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r9m
8:04 PM
@KhallilBenyattou if that is Barney the TRex .. he's high ! :P
 
@DavidWheeler What is the most verbose (as in overly explaining concepts for too long) introductory abstract algebra book you know
 
r9m
@Chris'ssis hello :) how are you? :)
 
@r9m Hello! Long time I haven't seen you. How are you doing? :-)
@r9m Not bad. Very creative these days.
 
r9m
@Chris'ssis :) stayin alive :D
@Chris'ssis :D :D
 
@r9m Great! :-)
@r9m These days I created a very special problem. Actually I created more, but one is particularly divine.
 
r9m
8:10 PM
@Chris'ssis may I see them? :)
 
@r9m Take it.
 
r9m
@Chris'ssis got it
 
@r9m Don't post any answer related to it here, but let me know if you like it and know how to do it. :-)
 
r9m
@Chris'ssis okay :) I'll try and mail if I have some idea :D
 
@r9m This is prepared for a magazine. :-)
@r9m It's one of the most beautiful things ever!
 
r9m
8:13 PM
@Chris'ssis cool :) !!
 
@r9m I cannot describe it in words. I almost cried at its beauty. Honestly.
 
r9m
@Chris'ssis hmm ... so this is another prof killer :D .. okay !!
 
@r9m :D
@r9m I saw you are deeply involved in the inequalities. I was thinking you might be possibly upset since you didn't come here at all.
 
Hello!!! When is the expectancy of max{n-q,q-1} , where n is a fixed number and q is in [0,n] , so max{n-q,q-1} is in [n/2,n], equal to $\frac{\frac{n}{2}+n}{2}=\frac{3n}{4}$? @quid @DavidWheeler @robjohn @DanielFischer @ThomasAndrews
 
r9m
@Chris'ssis nope ... I only wrote those answers because I didn't have to think much or invest a lot of time .. I am busy with my studies (got to make sure I pass the courses :P)
 
8:17 PM
Anooying, @evinda. Don't tag so many people. If you want lots of people to look at your question. post it on the main site.
 
@r9m You mean you are busy with teaching students? :D
 
I am sorry... @ThomasAndrews
 
r9m
@Chris'ssis no :P .. I am busy with my end sem studies .. there's a lot on my plate :P
 
@Chris'ssis I'm having a meeting tomorrow morning with my academic advisor (who is also a math professor), during which I will share my notebook filled with all of the math work I've been doing independently for the past several months. I'm both terrified and excited
 
@r9m One more thing about that problem, not to forget. It has infinitely many solutions.
@teadawg1337 Great! Star! It's hard to find these days someone that really appreciates your work.
 
8:19 PM
Hell @evinda, I am sorry I do not know right away @evinda and I need to finish a presentation urgently :(
 
A ok.. @quid What presentation?
 
r9m
@Chris'ssis I am ultra nervous with an interview I gave a few days ago .. they asked me 4 simple questions and that was it ,, it was so short that it made my head spin ,, I am confused and not sure why did they ask me just 4 questions while I heard they asked other interviewees more questions :( ..
 
@evinda on my research work.
 
Aha... @quid
 
@r9m Once they asked me to describe how a bycicle works ... (I mean the mechanism inside)
 
r9m
8:22 PM
@Chris'ssis :P lol ... and what was your answer ?
 
I felt in hell a few seconds and then I found a way to explain how that would work. Well, it's hard to say that in English. :-)
 
r9m
@Chris'ssis I see .. okay :) I think I read it a long time ago in my physics text book .. :P
 
@r9m I never imagined to hear such a questions, but well, it happened.
 
@Chris'ssis The thing is, though, this will be the first time I've shown anyone my independent work...
 
@teadawg1337 I'm not that secretive. It's not bad to cooperate with some people you may trust. Also here I posted enough of my work, but I trusted the guys here. :-)
 
8:26 PM
what up boys?
 
@teadawg1337 Doing reserach is the most amazing activity I've ever done. The satisfaction of getting lovely results is hard to describe ... :-)
 
why
it's so boring
:P
 
r9m
@teadawg1337 my profs usually mildy scold me when I don't attend their lectures and my solutions in the final exam has little to do with the materials taught in class .. (independent ideas) .. but I can see they are always secretly pleased :P .. but for the same reason I have never got a full grade in any of the papers :P lol
 
Can anyone offer a little helping hand? just a little clarification question regarding linear program standard forms...
 
