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Yeah, that picture is pretty reminiscent of what you were up to, isn't it? :)
 
Eh
But now you know:
Give me a closed set $[a,b]$
and a covering of it $F$
or I will kill one hostage every hour
 
12:13 AM
$[a, b]$ and $F$
 
12:32 AM
I'm cooking up a good one for you, @columbus8myhw. You'll never want to look at open covers again
 
Holy crap @DavidW is back!
 
@Incurrence He got a new job and moved to a new state
 
12:54 AM
Hey guys I am stuck on very last part of a question if someone could give me a push.
 
Hello, @columbus8myhw . I want to play a game. Up until now, you have spent your life among the dead, piecing together their final moments. By the time you read this, you will have one minute to find a finite open cover of $[0,5]$ among this cover $F$...
 
If p is odd prime, then any divisor of Marsene number $M_p$ = $2^p$ - 1 is of the form 2kp + 1. If q is a divisor of $M_p$ then qk = $2^p$ - 1, so we must have $2^p$ = 1 (mod q), so we must have ord(2,q) | p and by fermat little theorem $2^{q - 1}$ = 1 (mod q) hence we must also have ord(2,q) | q - 1.
any ideas?
 
@Robjohn Do you believe me on the chat functionality now there are 64 people here
 
@Incurrence can you help me with this little push ?
 
With seen 20 hours ago on quite a few
@KarimMansour I don't know Marsene numbers
 
1:01 AM
you don't need to know them just group theory
Mp is of the form $2^p - 1$
I want to reach the conclusion q - 1 = 2kp so we have q = 2kp + 1
 
Hi @DavidWheeler
@KarimMansour are you doing number theory?
 
yeah
my last assignment !
4 questions left then that is it for the assignment
I am almost done with 1 question but can't reach the conclusion
 
Is the question related to marsene primes
 
yeah
but you don't need to know marsene primes
to do it
 
1:17 AM
Oh....okay
 
quick question what does it mean to have a linearly independent eigenvector?
if I have an eigenvector like [0 0 ] does it still count as an eigenvector? I think so?
 
no any zero vector is not linearly indepedent
 
so all i have is one eigenvector... v_1 = [ 1 1], v_2 [0 0] since there is a 0 0 eigenvector, there is just one eigenvector?
 
ok
 
1:32 AM
@Incurrence nvm
 
Thanks @AndreyKaipov for the clue for my question
 
@robjohn Do you agree haha?
 
Do you ever try to search "math se" to get to this website but you accidentally type "maths e" and it just tells you about Euler's constant
Also, the thing I was referencing: xkcd.com/857
 
@Incurrence I am asking in the moderator's chat and they are just noticing it for the first time, too.
 
@robjohn Awesome
 
1:42 AM
Equivalent systems of linear equations have exactly the same solutions. @Incurrence can you help me with this?
I have to prove they are homogeneous
 
@Rememberme I have no time until Thursday sorry
Exam is tomorrow and I am stressed as hell
 
2:25 AM
does anybody want to check a proof for me my proof ?
 
@Robjohn Do derivatives of $z\in \Bbb C$ in general agree with the same rules of derivatives of $x\in \Bbb R$?
Meaning I can take derivatives as if I am dealing with $x$ in the most part
 
Fine@Incurrence
 
@Rememberme Sorry Sayan
 
0
Q: proof verification(number theory)

Karim MansourI proved the following if p is an odd prime, then any divisor of Mersenne number is of form 2kp + 1. Proof: If q is a prime divisor of $M_p$ then qk = $2^p$ - 1 $\rightarrow$ $2^p$ $\equiv$ 1 (mod q). gcd(2,q) = 1 since q can't be two since 2^p - 1 is always odd, so its prime divisor can't b...

 
No prob
 
2:33 AM
if someone could check my proof that would be great !
 
@Incurrence as I remember, most derivatives act similarly, but keep in mind that for differentiation on the real line we had $\lim\limits{h\to 0}$ where we just have to check that the limits agree as $h$ approaches zero from the left and from the right, but in the case of complex differentiation, $h$ might approach zero from all sorts of directions.
 
