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12:07 AM
I just completely screwed up on that edit...
Had to correct it three times....
 
12:32 AM
Wait, why did I jump straight to 301 rep... Is it because of my other accounts on SE?
As in, associated accounts
 
12:51 AM
hi
 
Hello, @beginner
Nobody saw that, right? Good.
 
i am trying to understand what fiber means? i have that fiber is the preimage of a mapping
 
@teadawg1337 I can see everything. ;)
 
does that mean i can think of it as a line that goes from x to y for the element that the mapping works on??
 
@MikeMiller cowers in fear
 
12:54 AM
like if $f(x)=x$, is there a line from x to y for all x and y, so that means i have one fiber for all x and y??
i cant render math on my mums tablet :(
they banned me from the computer for playing minesweeper :(
 
But minesweeper is a fantastic game....
 
im not allowed any games until highschool they said
 
ah, but high school is when you're supposed to stop having fun
 
it will ruin my brain i was told :(
but now i am stuck on my textbook i borrowed and now i cant do anything!
 
@Mike High school was a blast for me, since it was the first time I really enjoyed mathematics. Also, I loved physics
Although, the social aspect is a complete nightmare...
 
12:58 AM
is a fiber not normal in math? cause my brother said zorich is by a russian mathematician so it may be weird
 
@beginner Zorich is a fantastic read, I highly recommend it
 
my friends think my math textbook is weird so i don't tell them now
i just build forts in the bush with them and do math by myself
i am trying to do zorich but stuck on page 22 stuff
teadawg can you tell me what a fiber is?
 
I'm still a student myself, I'm not very well-versed in set theory quite yet
 
oh is it set theory? i think it is functions
@mikemiller can you tell me what a fiber is/does?
since you see everything?
i will ask it on the stack exchange
 
:(
 
1:08 AM
@Twink what's wrong?
 
@teadawg1337 A question the nobody answered :(
 
The first question I asked on this site didn't receive an answer for a while
In fact, I ended up answering the question myself :P
@Twink My apologies, didn't recognize your name; thought you were new here :(
 
1:32 AM
Well, I feel like an arse
 
If $T$ is an invertible matrix, and for matrices $B$ and $C$, and we have that $\|TA-TB\|<\epsilon$, can we say that $A$ and $B$ are close to each other in some sense?
Intuitively the above shows that $TA\approx TB$ and multiplying by $T^{-1}$ we have that $A\approx B$...
 
1:48 AM
i get it now teadawg. a fiber is all the x that yield a specific y after a function!
@teadawg1337 so $f(x) = x^2$ if we want the fiber of $y=4$, we get fiber (-2,2) because both go to $y=4$!
or i mean to say that both $-2$ and $2$ lie on the fiber of $y=4$
 
2:07 AM
@beginner Awesome, I'm glad you got an answer for your question!
 
2:18 AM
Ugh, that feeling when you're about to post a really good question you're sure is original, but you search and find an identical question... I almost posted a duplicate of this
Oddly enough, I got the correct answer but I was going to ask for a verification
 
 
1 hour later…
3:46 AM
holds a cricket up to his screen and transmits its stridulation through the internet
Y'all are boring tonight... Edit: Oh no, I just said y'all... I'm turning into a southerner
 
4:06 AM
@DavidH Did you see my answer to my own question on the Basel problem? I figured out the polylogarithm identities and such :)
 
@teadawg1337 Woohoo! I honestly have that about the problem since I posted my answer. I've been spending most of my stackexchange time working on some very frustrating and distracting problems. But I'll take a look now!
 
anybody want to guide me to an understandin of Galois theory based on my vague understanding of Galois' paper?
As I understand it you can motivate a group by thinking about expressing a polynomial in terms of it's roots
f(x) = x^n + a_1x^{n-1} + ... + a_n as f(x) = (x - x_1)...(x - x_n)
permuting the roots leaves the polynomial invariant.
We can attach each permutation of the roots to a "primitive element" X_1 = a_1x_1 + ... a_nx_n (so X_2 = a_1x_2 + a_2x_1 + ... + a_nx_n for example) but because this is not symmetric, like the polynomial is, we simply form a symmetric polynomial out of the
 
