« first day (1598 days earlier)      last day (3417 days later) » 

12:11 AM
@alizter You should be sleeping.
 
Hehe, I was downvoted.
Three questions in a row.
@JasperLoy Maroon suits you.
 
@JasperLoy The time only applies when I am tired.
 
@PedroTamaroff I call it brown, lol.
@Alizter But you have a cold, lol.
 
@JasperLoy It is maroon,.
Not brown.
 
@PedroTamaroff Hmm, OK. But when I used GIMP to create it, I typed 'brown'.
 
12:25 AM
@JasperLoy GIMP is bad.
So bad.
 
Not as bad as living in my country.
I am thinking of who I should email. I feel really sad and I would like to talk to some people, other than my real life friends.
Anyway, next week I will meet a friend and a prof. And next next week I will meet another friend and another prof.
 
@PedroTamaroff yo ho, it is duplicate time.
6
Q: Closed form of a Definite Integral

Nicholas PipitoneI attempted to integrate the following function from a practice problem in my Calculus textbook: $$\int_{0}^{\frac{\pi}{2}}{\frac{1}{1+\tan^\sqrt{2}(x)}} \ {\rm d}x$$ I failed to find an indefinite integral, and I am assuming getting an indefinite integral is simply impossible. Using Wolfram|Al...

 
@N3buchadnezzar Good.
 
12:40 AM
Is this site really a good thing? Hmm.
Has it really helped people who are studying math?
 
@JasperLoy I definitely helped me, Jasper.
 
@JasperLoy What is the point of living. What is love. Why did I eat so much puddi puddi?
 
@N3buchadnezzar That's not the point, lol.
 
@N3buchadnezzar Silly. Puddi puddi ate itself.
 
@PedroTamaroff Hmm, sometimes I think you might be better if you did not use this site.
 
12:42 AM
@JasperLoy Each time you refrain from voting to close a duplicate a puddi puddi eats herself.
 
@JasperLoy Shun the nonbeliever!
 
Voting will begin in a few hours for the election.
 
@PedroTamaroff Btw why do you help people here?
 
@N3buchadnezzar Because I like doing so.
 
These days I come mostly to chat.
 
12:46 AM
I mean sometimes I feel likeI am talking to a wall. I try to help someone, but they are just interestedin the answer. I am not sure this is completely bad. Not everyone need a large portion of maths in their lives. alas helping someone with very little motivation for something you love can be very demotivating
 
This help thing is becoming a very deep question...
I am going to sleep now.
I hope I find divine guidance in my dreams.
 
@JasperLoy Vote to close :p
 
That was a good opportunity not only for a "close" vote but also to downvote the original PSQ.
Two birds with one stone.
OK, it was really two stones I guess.
 
@Bungo I am thinking about making a post on meta for common duplicate questions
So it will be easier to look up duplicates. Especially integrals completely written in tex can be difficult to near impossible to finding dupes
 
@N3buchadnezzar: It could be useful if it doesn't become such a long list that it is hard to find things.
I have spent 20 minutes reading statements from candidates for the moderator election, which is about 15 minutes more than I spent on the actual U.S. election in November.
 
1:10 AM
@Alec: FYI, it's not "out of print"—they print it "on demand." shrug ... I wrote books for the same reason I devoted myself to teaching: I wanted to help people learn, but my books are definitely not for everyone (or even for 1% of everyone) :)
@Mike: One fifth of scotch, or one fifth of the exams?
 
@TedShifrin it may as well be out-of-print, the price is that high!
@TedShifrin seriously, wouldn't it be awful if someone leaked a PDF
 
I don't want to hear about that.
Given the prices of most books in the US, it is competitively priced.
 
The US bookstore is awful though, and they're clamping down on ability to sell second hand, it's awful.
But anyway @TedShifrin it'd be awful if someone with a PDF leaked it
AWFUL
 
Alec, f*** off.
 
As in - it keeps me awake at night
@TedShifrin what have I said?
/me is confused
@TedShifrin you don't get what I'm saying do you?
It'd be awful if people were to google your book and find a PDF.
I just had a coughing fit
@TedShifrin on a serious note I could not believe MIT has erased all of Walter Rudin's great work.
They've totally expunged all his open courseware stuff, which I adored. It's a real shame and quite tragic TBH. They were great contributions.
 
1:29 AM
@Ted Three fifths of exams now - the one fifth of scotch comes after!
 
