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07:03
@anon Let me see if I get this straight. In $R[x]$, $(I)$ is $\{\sum i_j x^j:i_j\in I\}$. On the other hand, $(I,x)$ is $\{i_1+\sum( i_j+r_j)x^j:i_j\in I,r_j\in R\}$, yes?
might as well just say {i+xp(x) | i in I, p(x) in R[x]}
@anon Yes, sure.
R[x]/(I,x) should be "obviously" iso to R/I
@anon Yes.
That was my hunch.
@anon So right, two elements in $R[x]$ are identified $\iff $ their constant terms' difference is in $I$. This gives the same structure as in $R/I$.
The $xp(x)$ kills everything else.
indeed, the constant coefficient projection map R[x]->R/I has kernel (I,x) and is surjective
07:09
@anon =)
Thanks, man.
hi, is there anyone good in ring theory?
@anon But we don't need maximality of $I$ here right? We've proven $I$ maximal in $R$ iff $(I,x)$ maximal in $R[x]$.
yes to your second sentence, wat to your first
nesreen, just ask
@anon LOL, nevermind.
if R is reduced ring which is not an integral domain, is this implies R has at least 2 minimal prime ideals?
07:22
what kind of ideals? (or is the ring commutative?)
we know if R is not an integral domain< then R has a zero-divisors more that {0} and any minimal prime ideal will be contained in Z(R)
ya< commutative
I want to know why at least 2????
hmm
I'm trying to think of a series to compare with $\sum \frac{1}{n \sqrt (n)}$ to show that it converges
Any ideas?
@Crake $$\int_1^\infty x^{-3/2}dx$$
or use an integral test
07:28
@anon Dibs.
lol
ya, that seems much easier
Anyway, can you think of a series?
"yes, that way is better. anyway, back to my way..."
lol no
I'll actually do it "your way"
but well.. I'm intersted
07:33
@Crake Not really. $p$ series are usually dealt with integrals.
At least, if you don't want to sweat much.
@PedroTamaroff Alright thanks
@anon isn't the intersection of the set of all prime ideals in reduced ring={0}?!
quoting the answer one your question
how is that?
but yes, according to wikipedia, the nilradical is the intersection of all nonzero prime ideals (hence all minimal prime ideals), and a commutative ring is reduced if and only if its nilradical is trivial. however I am not well-versed in commutative algebra enough to know why the intersection claim holds.
*on your question
07:41
ya< that's right!
I was know it but I couldn't write it
thans
$2$
How does one run startjax?
start chatjax
How do you guys view mathjax in this chat?
follow the instructions
which part of them are you failing to do or understand?
there aren't any instructions
I added it to the bookmark
and clicked on it
I'm a moron
07:52
you have to click on it while you're in this tab
I was clciking on it but not while on the chat
ya
$2$
ah good
$e^{i\pi}+1=0$
$\left \vert \sum_{n=1}^{\infty} \dfrac{e^{in\theta}}{n\sqrt{n}}\right \vert \leq \sum_{n=1}^{\infty} \dfrac1{n\sqrt{n}}$
sweet
Why is that true?
Isn't it always the same (=)
why would it be?
certainly |1+i| and 1+1 are different for example
this is just the infinitary version of |a+b+c+...|<=|a|+|b|+|c|+...
Well, I may be thinking wrong but isn't the modulus of $\dfrac{e^{in\theta}}{n\sqrt{n}}$ always $\frac{1}{n\sqrt(n)}$
07:58
correct
So, you're doing the modulus of the sum
just like the modulus of i is 1, but |1+i| does not equal 1+1
ya
I see
even more pointedly, |1-1| certainly does not equal |1|+|-1|
lol
I was not looking at the sum
This chatjax is awesome btw
08:01
you can use double dollar signs for large centered equations that take up their own line, or put \displaystyle in front of equations to make them just as large but inline
e.g. $\sum$ vs. $\displaystyle\sum$ vs. $$\sum$$
 