4:23 AM
@DavidH As far as I know, this can't be simplified using known identities: $$\frac12\ln^2\left(\frac{\sqrt5-1}2\right)+\frac12\ln^2\left(\frac{\sqrt5+1}2\r‌​ight)=\ operatorname{Li}_2(\sqrt5-2)-\ operatorname{Li}_2(3-\sqrt5)+\frac12\ operatorname{Li}_2(4\sqrt5-8)-\frac12\ln^2(3-\sqrt5)-\frac12\ln^2(\sqrt5-1)+\ln(‌​\sqrt5-2)\ln(3-\sqrt5)+\ln(\sqrt5-1)\ln(3-\sqrt5)$$
Ugh, I give up on editing that... It's the third equation from the bottom
 
4:55 AM
I shouldn't be staying up this late anymore, goodnight nobody (or possibly someone who reads this in the not-too-distant future
 
5:13 AM
lol goodnight @teadawg1337
 
5:25 AM
@teadawg1337 goodnight Teadawggg
 
6:14 AM
Anybody around?
@robjohn I'm still struggling with bit.ly/1vLjwQf
If we have $0 \le \theta \le \pi$ and having knowledge of $F(\theta)$
 
woah that looks hard
 
@beginner It is hard :D
@beginner Indeed, I think, there is a simple method, but I don't find it, I just feel it and beat around the bush!
 
i wish i knew how to do iteration!
what does "show that $u,v$ are unique up to multiples of $b$ and $a$" mean?
for $au+bv=1$
 
6:39 AM
@beginner Show that if $cu+dv = 1$ as well, then $a-c$ is a multiple of $v$ and $b-d$ is a multiple of $u$.
 
oh i think i get it thanks mike ill see if it makes sense
$u(a-c)+v(b-d)=0$?
 
that's true, yeah
 
what does it mean to show it is unique?
get u and v by themselves and show that the a,c choices change it?
by the way i heard 1 isn't prime?
 
Ugh, why is someone purposefully editing my posts and making them worse... I've had to reject two anonymous edits within the past 10 minutes
 
didn't you go to sleep?
 
6:50 AM
Yup. My brother was screaming, I assume he woke up from a nightmare
Now I can't fall back asleep
 
so you came back to help me :))
is this acceptable to show that $n\gt 1$ is either prime or has a prime factor $
$\leq \sqrt{n}$
assume $n=ab$ where $a,b\gt \sqrt{n}$ then $n\lt ab$ contradiction
thus $n$ has a prime factor $\leq \sqrt{n}$
oh wait that would either mean $a,b \leq \sqrt{n}$ or $n\ne ab$
just change that one line to assume $n=a_1 a_2 \dots a_i$ where $a_1 a_2 \dots a_i \gt \sqrt{n}$
and same thing, is that good? i cant see latex on my mums tablet so sorry if bugs
does $(2n)!$ simplify to something?
something else i mena
mean
 
Ugh, these edits... Someone's targeting me, I swear...
I'm heading back to bed, someone reject any edits made to my posts please.
 
ok i will learn how :)
 
They're gonna be awful, and @beginner you need 2k reputation to review edits
 
@beginner No, it is all not good.
 
7:01 AM
ok okay, i cant do it for atleast a few weeks then, hopefully i will get 2k soon
what is not good will hunting?
 
@beginner Your proofs.
 
oh they are wrong?
the one above?
or the renderings
 
Yes. Your proofs.
 
oh what is wrong
i get the prime factors and set them bigger than $\sqrt{n}$ and then that must be bigger than $n$ so contradiction?
 