@AlecTeal: Surely you don't mean Walter Rudin when you say Walter Rudin.
The former was affiliated with the University of Wisconsin, and so was the latter.
 
1:47 AM
@Bungo - sorry, Dyslexic moment, Walter Ludwin!
 
2:06 AM
@Studentmath I was busy all day yesterday and today so I did not yet get a chance to calculate the number of ways to make a connected graph out of 5 vertices using 4 edges, 5 edges, 6 edges. How would you make it using 4 edges...would this just be 0?
Can someone help me with this?
A ladybug is walking at random on a hexagon. She starts at vertex A. Every minute, she moves to one of the two vertices (chosen at random) adjacent to the one she's currently on. What is the probability that, after 10 minutes, she is back at A?
@Studentmath Hello!
 
can someone explain the me what the double integral of xy dx dy looks like from 0 to 1 for both integrals
 
@Studentmath: You still there?
*bump Iss anyone still active on here?
 
2:24 AM
im here
 
2:40 AM
I did the thing. I answered a question in meta meta.math.stackexchange.com/a/18803/18908 ^^
 
3:02 AM
Duplicate
8
Q: Integrate: $\int_0^\infty \frac{\log (1+x)}{1+x^2}dx$

Mula Ko SaagCan this integral be solved with contour integral or by some application of Residue theorem? $$\int_0^\infty \frac{\log (1+x)}{1+x^2}dx = \frac{\pi}{4}\log 2 + \text{Catlan constant}$$ It has two poles at $\pm i$ and branch point of $-1$ while the integral is to be evaluated from $0\to \infty$....

please close =/
 
r9m
@N3buchadnezzar the limits are different :O
 
Anybody have a copy of Apostol' Mathematical Analysis?
 
r9m
@TimRaczkowski hard or soft copy ? what are looking for ? :)
 
3:17 AM
Doesn't matter. I've got a photocopied electronic version and wanted to verify a problem.
 
r9m
@TimRaczkowski which problem ? page number plz :)
 
p.211 problems 8.9 and 8.10.
OK so 2 probs. ;)
 
r9m
@TimRaczkowski do you want me to write the two problems down here ?
First one: $|a_n| \le 2$ and $|a_{n+2} - a_{n+1}| \le \frac{1}{8}|a_{n+1}^2 - a_n^2|$
 
In 8.9, there's a multiplicative constant which appears to be 1/8.
 
r9m
@TimRaczkowski yes !
 
3:25 AM
How about the fractional exponent in 8.10.
The one for L.
 
r9m
@TimRaczkowski Second one : $a_1,a_2 \ge 0$ ... $a_{n+2} = (a_na_{n+1})^{1/2}$ and $L = (a_1a_2^2)^{1/3}$
 
OK. Got it. Thanks.
 
3:45 AM
 
4:11 AM
@Integrator That bumping was very annoying.
 
@PedroTamaroff People here, will do anything for hats!
 
r9m
@Integrator oo ! there's a Hat for that ? :D
 
267
Q: Winter Bash 2014 Secret Hats

KevinSo far, I've unlocked one of the secret hats. I'm wondering what other secret hats are out there and how to earn them?

 
r9m
@Integrator MOG !! :D Thanks ! ;)
 
@r9m Now I just need waffles!
@r9m No way to get eureka!
 
r9m
4:17 AM
I just looked up .. MOG means a moggie , Mother of the groom :P I meant to say Mother of God :P
 
@N3buchadnezzar Unless I'm overlooking something, that old question is not a duplicate. The limits of integration are different from the one you linked to on meta. The two can be related, though, as shown by answers to that question. Another question was closed as a duplicate of that question just today.
 
4:33 AM
@TedShifrin, I very much dislike the fact that Athena won't let us check final grades until December 23.
 
@KajHansen Yikes. Christmas present, eh?
 
Tell me about it @PedroTamaroff. The old system was better in so many respects.
 
@KajHansen I should look at my system. I sat for some finals this semester.
 
I got an S in "Teaching College Mathematics"
 
@MikeMiller S as in "Super"?
 
4:40 AM
as in sicknasty
 
@MikeMiller, I unlocked Naruto without trying ;)
 
Or S as in "Produces a difference in air pressure that pushes objects inside a hole"?
 
