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17:08
@Sawarnik ??
@Sawarnik oh
A hippalectryon (or hippalektryon, from Greek ἱππαλεκτρυών) is a type of fantastic hybrid creature of Ancient Greek folklore, half-horse and half-rooster, with yellow feathers. The front half is that of a horse, the rear half a rooster's wings, tail and legs. The oldest representation currently known dates back to the 9th century BCE, and the motif grows most common in the 6th century, notably in vase painting and sometimes as statues, often shown with a rider. It is also featured on some pieces of currency. A few literary works of the 5th century mention the beast, though no myths relat...
@Hippalectryon are you sure I answered something like that?
@MikeMiller uh ?
@MikeMiller Oh no sorry
@MikeMiller it was @RossMillikan
@MikeMiller No idea why i made that mistake :c
What^ are you doing?
What part
All the @'s?
17:13
Uh... talking to Mike ?
xD
only needed one to see the message :)
@Studentmath Hey ;)
17:34
@Hippalectryon Ah.
What prerequisites must I know to understand Bezout's theorem (for curves)?
I imagine Hartshorne has a proof in the first chapter of his book
What do I need to understand Hartshorne?
some algebra
why don't you just find a proof online and try to read it?
The proof seems long and I don't want to find something that would take months to understand in the middle of it.
17:48
it uses some fancy stuff, but it develops it all
(as far as I can tell from a glance)
It does, thanks. I've always wondered: for equivalence relations, is $\text{set}/ \sim$ directly related to the quotient category at all? I see intuitively why the symbol $/$ is used, but not properly.
The pdf reminded me.
18:09
Find the closed form of $$\sum_{i=1}^{\infty} \sum_{j=1}^{\infty} \sum_{k=1}^{\infty} (-1)^{i+j+k+1} \frac{H_{i+j+k}}{i j k (i+j+k)}$$ (newly created)
@Chris'ssis You really want to kill us uh xD
And then attend the generalization $$\sum_{k_1=1}^{\infty} \sum_{k_2=1}^{\infty}\cdots \sum_{k_n=1}^{\infty} (-1)^{k_1+k_2+\cdots +k_n+1} \frac{H_{k_1+k_2+\cdots +k_n}}{k_1 k_2\cdots k_n (k_1+k_2+\cdots +k_n+1)}$$
@Chris'ssis That's cool :D
@Hippalectryon The art is meant to feed the soul. This is an art.
2
@Chris'ssis YES
18:13
@Hippalectryon :-)
@Chris'ssis Too bad i'm not good enough to solve those :/
@Chris'ssis If you have some time, send me the solution one day :)
OK
There is a mistake in denominator. (without "+1")
Ah yeah
@Chris'ssis The first one $\sum_{i=1}^{\infty} \sum_{j=1}^{\infty} \sum_{k=1}^{\infty} (-1)^{i+j+k+1} \frac{H_{i+j+k}}{i j k (i+j+k)}$ reminds me of Quaternions :D
@Hippalectryon An American mathematician told me once some things about quaternions, but I forgot ...
@Chris'ssis Nah it's just the $i,j,k$ stuff that made me say that :) nothing complex (haha such pun)
18:21
@Hippalectryon :D
wow, I was recently looking at one of Arturo Magidin's answers and boy does he know how to write well.
@Hippa Hamiltonian quaternions freaks me out.
@BalarkaSen Lots of things in maths freak me out
Anyone else having problem loading the chat?
@Hippalectryon Like?
mathematicians freak me out
2
18:28
Haha, @Charlie
@BalarkaSen yes
@BalarkaSen Like, the possibility of making an exercise on a polynomial whose roots are the Vapnik-Chervonenkis dimension of a hypergraph of the set of coefficients of the rank n multinom formula :3
I don't even wanna think about it
-___________-
@Hippalectryon I don't even get what you're talking about.
@BalarkaSen What part
Vapnik-Chervonenkis dimension, hypergraph and how it relates to polynomial roots.
Did you made that up?
18:35
@BalarkaSen Let $S$ be a set with $n$ elements.

Let $P(S)=\{X\mid X\subseteq S\}$

Let $H\subseteq\mathcal{P}(S)$ (hypergraph with edge set $S$).

Let $H_{|U}=\{U\cap A\mid A\in H\}$

Let $\dim_{\text{VC}}(H)=1+\max\{|U|\mid U\subseteq S\text{ and }H_{|U}=\mathcal{P}(U)\}$
The only relation of graphs with polynomials roots I know of come from an older version of dessin d'enfant by Klein.
@BalarkaSen $\dim_{\text{VC}}$ is the Vapnik-Chervonenkis
@Hippalectryon What's $X$?