2 hours later…
10:21
morning all
Someone in the same timezone!
$$\sum_{n=1}^\infty \frac{(1+i)^n}{2^{n/2}cos(in)}$$
yep, cool. Seems about right. How would you prove that it converges?
it being the series above
Ratio won't take me anywhere
why do you think it converges?
Oh well, I worded it wrong
I want to test for convergence
It may converge or it may not
hint: how big or small are the summands?
you will see each summand has abs val >1
and you need summands to tend to 0 for convergence
10:37
Might it help noting that this converges iff

$\sum_{n \ge 1} e^{i \pi n/4}sech(n)$

converges?
I don't see the point in that
oops, cos(in) has an i in it
I think I solved it using the ratio test
Well, I did, unless I made a typo somewhere
alyosha has it right
the i in the cos(in) is annoying, so turn it into the series he posted
You can rearrange it into

$\sum_{n \ge 0} e^{i \pi /4} (sech(8n+1)-sech(8n+5))+e^{i \pi / 2} sech(8n+2)-sech(8n+6))+e^{i 3 \pi /4 } sech(8n+3)-sech(8n+7))+e^{i \pi}( sech(8n+4)-sech(8n+8))$
Sech addition formulae may simplify that
The differences in sech shrink really fast, so it's not too surprising that it converges
just use comparsion with geo series, as sech~exp/2
10:51
I suppose so
 
3 hours later…
13:28
I have noticed many math presentations that look something like this (cs.illinois.edu/~heath/scicomp/notes/chap09.pdf). Does anyone know what tool they use to generate these (does not look like Powerpoint) Thanks
 
2 hours later…
leo
leo
15:53
@Amzoti beamer
16:15
@leo: Thank you!
16:46
Can someone help me prove how the dual of the dual norm is the original norm?
17:31
I've actually been trying to do this for sometime now
but I'm stuck
Is $U(2^n)$ another way to say $(\mathbb{Z}/2^n \mathbb{Z})^*$?
@EnjoysMath Thanks :-)
HOw is the double dual the original norm?
@Alyosha yech... there are parentheses missing...
That's the hyperbolic yec function
17:50
rhetorical exaggeration ^
;-)
Hi @OldJohn how are you?
@skullpatrol Hi - recovering slowly from a nasty kidney stone problem, but otherwise fine, thanks - and you?
@OldJohn I'm sorry to hear that...I'm ok thanks.
17:55
math has taught me an important lesson....
modesty.
without modesty i will not succeed in this field.
are you being modest?
:D
unfortunately not. i will have to take up buddhism soon.
i don't understand what the notation $(x_n)_{n \in \mathbb{N}}$ means. Someone asked the teacher the other day, I nodded up and down like I thought I knew it - and now I don't.
@DonLarynx Looks like a fairly standard notation for a sequence
but the teacher wrote it like this
$(x_n)_n$
although some would prefer curly brackets
18:02
Similarly, scroll to "definition in terms of limits and sequences" en.wikipedia.org/wiki/Continuous_function
is that notation for a subsequence?
I can't see anything on there that looks like a subsequence
a subsequence should look more like ${x_n}_m$
so what does $(x_n)_n$ mean
Have you looked at this?
not sure I have ever seen that notation, exactly
$(x_n)_n$ is bad notation for a sequence of sequences. Should be $(x_{m,n})_m$
18:09
Greetings
@KevinDriscoll :-)))))))
@Chris'ssis Yes, you can mix chocolate and peanut butter. Glad I could help.
@EnjoysMath Here too.
Let $f:\mathbb{R}\rightarrow \mathbb{R}$, $F$ - its antiderivative, $e^{x-F(x)}=F(x)$.
Compute