@KajHansen I guess your answers are just worse than mine :)
 
lol'd @PedroTamaroff
LOL
 
r9m
5:00 AM
wha ??!!! I can't have Fascinating and Fascinating Maam together ? :O
This guy is hyped up !! HYpe level
 
r9m
5:32 AM
@MikeMiller can I ask you to do me a favor regarding hats ? ;) :D
 
 
2 hours later…
7:41 AM
@epimorphic: nice aurora...
 
8:04 AM
Glad you like it! :)
It's from a cg wallpaper I found online.
@robjohn Was it enough to turn that frown upside down?
 
@Chris'ssis yep caught that. Thanks Chris's sis :)
 
Hello @AlecTeal !! Are you familiar with the programming language C ??
 
8:28 AM
Hello @MaryStar !! Are you familiar with the gum brand Orbit ??
 
Are you guys familiar with the Khan academy?
 
8:44 AM
OK you guys I've got a stupid question
 
Don't expect a stupid answer :P
 
@AlexanderGruber OK
 
Why @BalarkaSen is ted ignoring you?
 
It's because I said algebra > geometry.
:P
 
so I've got that $\forall \epsilon > 0$, the measure of $\{a:f_n(a)-f(a)|\geq \epsilon \}$ is $0$. I've also got that for some subsequence $n_k$, $f_{n_k}(x)\rightarrow f(x)$ almost surely.
also $f_n(a)$ is monotonic for all $a$
 
8:49 AM
Yuck.
 
what i would like to do is say, "then $f_n(a)\rightarrow f(a)$ for all $a$ for which $n_k$ converges"
which would be nice, except that I need to establish that $f_n(a)$ does actually converge at those points, and I don't see how the first condition with the measure gives that to me.
 
9:08 AM
Does $\displaystyle \int \frac{\ln\left|e^{2x} + 1\right|}{1 + \sin2x}\mathrm dx$ have a closed form?
I've occupied 8 pages so far in search for it... By-parts is an ugly tool on this beast.
 
9:35 AM
Greetings
Here is a question in Ramanujan-like fashion
$$\int_0^1 \frac{1-x+\log(x)}{1-x+x\log(x)} \ dx$$
 
9:47 AM
Greetings
 
Haha
@Chris'ssis What were you doing in election room? You made laugh :D
 
A harmless error :-)
 
@Venus I entered the wrong room :D
 
LEL
 
@Chris'ssis You just made my day :D
 
9:50 AM
@Venus :-)))))
 
Morning @DanielFischer
 
@Chris'ssis This integral is tough: $$\int_0^1\left(\frac{1}{\ln x} + \frac{1}{1-x}\right)^2 \mathrm dx$$
 
Morning @Balarka Sen.
 
@Venus I asked that in the past. It's not tough at all.
 
@DanielFischer Good morning!!!

Can you help me at this function: http://stackoverflow.com/questions/27562431/insert-new-node-in-tree?noredirect=1#comment43552079_27562431 ??
 
9:53 AM
@DanielFischer Guten morgen...
 
Guten morgen!!! :D
 
@Chris'ssis I saw it you just edited it
 
@Venus :D
@Venus I think I can come up with an incredibly fast solution.
 
@Chris'ssis I'm thinking of Feynman's way
How?
 
@Venus What happens if you integrate over $[0, 1]$ with respect to $s$?
In mathematics, the digamma function is defined as the logarithmic derivative of the gamma function: It is the first of the polygamma functions. == Relation to harmonic numbers == The digamma function, often denoted also as ψ0(x), ψ0(x) or (after the shape of the archaic Greek letter Ϝ digamma), is related to the harmonic numbers in that where Hn is the n-th harmonic number, and γ is the Euler-Mascheroni constant. For half-integer values, it may be expressed as == Integral representations == If the real part of x is positive then the digamma function has the following integral representation...
 
9:58 AM
@Chris'ssis That's my Feynman's way to evaluate your integral
 
@Venus Then you can write our integrand by using a double integral.
@Venus OK. Still, I'm thinking of a faster way.
 
@Chris'ssis I love sos440's approach
@Chris'ssis Please elaborate
 
@Venus Let me see if it works ...
 
10:15 AM
@epimorphic I am also an astrophotographer, so I enjoy pictures of atmospheric and astronomical targets (all nature photography can be beautiful).
@Venus I just got downvoted on my answer to that question :-(
I asked for a comment, but most likely will not get one.
 