@r9m is "I can see they are always secretly pleased" a nice way of saying "They are always glad to learn from me."? :D
 
r9m
8:31 PM
@Chris'ssis not at all :P lol ... just means my efforts are perhaps appreciated :P
 
@r9m :D
 
I love doing independent math work as well, but this is the first time I've truly stepped out of my comfort zone... If the professor doesn't appreciate my work, then I don't know where I'll find future motivation. I'm completely terrified
 
@teadawg1337 You mean that if $X$ doesn't appreciate your work then your work isn't valuable? I would be very careful about that.
 
r9m
If I become a prof some day (pun) .. I'll keep students who disregard my methods and work on their own at the problems closer .. (keeping friends close but enemies closer :P lol)
 
I don't think that there is a possible event, or a person ever that might be able take my self-confidence away and the strong belief in my work. I don't say I'm special, but I prefer to put trust in myself and be honest with me. I'm usually the toughest person with myself. As regards my work, I don't ever expect someone to be nice with me, but to be just honest.
I also strongly believe in constructive opinions.
 
r9m
8:37 PM
^ :)
 
@r9m :-))) I think I'd also like to have such a student in my class (if I were a professor).
 
if we have some subgroup H < K < G such all of them are finite
and K is normal
is |KH| = |K| ?
 
r9m
@Chris'ssis hehe .. that just proves you never had anyone like me in your lectures :P lol (just joking)
 
If the meeting goes extremely well, then I would have more motivation to continue my work and I'll probably be more productive doing it. I just worry about the opposite scenario...
 
8:41 PM
@r9m lol, I'm just a self-educated one. :-)
 
r9m
@teadawg1337 don't worry before the meeting has happened .. you'll have lots of tangible reasons to worry about after the meeting .. (i.e. the true nature of the problem will be revealed after the meeting ... no point in worrying now)
 
what meeting
@DavidWheeler or @BalarkaSen still here?
 
:D
I have a question
I am solving this question in algebra want to verify if my reasoning is correct @DavidWheeler
 
is this gonna be like that Taylor Swift video with the uber-long introduction?
 
8:47 PM
If K is normal subgroup of G of prime index p , then H <= either 1) G = HK ^ [H: K intersect H] = p OR 2) H <= K
so
we use second iso theorem
 
@Chris'ssis funny story
 
lool
|H| / (K $\cap$ H) = |KH| / |K|
since K subgroup of G
we have |G| = |K| * [G: K] = |K| * p for some prime p
hence |K| = |G| / p
hence |KH| / |K| = |KH| * p / |G|
since |G| doesn't divide p then it must either divide |KH| * p or it must divide |KH|
so far so good @DavidWheeler or do you notice any logical inconsistenties?
 
|G| obviously divides |KH|*p
and it obviously DOESN'T divide |KH|.
 
@teadawg1337 Be very careful about what a bad scenario might be. Mathematics is hard, there is a long way to go for getting good result, there is a price to pay, there might be moments when you think you're alone and maybe it's hard for you to continue, but it's important to never give up. Don't expect miracles. Just do your best.
 
yes
I want to reach the conclusion
that for the case that it divides |KH| then KH = G
 
8:54 PM
Your original statement you want to prove is confusing to me because of formatting
 
I mean I see the result in my mind but want to be careful with my argument
why ?
 
Re-state what you are trying to prove.
 
If K is a normal subgroup of a prime index p, then for H subgroup of G either
1)G = HK AND [H: K $\cap$ H ] = p OR
2)H is subgroup of K
 
@Chris'ssis A price to pay??
 
@teadawg1337 A huge amount of effort, time ...
 
8:58 PM
Do you have an idea about the following?
0
Q: Expectancy of $max \{n-q,q-1\}$

evindaWhen is the expectancy of $max \{n-q,q-1\}$ , where $n$ is a fixed number and $q$ is in $[0,n]$ , so $\max\{n-q,q-1 \}$ is in $[\frac{n}{2},n]$, equal to $\frac{\frac{n}{2}+n}{2}=\frac{3n}{4}$?