$$\lim_{x\to\infty}\frac{f(x)}{x}$$
4
I try to find an elementary proof. (it was given on a high school contest)
For what grade or age?
@skullpatrol $11$th grade
18:15
"We now show that a is $a$ product of irreducibles. If $a$ is irreducible, we are
done. Otherwise let $p_1$ be an irreducible such that $p_1 | a$. Then $a = p_1 c_1 $. If $c_1$ is a unit, we are done. "
How did they get that they're done?
That's a tricky one
Thanks I understand it now
np
9 mins ago, by skullpatrol
@EnjoysMath Here too.
What was that @Don
:D
This is so wrong
In abstract algebra, a non-zero non-unit element in an integral domain is said to be irreducible if it is not a product of two non-units. Irreducible elements should not be confused with prime elements. (A non-zero non-unit element a in a commutative ring R is called prime if whenever a | bc for some b and c in R, then a|b or a|c.) In an integral domain, every prime element is irreducible, but the converse is not true in general. The converse is true for UFDs (or, more generally, GCD domains.) Moreover, while an ideal generated by a prime element is a prime ideal, it is not true in gener...
Look what they say about primes
isn't that wrong?
$2$ is a unit..............
@EnjoysMath If $c_1$ is a unit, then $a = p_1$.
18:30
@skullpatrol 8 = 4*2.
Okay, if $a = p_1 c_1$ and $c_1$ is a unit, then $aR = p_1 c_1 R = p_1 R$
but I don't get that $a = p_1$
why are you telling me this?
lol
so is ZFC a class?
like do you take it during grad school?
Hi @skull
Hi @Charlie
@skullpatrol how are you doing?
18:36
@Charlie Fine thanks, how are you?
@skullpatrol fine, fine
@Charlie :-)
@skullpatrol :D
0
Q: In the proof that in a PID, every non-zero non-unit is the product of irreducibles...

Enjoys MathIn proving that all non-zero non-units of a PID are a product of irreducibles, theres: "We now show that $a$ is a product of irreducibles. If $a$ is irreducible, we are done. Otherwise let $p_1$ be an irreducible such that $p_1|a$. Then $a=p_1c_1$. If $c_1$ is a unit, we are done. " How...