@Hippalectryon @MikeMiller In the next version (14.12.2) review task counts should auto-refresh every 100 seconds. Also, I added individual "reload" links to the question links, so those can be updated as needed, without reloading the whole page. Let me know if anything goes wrong with this.
 
@robjohn What? I didn't understand your answer but I didn't downvote it
 
@DanielFischer Do you maybe have an idea??
 
@Venus I didn't think you had. I was just disappointed since it seems baseless.
 
@robjohn I know that maybe someone is jealous of you.
@robjohn I've just seen this integral on internet $$\int_0^1 \frac{1}{\left(1+x^2\right)\left(e^{x}+1\right)}\, dx$$
Here is my approach
\begin{align}
\int_0^1 \frac{1}{\left(1+x^2\right)\left(e^{x}+1\right)}\, dx&=\int_0^1\int_0^\infty \frac{e^{-xy}\sin y }{e^{x}+1} dy\;dx\\[7pt]
&=\int_0^\infty \sin y\int_0^1 \frac{e^{-xy}}{e^{x}+1} dx\; dy\\[7pt]
&=\int_0^\infty \sin y\int_0^1 e^{-xy}\sum_{k=1}^\infty(-1)^{k-1}e^{-kx} dx\; dy\\[7pt]
&=\int_0^\infty \sin y\sum_{k=1}^\infty(-1)^{k-1}\int_0^1 e^{-(y+k)x}\; dx\; dy\\[7pt]
&=\int_0^\infty \sin y\sum_{k=1}^\infty(-1)^{k-1}\frac{1-e^{-(y+k)}}{y+k}\; dy\\[7pt]
\end{align}
I'm stuck in the last expression
@robjohn Do you have an idea to evaluate this one?
@M.N.C.E. Hi. Could you help me to evaluate the above integral? Hehe
 
10:29 AM
@Venus Hi. I will give it a try later. At first glance it might not have a simple closed form though.
 
@M.N.C.E. OK, take your time. Me think so it has no CF
 
10:46 AM
@Venus I just got an idea... give me a bit to investigate.
 
@robjohn OK
 
@DanielFischer
 
@BalarkaSen ?
 
10:59 AM
@DanielFischer That problem is too hard.
 
Which?
 
@Nick May I believe there could possibly be a closed form?
 
Given a chain of Galois covers $X \to Y \to Z$ an element of $Aut(X, Z)$ induces an element of $Aut(Y, Z)$.
$Aut(A, B)$ being the Deck transformation group.
Equivalences of $A$ absorbed by covering map $A \to B$
 
@Chris'ssis ... I am sorry but I do not know. It's ofcourse beyond me.
and wolfram can't express the indefinite integral interms of known mathematical constants or functions.
@Nick Mathematica yields $$\int_{0}^{\pi/2} \frac{\ln\left(e^{2x} + 1\right)}{1 + \sin2x}\mathrm dx\approx1.830481056481482\cdots$$ — Integrator 26 mins ago
I just need to know to get there.
 
@BalarkaSen What was a Galois cover again?
 
11:06 AM
$p : E \to B$ is Galois if $\pi_1(E)$ is normal in $\pi_1(B)$ under applying the functor $\pi_1(p)$
 
Because we know $$\int_{0}^{\infty}\dfrac{\sin{kx}}{e^{mx}+1}dx=\dfrac{1}{2k}-\dfrac{\pi}{2m}csc {\dfrac{k\pi}{m}}$$ and $$\dfrac{1}{1+\sin{x}}=\sum_{n=0}^{\infty}(-1)^{n}(\sin{x})^n$$ — math110 7 mins ago
@Chris'ssis: Is the method associated with doing this integral going to be complicated?
 
@Nick At the moment I don't see any natural way of computing the integral. It might involve some ugly hypergeometric series.
 
@Chris'ssis: I was afraid of that. Someone I knew IRL was cooking up an elegant method to evaluate it before he croaked.
 
@BalarkaSen Okay. So you look at an $y\in Y$, and all $x\in p^{-1}(y)$, then $p(h(x))$. If you can show that $p(h(x))$ is independent of the $x\in p^{-1}(y)$ you chose, then everything is fine, you have your push-down. If not, what are the problems?
 
@Nick to croak = to die?
 
11:15 AM
@Chris'ssis Yes, this integral is that deadly. Nah, I'm just kidding. It was cardiac arrest. The guy was like 90.
 
