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user19161
10:00 PM
@jimmy First, ask yourself why you want to get a degree in math.
 
They offer night courses, and the quality of education -- although it varies -- is not that different than a typical university (for the first two years)
 
@JimmyHoffa okay, for that you have part time registration I guess, where you can register for courses on a per credit basis. or you can go to community colleges.
 
Doing the equivalent of 2 years of undergrad at a comm. college might take 3 or so years.
 
@JacobBlack because after 12 years as a software engineer I've finally realized all the mathy parts are what have always made me interested in it. But I have a wife and kid and full-time job to manage.
I'm just probing the possibilities to see if it's even useful for me to bother
 
Then, for your upperclass years (3rd/4th years), I'd say you can do 2-3 classes, part time, and finish that within a 4 year period.
 
user19161
10:01 PM
@JimmyHoffa My answer is, study math on your own if it is just interest.
 
@JimmyHoffa, maybe you can get it from openuniverty
 
user19161
A degree is for you to find a job, not for you to study math.
 
@JacobBlack A degree in math is not for anyone to find a job with ;P
 
Hi there...
 
hi
 
10:02 PM
@user58512 hey how are you?
 
user19161
@JimmyHoffa But anyway, that was my serious answer to you. =)
 
I had a bad day but I cheered up from this chat
how are yhou
 
@JacobBlack You think this idea of mine is basically impossible then?
 
fine thanks
 
@JimmyHoffa Where do you live?
 
10:03 PM
@user58512 didn't you enjoy carneval?
 
user19161
@JimmyHoffa I don't know, you need to find out from the college. But my answer still stands.
 
I don't know anything about it
 
Denver
 
Oh have a look outside
Isn't @charlie living in brasil?
 
10:05 PM
@user58512 you meant carneval or a certain question?
 
The best thing to do would be to contact them, maybe an admissions officer, and ask about options
 
carneval
 
Your situation is not unique
This is something they should be equipped to answer.
 
me neither...
 
Sometimes, a university will grant waivers to certain early requirements to older students
 
10:06 PM
@user58512 so why did you have a bad day?
 
@EdGorcenski interesting. Thanks for the tip, perhaps I will call them up.
On a side note, @JacobBlack you suggest I just study on my own; is it really that easy to study math on your own with no instructors to let you know of your correctness?
 
my course is not going well
and my time atuni is not good
 
@JimmyHoffa It can be done to varying degrees of success. Early things, say through Differential Equations, yes.
 
which one are you taking currently?
 
There are 2 things I think would be hard self-learn, a foreign language (unless immersed) and math. Both for the same reason: With no outside parties there's very little certainty of correctness
 
10:08 PM
I mean my whole course
 
ahh
don't worry
 
I'm very worried about money
 
I like him
 
the land lord says the rent is late
gotta get some wine and a cigarette...
@user58512 this always cheers me up...
the chords fit nice to this one:
 
10:20 PM
oh this is good!
I haven't heard this in a long time
 
we used to mingle both up with the band...
How does anon's comment fit to your answer...
 
I think he's just being pendantic about terminology
 
@draks... yes
 
@user58512 ah ok
@Charlie where from brasil are you?
 
@draks... southeast
@draks... why?
 
10:25 PM
@Charlie isn't there carneval as well right now?
 
@draks... yes, it is
 
@Charlie and you're chatting on M.SE?
 
@draks... yes, of course
i don't samba
 
@Charlie very good
 
i'm not very fond of it
 
10:26 PM
oh you don't?
 
@draks... no
 
you're from brasil...
 