@JMoravitz Yep
 
@Incurrence I recommend looking over this article which does a pretty good job of explaining the differences between the real and complex cases.
 
@JMoravitz what do you think of my proof ??
 
I'm not familiar with the notation "ord(2,q)" @Karim
 
2:41 AM
@JMoravitz Nor am I
 
I think in general, if we have a real-valued function $f$ that has an "analytic counterpart", then we can just pretend we're taking the derivative of $f$, and plug in complex values. Is that what you're getting at, @Incurrence ?
 
I believe so, but I won't commit to that until after I have read the following two(small) chapters
 
Can someone help me with a bit of linear algebra
 
yes
 
@JMoravitz ord(2,q) is the smallest integer k such that 2^k $\equiv$ 1 (mod q)
 
2:47 AM
Equivalent systems of linear equations have exactly the same solutions. @eoin I have to prove that they are homogeneous and have the same soloutions
 
You have to prove that equivalent systems are homogenous?
 
@Eoin I have to run....can you please ping the hints..
 
Could you ask that again, I think I have it wrong.
 
@Karim ah, then yea, trivially ord(2,q) will be p then (since 2 is of order p, prime, in $\mathbb{Z}_q^\times$)
it looks fine to me, and Andre showed where the necessity that p is odd comes into play
 
This is a theorem whose proof is not given:-
I have to prove that equivalent systems have the same soloution basically they are same and homogeneous @Eoin
 
2:51 AM
What's your definition of 'equivalent system'? Mine is basically that they have the same solution set...
 
Mine is that one equation can be written as the linear combination of other equations in the other system @pjs36
 
yeah
thank you @JMoravitz
 
@LutzL I'm trying to decypher your reply to my question earlier, but I'm still having a bit of trouble figuring out exactly what to do. Aside from the central difference formulas, most of the formulas you have displayed are going over my head a bit and thus I'm struggling to apply them to my problem. Can you maybe just construct the u(t) vector and I'll see if I can pick it up from there? That's where I'm stuck at the moment.
1
Q: Need help with a Crank Nicholson Method example problem.

user3330644I have an exam coming up and the professor released the sample test containing a Crank Nicolson question. I was out of town for those two lectures, so I missed the information. Even though I have acquired the notes, the professor didn't do an example problem, which is the best way I learn a new...

 
if you have a linear system $A$ in $n$ variables, and say you have a solution $x_1,x_2,...,x_n$. If $B$ is another linear system that is equivalent to $A$ then it can be formed by linear combinations of equations in $A$, and then $x$ is again a solution.
If they are homogeneous, then it still holds. I'm not sure if this was what you were looking for still though.
@rememberme Maybe you wanted, say $Ax=0$ is a matrix presentation of our homogeneous system. If $B$ is equivalent to $A$ then there is a matrix $E$ of elementary transformations such that $EAx=E0$, thus $EAx=Bx=0$ and $x$ is a solution to $B$ as well.
The only other thing I have to say is perhaps it would be useful to write out an arbitrary system and an arbitrary system of linear combinations of that system. It is just a long process of bookkeeping.
 
3:13 AM
anyone know how to do brackets in latex
 
3:23 AM
Which brackets? @usukidoll
 
if I have $2^{2^{n + 1}}$ $\equiv$ 1 (mod q), then ord(2,q) | $2^{2^{n + 1}}$ does it imply that ord(2,q) = $2^{n + 1}$ ?
if so why I don't see it directly
 
Most brackets you just type, (, ), [, ], \{, \}, \langle, \rangle
$(, ), [, ], \{, \}, \langle, \rangle$
 
knows all the different kinds of brackets... No, that's a lie. I only know a few :(
 
@DiscipleofBarney do you see it?
 
See what? @KarimMansour
 
3:29 AM
the above on my question why is ord(2,q) = $2^{n + 1}$?
 