@DanielFischer I know, and I am having trouble showing that.
But let me think.
I have to show that if $p(x_1) = p(x_2)$ then $p(h(x_1)) = p(h(x_2))$
 
@Nick However I found an interesting series while workign on it ...
 
@Chris'ssis Cool. I knew this was up your alley :D This integral is 8 years old btw.
 
@Nick It's younger than me, that's clear. :-)
 
We know that $r(h(x_1)) = r(h(x_2))$ as $r \circ h = r$.
So $q(p(h(x_1))) = q(p(h(x_2)))$
This is getting more and more complicated. Hehe.
So $p(h(x_1))$ and $p(h(x_2))$ lies over the same fiber in $Z$.
 
11:35 AM
@Nick If you know to compute this one then you're done $$\int_0^{\frac{\pi }{2}} \frac{1}{\left(e^{2 x}+1\right) (\cot (x)+1)} \, dx$$
 
@DanielFischer When we print the values of a binary search tree using in order traversal are the values printed in an ascending order??
 
@DanielFischer Should I consider looking at small evenly covered nbhd $U_y$ around $y \in Y$?
 
@Chris'ssis Hint Please.
 
Choosing such an nbhd and considering two fibers $x_1$ and $x_2$ over $y$ gives unqiue disjoint nbhds $U_{x_1}$ and $U_{x_2}$ around $x_1$ and $x_2$ respectively which lies in the fiber over $U_y$
$h(U_{x_1})$ and $h(U_{x_2})$ are also disjoint.
 
@Chris'ssis lol, I don't have tools to do that. All I know is by-parts and it's a dead end for these.
 
11:42 AM
Gah I have no idea how to do this.
 
@Nick Neither do I.
 
@Chris'ssis Hopefully this has been interesting :D
 
Morning all!
There are so many scary integral. :0
How've you been, @Nick?
 
Integrals are stupid.
 
Hi pal :D
 
11:48 AM
How are you doing, @skull? ^_^
 
Fine thanks @KhallilBenyattou how's it going?
 
I kinda agree but not totally, @Balarka. Integrals are fairly fun to do as they don't follow an algorithmic approach (like differentiation) and you can show off when you've cracked a hard one, but if they can't be used in other fields, they are pretty much just sitting there taking up space in the garage of mathematical tools.
It's going quite well, @skull! I just woke up after 10 hours of sleep so I'm feeling pretty refreshed (although I thought I'd feel tired because of oversleeping, but apparently not)!
I'm glad you're doing ok!
 
Sleep helps the brain to recover.
 
@KhallilBenyattou A bit disoriented. I'm closing my books now. I've given up on everything... Just for a couple of hours. I think I'll utilize this time to learn how to type in the way of a typist.
@skullpatrol Yeah, remember Wolverine in X2. He got shot in the head. A bit of sleep is a miracle.
 
How to type in the way of a typist, @Nick?
 
11:54 AM
An all natural miracle @Nick :-)
 
Also, don't get yourself down, @Nick. We all hit roadblocks. What matters is how you orient around them. ^_^
 
@KhallilBenyattou Yeah, I touch type. I'm learning how to type properly at typingweb.com
 
Oh, I always need to find the first key, but after that, I can type the rest out without any problems, @Nick.
Does that count? I feel like it doesn't.
 
@KhallilBenyattou I'm one of those people who look at the keyboard rather than the screen wile typing. Ooh, and it seems I can type at a rate of 9wpm. Whoopie!
 
12:01 PM
@Nick In general these questions don't work like: "OK, I saw it and now I'm writing up the proof". One needs to do some research ...
 
I tried this test and got 72wpm, @Nick. Try it out!
 
@BalarkaSen If the integrals are stupid things you cannot approach, what could be said then about smart things? :-)
@BalarkaSen Maybe you ignore stupid things, and only admire the smart ones. I agree with that. ;)
 
I don't understand what you mean, @Chris'ssis.
 
Neither do I.
@Khallil I got 52wpm.
 
@BalarkaSen why would you say algebra > geometry to an algebraic geometer? That is like saying your father > your mother. The prof knows, first hand while writing his books, the intimate relationship between the two
 
12:05 PM
Nice! I take it you didn't make any mistakes either, @Balarka?
 
