do it.
@Balarka pick a number in random from 1 to 100
 
I am gonna do an all nighter
 
Why ? @BalarkaSen
 
too much homwork
and cambridge organ scholarships don't grow overnight so need to practice
 
8:28 PM
let me get my head round that one.
 
It's not supposed to be easy but... If you manage to do it quickly, i'll have a harder one :)
 
@Hippalectryon 3 is half between 2 and 4. 5 is not so half between 3 and 9. Therefore the former is a larger bit.
 
@Alizter What kind of proof is that ??
 
heh
 
you get me though?
 
8:34 PM
it's not a proof.
 
i never siad it was
just ituitional reasoning
 
The point is to find a proof
It's the hard part
 
your ituition is better than our intuition.
 
proofs start with ituition
:-)
 
8:36 PM
@IceBoy Got a proof of what you're saying ?
:P
 
no :(
 
The best I have is $$\log_2(3) \log_3(4) \log_4(5) = \log_2(5) > \log_3(5)$$
Not sure how to push it though
 
Want a hint ?
 
no
 
8:39 PM
Good :D
 
no
 
Good answer
I like giving false hints anyway
 
@Hippalectryon A kid asked me an apparently weird question these days, you reminded me of this by posting that logarithm question above.
If $a, b, c \in (1,\infty)$, then prove that

$$a^{\large\log_b(c)}+b^{\large\log_c(a)}+c^{\large\log_a(b)}\ge a+b+c$$
 
@Chris'ssis Want some geometry ?
 
got it
 
8:40 PM
oh @Alizter!
 
taylor series
 
@Hippalectryon Can you do it?
 
and cauchy product
 
nah
 
with special attention to terms
should get the job done
 
8:40 PM
i am looking for an elementary solution. it's easy to do it by taylor
just evaluate taylor of both for some 10 terms or somewhat
 
@Chris'ssis Let me get my slate :/
 
ze question is $3^{\log 3}>5^{\log 2}$
 
OK :-)
 
well $5^{\log 3}>5^{\log2}$
 
yes, so?
 
8:44 PM
but no
 
exponentiation will only cut off your errors
try working at the logarithmic level
 
@Chris'ssis Wow that's funny $a^{\large\log_b(c)}+b^{\large\log_c(a)}+c^{\large\log_a(b)}=c^{\large\log_b(a)}‌​+c^{\large\log_a(b)}+b^{\large\log_c(a)}$
such permutation
 
@Hippalectryon Ah, yes ...
 
I kinda feel like it's useless though xD
 
@Hippalectryon eh?
you have just shifted terms.
 
8:47 PM
Shifted ?
 
It's as ugly as before
 
@Studentmath That's what I was saying
 
@Hippalectryon $b^{\log_c(a)} = b^{\log_c(a)}$
 
I didn't change this term
 
@Alizter $3 > 2^{3/2}$.
 
8:48 PM
$c^{\log_a(b)} = c^{\log_a(b)}$
 
Only the first
xD
 
oh ah
 
@Hippalectryon but wait, he gave me another one that says that $$(a^a b^b c^c)^2\ge a^{b+c} b^{a+c} c^{a+b}$$ for all $a,b,c>0$. This one was an A-bomb to me.
 
that's easy
or it might not be
i stand corrected
 
@BalarkaSen @Chris'ssis Solve $\sin^7(x)+\frac{1}{\sin^3(x)}=\cos^7(x)+\frac{1}{\cos^3(x)}$
On $\mathbb{R}$
 
8:51 PM
@Hippalectryon Sure, after you give me some support on that inequality above. I'm focused on that now.
 
assume WLOG $a < b < c$. then $a^{b + c} b^{a+c} c^{a + b} = a^b a^c b^a b^c c^a c^b > a^a a^a b^b b^b c^c c^c$
no it is easy @Chris'ssis
 
^ what he said
 
patently obvious.
and now i am going to get some sleep.
 
@BalarkaSen sleep is for nonloser people
 
@BalarkaSen What???
 
8:54 PM
hence why I am not sleeping
 
@Chris'ssis it is obvious, that's what i meant
 
-_______________-
 
throws table at @Hippa
 
@Alizter lots of people are not sleeping waiting for iPhone 6
 
iPhone 6+ is awfully big
awfully
 
8:57 PM
What?
 