Everything. But I don't have time to write a proof for you now. You should think about it more and ask on the main site if needed.
 
7:03 AM
ill ask them now
thank you william
 
7:17 AM
Apparently, the automatic spam filter has caught scent of the edits being made to my posts. Good.
I won't go into details on the expletives used in a couple of them... I'm heading to bed, hopefully the spam filter keeps doing its job well
 
it was correct someone thinks @will
hey again @tea goodnight again :)
 
@beginner g'night
 
@will if he said it is right, then why did you say that everything was wrong? surely it must be atleast half right in your eyes if he thinks it is right
 
7:59 AM
oh thanks @mike i think i get it, i need a=c, b=d, so they are the same, so it is unique
like linear independence
 
8:57 AM
Not much chat here these days. I remember two months ago the chat would be really active at this time
 
Due to finals perhaps
 
Maybe, I miss people xD
My girlfriend is gone for a week(and then back for a week, gone for a week repeat for 12 weeks). I haven't seen or spoken to someone in real life today and it is 7PM
and I got up at 4AM
 
The internet is here for you!
 
I guess now you are here xD
What subjects are you doing this semester?
 
Real analysis and ODEs
 
9:04 AM
Real analysis like Rudin's real(and complex) analysis?
 
We're working from Rudin's PMA
 
Oh okay, cool, I did an Analysis course using PMA, but I was still in engineers mindset at the time, I did poorly
Scraped a pass
 
Yeah, that could be rough
 
I wish I could take it again now with the proper mindset, but oh well, I can only hope to do well in the rest of my courses. Grade release is in 5 hours for me
 
Hopefully things come out well
 
9:50 AM
If $f\in L^1$ is it common notation to write $||f||$ in place of $||f||_1$?
 
@Committingtoachallenge I think if you want to read so many other books, you should just forget about my 12 holy books. It does not make sense to read so many books with so much overlap. I won't comment on your books again.
 
@WillHunting The overlaps are great :). Everyone of my first books uses different notations and it works together beautifully. The exercises are all from different perspectives I love it
@WillHunting In regards to spying from your government, do you want me to modify my statement about you? Here
@WillHunting Also, you don't seem as happy as you once did, anything changed in your life??
 
@Committingtoachallenge I cannot even click on the link because it is at the end of the line exactly, lol.
 
Under where the idea started I reference you
 
10:02 AM
@Committingtoachallenge Just don't mention my location anymore in this chat, that will do.
 
I live in Antarctica. I catch fish to eat every day.
@Committingtoachallenge Never mind. Just stop talking about it.
 
So why aren't you as happy these days?
 
I never was happy. I will be unhappy until the day I recover completely from my mental illness, which is targeted at the first day of 2016 currently.
 
@WillHunting Do you have an estimated probability of success?
 
10:08 AM
@Committingtoachallenge No, I do not. But because of the way my methods work, if I don't make it, it will be delayed by a year. So the next target will bw 2017, then 2018, and so on.
@Committingtoachallenge Thank you for mentioning the 12 holy books on your blog. Many people in the world will now benefit, lol.
 
@WillHunting Did you have a new years resolution this year, or something of the sort?
@WillHunting xD. I thought I should attribute your specific version, since it is probably better for the general person
 
@Committingtoachallenge Not really. I just hope to get well asap and then go to grad school.
@Committingtoachallenge Yes. But you should actually state the titles properly. Some of them are not the full titles.
 
@WillHunting Was it just Weinstein to be fixed?
 
@Committingtoachallenge Marsden and Weinstein: Calculus I, Calculus II, Calculus III; Cohn: Classic Algebra, Basic Algebra, Further Algebra and Applications; Rudin: Principles of Mathematical Analysis, Real and Complex Analysis, Functional Analysis; Lee: Introduction to Topological Manifolds, Introduction to Smooth Manifolds, Riemannian Manifolds
 
10:17 AM
@Committingtoachallenge Which grad school are you applying to?
 