@draks... I feel like an ET for not liking it
 
@Charlie ET calling home (M.SE)
 
yes...
I'm a weirdo
 
10:28 PM
you're a nerd
 
@draks... yes
 
to admit: me too
 
NO
 
that's straight
Hey have a look at this one:
0
Q: Construction of Hadamard Matrices of Order $n!$

draks...I'm trying to get a hand on Hadamard matrices of order $n!$, with $n>3$. Payley's construction says that there is a Hadamard matrix for $q+1$, with $q$ being a prime power. Since $$ n!-1 \bmod 4 = 3 $$ construction 1 has to be chosen: If $q$ is congruent to $3 (\bmod 4)$ [and $Q$ is the c...

@Charlie you're a rebel...
 
@draks... yes...in a very ackward way
 
10:33 PM
In algebraic topology: can a retraction map induce a homomorphism between fundamental groups of the original space and the retract if the retract is not a deformation retract?
 
@draks... I don't like beaches either
 
@Charlie at least you seem to have one
to not like it...
 
@draks... yes...
 
Do you like snow?
 
@draks... i never saw snow, actually
 
10:35 PM
really, just look outside my window...
 
@draks... oh really!
 
heaps of it...
 
cool
 
Here I will use $\mathbf{N}=\{1,2,3,\ldots\}$. If $(f_n)_{n=1}^\infty$ is a sequence of functions from $\mathbf{N}\to\mathbf{N}$ and I define $g(n)=n \prod\limits_{i=1}^n f_i(n)$, then can I say that $\forall m\in\mathbf{N}$ $\lim\limits_{n\to\infty} \frac{g(n)}{f_m(n)}=\lim\limits_{n\to\infty} n \prod\limits_{i=1,i\neq m}^n f_i(n)=\infty$?
 
indeed
 
10:36 PM
@draks But you are on a moon of jupiter... how do you have snow?
 
let me tell you: space is cold...
behind the snow line, of course
enceladus is full of it
 
Enceladus
isn't it in Saturn?
 
:This article is about the moon of Saturn; for the mythological giant, see Enceladus (mythology). Enceladus is the sixth-largest of the moons of Saturn. It was discovered in 1789 by William Herschel. Enceladus seems to have liquid water under its icy surface. Cryovolcanoes at the south pole shoot large jets of water ice particles into space. Some of this water falls back onto the moon as "snow", some of it adds to Saturn's rings, and some of it reaches Saturn. Because of this apparent water at or near the surface, Enceladus may be one of the best places for humans to look for extraterres...
 
Alas you are on Mimas!
 
it's not that far
 
10:39 PM
Space is so sexy
 
nice trip to go boarding
@Charlie are you hitting on me?
 
@draks... no
 
:-(
 
@draks... is your name Space?
 
to answer my earlier question:"No known zeros with order greater than one are known. While the existence of such zeros would not disprove the Riemann hypothesis, it would cause serious problems for many current computational techniques (Derbyshire 2004, p. 385)."
 
10:40 PM
not yet?
 
@draks... haha
 
so no one has proved the zeros all simple
 
@draks... I have to go, see you!
 
@Charlie bye, go samba...
 
@draks... no
 
10:42 PM
I had to try...
 
@draks... ok
 
@user58512 wait where does this belong to?
 
the zeros of zeta
 
@user58512 ahh...
@user58512 do you care up them?
 
hm?
 
10:45 PM
hey it's 23:45
 
wait until 23:45:67
 
@user58512 why did you post that?
@user58512 I missed that
 
because I asked it earlier and no one answered
 
@user58512 really? in a question or on chat?
 
chat
 
10:47 PM
gimme a link. I forgot that totally
must be the wine...
 
1 hour ago, by user58512
just curious, are the zeros of zeta all multiplcity zero?
 
ahh, I wasn't here then...
 