Do you have any reason to believe that?
(is $n$ something special?)
 
its related to this question I am solving something related to this
6
Q: A theorem about prime divisors of generalized Fermat numbers?

pedjaA theorem of Édouard Lucas related to the Fermat numbers states that : Any prime divisor $p$ of $F_n=2^{2^n}+1$ is of the form $p=k\cdot 2^{n+2}+1$ whenever $n$ is greater than one. Does anyone know is there some similar theorem for generalized Fermat numbers: $F_n(a)=a^{2^n}+1$ ? I've b...

but I don't see it at all how ord(2,q) = $2^{n + 1}$ I mean that would work yes if it is prime
but $2^{n + 1}$ isn't prime
 
Let $\textbf{F}(x,y,z) = (xyz, x^{2} + yz)$ for $(x,y,z) \in \mathbb{R}^{3}.$ Find the first-order \hphantom{stuff}approximation of $\textbf{F}(x,y,z)$ near the point $(1,2,3).$
Any help would be appreciated.
 
Well I am pretty sure you are missing some informations, and I don't really feel like wading through some other question at the moment, I am doing some things @KarimMansour
 
its okay @DiscipleofBarney :)
I asked author and he answered it very weird because I asked a question and he answered it few moments ago lol.
 
3:45 AM
Ah okay, that is good @KarimMansour
 
 
2 hours later…
5:22 AM
is there different algorithms for binary powering mod n?
 
I don't know much about binary powering mod n, but how is someone suppose to give you a different algorithm when they don't know what algorithm you are using? @KarimMansour
 
our prof told us to know the definition of binary powering mod n
not the definition but briefly describe it
 
why are there sooooooooooo many people in the room today
 
but we didn't learn about it and he didn't even mention it, so I don't know if there is different versions of it @DiscipleofBarney ?
 
Well you never gave the version you know so how is someone suppose to give a different version? @KarimMansour
 
If we answer other's question, we click the button "post uor answer" But before this we can find "community wiki" under the answer box. What is this ? If we click, what happen ?
 
I really dont like algorithms
 
@HKLee Community wiki is sort of giving up any "ownership" of the question, so you don't get reputations if someone votes and it people feel less squeamish about editing or adding to community wikis. Some of the main uses is for giving opinion based questions and don't feel like you should get reputation, or if a question is answered in comments and you decide to move the ideas of comments to an answer.
Makes it more of a wiki basically
 
@ Disciple : Ah.. Thank you for your kind reply.
 
5:52 AM
@MikeMiller Install Tampermonkey extension, which manages such scripts.
 
6:08 AM
@pizza I have Tampermonkey but i don't know how to use it
 
6:29 AM
@Robjohn Any update now that the room is at 70 xD
My theory is that now there is a long duration it takes to kick you(already shown to be longer than a day), or you now must click the leave button to leave the room
 
@Incurrence any idea why are there soooooooooooooooo many people in the room?
 
Read my theory above
 
Hi @EricStucky
yup i think the same thing.....is it because of the new update
Can someone here help me with linear algebra
 
6:53 AM
@Balarka are you online???
 
@pizza Thanks, works perfectly.
 
@Rememberme Just post the problem, maybe someone will help if they have seen the problem...
 
Can someone please help me what is wrong with my question.
http://math.stackexchange.com/questions/1240936/solving-a-recurrence-without-initial-conditions
 
7:17 AM
@DiscipleofBarney can you help me
 
@Rememberme Post the problem/questions, describing what you need help with, then people can decide whether or not people want to help with whatever you problem is.
 
Mine is a recurrence, I have provided an answer, but I would like to know if it is correct.
 
Ok fine...... the question is:-
Prove that two system linear equations which have the same solutions are equivalent...?
 
Huy
Isn't that the definition?
 
thats given as a theorem in hoffman kunze
not proved
Any ideas @Huy
 
7:29 AM
@Rememberme Why don't you give the definition of equivalent systems?
 
Equivalent systems are the systems in which the linear combination of one system can be derived from the linear combinations of the other system
 
It is proved in Hoffman and Kunze, just above the statement of the theorem
 
@DiscipleofBarney i am still in section 1.2
 
Like I said it is proven above the statement of the theorem
 
OK i didn't really see that.......
 
7:40 AM
@Incurrence Yes. If the function can be written as a power series (which includes polynomials and most common smooth functions) then they are called analytic and obey all the usual rules.
 