@WillHunting I have no idea, I have one more year of undergraduate studies and then I will likely work as a sell out(I mean no offense to anyone) for a year or two while studying more on my own, then grad school
 
@Committingtoachallenge Good luck. Maybe we will meet in grad school, lol.
 
@WillHunting maybe ;). Probably Grad school in 2018
 
@Committingtoachallenge What's a sell out?
 
@WillHunting Application Mathematician(Working Mathematician)
 
10:27 AM
Why I can edit other users answer without waiting peer review? I don't get +2 too
Is there an error to the system?
 
@Venus Because you got 2000.
 
@Venus You now have too much reputation for that :-)
 
@WillHunting What did you mean?
 
@Venus Because your reputation has reached 2000.
@robjohn Today, I learnt when Thanksgiving and Black Friday are.
 
@WillHunting So I can edit without waiting peer review. I don't know that
 
10:29 AM
@WillHunting Ah, that was last weekend (Thursday and Friday) :-)
 
@Venus Now you know. You seem to be a non-native speaker of English.
@robjohn I will be studying math in 30 days from now. I am excited.
 
@WillHunting You are not studying math now?
 
@robjohn No, I am not. I have to start on 01 Jan due to some weird reasons.
 
@robjohn Thanks for the info about the privileges
@WillHunting How did you know without hearing my accent?
 
@Venus Because you said 'Why I can do this?' instead of 'Why can I do this?'
 
10:37 AM
@WillHunting I forgot to swap the subject & the auxiliary verb, haha
 
@Venus 'Why can I do this?' is the form a question should take. 'Why I can do this' would not be a question or a complete sentence, but may be used as a title for example.
 
@WillHunting The structure of the sentence in English is very different with my country
 
@Venus Where do you live, if it is not a secret?
 
@WillHunting Serawak, Malaysia
@WillHunting Why would that be a secret?
 
@Venus Some people don't want to say their location because there may be people spying on them. Especially if they are from Iran or North Korea, lol.
@Venus Are you happy living there?
 
10:42 AM
@WillHunting There's no internet connection in North Korea for the ordinary people
@WillHunting My friends make me feel happy here
 
@Venus Thanks for telling me. I do not know so much about that country.
@Venus OK. I am very unhappy here. I will keep my location a secret. However, I can tell you I am not from Iran or North Korea, but it is bad enough here.
 
@WillHunting I know that. It's impossible you're from North Korea. I don't believe either you're from Iran
 
@Venus What is your native language?
 
How come you say bad living there?
 
@Venus Well, I think I must keep that a secret too.
 
10:47 AM
@WillHunting Malay, sometimes English.
 
@Venus OK. I speak English and Chinese.
 
@WillHunting I won't ask then
@WillHunting I'm a half Chinese
 
@Venus Can you speak Chinese? I learnt English and Chinese in school for 10 years, lol.
 
@WillHunting Unfortunately not
 
@Venus I once liked a very beautiful Malay girl, lol.
 
10:51 AM
@WillHunting That must be a Chinese-Malay girl
 
@Venus No, totally Malay, lol.
 
@WillHunting Seriously? A Moslem girl?
 
@JasperLoy $\iff$ @WillHunting ?
 
@Venus Yes. At my workplace. But I left soon so I did not really do anything about it.
 
Wait!? How can you interact with a Malay girl?
 
10:54 AM
@Integrator Yes, indeed.
@Venus Why not?
 
@WillHunting Cool!
 
@WillHunting You said at your workplace. I guess you're either from Malaysia, Philippine, Singapore, or Indonesia.
 
@Venus Maybe, lol.
 
@WillHunting Obviously you're not from Malay since you said you only speak English & Chinese
 
@Venus LOL. Maybe you're right. Are you an undergrad math major?
 
10:59 AM
@WillHunting Nope
 
hello. Anyone here comfortable with algebraic geometry?
 