I don't think so! I searched about that a while ago and it is also unproved
 
$$
\begin{align}
\int_{-\infty}^\infty\frac{\arctan\left(e^{-x}\right)}{\cosh^{2q}(px)}\,\mathrm{d}x
&=\frac\pi4\int_{-\infty}^\infty\frac{\mathrm{d}x}{\cosh^{2q}(px)}\\
&=\frac\pi{2p}\int_0^\infty\frac{\mathrm{d}x}{\cosh^{2q}(x)}\\
&=\frac\pi{2p}\int_0^\infty\frac{\mathrm{d}\sinh(x)}{\cosh^{2q+1}(x)}\\
&=\frac\pi{2p}\int_0^\infty\frac{\mathrm{d}u}{(1+u^2)^{q+1/2}}\\
&=\frac\pi{4p}\int_0^\infty\frac{u^{-1/2}\,\mathrm{d}u}{(1+u)^{q+1/2}}\\
&=\frac\pi{4p}\int_0^1t^{-1/2}(1-t)^{q-1}\,\mathrm{d}t\\
3
 
@draks..., very interesting answer!
 
10:53 PM
@user58512 acceptable indeed
@robjohn doesn't render for me...
 
@draks... Is there a link at the bottom?
 
@robjohn that doesn't render for me as well...
 
@draks... Do you have ChatJax installed?
 
sure
 
click (see full text)
 
10:55 PM
Oops, I just murdered MATLAB
 
@user58512 that was the link
 
@EdGorcenski shit happens...
 
I just tried to iterate over all permutations of 60 indices.
 
don't do that...
ask matlab "whynot"...
 
how can i prove that $\sum_{k=1}^{\infty} k^3/e^k$ converges without using the ratio/root test. only with the integral comparison, and limit comparison tests.
 
10:56 PM
Hi all (again)
 
@Chris'ssisterandpals and pals?
 
@Chris'ssisterandpals hi pals...
 
@robjohn: hi!
 
@Eric, easily k^5/e^k < 1 for sufficiently large k
@Eric, so k^3/e^k < 1/k^2 which we know converges
 
@robjohn: sister here. I refer to some colleagues of mine from that I take some of the problems.
@robjohn: I really need your help now :D
 
10:58 PM
@draks... rerun ChatJax and then click on the (see full text) link
 
@robjohn how does that work? are the names cut from a certain length on?
@robjohn i did
 
@user58512 Is that really enough? I know it is true that $k^n/e^k$ will converge for sufficiently large $k$
 
user19161
@user58512 Don't worry about money. Since you can't back out now, there's no use worrying. Just get the course over and done with and move on. No use worrying. Just relax.
 
@user58512 that is a big open problem !!
 
@user58512 Can I do this:
 
10:59 PM
@draks... Only a certain number of a multiple line comment are shown unless you click that link
 
user19161
@Chris'ssisterandpals It's always the sister here, I know.
 
@Chris'ssisterandpals for what?
 
@robjohn: I'm in trouble with math.stackexchange.com/questions/300542/…. I think I found a problematic case $x_{n}=\frac{1}{n}, y_{n}=\frac{\sin n }{n}$.
 
Since $\lim_{k \to \infty} (k)^{1/k} = 1$ we get
$k^5 < e^k$
$(k)^{5/k} < e$
 
@robjohn ahh I opened a new tab, that didn't work.thx
 
11:00 PM
@robjohn: for those sequences the limit doesn't converge as expected. I checked that with Mathematica. This is really weird.
 
goodbye
 
@JacobBlack, thanks - if I just keep working hard for the rest
 
bye @mick
 
bye @draks...
 
@mick, the simple zeros? seems quite interesting
 
11:01 PM
@user58512 yes but HARD !!
 
Uh oh. Is this question an invitation to a political flamewar in the comment thread?
 
@Eric, I don't think you understood my idea: let's say a_k < 1/k for all k, then a_1 + a_2 + a_3 + ... < 1/1^2 + 1/2^2 + 1/3^2 + ...
 
user19161
@HenningMakholm I know about the free spending but it can't be that much...
 
@user58512 almost as hard as Riemann hypothese !!
 
@mick, it seemed to me like something that would be known by basic complex analysis, but I was wrong
 
11:03 PM
@user58512 oh simple sorry
that is known
i thought critical
 
@mick which question are you talking about?
 