8:31 AM
@Robjohn

$$\sin z = 6$$
$$\frac{e^{iz}-e^{-iz}}{2i}=6$$
$$e^{iz}-e^{-z}=12i$$
$$u^2 - 12ui - 1=0$$
$$\frac{12i\pm \sqrt{-144+4}}{2}$$
$$e^{iz}=u=6i\pm \sqrt{35}$$$$iz=\ln(6i \pm \sqrt{35})$$
$$z=-i\ln(6i+\sqrt{35})$$

Where was I meant to insert the fact that it is multivalued? The first step?
Where do I put in the $2\pi k + \frac{\pi}{2},\quad k\in\Bbb Z$
 
@Incurrence nvm, you asked for sin :-)
@Incurrence Isn't that $i\sqrt{35}$?
 
Ahhh
I don't think so
Oh
Yes
Woops
$$e^{iz}=u=6i \pm \sqrt{35}i$$
$$iz =\ln(i(6\pm\sqrt{35}))$$
$$z=-i\ln(i(6\pm\sqrt{35}))$$
 
@Incurrence You need both $i(6\pm \sqrt{35})$
@Incurrence You can make use of the fact that $\log(i(6\pm\sqrt{35}))=\frac\pi2i\pm\log(6+\sqrt{35})+2k\pi i$
 
ahhh
hmm
I don't see it sorry
Just the log(ab)=log a + log b, rule and I am missing it perhaps?
 
So the solutions are $\frac\pi2\pm i\log(6+\sqrt{35})+2k\pi$
@Incurrence $\log(1/a)=-\log(a)$
 
8:43 AM
How did you do that step above my hmmm though?
 
$(6-\sqrt{35})(6+\sqrt{35})=1$
 
There they go!
Why did that happen?
56 were just kicked
 
They were some of you accounts and you left all of them :D
 
@DiscipleofBarney Did you see it happen?
@DiscipleofBarney Looked awesome seeing 56 pictures flying all over the screen
Thanks heaps @Robjohn
 
8:48 AM
brb
 
9:18 AM
How are those equal ^
Shouldn't the $-i$ get rid of the $i$ term for the $(2n+\frac12)\pi i$?
 
@Incurrence hey
 
Hello
 
@Incurrence something is weird there. The $-i$ was just dropped.
47 mins ago, by robjohn
So the solutions are $\frac\pi2\pm i\log(6+\sqrt{35})+2k\pi$
 
Okay that's what I thought, thanks!
@robjohn Is that somehow a valid thing to do?
 
@Incurrence only against a factor of $0$...
 
9:33 AM
Okay ahaha
 
In other words, no.
 
Hello everyone. First time in this chat room :)
 
Welcome
 
thanks incurrence
I have question regarding worst-case (max-min) optimization.
any guru of worst-case robust optimization here... or any tips how to find them here
 
Usually you just ask the question and hope someone answers in this room
 
9:35 AM
@Mohsin algorithms?
 
no it's not algorithmic... it's more conceptual
I have received a response on my question on the forum... But it is always nice to have a conversation... it is always fast :)
 
@Mohsin where is the question?
 
1
Q: Non-uniqueness of worst-case (max-min or min-max) optimization

MohsinI have a worst-case optimization problem, where i want to maximize the minimum from the uncertainty set (uncertainty is given as an ensemble of 100 realizations, so an ensemble based approach). It is a typical max-min problem. What i am trying to understand is: for different inputs, i have a dif...

Here it is :)
 
9:50 AM
@DanielFischer Hi, I don't understand the following 'fact' : why when we have a holomorphic function on a domain we have $\sup\{\vert h(z)\vert: \vert z \vert=\alpha\}=\sup\{\vert h(z)\vert: \vert z \vert\le\alpha\}$ ? I don't see the relation between this and the maximum principle.. Thanks
 
@Gato What does the maximum principle state, in the formulation you know?
 
10:11 AM
@Gato That is, the maximum principle on the disk $|z|\le a$
 
10:27 AM
Wow I have been finding so many errors in the solutions after that one
 
Please if anyone has a good explanation to my questions, post it here with my name tag... many thanks in advance :)
 
10:40 AM
@robjohn Are you sure there is not some strange $\log$ rule? They did something similar again
 
@Incurrence who did this?
 