Greetings
 
hi
 
@Venus I wonder if you know a brilliant way to finish $$\int_0^{\pi/2}\log(\sin(x))\tan\left(\frac{x}{2}\right) \ dx$$ (high school level approach)
 
@Chris'ssis Why did you ask me? I'm no good at integral by the way :D
 
11:07 AM
I have a question about computing resultants of two polynomials
if anyone is up for it, let me know
 
@Venus The integral I showed some days ago was a bit wrong, it had a wrong upper limit, that is $1$ instead of $1/2$.
 
@Chris'ssis Using $\displaystyle\tan\left(\frac{x}{2}\right)=\frac{\sin x}{1+\cos x}$ and taking $t=\cos x$, we get $$-\frac{1}{2}\int_0^1\frac{\ln(1-t^2)}{1+t}\,dt$$
@Chris'ssis I know, I saw your chat with robjohn
 
@Venus Yeah, that works. Let me show you my way.
 
@Chris'ssis Yes, please
@Chris'ssis You always say "trivial", I hate that word :D
 
@Venus :-)
 
11:15 AM
@Chris'ssis Do you have any slight interest outside of integrals and series etc?(Honest question)
 
@Venus Anyway, I'm preparing to write up the proof to $$\int_0^1 \frac{\log(1-x) \log(x) \log(1+x)}{1-x} \ dx$$ that is related to that wrong integral I showed to you.
 
@Chris'ssis I think we can use Beta function for the integral which you have shown to me
 
@Venus Yeah, it can be approached in many ways.
@Committingtoachallenge "Integrals, series and limits"
 
@Chris'ssis Yeah I mean outside of them
 
@Committingtoachallenge Of course I have.
 
11:18 AM
@Committingtoachallenge Yes, @Chris'ssis likes running, lol.
 
@Chris'ssis What fields?
@WillHunting I meant other math fields xD. Like Coding theory or Geometry etc
 
@Committingtoachallenge Geometry, yes.
 
I get my grades released in two hours and a half :)
 
@Committingtoachallenge I also like inequalities, probably the hardest math area ...
@WillHunting :-)))
 
@Chris'ssis Probably not the hardest, but it could be one of
Inequalities are relatively accessible
 
11:24 AM
To understand me better ... let me find a problem given some time ago in American Monthly that I solved in 4, 5 different ways, one using clever inequalities.
You might like to see the power of inequalities ... (and at the same time realize it's not that easy to go there)
Until I find my proof there, give a try to this one
Test for convergence, in an elementary way, the following series $$\sum_{n=1}^{\infty} \left(\frac{(2n-1)!!}{(2n)!!}\right)^2$$
@Committingtoachallenge ^^^
 
Again, I know this is a shot in the dark, but... in case anyone can give a hint... it'd be massively appreciated :) math.stackexchange.com/q/1048049/173397
 
@Chris'ssis I didn't say I had any experience with such problems. My point was merely that highschool student can become quite proficient at solving these sort of problems, yet essentially no students are capable of touching problems in topology etc
E.g. I have a lecturer who could trivially solve these problems and cares little for them
 
@Committingtoachallenge Here is my belief: I'd be on top no matter what I attended. if I studied topology I'd be definitely on top, but for doing that I need attraction and love to it since without them one cannot get top results.
 
11:39 AM
@Chris'ssis Comments like that are why I believe you are narcissistic.
 
@Committingtoachallenge I'm very self-confident, I'm trained to be like that.
 
Or maybe comments like your current About me: "A Romanian that one day is going to be like Ramanujan or far beyond."
 
@Committingtoachallenge Yeah, I don't believe in the limits imposed by the other. I mean our duty is to break all the limits previously set by the others.
 