@user58512 no i understood that, obviously the sum holds too
thanks
 
@draks... question here by user58512
bye guys
 
still this one. ok...
 
it was just idle curiousity, I don't if it has any implications
well it says it has implications for computation
 
user19161
11:05 PM
Curiosity is the start of all mathematics.
 
i thought logic is the start of all mathematics
 
user19161
Also, you misspelled the word.
 
@Chris'ssisterandpals You can get that to converge to any number you want.
 
@robjohn: then the answers provided are wrong?
 
@mick, homework problem. Show the zeros of zeta are (i) simple (ii) irrational (iii) *have real part is 1/2. Starred question is particularly difficult.
 
11:08 PM
@user58512 how close do the roots get?
 
@draks..., I think the zeros near T are separated by c/log T uniformly or something
 
user19161
@jonas and @old appear together, how romantic!
 
@robjohn: actually, the answer that I received as good is wrong (from my friend). I was told that the answer is $0$.
 
Yes.
 
@draks... yeah in davenport it gets that there are O(log T) zeros with T-1 < Im(\rho) < T.
well that imply it :/
 
11:13 PM
@Chris'ssisterandpals it is...
 
they might all be clumped together hm
 
i would bet they will
 
good question but I have to sleep now
bye
 
user19161
@user58512 Good night!
 
@robjohn: what do you mean? I'm too tired right now. :)
 
user19161
11:15 PM
@Chris'ssisterandpals Then good night!
 
@Charlie hey I added some (a) "Space" just for you...
 
@JacobBlack: you're right. (soon :))
 
user19161
@Chris'ssisterandpals Always. =)
 
@Chris'ssisterandpals Hello!
 
@user58512 bye
 
11:18 PM
@OrangeHarvester: hi! How are you? :)
 
@Chris'ssisterandpals I am good. What about you?
 
user19161
I never ask "How are you?"
 
@OrangeHarvester: I'm going crazy. (in trouble with a limit)
 
@JacobBlack what do you ask?
 
user19161
Because I don't know what the other person is supposed to say.
 
11:19 PM
@Chris'ssisterandpals Tell me. I can try.
 
user19161
@draks... I ask "Will you marry me?"
 
@JacobBlack no
 
@draks... He asks.
 
@Chris'ssisterandpals I posted an answer
 
@robjohn: (+1) but how about setting $x_{n}=\frac{1}{n}, y_{n}=\frac{\sin n }{n}$? The limit doesn't exist.
 
user19161
11:21 PM
@robjohn Your answer has grammatical errors. =)
 
@draks... oh how cute
 
@Charlie sexy, isn't it?
 
user19161
I also never say "Hi".
 
@Chris'ssisterandpals It doesnt matter, as long $x_n$ goes to $0$
 
@robjohn: OK. Thanks.
@robjohn: Mathematica let me know that the limit doesn't exist. Anyway. I let things this way.
 
user19161
11:27 PM
I never used any of the three Ms.
 
@JacobBlack you are forcing the question: 3Ms?
 
@robjohn About your answer, may I know if that was your thought process? For example, how did you think of those $2\sqrt{3}$ and using A.M.-G.M. from the outset? Or did you do it otherwise and this was how you presented it?
 
user19161
@draks... You mean what are the 3 Ms?
 
yes tell us..
 
user19161
Mathematica, Maple, Matlab. QED.
 
11:29 PM
really? they are great...
 
@OrangeHarvester Pretty much as it is there. I divided the top and bottom by $x_ny_n$, then you have $|x+1/x|>2$, so you just need to adjust that using AMGM
 
user19161
@draks... I only use the calculator.
 
@robjohn Ahh, I see. Nice. :-)
 
user19161
In my undergrad math course, I almost never had to use a calculator either.
 
@JacobBlack especially the symbolic toolbox (Maple IF) in Matlab is ever so cool...
 