@robjohn A professor
He just moves the $i$ to the top without adding a minus as I would expect
 
@Incurrence there, they dropped a $-1$
 
:\
Then their proof that $\tan^{-1}z = \frac i2 \log \left( \frac{i+z}{i-z}\right)$ doesn't even work
Which is weird, since every prior step seemed to work
@robjohn There is a third error, but it gets ignored right at the start(the cos has a minus), but otherwise I can't see how they derived it wrong
Ohhhhhhhhhhhhhhh
They didn't inverse the fraction
They meant to absorb the $-1$ into the fraction
But they didn't
$\frac i 2 \log \left( \frac{i+z}{i-z}\right)$
Sorry!
If I am ever a professor, I am going to triple check my damn notes I put up
 
10:58 AM
Well there is your problem.... your actually reading notes a prof puts up.
Although, the mistake catching might be a good exercise
 
@DiscipleofBarney Yeah, I usually never do, since the textbook is usually better(unless they do some nice typed notes)
But these are answers to practice questions
 
Studying for a test, pfft
 
I found out about it only 6 days before hand, and I had to work on my functional assignment for most of the time prior :|
So really I had yesterday and today
and I had badly neglected this class :\
Was 30 hours study deficit yesterday
(after spending a little over 10 hours studying)
 
Huy
11:20 AM
Are you committing to a challenge, @Incurrence?
 
Indeed
 
Huy
Why the namechange?
 
Don't like classmates knowing my username
And I changed it immediately after my friend walked away from my uni computer
 
Huy
What is the problem with classmates knowing your username, for you?
 
Potentially judging me for asking trivial questions here
 
Huy
11:22 AM
Why does it matter to you what they think of you or your questions?
 
They are less likely to ask me for help if they think I am bad at math
 
Huy
Do you need them to ask you for help?
 
and usually classmates only help you if you help them, so I want them to come to me
If they think I am bad at math, they won't collaborate with me
 
Huy
Over here everyone just helps whoever asks, nobody cares whether the other person is good or bad at math.
 
Indeed
Oh I thought you meant on MSE
You live in a good math community haha
 
Huy
11:24 AM
I don't know, I've never really been part of the community, but when I have a question, I have no problem asking someone and they don't have any problem giving a good answer.
 
People here are happy to help you frequently, but only if you help them frequently
I have had 'friends' stop talking to me after I have asked them for help three times
Without them coming to me
Whereas I have helped some people more than 10 times in a row without caring
 
Huy
@Incurrence: I see. I slightly remember similar behaviour taking place in high school, but definitely not at uni. There's other things to worry about at uni, no need to produce more drama, imo.
 
Indeed, now I usually just stick to working with the people who actually seem interesting in learning, rather than their grades
 
Huy
I'm interested in my grades as well, but my grade doesn't get any better if I don't answer questions from other students. At least I would think so. :P
 
@Huy Haha, I mean in a competitive sense. Since often grades are scaled against eachother, people will try to use diplomacy to get the maximum amount of help from people with minimum gain for others.
 
11:32 AM
So you are saying I should stop helping... :)
 
Huy
After the first two years, over here I don't think grades are scaled against each other anymore.
 
@DiscipleofBarney You aren't in my class I hope. I have actually considered this
@DiscipleofBarney None of my current classes are scaled for the record
@DiscipleofBarney You do live in the US right lol
 
@Incurrence Well I am in your class
I am you....
 
lol, not funny hahaha
 
ADG
hello
 
11:35 AM
@DiscipleofBarney You really aren't in my class though right. You claim to be in the US, but have Australian sleep times based on your chat activity graphs
 
You have multiple chat personality disorder
 
ADG
me?
 
No Disc
@DiscipleofBarney Can you tell me your IP?
 
Mew
sup peepls
 
Nothing much, just freaking out
 
Mew
11:38 AM
anyone smart on who can help me
freakin out about what mate
 
This guy is stalking me IRL
 
Mew
u full serious mate?
 