@Chris'ssis You are a bit late though aren't you? You are 25, and ramanujan was the same age when he was doing world renown things was he not?
With little of your resources
 
@Committingtoachallenge The point of my comment is mainly related to the need of breaking the limits, there is no need to always live in the shadow of the others.
Wouldn't you like to be far better than your professors? Maybe not, but some would like. It's just an example, and you can go further thinking like that.
 
11:46 AM
@Chris'ssis I don't think anyone alive today will surpass Nikola Tesla
If you base so much of your life around beating people, you won't enjoy it.
 
@Committingtoachallenge No, this is what I was preparing to write and clarify. It's not for the sake of beating the others but only for knowledge.
 
So it is a hope, not a goal?
 
@Chris'ssis Are you 25? I thought you're a lecture
 
@Committingtoachallenge It might be a consequence of a thinking style.
@Venus As I previously said, I have absolutely no background in math, I'm self-educated.
 
@Venus She is an accountant
Although her background as I have researched is inconsistent(self-contradicting)
 
11:53 AM
@Chris'ssis Really? Wow! You haven't said "no background in math" to me
 
@Venus Yeah, I told you in the past. Maybe you forgot (or?).
 
@Committingtoachallenge An accountant?? That makes me amazed
 
@Venus Depending on the story she is giving yes
 
@Committingtoachallenge You mean you don't believe I'm an accountant. I definitely do not lie (as regards this point).
 
@Chris'ssis When?
 
11:55 AM
@Chris'ssis So is your brother right?
 
@Venus On this chat I think.
 
@Chris'ssis You are an accountant in your final year of school :)
 
@Committingtoachallenge lol, not really :-) I graduated from financial-accounting some time ago.
 
Feb 11 '13 at 12:21, by Chris's sister
@Theorem: I'm in the penultimate year in high school and my borther is a prospective accountant.
32 secs ago, by Chris's sis
@Committingtoachallenge lol, not really :-) I graduated from financial-accounting some time ago.
 
@Committingtoachallenge lol, maybe that was my brother that sometimes says jokes. He uses my account once in a while. :-)
 
11:57 AM
oh yeah ;)
@Chris'ssis There are other funny things, but I want to see more contradictions xD
 
@Chris'ssis A friend of mine gave me this one today
$$\int_0^{\pi/2}\frac{\sin^2x\tan^{-1}\left(\cos^2x\right)}{\sin^4x+\cos^4x}\,dx$$
 
@Committingtoachallenge But a nice proof don't you wanna see?
 
@Chris'ssis ;)
 
@Committingtoachallenge ^^^
@Venus OK, I'll look at that.
 
12:18 PM
@Venus Done
 
@Chris'ssis Really? How?
 
$$1/Sqrt[2] (1/64-I/64) ([Pi] ((8+8 I) ArcTan[2]+6 ArcTan[Sqrt[1+I]]-6 ArcTan[(1+I)^(3/2)/Sqrt[2]]+(4+4 I) Log[5]+3 Log[1-Sqrt[1+I]]-3 Log[1+Sqrt[1+I]]-3 Log[1-(1+I)^(3/2)/Sqrt[2]]+3 Log[1+(1+I)^(3/2)/Sqrt[2]])+2 I (ArcTan[Sqrt[1+I]] (Log[4]-8 Log[I Sqrt[1+I]])+ArcTan[(1+I)^(3/2)/Sqrt[2]] (Log[64]-8 Log[-(1+I)^(3/2)])+Log[2] Log[1-Sqrt[1+I]]-4 Log[I Sqrt[1+I]] Log[1-Sqrt[1+I]]-$$
$$Log[2] Log[1+Sqrt[1+I]]+4 Log[I Sqrt[1+I]] Log[1+Sqrt[1+I]]+3 Log[2] Log[1-(1+I)^(3/2)/Sqrt[2]]-4 Log[-(1+I)^(3/2)] Log[1-(1+I)^(3/2)/Sqrt[2]]-3 Log[2] Log[1+(1+I)^(3/2)/Sqrt[2]]+4 Log[-(1+I)^(3/2)] Log[1+(1+I)^(3/2)/Sqrt[2]]))$$
$$\approx 0.29939724665821598804991831682348419888733311616530689684167804324416\
86999841095892401842889137800962$$
@Venus I used my research. The last expression can be simplified further.
 