A hypergraph with edges set $S$ is just a subset of the power set of $S$
The zero locus of the polynomial?
18:35
Uh no
Are you not used to the $\{\text{element}\mid\text{condition}\}$ notation ?
Oh.
Missed that.
So $X$ is defined that way.
Basically, the power set of $S$ is the set of all the sets subsets of $S$, ie all the $X$ such as $X\subseteq S$. Hence the line $\mathcal{P}(S)=.......$
I know, I know.
@MattN. Huehue
@MattN. We should tell @TedShifrin :D
@Chris'ssis what's that strange $F$ ?
And what's a hypergeometric or generalized hypergeometric function ?
I need to do some jogging ... back later on.
@Hippalectryon You know more than it seems at first sight. ;)
18:47
what happened to Jasper Loy?
Did he delete his account again ?
I don't know
probably
@bana he said he'd never come back
18:56
@bana you do know it's not the first time he's done that (probably not the last), but he doesn't give any reason...
@G.T.R He says anything.
I see.
@G.T.R Ils sont quand tes oraux ?
Dans 2 semaines
19:25
@G.T.R Tu me donneras tes exercices :)
english sucky soccer. Go ARgentina
19:39
Hi all! Here you may find a problem in the context of calculus of variations. Reproducing Kernel Hilbert Spaces are also included. Any help would be very very desired!
20:00
welcome
@Hippalectryon A particular case, $n=3$, below
@Chris'ssis Mind making a correction ? :D
@Hippalectryon Go ahead. What is about?
@Chris'ssis I meant, about how to find that rseult
anyone know about topological cones or asymptotic cones?
20:14
I've just developed a novel avocado processing algorithm. I call it the triptych peel, it allows one to both remove the spoon requirement from the process, while increasing the recovered fruit material efficiency to nearly 100%.
Hi all
I asked 2 new questions today :)
@mick hi
hello mick, nice to see someone with such a good name around :D
3
Hi charlie ! Like your new haircut ! @C
@Charlie
@mick thank you
20:16
@Charlie your welcome my dear
@mick did you make birthday already?
yes
how old?
I gotta go in bath guys
bye
look at my questions :p
avadado
20:17
maybe later
bye
-_-
@ComTruise it's been like two years he claims to be 14
sometimes this math is too much for me
frying my brain
mmm fried brain, got any acid? integration problems?
Thanks God! Another titan is dead! :-)
20:23
I tried to take a nap and had nightmares about math
who died?
@Chris'ssis england?
@Charlie No. I'm referring to a multiple integral that comes from my research.
@Chris'ssis ah
@Hippalectryon Just give it a try. You'll have a lot of fun! I guarantee you that! :-)
@Chris'ssis Don't tell me that >:o i'm a newb wit those kind of integrals
20:25
Hippopatamus
@ComTruise >8c
hungry as a Hippo
@ComTruise
hippopotamus are violent creatures
whaaaaaaaaat
20:28
is Hip Pop sortof like Hip Hop Pop?
but they look sweet
@ComTruise They can kill you
@ComTruise they're thirsty for blood
@ComTruise if you get near babies you can be chased after
20:29
@Charlie They run FAST
@Hippalectryon yeah bitches
@ComTruise And they don't stop running after you until you're really far away
Wanna be crushed to death by a hippo ? :D
no I just want to hug one
20:33
i remember that, Fantasia? Dumbo?
@ComTruise fantasia
baby Hippos look like pigs kindof
anyone here versed in primes and can help me determine if I'm onto anything new?
@Hippa: Speaking of voting irregularities. I think they took away some of your votes for me, too ... "Serial voting reversed."
Hi @Charlie @Com
20:37
@TedShifrin It was like 5 or 6
serial voting? lol
@TedShifrin And i do that often
Well, they took away points when they restored the ones from the pissed-off downvotes.
@TedShifrin hello theodore
@Charlie Can you speak Minionese?
20:39
Oh, and @Hippa ... Thanks so much ... Your meme is now viral among my students on FB. GRR.
@Sawarnik oppa massara
@TedShifrin How ? :D
@TedShifrin I have seen it in FB too, and not from one of your student. :D
Oh great @Sawarnik.
20:40
@TedShifrin hmmm, I wonder if the mathematical e-books have a good selling volume ... (in general I mean)
huh? @Chris'ssis
@TedShifrin I saw you have some books on Amazon, don't you? @Hippalectryon pointed out that in a picture.
Not e-books, @Chris'ssis.
@TedShifrin Ah, they are not e-books ... OK
There's a free .pdf book on my webpage.
4
20:42
@TedShifrin Don't worry i can't find you bu searching the image on google yet :D
2-1
Uruguay!
you guys love to have someone to worship....