His username is @DiscipleofBarney
 
Mew
someone from stack exchange?
fk dat sht man
u shud call da police
 
wut u wnt mew?
wut u nd hlp wif?
 
Mew
11:40 AM
help me solve this equation
 
mybs i cn help
 
Mew
e^x - x = 0
 
did u wolfrmelpha et?
 
@BalarkaSen oh yeah, i forgot about that, thanks for the reminder
 
Mew
wtf is dat
 
11:40 AM
@DanielFischer (some problems with internet ) The principle is stated as if $\vert f\vert $ has a local maximum at $a\in U$ then $f$ is constant in a neighborhood of $a$.
 
i'll think about it
but got plenty of lectures today
 
@Mew Did you chuck it in wolframalpha.com
 
Mew
i'm just trying to solve the problem of log(log(log(log(log(.... log(x))))
mate i'll do dat now
 
@Incurrence No
 
@DiscipleofBarney Are you really in the US?
 
Mew
11:42 AM
na mate
 
Who knows? Do you have any US personalities?
 
Mew
Incurrence, i plugged into wolphmatic
 
@Gato Okay. So if $f$ is not constant, then $\lvert f\rvert$ has no local maximum in the interior. Now, $\{ z : \lvert z\rvert \leqslant \alpha\}$ is compact, so what does that imply about $\lvert f\rvert$?
 
Mew
it said the solution was - Wn(-1)
so wtf is dat
 
@Mew bro it says at the bot right of the result did ya chk dat?
 
Mew
11:44 AM
mate it made no sense
 
@DiscipleofBarney Can you tell me please I am so confused
 
@Incurrence Tell you what?
 
@DiscipleofBarney If you are actually in the US
 
@DanielFischer The supremum is reached on 'this' set.
 
@Gato Right, it attains a mximum somewhere on $\{z : \lvert z\rvert \leqslant \alpha\}$. Now there are two possibilities, either $f$ is constant, then the maximum is attained at every point, or $f$ is non-constant, and then the maximum must be attained - where?
 
11:47 AM
@Incurrence I leave that as an exercise to the reader...
 
Mew
University of queensland?
na mate i work
are you at UQ incurrance?
Is anyone smart on who can help me?
 
@DanielFischer On the boundary of $\{z : \lvert z\rvert \leqslant \alpha\}$, witch is what I ask for.
Thnaks
 
Welcome
 
Mew
when does e^x = x
does it ever?
 
@Mew Graphing works well here. Also that the former increases much faster than the latter.
 
Mew
11:52 AM
how do i graph complex numbers
 
$e^x = x$?
 
Greetings
 
Mew
sup sis
 
ADG
anyone?
 
Anyhow, you can use series expansions or whatever to see why that never holds true.
 
11:55 AM
$$\lim_{(x,y)\rightarrow (0,0)}\frac{x^2 y^2 (\log (x)-\log (y))}{x^2-y^2}$$
 
Mew
i can't see the latex
for some erason
 
ADG
see chat at latex on right side ------->
 
Mew
where
 
ADG
$\LaTeX$ in chat
first starred message
 
Mew
dude it's working now
thanks
 
 
2 hours later…
2:17 PM
@MikeMiller well, surely that's false? consider the inclusion $M_2 \hookrightarrow M_3$. (i was going through some old conversations and stumbled upon this message, and it felt wrong)
 
@Chris'ssis Hold on... at first I was thinking about $\frac{xy(\log(x/y))}{x^2-y^2}$
 
@robjohn OK :-)
 
@Chris'ssis This is $\frac{xy(x/y)\log(x/y)}{(x/y)^2-1}$ which does go to $0$ since $0\le\frac{u\log(u)}{u^2-1}\le\frac12$
 
@robjohn It seems so.
 
2:35 PM
@robjohn Good.
 
Hi @robjohn
 
@Gigili hi there. Haven't seen you in a while
 
Do you have time to take a look at my question here?
@robjohn Yeah, I'm busy preparing for my thesis defense
Oh never mind, got it!
Thanks anyway
 
@BalarkaSen What inclusion?
 

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