12:46 PM
@Chris'ssis Jesus. What's that?
 
(1/(8 Sqrt[2]))\[Pi] (-ArcCot[(Csc[\[Pi]/8] (1 + Re[Sqrt[1 + I]]))/2^(
1/4)] + ArcCot[Sec[\[Pi]/8]/2^(1/4) - Tan[\[Pi]/8]] +
2 ArcTan[2] - ArcTan[1 + 1/Sqrt[2] + Sqrt[-(1/2) + 1/Sqrt[2]]] -
ArcTanh[2/7 Sqrt[1 + 5 Sqrt[2]]] +
Im[2 ArcCoth[3/Sqrt[(8 + 2 I) - (4 + 2 I) Sqrt[2]]]] + Log[5] -
Re[2 ArcCoth[3/Sqrt[(8 + 2 I) - (4 + 2 I) Sqrt[2]]]])
@Venus a simplified form above.
 
What a closed-form :D
 
@Venus Yeah, it looks pretty exotic, but I'm used to them.
 
@Chris'ssis OK, thank you so much ^^
 
@Venus I wanna show you something delightful.
 
12:52 PM
@Chris'ssis Yes please
 
@Chris'ssis $$\displaystyle{\int_{0}^{\infty}\frac{dx}{\left [ x^4+\left ( 1+\sqrt{2} \right )x^2+1\right ]\left ( x^{100}-x^{99}+\cdots+1 \right )}=\frac{\pi}{2\left ( 1+\sqrt{2} \right )}}$$
how?
 
Some months ago I evaluated in closed form $$\int_0^1 \int_0^1 \int_0^1 \left\{ \frac{x}{y}\right\} \left\{ \frac{y}{z}\right\} \left\{ \frac{z}{x}\right\} \ dx \ dy \ dz$$ where $\{x\}$is the fractional part of $x$and sent it to the author from that I know I was the first one that sent to him a closed form. He considered it an open problem. Now, based upon this I create another question
$$\int_0^1 \int_0^1 \int_0^1 \left\{ \frac{x y}{z}\right\} \left\{ \frac{z x}{y}\right\} \left\{ \frac{y z}{x}\right\} \ dx \ dy \ dz$$
@Venus ^^^
 
@Chris'ssis Nice joke! :D
 
Any ideas on my integral chris'ssis? Good job on the fractional part integral though
 
I give up already
 
12:57 PM
@UserX This is not hard.
 
I'm not sure if we're in the same page
 
@Venus It's not a joke at all :-)
 
@UserX Your integral seems familiar to me. I've seen a similar one here recently
 
@UserX is it useful to let $x\mapsto 1/x$?
 
@Venus maybe. I didn't find it on the site though,
 
12:59 PM
@UserX Wait. I think I found it
 
I guess I'll use $\sum_{n=0}^m (-1)^n x^n= \frac{(-1)^m x^{m+1}+1}{x+1}$
yea ok I simplified it to $\int_0^{\infty}\frac{x+1}{(x^4+(1+\sqrt 2) x^2+1)(x^{101}+1)}\mathrm{d}x$
 
@UserX Maybe you had in mind this one
31
A: The Wicked Integral

Omran KoubaIndeed let $$ I(n,a)=\int_0^\infty\frac{dx}{\sqrt{x}(1+ax+x^2)(\sum_{k=0}^n(-x)^k)} $$ The change of variables $x\leftarrow 1/x$ yields $$ I(n,a)=\int_0^\infty\frac{(-1)^nx^{n+1}dx}{ \sqrt{x}(1+ax+x^2)(\sum_{k=0}^n(-x)^k)} $$ Thus $$ 2I(n,a)=\int_0^\infty\frac{1+x}{\sqrt{x}(1+ax+x^2)}dx= 2\int_0^...