2
this guy keeps asking questions like this math.stackexchange.com/questions/840019/…
@robjohn you're too silent there :D
@Chris'ssis worship @Chris'ssis
20:46
@Hippalectryon lol, how is that? :-)))
I wonder if suspensions alter people's behavior once they're reinstated. I would tend to doubt it.
@Chris'ssis You are my integral god :3
they comeback like this:
http://i245.photobucket.com/albums/gg72/heresjohnny_05/The-Shining---Heres-Johnny-Poster-C.jpg
you're not an atheist, @Hippa? :)
20:47
@TedShifrin notban ????
I should quit typing on the iPad :(
@Hippalectryon No, I'm not at all. I don't appear amongst those gurus Ron was talking about in the picture you showed me yesterday. :-)
@TedShifrin Can't be one since i've met @Chris'ssis 's integrals :3
@Chris'ssis Secret god then :P
@Hippalectryon lol :D
@Hippalectryon by the way, robjohn is a model to me. I learned a lot of things from him. He had a lot of contribution to my development.
@Chris'ssis He's my second god :D
20:52
@Hippalectryon Well, he is a math god.
@Ted issue's been dealt with.
Thanks, @Alex. We saw. I wonder if suspension modifies behavior. The computer restored those points but took away others for serial upvoting!
I can't say much but thanks for informing me, you were the tip of the iceberg.
@Hippa: Clearly I am a math satan.
2
@TedShifrin Stop giving me new meme ideas :3
20:56
I want to be the iceberg, not just the tip!
4
@TedShifrin Welllll, you were half the iceberg anyway. It was really more like two icebergs. That makes you an iceberg.
You clearly have way too much free time, @Hippa. I need to give you more problems.
Silly @Alex
@TedShifrin :c i still haven't solved @G.T.R 's $P-aP'$ one
I haven't thought about it.
@TedShifrin What happened there btw is that invalidating votes usually means simply wiping the serial vote history between two users, so he must have just upvoted you 3 times at some point. I'm not sure it's possible to do it more precisely.
21:07
ohhh, interesting, @Alex ... Thanks for that. Yeah, we exchanged friendly words (when I corrected him on something) a year ago, but that won't likely happen again.
I want to complain about serial starring of my comments in here!!
Someone should write a script, someday, to go through the star history and see who's gotten the most stars total, the most stars for a single comment, etc.
We could hold an awards ceremony.
Or, more likely, a disawards ceremony.
@r9m Bye.
2:40 now.
r9m
r9m
@Sawarnik ah .. okay ,, gdn8 :P
howdy @r9m
Salut, @Gabriel ... Tu vis toujours!
21:13
@Ted salut, oui encore en vie
r9m
r9m
@TedShifrin hello professor :D .. how was your day ? :)
@skull: You should spend more time on math and less time on prattling.
@G.T.R t'as vu mon meme ? :D
Tout le monde l'a vu, @Hippa.
@TedShifrin How did your student even find it ?
21:14
Kaj was in here all yesterday evening, right?
anyone want to help with a quick opinion on prime psuedo-code
Don't know anything to help, @AlexL.
@TedShifrin Is Kaj a student of yours ?
@DanielFischer I can't prove your statement. The best I come up with is $\displaystyle \int_a^x(x-t)f''(t)dt\geq 0$ but what then ?
@ted
Thanks tho
21:15
Not at the moment, but he's taken 3 courses from me, @Hippa. I think that's it for him.
daccord
He came in here because I asked him about @Pedro's Ramsey theory question, @Hippa.
hi @skullpatrol
@TedShifrin The vertices thing ?
21:16
@Charlie hi
@G.T.R If $a$ is a local minimum and $f''(a) \neq 0$, then $f''(a) > 0$. If $f$ is twice continuously differentiable, you have $f''(x) > 0$ on some $(a-\varepsilon,a+\varepsilon)$. But $f'' > 0$ implies that $f'$ is strictly increasing, and that implies that $f$ is strictly convex.
@skullpatrol how are you, mister?
@Charlie Fine thanks, how are you?
@AlexLieberman Prime pseudocode? What's the goal?
21:19
@skullpatrol I'm fine thanks
Find primes one by one up to N - faster.
Miller Rabin
to find primes
@Charlie Ooookaaay....any news from the world scam I mean cup?
@AlexLieberman Sieve of Eratosthenes, enhanced, is the fastest way.
Have you seen my code? Anything enhanced like it?
21:22
@skullpatrol a bunch of guys from chile tried to inavde the staium withou tickets
oh dear, @Ted, that top comment could be truncated... poorly
Lots of me can be poorly truncated ...
just the tip ?
@G.T.R He definitely said "not just the tip"...