 
OK, i think what I would do is assume $H$ is NOT a subgroup of $K$
 
no
one can reach that conclusion in the middle
if |G| | (|KH| * p)
 
I'm not sure if any popular math discussion has led more people astray in my lifetime than the $1+2+3+\cdots =\frac{-1}{12}$ thing.
 
and then show when that happens we MUST have $G = HK$
 
Is there a new one today, @ThomasAndrews?
 
9:00 PM
This weird question wants to use $1+1+1+\cdots =\frac{-1}{2}$. math.stackexchange.com/questions/1211949/…
 
that is the case if |G| | |KH| I think
 
It's leading to a whole new generation of cranks.
 
...
I have no words for that @ThomasAndrews
 
Ok, sure, if $|G|$ divides $|KH|$ it's clear $G = KH$.
 
It can be fixed fairly easily (if they pay attention to responses...) by pointing out that you can't do algebraic manipulation with things that are ill-defined. There are, of course, fancy explanations like Banach limits by which these things make sense; I don't really think one should mention them in these discussions.
 
9:03 PM
yes
for case II
|G| | |KH| * p AND |G| doesn't divide |KH|
 
so, assume not, so that $|G|$ divides $|KH|\ast p$ but not $|KH|$
 
yeh
I mean I see the result but I don't see the middle conclusion to reach that H must be subgroup of K
 
Let's look at it another way: we have $p$ cosets of $K$ in $G$, right?
Now $HK \leq G$, so we have $|G| = n|HK|$.
 
good night, @Mike
hi @teadawg, @Thomas, @DavidW
 
Divide both sides by $|K|$, and we get:
$p = [G:K] = n[HK:K]$
 
9:16 PM
Heya @Ted, opinions on my meeting tomorrow?
57 mins ago, by teadawg1337
@Chris'ssis I'm having a meeting tomorrow morning with my academic advisor (who is also a math professor), during which I will share my notebook filled with all of the math work I've been doing independently for the past several months. I'm both terrified and excited
 
So the index of $K$ in $HK$ is a divisor of $p$.
 
I just saw that in the star list. Remember that at junior colleges, many math professors don't have Ph.D.'s in math.
Do you know the background of this particular person?
 
So $[HK:K] = 1$ or $p$.
@KarimMansour when you have a prime number in a problem, you want to find things that divide IT.
 
back sorry was making food reading what you wrote @DavidWheeler
 
hi @Ted
 
9:20 PM
@Ted She got a double Bachelor's degree in Math and English from the University of Michigan, and her Master's in Mathematics from... Bowling Green State
 
filling out my income taxes...
 
I already paid the taxes I owed, @Mike ... Enjoy.
 
@DavidWheeler |HK| = |H| * |K|?
 
Finished federal already. Just filled out the forms instead of dealing with software crap. State's what's left.
 
no nvm
you said divide both sides
 
9:21 PM
OK, @teadawg .... Michigan is good, and Bowling Green is ok. But you're doing stuff that's pretty recondite. So don't be upset by a neutral reaction (like you'd get from me :)).
By all means express your enthusiasm and passion for mathematics, @teadawg.
 
Yeah I see that is another way to provde that intersection is p but the problem is for the second case
 
hi @Kaj
 
that is H subgroup of K
 
Hey hey
 
@Mike: I saw you all had fun on the mountain yesterday ...
 
9:22 PM
I mean if H is subgroup of K yeh then result will hold however I want to reach that as conclusion
 
Highest point in LA county, @Ted.
 
Not quite the high Sierra :)
 
Hard as hell. We were not nearly pessimistic enough in our guesses as to how long it would take.
 
How much of the height did you drive up, @Mike?
 
@TedShifrin, do you know if it's possible for someone to get both the regular and the applied math degree at UGA?
 
9:24 PM
I do know, @Kaj, and the answer is NO.
 
Some. The actual trail we took was about four miles horizontal, a little over 1/2 mile vertical.
 
That already came up with one of my advisees.
 
Or at least, that's what it officially was. I don't buy it.
 
@TedShifrin I've gotta start somewhere, right? Maybe she'll direct me towards someone who would know more about the stuff I'm working with
 
That's unfortunate. I'd only need something like 4 more classes for that.
 