@BalarkaSen @KhallilBenyattou sometimes I'm not easy to understand like my integrals. :-)))))
 
@KhallilBenyattou Nope.
@skullpatrol It was just a leg-pulling.
 
@KhallilBenyattou = @BalarkaSen?
 
Yup, I tried it again and got a really obscure looking text and got 66wpm with 2 mistakes but I swear it said fixt and I typed it, but it said fist afterwards. :/
 
No, @Chris'ssis.
 
12:06 PM
As far as I know, we aren't the same person, @Chris'ssis.
 
I'm not a naruto fanatic.
 
Is there something that's making you suggest that, @Chris'ssis?
 
Yes, @Khallil. We both disagree with her :P
 
@KhallilBenyattou I was saying something to @BalarkaSen above and you answered me back.
@BalarkaSen lol :-)
 
That must be it, @Balarka. ^_^
If everyone agreed on the same things, the world would be a very boring place.
 
12:08 PM
@KhallilBenyattou Agree.
 
*Disagree
:P
 
@Chris'ssis = @KhallilBenyattou
Since they both agreed with each other.
 
Confirmed, just like the Illuminati, @Balarka.
 
LEL
 
12:09 PM
@BalarkaSen lol, he disagreed then ...
 
You should show him how much you respect his experience before trying to kid around with him, right @BalarkaSen?
 
@skullpatrol I know. I apologized for saying that yesterday.
 
I tried for you.
 
Well, I guess I'll just have to wait. Thanks for that anyway.
 
12:11 PM
np pal
Live and learn.
 
He isn't really that offended I guess.
Just having a bit of fun giving me a hard time :P
 
@MaryStar Inorder traversal means you first print the values in the left subtree, using inorder traversal, then the value of the root node, then the values in the right subtree, using inorder traversal. What are the criteria for a value to be in the left resp. right subtree?
 
Good morning guys
(and gals)
 
Hi pal
 
@DanielFischer In the left subtree there are the elements that are smaller the element of the root and in the right subtree there are the elements that are greater than the element of the root. That means that printing the elements while traversing using inorder traversal we et the elements in ascending order, right??
 
12:15 PM
@Chris'ssis Good Luck, I'm not expecting any elegant solution. Any solution will do.
 
@BalarkaSen Think about what happens to the points first. And don't forget that the covers are Galois, that may be needed to get a well-defined map on $Y$.
 
Hmm.
 
@MaryStar Right.
 
I think I have an idea.
 
@Chris'ssis Yes, well, it would be nice if an answer could be found before Christmas. So, it's now a race between you and another 11 guys. Good Luck on helping santa :D
 
12:18 PM
@Nick I miss the proper tools for that integral (at the moment).
 
@DanielFischer Ok!! Thank you!!!
 
@Balarka @Chris'ssis
 
Hahahah
 
12:19 PM
@KhallilBenyattou We have too much free time.
 
@KhallilBenyattou hahahaha, nice!!! :-)))))
 
I have exams in 2 weeks.
Pahahahaha!
A good worker creates time, @Nick. =P
 
@KhallilBenyattou TimeLord!
@Chris'ssis Well then, talk to you on Christmas Eve then. :D
 
@Chris'ssis do you celebrate christmas? :)
@Nick Sup, how have u been
 
@TheArtist Yeah, I'll do it with my dogs. We'll eat some cake together. :-)
 
12:25 PM
@Chris'ssis you have dogs? I have one :)
 
@TheArtist Yeap. :-)
 
@DanielFischer OK, let $p(x_1) = p(x_2)$ for $x_i \in X$. Then $r(h(x_1)) = r(h(x_2)) = z$ where $r : X \to Z$ as $r \circ h = r$. This means $p(h(x_i))$ are in the fiber of $z$. Now $q : Y \to Z$ is a Galois cover, that means the Deck transformation group $Aut(Y, Z)$ acts trasitively on the fiber over any point in $Z$. This means there is a $h' \in Aut(Y, Z)$ such that $h'(p(h(x_1))) = p(h(x_2))$.
 
@Chris'ssis I'm your dawg, homie. Cake for me?
 
@Chris'ssis I have a golden retriever. By dogs, you meant plural ? :) how many dogs?
 
@Nick :-)
@TheArtist Now I only have 3.
 
12:27 PM
@BalarkaSen Looks promising.
 
@TheArtist occupied.
 
That $h'$ is our desired element induced from $h$, @DanielFischer, right?
 