11:30 PM
@JacobBlack And you are worse off with it. I feel my experience with checking out my answers on computer helped me a lot about developing intuition. In high school, I did not have the M's so I would use C, C++. In college, I used mathematica + matlab a lot.
 
user19161
@OrangeHarvester My brain is too powerful, face it bro!
 
@Chris'ssisterandpals My answer doesn't require that $y_n\to0$
 
@JacobBlack Meh.
 
@JacobBlack answer this math.stackexchange.com/q/300635/19341, power brain...
 
@draks... yes
 
user19161
11:32 PM
@OrangeHarvester Bleh.
 
@JacobBlack where?
 
user19161
@robjohn It's better now, though I would write it differently myself, using "AM-GM inequality" and a full stop even after display math.
 
I do not put fullstops when writing math, only write them in text.
 
@OrangeHarvester me neither, I put three...
 
user19161
I avoid display math like the plague.
 
11:35 PM
@JacobBlack A full stop? even if it is the middle of a sentence?
 
user19161
I like my writing to take up as little space as possible.
 
@JacobBlack I never used displaymath and I am not even sure why is it needed.
 
user19161
@robjohn No, at the end of sentences.
 
user19161
@OrangeHarvester To highlight equations and if it is hard to read if too small.
 
@JacobBlack Never had that problem. :-/
 
11:36 PM
@draks where this name came from?
 
@robjohn why do you use $$ around align environments? Just curious.
 
@Charlie an abbreviation. What about yours?
 
@draks... Chaplin
 
@draks... Chaplin.
 
user19161
Jinx.
 
11:38 PM
@JacobBlack you are not allowed to call jinx, you did not speak.
 
@OrangeHarvester I thought I noted that some things worked funny if a math envronment wasn't put in math mode.
 
@robjohn Okay. Will keep a lookout about that.
 
user19161
My LaTeX is too bad, I am not qualified to speak.
 
@OrangeHarvester You mean the $$...$$?
 
@robjohn Yes.
 
user19161
11:41 PM
But guys, note that the double dollars will probably be deprecated in LaTeX 3.
 
@OrangeHarvester I should be glad I haven't read any of your answers. Things are hard to read without displaymath
 
@JacobBlack It is already deprecated I though.
 
user19161
@OrangeHarvester It still works.
 
user19161
It may not work anymore.
 
@Charlie why do you ask?
 
11:42 PM
@JacobBlack Yes. deprecated means it will still work, just that you are not encouraged to use it and that it may not work in future versions.
 
@JacobBlack Okay, so we will use \[...\]
 
@robjohn I used align/equation everywhere. And $...$ for inline.
 
user19161
@OrangeHarvester I used it in the other sense.
 
@JacobBlack There is no other sense!
 
user19161
@OrangeHarvester Check JL dictionary.
 
user19161
11:43 PM
However, note that single dollars will probably remain working.
 
@JacobBlack I prefer software development seminars and books and etc.
 
@OrangeHarvester In my papers before coming to MSE, I used \begin{equation}...\end{equation}
 
@JacobBlack Single dollars is the only way to display inline math right?
 
user19161
@robjohn That is what I prefer too, and they get numbered as well.
 
@JacobBlack I am pretty sure of that
 
user19161
11:44 PM
@OrangeHarvester No.
 
@JacobBlack There is another I know [...] I think but I do not use that.
I think it is confusing.
 
user19161
Use backslash followed by parenthesis instead of square bracket for inline.
 
Does anybody know how Williamson's or Turyn's kind of construction for Hadamard matrices work? Answers are welcome here:
 
@robjohn Yes, but now I prefer not to use equation, and use align instead.
 