9:24 PM
They aren't separate degrees @Kaj; there's only an area of emphasis.
 
Ohh, I see
 
It's irrelevant. Do the regular degree, @Kaj, and take whatever courses you want.
 
If $[HK:K] = p$, then that means that $|HK| = [HK:K]|K| = [G:K]|K| = |G|$, and $HK = G$.
 
Amusingly, I could have gotten the applied and pure emphases in my degree if I'd asked for them.
 
9:25 PM
And that would have made what difference, @Mike?
If you want a real-world job, maybe the applied means something. Otherwise, irrelephant.
I hope I didn't upset @teadawg with my frankness ...
 
No idea, @Ted. I didn't ask for the applied emphasis.
 
Can anyone recommend a 'gentle' introduction into both projective geometry and (more specifically) homothety a secondary school (Olympiad) student?
 
Otherwise, $[HK:K] = 1$, that is $HK = K$, so $H \leq K$
 
No, not at all @Ted
 
9:26 PM
@TedShifrin: He responded to you. Not sure if you saw that.
 
Oh, I missed it ... Thanks, @Mike @teadawg.
 
ohh I see
okay good thanks alot @DavidWheeler
 
You should look at the UT website, @teadawg, and look for people doing special functions.
@Gridley, an Olympiad-level secondary student is better than most college math majors.
 
@Ted University of Tennessee, right? I'm not driving out to Texas in my spare time :P
 
correct, @teadawg ... silly boy.
 
9:28 PM
what is olympiad level secondary student
 
Nice, done with taxes. Cali tax was painless after the federal.
 
Mine will be two states next year, @Mike, hence much painly.
 
That's where I plan on transferring to finish my undergraduate degree
 
Hence my suggestion, @teadawg ... I didn't just fall off the turnip truck :)
I know a few people at UT, but more in geometry, so it won't help you.
 
Well, @Ted, I've only got two sources of income: stuff on a W2 and stuff on a 1098T. Even if I'd moved here from a second state it wouldn't have been hard to consolidate the two. I bet yours has a lot more, and hence will be much more tear-inducing...
 
9:30 PM
Yup, life was trivial when I was a grad student, @Mike.
@Gridley: The first thing that comes to mind is a book I've recommended a lot on here for geometry. It's a bit sophisticated, but I'll bet you can handle it. Look at Pedoe's Geometry: A Comprehensive Course. There's also Hartshorne's introductory book.
And, of course, Coxeter ...
OK, I have food to cook ... See you all later ...
 
Bye.
 
Cya @Ted, I'm looking through the 79 pages of names on the UT website
 
@TedShifrin how many problems did you create in your activity so far? (approximatively)
 
@TedShifrin Thanks, I'll look into it. The only Geometry book I have used so far is one by the British Olympiad called 'plane Euclidean geometry' which goes up to inversion. Most of my Olympiad experience comes from actually trying (and sometimes failing) past questions, by sophisticated do you mean theory intensive?
 
Hello all !
 
9:41 PM
Hmm
 
I asked a new question
0
Q: Are there infinitely many Thâbit ibn Kurrah cousin primes?

mickPositive integers of the form $3 * 2^n - 1$ are called Thâbit ibn Kurrah numbers. and if those numbers are prime they are called Thâbit ibn Kurrah primes. Now if for a fixed positive integer $n$ , $3 * 2^n - 1$ and $3 * 2^n - 5$ are both prime then we have a Thâbit ibn Kurrah cousin prime. Con...

help would be appreciated :)
 
@Chris'ssis .... hmm
 
@AlecTeal ?
 
bye
 
9:49 PM
bye
@r9m when you have more free time work on that, you won't regret any minute spent there. I'm planning to develop another similar integral, slightly modified, but that again is very special for some reasons. :-)
@r9m that article with Au-Yeung series is going to be published soon. Unfortunately, to publish an article takes so incredibly much. I have more stuff to be published but all goes terribly slowly. I need that stuff for my book.
 
Is there someone familiar with the Knapsack problem??
-1
Q: Properties of the Knapsack versions

Mary StarThere are two versions of the Knapsack problem, the integer and the fractional one. The difference between the integer and the fractional version of the Knapsack problem is the following: At the integer version we want to pick each item either fully or we don't pick it. At the fractional versio...