 
@Chris'ssis well, I can't use that to solve mine :P maybe I can use the methods
 
@UserX Yes, that one!
@UserX Set $x=t^2$, you'll get your integral
I've just tried it
 
@Venus I'll get my integral how? in a form that I can use that general result?
 
1:13 PM
Does Ve=Ch ;)?
 
@UserX We have $$I(n,a)=\int_0^\infty\frac{dx}{\sqrt{x}(1+ax+x^2)\displaystyle\sum_{k=0}^n (-x)^k}$$
 
brb in 5 mins. I'll see them in a sec Venus
 
@UserX Sub $x=t^2$, we will get
$$I(n,a)=\frac{1}{2}\int_0^\infty\frac{dt}{(1+at^2+t^4)\displaystyle\sum_{k=0}^n (-t)^{2k}}$$
 
k I'm back
 
@UserX I meant $$I(n,a)=2\int_0^\infty\frac{dt}{(1+at^2+t^4)\displaystyle\sum_{k=0}^n (-t)^{2k}}$$
@UserX Sorry!! I made a mistake
 
1:23 PM
@Venus the sum changes to $\frac{(-t)^{2m+2}-1}{t^2-1}$ now...
 
@UserX Try to use Omran Kouba's answer. I bet your integral can be done in similar fashion
 
@Venus I will do that but I also bet there's a simpler way :P
 
I wish I was sleeping x.x
 
@Committingtoachallenge Sometimes I have the feeling you spying on me.
 
lol
176
Q: How do I draw a pair of buttocks?

Simpleton JackI'm trying to develop a function which 3D plot would have a buttocks like shape. Several days of searching the web and a dozen my of own attempts to solve the issue have brought nothing but two pitiful formulas below. They have some resemblance to the shape I want, though not quite. Could you...

 
1:33 PM
@Chris'ssis Why is that?
 
@Committingtoachallenge you ask me many personal questions. :-)
 
@Chris'ssis ;)
 
@Committingtoachallenge Don't worry, I'm an open-mind person. ;)
2
It's OPEN-MINDED I think. :D
@Committingtoachallenge what is the correct form?
 
@Chris'ssis open-minded :)
 
I wonder, can I ask this question on meta.mse: "are there users on MSE getting married or falling in love?"?
3
 
1:42 PM
@Committingtoachallenge hehe, I guessed something wasn't correctly written ...
@Venus lolllll - You have my upvote if you ask that!!! :-)))))
 
@Venus Yes, it has happened many times on the SE network. I am largely responsible for it.
 
@Chris'ssis But I would get so many donwvotes for that :D
@WillHunting You have a burden of proof by saying that :P
 
@Venus Let's see, I'm very curious about that.
 
@Chris'ssis Okay, I'll try but @WillHunting can you help me to arrange sentences into a good question about this topic? You know, I'm not good at English
 
@Venus I can. But you cannot be serious about asking this, lol.
 
1:46 PM
@WillHunting please give her the needed support for the best question I ever saw on meta.
 
@WillHunting If you help me, I'll ask it
Seriously, no kidding
 
:D (thumb up)
 
@Venus OK. I will help you, lol.
But note that such a question is very likely to be deleted soon, lol.
 
@WillHunting Why should it be deleted? I don't see the logical reason for such a decision. I mean doing math so much, being so addicted to this site, you might miss some important aspects of life. :-))))))
 
@Chris'ssis Because it has nothing to do with math? LOL.
 
1:49 PM
@WillHunting It's just another way of talking about math addiction amongst other things.
 
But meta is supposed to ask about MSE right?
 
7 minutes until grade releasssee
 
I am going to take a nap.
 
@WillHunting Where's the question?? ^^
 

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