1
Q: Construction of Hadamard Matrices of Order $n!$

draks ...I'm trying to get a hand on Hadamard matrices of order $n!$, with $n>3$. Payley's construction says that there is a Hadamard matrix for $q+1$, with $q$ being a prime power. Since $$ n!-1 \bmod 4 = 3 $$ construction 1 has to be chosen: If $q$ is congruent to $3 (\bmod 4)$ [and $Q$ is the c...

 
11:45 PM
@JacobBlack \[\huge{\text{HA!}}\]
6
 
user19161
And this one people don't know: there actually is a \begin{math} and \begin{displaymath}
 
@JacobBlack That is two strokes, I would rather prefer Shift+single stroke
 
@JacobBlack You can't see that I used \\[...\\] to do the HA!
 
@draks... curious
 
user19161
@OrangeHarvester Exactly. I asked a member of the LaTeX 3 team about why double dollars would be deprecated and he said it was for technical reasons.
 
11:47 PM
@Charlie good starting point...
 
user19161
Of course, all this still works in plain TeX.
 
user19161
@robjohn Yeah, formatting here is terrible, one has to care about TeX and other factors.
 
@Charlie what are you working on recently?
 
@JacobBlack Double dollars were hard to parse as well I think. Also, inside the verbatim environment, you have to use \(...\) or \begin{math} \end{math}
 
user19161
@OrangeHarvester I would never use that environment.
 
11:48 PM
@JacobBlack And why would that be?
@robjohn I starred it.
 
user19161
@OrangeHarvester No reason to use it.
 
@JacobBlack Writing about LaTeX code in LaTeX?
 
user19161
@OrangeHarvester I will never write about LaTeX.
 
user19161
I am only a mango.
4
 
@JacobBlack ahh, then you are off the hook.
 
11:49 PM
@OrangeHarvester Yeah. I was just commenting that one can see the TeX but not the surrounding environment delimiters.
@JacobBlack with just the right papaya?
2
 
user19161
This chat is very healthy: it is full of fruits.
 
@JacobBlack if you don't like it, you can put some strange html code in between to break it, see here:
1
Q: Invisible $\LaTeX$ error: $\mathrm{Im} \left( \right) $ vs. $\mathrm{Im} \left( \r ​ight) $

draks...The following tex works fine, when I write it myself: $\mathrm{Im} \left( \right) $, but when I copy the same tex from here it doesn't work: $\mathrm{Im} \left( \r‌​ight) $ Why is that? And you might have noticed the different look in title and post. It was just a copy-paste. Are they rende...

 
@draks... nothing
 
user19161
@OrangeHarvester But really I am not interested in TeX itself.
 
@JacobBlack Okay.
 
user19161
11:53 PM
@OrangeHarvester I am only interested in mathematics and women.
 
@draks... but abreviation of what?
 
@JacobBlack why don't you have a tex.stackexchange.com account?
@Charlie my name...
 
@draks... don't you say?!
 
user19161
@draks... I deleted it.
 
@draks... dra is the first three letters of your first name and then k and s are the first letters of middle name and surname?
____
 
11:55 PM
@Charlie no, can't I just be draks ... for you?
 
@draks... hmm...
 
@Charlie what's your first name?
hey no let's keep
 
@draks... Can't i be just Charlie for you?
 
sure charlie...
 
user19161
3
A: Does this prove that there are no infinite sets?

Asaf Karagila$\infty+1=\infty-1=\infty$. You can't cancel it out. $$\Huge\text{Infinity is not a finite number.}$$ You can't apply the rules of finite arithmetics to infinite sets.

 
user19161
11:57 PM
HAHAHAHA
 
@draks... ;)
 
user19161
@Charlie Is it a secret?
 
@JacobBlack said Jason Bourne...
 
user19161
@draks... Yes, and many others.
 
@JacobBlack I've heard of Jasper, but I never met him...
 
user19161
11:59 PM
@draks... Have you met Jacob or Jason?
 
people who don't define their notation don't deserve to get answers ...
 
@JacobBlack Are you Joe Black's ugly little brother?
 

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