 
I have no idea!
Can anyone help me with an easy proof?
Prove that
$\lim_{(x,y) \rightarrow (0,0)}\frac{sin(2 x + 2y) - 2x - 2y}{\sqrt{x^{2} + y^{2}}} = 0.$
 
10:05 PM
L'hopital rule
and your done
 
Are you sure?
 
Alright. Sweet. Thanks.
 
Can someone please take a look at this problem? I need to find a counterexample: math.stackexchange.com/questions/1212047/…
 
@DanielFischer Could I ask you something about an algorithm?
I want to write a $O(n \lg k)$-time algorithm in order to merge k sorted lists into one sorted list, where n is the total number of the elements in all the input lists.

I thought the following:

We put in a min-heap of length k the first elements from the k sorted lists.
Then, we heapify and delete the root , which will be the smallest element and put it at a new list, and at this position we put the second element from the list from which the smallest element came from. Then we heapify and continue the same procedure till the li
 
10:20 PM
@evinda If you can't find help here, maybe the cstheory stack exchange site can check your solution.
 
@ThomasAndrews If I want to write an algorithm, could I just write the steps that we have to follow?
 
Hello @AntonioVargas !! Are you familiar with the Knapsack problem??
 
In math class, that might be enough. In a computer class, you might or might not have to write either real code or pseudo-code.
 
@ThomasAndrews A ok... It is from a computer class.. Could you maybe take a look at my algorithm?
 
10:36 PM
I searched him up on Google and saw 'Evil' as a sub search.
Naturally, I went straight to the images!
It's been a while since I've been on here.
How've you been, @r9m?
 
It's interesting that a Google search for "is used in such generality" (with quotation marks) only turns up one result. Five very common words, and of the entire internet, only one page has those words in that order.
 
Google doesn't search everything
 
Hello @DanielFischer !! Do you maybe an idea for the following??
-1
Q: Properties of the Knapsack versions

Mary StarThere are two versions of the Knapsack problem, the integer and the fractional one. The difference between the integer and the fractional version of the Knapsack problem is the following: At the integer version we want to pick each item either fully or we don't pick it. At the fractional versio...

 
10:54 PM
I wouldn't call "generality" common.
 
at least a minor peerage word, eh what?
 
I tried L'Hopital's formula, but it doesn't work.
Does anyone know a different method for this proof? Prove that

$$\lim_{(x,y) \rightarrow (0,0)}\frac{sin(2 x + 2y) - 2x - 2y}{\sqrt{x^{2} + y^{2}}} = 0.$$
 
Google NGRAM, which searches books, has it occurring about 1/100th of the times that the word "general" does.
I'd evaluate it as $$\lim \frac{\sin(2x+2y)-2x-2y}{2x+2y} \frac{2x+2y}{\sqrt{x^2+y^2}}$$ The first is going to have limit $0$, so you need to show that $\frac{2x+2y}{\sqrt{x^2+y^2}}$ is bounded.
 
How did you get that first bit?
 
Whoops, that's how I'd aproach it, but I'm not sure the first statement is right - I think the first part is actually $1$. @buddhababe
 
11:03 PM
Oh, okay.
 
Oh, no, the first bit is $0$. You can see that by showing $\frac{\sin(h)-h}{h}\to 0$ as $h\to 0$. That is a case for L'Hopital. Or rewriting $\frac{\sin(h)-h}{h} =\frac{\sin(h)}{h}-1\to 0$.
 
Oh, okay. That helps.
 
@MaryStar I'm not, unfortunately.
 
Ok... No problem... @AntonioVargas
 
11:40 PM
anyone here?
 
Kalispera @KaliMa :)
 
@DavidWheeler still here?
If [HK:K]=p, then that means that |HK|=[HK:K]|K|=[G:K]|K|=|G|, and HK=G.
in your answer
I am looking back to it I don't see how did you get
 
Anyone here familiar with charpol methods for big matrices?
 
[G: K] |K| = [HK: K]|K| ?
I thought I understood that part but I don't
 
11:58 PM
Look, we established that $|G| = n\cdot|HK|$, right?
$HK$ IS a subgroup of $G$ (since $K$ is normal) and its order divides the order of $G$ by Lagrange.
 

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