@Charlie I see you were up late last night? Sorry I didn't have time to talk.
18:45
@skullpatrol no.problem. I coyldn't sleep,.i went to sleep about 3h30
@Charlie icic
@EnjoysMath, do you know if ZFC is a class?
like do you take it during grad school?
no, you read about it in the beginnings of all your books
or in the appendix
or you can study it independently, but I've never heard of a ZFC class
that's a horse of a different color!
oh
i thought it was significant....
more significant than naive set theory class for example
ask someone in school, I'm not
18:56
okay
In the topology of metric spaces the Heine–Borel theorem, named after Eduard Heine and Émile Borel, states: For a subset S of Euclidean space Rn, the following two statements are equivalent: *S is closed and bounded *every open cover of S has a finite subcover, that is, S is compact. In the context of real analysis, the former property is sometimes used as the defining property of compactness. However, the two definitions cease to be equivalent when we consider subsets of more general metric spaces and in this generality only the latter property is used to define compactness. In fact, the...
What is the point of the first paragraph?? After "If a set is compact, it is closed."
@skullpatrol :D
@EnjoysMath where you said you are from?
where does it say "If a set is compact, it is closed." I don't see it anywhere
it's further in the actual wikipedia page
and the point is to prove that claim
19:00
Right after: Proof. @EnjoysMath
I guess I just don't understand compactness yet
at least topologically
I can understand it sequentially, but my teacher has only covered it topologically, so I see no purpose in remembering it sequentially
:(
you will when you prove things with sequences in a compact space
@Don, did you know that a set is closed iff it contains all of its limit points?
How's the convergence test that you do $a_n^{1/n}$
?
called
I forgot the "called" lol
yes...
the root test @Crake
what's the first part of the paragraph you don't understand
19:04
Where "compactness" was used
oh yes, thank you
every open cover has a finite subcover
so what?
compactness isn't used until the second paragraph
oh wow lol
you lie
:D
19:05
So in the first paragraph they are just rambling?
I really hate pedantic writings
No, they're preparing to use those remarks
You'll have to read through it again and again until it makes sense
skip to the part where they construct $C$
yeah
i just did
then go back and read par 1
it made more sense
thnx :)
Since you're in a metric space, you can "choose open neighbordhoods small enough" That's a huge point because you can't do that in just any topo space
In that article limit point = accumulation point I think
19:12
Okay so in the first paragraph they show that the cover must be closed.
That's the only way it will cover the set
True @EnjoysMath?
No, what they show is stated in the first sentence
that's exactly what i just said, intuitively. can you re-read it?
@skull :?
@Charlie :/
correctomundo it sounds like. I will be busy rolling one
19:21
Am I correct to assume that there are no solutions to this congruence? $$4x\equiv 21\mod{46}$$ because $(4,46)=2$ and $2\nmid 21$?
I think $$\lim_{x\to\infty}\frac{f(x)}{x}=0$$ (hoping everything is fine in my reasoning)
@skullpatrol @Charlie how do you two know each other?
Correct @agent154
@DonLarynx OK, thanks
@Chris'ssis Only if $f(x)$ is a polynomial with degree 0
I hate myself I wasn't able to compute it without pen and paper and spent 20 minutes on it.
19:36
@Chris'ssis What is $f(x)$?
@DonLarynx Wouldn't it also be true for any polynomial with degree less than 1, such as $x^{1/2}$? Or how about logarithmic functions that grow slower than a polynomial with degree $1$?
@agent154 you don't need to find that in order to compute the limit.
@Chris'ssis It matters in this case, because if $f(x)=x^2$, then $\lim_{x\to\infty}\frac{x^2}{x}=\infty$
@agent154 don't you see the problem on the right side of the screen?
Oh, I didn't see that
@agent154 hehe, no pb.
19:40
@Chris'ssis you mean my attempt at a sol'n?
@DonLarynx no. Don't you see my entire problem on the right side of your screen?
Oh I see it
@DonLarynx ok :-)
HINT: $F(x)= e^xe^{F(x)}$.
Ahh! I'm trying to edit that but it won't let me
I'm wondering if my work is right here... $$2^{581}x\equiv 1\mod{19}$$

Since $(2,19)=1$, we have $2^{\varphi(19)}\equiv 1\mod{19}\Rightarrow 2^{18}\equiv 1\mod{19}$ by Euler's Theorem.