@BalarkaSen set $a=1, b=2, c=3$
 
I don't keep with these things
 
iPhone 6 line-up^^
 
@Chris'ssis yes?
your inequality holds and the proof is this
7 mins ago, by Balarka Sen
assume WLOG $a < b < c$. then $a^{b + c} b^{a+c} c^{a + b} = a^b a^c b^a b^c c^a c^b > a^a a^a b^b b^b c^c c^c$
 
8:59 PM
@BalarkaSen $$(1^1 2^2 3^3)^2\ge 1^{2+3} 2^{1+3} 3^{1+2}$$
 
yes, so what are you trying to imply by that?
 
$$11664\ge 432$$
 
i don't get you.
 
@BalarkaSen @Chris'ssis wrote it backwards
 
@BalarkaSen Shouldn't it be the other way ?
 
9:01 PM
it is a counterexample
 
throws table at himself
my bad.
 
yes you're bad
oh I am so punny
 
there you go @Chris'ssis
or no.
 
how is that a proof?
 
i am screwing up. it's a false proof.
 
9:03 PM
Proof by intimidation
 
@Chris'ssis Apologies, seems like it's a nontrivial problem after all.
throws a second table at @Hippa
 
@BalarkaSen No worry. :-)
 
I will be back, inequality! You can't run forever.
 
Balarka Gump
 
taps fingers at the third table
 
9:05 PM
@Hippalectryon Don't you work on my inequality? To quote Ted, I'm shocked! :-)))))))
 
@Chris'ssis I'm working on the first one
and trying to follow the chat
 
Ahhhh sonata IV II movement.
 
^ unexpected
 
It's like going bach in time
 
Bach decomposing... in his coffin... he composed to much when he was alive
 
9:08 PM
@Hippalectryon I think he is long past that stage
 
Too bad. I'd gladly have ordered a Bach Burger. 100% Bach meat.
 
who is the most famous french composer?
 
No idea who the most famous is, we have so many.
@Chris'ssis ??
 
@Hippalectryon :D
 
@Chris'ssis Find my age and I tell you :P
 
9:12 PM
@Hippalectryon 61
 
@Hippalectryon who is your favourite?
 
Hey I told you, I'm not Ted :P
 
@Hippalectryon :D
 
@IceBoy No idea, i stopped listening to classical 'orchestral' music long ago. But I like most of them.
 
icic
too bad, since you have so many to choose from
 
9:14 PM
@Chris'ssis I'm obviously younger than you are
 
@Hippalectryon or not :-)
 
Or not not ?
@Chris'ssis Here a very important hint : i'm older than my younger brother, and younger than my older brother :-))))))))))))
 
lol, that's the hint! :-)
By the way, I'm learning math with Ted.
 
Also, I live. Somewhere.
And I'm not dead. Yet.
Wow i can't even find myself on google
 
9:30 PM
cowfe
 
cowfe ??
 
I laughed so well watching that part with Ted saying "I'm shocked, I'm shocked, I'm old". :-)))
 
Mhhmmm... that is a good cup'o joe.
 
Bathroom singing, also known as singing in the bathroom, singing in the bath, or singing in the shower is a widespread phenomenon. Many people sing in the bathroom because the hard wall surfaces, often tiles or wooden panels, and lack of soft furnishings, create an aurally pleasing acoustic environment. The multiple reflections from walls enrich the sound of one's voice. Small dimensions and hard surfaces of a typical bathroom produce various kinds of standing waves, reverberation and echoes, giving the voice "fullness and depth." This habit was reported (with an attempt of explanations) centuries...
I was referring to this kind of singing. :-)
 
9:42 PM
@Chris'ssis: You laughing at me like @Hippa?
 
Uh I see :)
@TedShifrin You gotta see this masterpiece
 
@TedShifrin I laughed so so so well watching some part of youtube.com/…
:-)))
 
You clearly have way too much free time, @Hippa. What happened to your monitored studies, @Hippa?
Well, @Chris'ssis, students learn a lot more when they have fun.
 
@TedShifrin I'm studying. I had a 4h math exam today. Mind helping ?
 
@TedShifrin True! By the way, that was a great lessson, I liked that (especially the tricks to the last integral). :-)
 
9:43 PM
2
Q: Limit of the sum of $\gamma_k(x)=xf((k+1)x)-\int_{(k+1)x}^{(k+2)x}f(t)\mathrm{d}t$

HippalectryonLet $f$ be a continuous, decreasing function, with $\displaystyle\lim_{x\rightarrow\infty}f(x)=0$. Let $\gamma_k(x)=xf((k+1)x)-\int_{(k+1)x}^{(k+2)x}f(t)\mathrm{d}t,\displaystyle x>0$. Let $\Gamma(x)=\sum\limits_{k=0}^{\infty}\gamma_k(x)$, suppose that $\displaystyle\lim_{x\rightarrow0} xf(x)=A...