@Chris'ssis wow , that's so cool
@Nick occupied collecting hats? :D
 
@TheArtist no... with stuff.
 
@TheArtist I'd like people be like dogs. The world would be far, far better!:-)
2
 
12:29 PM
I hate dogs. I'd rather cat.
 
@BalarkaSen I don't know if that's actually already it. I would have to think about it myself, I'm trying to make you do the thinking ;)
 
I like hamsters with big eyes that wear sweaters!
 
@Chris'ssis I totally agree, I think about this many times when I'm with my dog
@Nick okay
 
@DanielFischer You're being Mike-ish, but you're doing it better than Mike :P
 
@BalarkaSen You mean the next thing is to put you on ignore for a week?
 
12:31 PM
No!
I mean being completely unhelpful.
But yet helping.
 
Back in 20 min (I need to send a solution to a kind I'm tutoring)
 
Oct 23 at 22:31, by Mike Miller
I think I can get Balarka to solve most problems just by giving him a hard time and being completely unhelpful.
I mean that's mean, but it really does help.
 
@Jasper Hello there, I see you changed colors
 
@teadawg1337 Yes. It matches the Christmas tree trunks.
 
@BalarkaSen It's the senior school teacher policy. It works only on those who have patience and interest.
 
12:34 PM
haha
 
@BalarkaSen never tell someone they are being completely unhelpful while learning
 
@skullpatrol It was a compliment.
Look at the comment above.
 
@skullpatrol There was context, and in that context it was okay to say that.
 
okay
 
Yet, the Galois cover hint made me come up with $h'$, so that's a little less than being completely unhelpful.
:P
 
12:40 PM
 
I don't like pets.
 
@JasperLoy Blue Jasper now Red Hulk for holidays?
 
@Nick Yes. It's brown actually.
 
I think my construction gives a map $Aut(X, Z) \to Aut(Y, Z)$.
The kernel even contains $Aut(X, Y)$
 
12:45 PM
@JasperLoy So is the Red Hulk. Also, Maroon 5 would disagree it's brown.
 
All I have to show that there kernel is all of $Aut(X, Y)$
 
@Nick It's called brown in GIMP.
 
@DanielFischer So I guess the construction solves the problem for the time being.
 
Maroon 5 are fantastic, @Nick.
^_^
 
Unless it turns out that kernel is greater than $Aut(X, Y)$, this map should so.
 
12:47 PM
@JasperLoy screw gimp. That's too bright to be brown. I'll go fix it in the souce code. You'll see a change in the next version (hopefully)
Brown is the color of dark wood or rich soil. It is a composite color; in printing or painting, it is made by combining red, black and yellow, or red, yellow and blue. In the RGB color model used to make colors on television screens and computer monitors, it is made by combining red and green, in specific proportions. The color is seen widely in nature, in wood, soil, and human hair color, eye color and skin pigmentation. In Europe and the United States, it is the color most often associated with plainness, humility, the rustic, and poverty. It is also, according to public opinion surveys in Europe...
 
Thanks for all the help @DanielFischer. Just so you know, I personally believe you are a great teacher and by no means I wanted to offend you by saying "you're being completely unhelpful".
 
@BalarkaSen Don't guess, check.
 
Doing that right now.
 
@BalarkaSen I know how the "being completely unhelpful" was meant.
 
I want a short exact sequence $1 \to Aut(X, Y) \to Aut(X, Z) \to Aut(Y, Z) \to 1$ actually.
The right-exactness is what we want to show.
That'd be a pain in the neck, yeesh. it doesn't hold for groups at least. only fields and galois groups.
 
12:49 PM
@KhallilBenyattou gfy
 
"Good for you?", @Nick?
 
Back.
 
@KhallilBenyattou Yes, but gfy is an abbv. which can mean to opposite things. I wanted to leave you puzzled.
 
So what we want to show is that $p(x_1) = p(x_2)$ implies $p(h(x_1)) = p(h(x_2))$ only if $h$ is a deck transformation of $X \to Y$
Hello @bolbteppa
 
@Chris'ssis Welcome Back!
user image
5
 
12:56 PM
@Nick Nice :-)
 
So cute, @Nick.
So. Darn. Cute.
 
@KhallilBenyattou Thank you for you analysis, captain obvious.
 

« first day (1598 days earlier)      last day (3417 days later) »