$2^{581}=(2^{18})^{32}\cdot 2^5\equiv 2^5\mod{19}\Rightarrow 32x\equiv 1\mod{19}\Rightarrow x\equiv 3\mod{19}$ (by Extended Euclidean Algorithm)
I don't know if I can legally do those things... I'm fairly confident I can, but it wouldn't be the first time that I misunderstood a theorem
That is, can I reduce $2^{581}$ modulo 19 before solving?
19:50
I would suggest you use wolframalpha to double-check your work
especially with arithmetic
Since $2^{581}\equiv 32\mod{19}$, by my calculations assuming they're right, then I should have $32x\equiv 1\mod{19}$
Sorry, that should be $2^{581}\equiv 13\mod{19}$. I had that written down but got mixed up between other numbers.
WA seems to say that's correct
great :)
Can anyone help me find the dual of this optimization problem? $min_x 1/2 ||y-x||_2^2 + \lambda||x||_1 $
I understand how to compute the dual in a setting where there are constraints
but any hint on how to do this?
@Chris'ssis do you know about differential equations
@DonLarynx a tiny bit.
19:58
Take F(x) = y and F'(x) = y'
I have managed to get your equation in the form $y' = \frac{e^x}{e^y}(1-y')$.
But I haven't found a form for it yet
@DonLarynx I already solved the question without using differential equations. But thanks!
@DonLarynx OK, let's think together
@leo are you here?
leo
leo
@Charlie yeah
20:00
@DonLarynx What does imply the fact that $e^x=F(x) e^{F(x)}$? Think of it when $x$ is extremely large (it tends to $\infty$).
leo
leo
@Chris'ssis why are you no longer wise?
$x = F(x)$
@leo are you fine? :)
@DonLarynx $\lim_{x\to\infty} e^{x} \rightarrow \infty$. This simple fact forces F(x) to tends to $\infty$.
$x = F(x)$ @Chris'ssis?
leo
leo
20:03
@Charlie yes :-)
how are you?
@leo ah, great , I'm fine:D leo, do you have another communication device? Email, fb,etc?
leo
leo
@Charlie email, Skype, Viber?
@leo like that
leo
leo
yes, those
what can I do to make my head hurt less when studying topology? will it get better?
20:06
@leo hard to say. I thought that "wise" could annoy some people. :-)
@DonLarynx did you get my idea?
leo
leo
@Chris'ssis haha
I see
@leo ok :)
leo
leo
@Charlie which do you want?
@leo email is fine
leo
leo
@Charlie ok get ready
20:09
@leo ok
leo
leo
@Charlie
got it?
@leo thanks leo
leo
leo
@Charlie :-)
Hi guys. Can anyone help me with a convex optimization problem?
I have been trying to do this for sometime but I am not sure how to proceed.
leo
leo
@Alice we don't know until you ask :-)
20:13
:-)
I'm trying to compute duals actually.
Specifically for a norm.
I want to prove that the dual of a dual norm is the original norm.
How do I do that?
I'm new to mathematics stack exchange and mainly here because trying to learn more about convex optimization.
@leo
do you know?
leo
leo
@Alice I'm not your guy
:-(
anyone else?
leo
leo
:-(
@leo :)
leo
leo
@Charlie :-)
20:22
@Charlie do you know?
ask on the main
you should get a response fairly quickly, usually
you mean chat room or post the question on the website?
I did do that
But didn't get a response.
:-(
(I mean I posted the question on the website here: math.stackexchange.com/questions/540020/…
But didn't get any response.
oh yea thats what I meant
I can't understand the question myself lol but if its research based you could try math overflow
there usually quick to get back to you
@Alice no, sorry
If its homework or something I wouldn't ask it there.
20:27
Not all of it is homework.
Why wouldn't you ask it there?
Where else could I ask?
(Been stuck for hours).
@Alice Showing your work is a very good thing to do to get more help.
Well the site is usually for research based questions I think
But if you get rid of the numbering, and maybe show a little work, I think they might be OK with it
I don't see the harm of posting it I guess though, the worst they can do is ask you to remove it
@DanielR thanks for your suggestion.
@Ethan thanks. I'll show my work.
Hey guys, if $A$ is a linear bounded operator of finite rank (on a Hilbert space $H$) then is it true that $\overline{ \text{range } A} = H$?
why would it be?
seems like finite-dim subspaces of a hilbert space would be closed
20:40
is there an easy counter example to this?
I was thinking something to do with the kernel and range and the orthogonal decomposition of $H$
huh?
for example, take H to be comprised of infinite vectors of real numbers, (x1,x2,x3,...). consider the projection map onto the first n coordinates. then A has finite rank and its image is the set of all tuples with zero outside the first n coordinates; this image is closed and not dense.
Can someone help me with this problem? math.stackexchange.com/questions/541660/…
@anon, okay, so if the question is reworded to saying if $A$ was compact rather than finite rank, the justification would stay the same right as all finite rank operators are also compact.
@DanielR and @Ethan : Here is the improved : math.stackexchange.com/questions/540020/…
I showed my steps
mainly stuck on the last step
but would like to verify the rest is correct.
@Alice Great, good luck! I can't help you though, but there are probably plenty that can.
Also, your question was now bumped to the front page, since you edited it. This will hopefully draw more attention to it.
20:53
I hope so too!
Thanks a lot.
@DanielR :-)

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