I didn't manage to do that last question
 
Which lesson, @Chris'ssis? The one @Hippa loves was just polar coordinates.
 
That was the polar coordinates in double integrals ...
 
Yeah ...
 
We're getting to the subtleties of limits in several variables as we now video the first semester.
@Hippa: I see robjohn gave you a hint. He's almost always on the money.
 
9:46 PM
video? :D
 
@TedShifrin Tbh i don't really see how the link would lead me to the limit
He's giving an upper bound that isn't the limit
And I have no idea how $\gamma$ is supposed to appear
 
@Hippalectryon Why do you think no one comes up with a solution on MSE?
 
@Chris'ssis I never use MO
The site desc says it's for pro mathematicians if I remember well
@Chris'ssis Also, some awfully difficult problems have been solved on MSE
 
@Hippalectryon MO is more focused on research level questions, those related to new ideas, concepts in math. On the other hand, I'm sure many from there would be in trouble with some of the questions posted here.
 
Hi guys, I am Narasimham I am quite new here
 
9:51 PM
@Chris'ssis Where should I ask then ? (well, you'll probably find an answer in a few days :) )
@Narasimham Hello
 
Welcome
:-)
 
@ Chris' s sis hallo
@ Ted Shirin hallo Sir,
 
@Hippa: Are you given $\lim_{x\to 0} xf(x)=A$ or perchance $\lim_{x\to\infty} xf(x)=A$?
hi @Narasimham
 
hallo
 
9:53 PM
@TedShifrin it is definitely $x\to0$
 
@TedShifrin $\lim_{x\to 0} xf(x)=A$ is a given hypothesis for the question
 
@Narasimham Hi
 
your name is one letter different from a very famous mathematician's.
heya @robjohn
 
just know that there is thing called chat'
 
@TedShifrin howdy
 
9:54 PM
& we can dynamically interact
 
Well, I trust you, @robjohn. But it would appear we want to compare $f(x)$ to $A/x$ for large $x$ to get $\gamma$ in there. Presumably there's a change of variables.
 
Hi Rob John I see u r in CA
 
@Narasimham there should be no between @ and the first letter
2 mins ago, by Narasimham
@ Ted Shirin hallo Sir,
 
So, @Hippa, have you contemplated a change of variables?
 
@TedShifrin $x\rightarrow tx$ ?
 
9:55 PM
no
 
@TedShifrin You might get a solution using a change of variables, but I don't think it is needed (if you were thinking of $x\mapsto1/x$)
 
<--- feels stupid thinking about the problem with @robjohn in the room :D
 
@TedShifrin Have you looked at the questions that come before ?
 
Right Ted Of course it is a name Srinvasa Ranujan = 10^ 100, me 10^-100!!
 
of course not, @Hippa. Who do you think I am?
 
9:56 PM
@TedShifrin Casimir errno
 
no, no, @Narasimham ... It's Narasimhan to whom I refer.
not Gauss either, @Hippa.
 
Not that one
That one
:D
 
I was reading the Dudley's book, a very good book, and I found the following exercise but I don´t understand what exactly he ask If $(K,d)$ is a compact metric space and $u\in K$ show that for any finite $M$ and $\alpha\in (0,1]$, $\mathcal{F}=\{f\in \text{Lip}(\alpha,M): \lvert f(u)\rvert\ \le M \}$ is compact with respect to the $d_{sup}$,. There the $M$ is fixed as well as the $u$ or that means that $\mathcal {F}$ is pointwise bounded I don't understand what he's asking? Thanks
 
Understood, he was a genius, he was not educated in the routine system and that explains his insight..
 
@robjohn did you make some progress on @Hippalectryon's question?
 
9:57 PM
@JoseAntonio: No, it holds for fixed $M$ and all $u\in K$.
@Narasimhan was a very well-known researcher in several complex variables. I know some of his work and books.
 
@ Ted Shifrin. Saw your book I liked it , havy many things to ask and learn
 
@Chris'ssis I think so. I have had to work on things offline, so I haven't gotten things written down.
 
OK the $M$ is fixed and any $u\in K$. Thanks @TedShifrin
 
@robjohn: You need to retire, like me :)
 
@robjohn So, you have a solution?
 
9:59 PM
@TedShifrin unfortunately, I must trudge on...
 

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