@Mike that's a phrase in English right ?
21:24
I guess my code could be considered S-O-E, but I feel like its modified beyond standard S-O-E methods.
@AlexLieberman Is it readable pseudocode?
Surely
I mean I believe so - I have the code in Java and the pseudocode
Looking.
check out the miller-rabin algorithm. I think it is based on SOE with some modification
@GTR: Your $P-aP'$ problem ... $P$ had real roots, $a\in\Bbb R$? $a>0$?
21:26
@Ted the sign of $a$ is irrelevant, and $P$ has only real roots
$a$ is real, yes
I'm trying to think about it for a quadratic. The proof is not clear.
Oh, a typo isn't helping.
r9m
r9m
@G.T.R what is the problem statement ?
OK, so it's easy for a quadratic.
$P$ is a real polynomial with only real roots. Prove that $\forall a\in \mathbb R$, $P-aP'$ has only real roots.
r9m
r9m
@G.T.R ah okay ... is it useful to look at $e^{-x/a}P$ ?
21:31
@r9m I like this trick as much as you do, but no
LOL ...
I contemplated Rouché, and discarded him.
He's French, after all.
He's thinking, @AlexL. Don't ping him multiply :P
o
the first one wasnt a ping
Well, yeah, it was.
Oh, I wasn't certain if it worked haha
I didn't mean to be rude.
:P
Daniel is very conscientious, but I'm sure your code takes time to parse.
21:34
Yeah I suppose so. I've spent a good few weeks of sleepless nights on finding it.
Daniel is great.
2
@Ted the key is logarithmic derivative
I had just gotten there when you said that, @GTR. Growl.
This reminds me of Lucas's Theorem about complex roots.
@Ted good thought
:16173728 Too complicated. Just marking multiples is faster than your logic. You should not use a boolean[], but single bits, the smaller memory footprint more than compensates the small additional bit-fiddling work. And your sieve should not contain even numbers or multiples of $3$, that space usage costs time.
21:38
Hrm.
I've tried BitSet and its slower.
I thought they had fixed that meanwhile. Just use an int[] with every array entry containing 32 flags.
What do you mean my sieve should not contain even numbers or multiples of three?
It doesn't contain them.
@G.T.R How is log derivative helping ?
put a dollar in the kitty
I've tried storing all the primes one after another, but that takes more time I believe.
21:41
@AlexLieberman new boolean[maxNumber]; says it does. Your flag i corresponds to the number i.
Not sure I understand exactly, but I am trying hard to.
I add more space I understand - wasted boolean slots.
So the time complexity space wise isn't that good.
But the time complexity of the time to calculate seems faster...
Hmmm, I think I solved an open question (well, maybe it's too much to call it "open question" - and then my solution only offers a transformation to an integral that for the small cases can be easily evaluated)
@r9m
r9m
r9m
@G.T.R its not a single step problem .. you have to look at several cases .. :) we begin with $e^{-ax}P$ when $P$ has distinct real roots .. next comes the case of multiple roots .. :)
wth ... took me like minutes of retry and retry fail to post this one :(
@Chris'ssis What one ?
r9m
r9m
@Chris'ssis \m/ Height !!!! :D
21:48
@Hippalectryon @r9m $$\sum_{n_1=1}^{\infty}\sum_{n_2=1}^{\infty}\cdots \sum_{n_k=1}^{\infty} (-1)^{\large n_1+n_2+\cdots +n_k} \frac{H_{\large n_1+n_2+\cdots +n_k}}{n_1\cdot n_2\cdots n_k}$$
@Chris'ssis Send me how you did this one too if you get some time one day :)
r9m
r9m
@Chris'ssis cool .. its mentioned as open problem in Furdui's papers :-)
@r9m Well, it's easy in a way, but some more research in the area of the polylogarithm may reveal a nice finite series of expressing the result.
@DanielFischer got it
@Chris'ssis :o
r9m
r9m
21:53
posts another not bad meme pic @Chris'ssis
what are the H's?
@ComTruise harmonic number
Did you know AM-GM was mere concavity ? I thought there were no other proofs other than backward-forward induction
This is a nicer version though
@r9m @Hippalectryon$$ \sum_{n_1=1}^{\infty}\sum_{n_2=1}^{\infty}\cdots \sum_{n_k=1}^{\infty} (-1)^{\large n_1+n_2+\cdots +n_k} \left(\frac{H_{\large n_1+n_2+\cdots +n_k}}{n_1\cdot n_2\cdots n_k}\right)^2$$
r9m
r9m
@G.T.R ya .. log concavity (that is the only proof I knew for a long time .. until one morning I saw a proof by induction on M.Se :P )
21:59
hi

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