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12:07 AM
Anyone around?
 
12:21 AM
hey
 
@Link Any Linear Algebra knowledge there?
 
ugh, no, still in calc at intergerals
 
@Link OK.
 
but one question
for a projectile, how can you find angle at a certain time relative to the horizontal?
i have intial speed, angle, and range traveled
but i need it at a certain time
any ideas?
 
@Link There are nice eqns for that
 
12:27 AM
I have the basics
i know about x(t) = .5*g*t^2 + x_o*t+x
and basic equations
but I just need to know how to get the angle at a certain time
 
@Link Oh, the angle is just the slope at that moment.
 
but how would you get it?
 
@Link Say $dy/dx=p$. Then $\tan \alpha =p$
@Link Make a drawing.
 
okay
 
@Link You should get a concave parabola. Now make a tangent to it at $t=$ something.
Then slope of that tangent $=p=dy/dx=\tan\alpha$
 
12:29 AM
ah
okay
 
Where $\alpha$ is the angle it makes with the $x$ axis.
 
so one i get p
i just do arctan?
 
@Link Yes, just be careful about the quadrants.
 
and get angle?
thanks
:D
teacher's explanition was a bit difficult.
 
@Link Sometimes that happens.
 
12:39 AM
if A and B are modules, A+B can be given as a quotient of a direct sum, yes? is it $A+B=(A\oplus B)/(A\cap B)?$ Does this make sense?
 
@anon Man, I have a Lin Algebra mid term tomorrow, could you help me with something?
 
@PeterTamaroff surely you jest
 
@Thelonius That equality should be an isomorphism, and A+B only makes sense when A,B are submodules of something bigger.
@PeterTamaroff maybe
I really hate 9pm-5am work shifts.
 
@anon Well, I have done almost all this exercise, I'm almost done.
@Thelonius ?¿
@Thelonius Why?
@anon I have three subspaces.
 
@anon, yes i meant isomorphism. can we not say something in general, however? for any A and B? can't it be given as the quotient of a direct sum somehow?
 
12:43 AM
How is it not already "for any submodules A,B"?
 
$$\eqalign{
& S = \left\{ {x:{x_2} - {x_3} - {x_4} = 0} \right\} \cr
& T = \left\{ {x:{x_1} + {x_2} + {x_4} = 0} \right\} \cr
& H = \left\{ {x:{x_1} + 2{x_2} - {x_3} + 2{x_4} = 0} \right\} \cr} $$
All in $\mathbb R^4$.
 
@anon, i want an expression that is valid for all modules A and B. you seemed to imply that A+B only made sense if they were submodules of a bigger ring. obviously they both have to be R-modules for some fixed ring R, but i don't see what you mean about being submodules
 
Now, I need to find a subspace $W$ such taht $S^\perp+T^\perp\supset W$ and such that $W\oplus (S\cap T)=H$.
I have done this.
 
@Thelonius If $[A]$ and $[B]$ are two equivalence classes of modules, surely there are modules $M$ with the property that $X,Y\in[A]$ and $U,V\in[B]$ are submodules of $M$ but $X+Y$ and $U+V$ are neither equal nor isomorphic...
 
$$\eqalign{
& S = \left\langle {\left( {1,0,0,0} \right)\left( {0,1,1,0} \right)\left( {0,1,0,1} \right)} \right\rangle \cr
& T = \left\langle {\left( {0,0,1,0} \right)\left( {1,0,0, - 1} \right)\left( {0,1,0, - 1} \right)} \right\rangle \cr
& H = \left\langle {\left( {0, - 1,0,1} \right)\left( {1,0,1,0} \right)\left( {0,1,2,0} \right)} \right\rangle \cr} $$
And
 
12:47 AM
@anon what if $A$ and $B$ are finitely generated?
 
forgetting some \perps I believe
 
$$\eqalign{
& S \cap T = \left\langle {\left( { - 1,1,1,0} \right)\left( { - 1,0, - 1,1} \right)} \right\rangle \cr
& S^\perp = \left\langle {\left( {0,1, - 1, - 1} \right)} \right\rangle \cr
& T^\perp= \left\langle {\left( {1,1,0,1} \right)} \right\rangle \cr
& S^\perp+ T^\perp = \left\langle {\left( {1,1,0,1} \right)\left( {0,1, - 1, - 1} \right)} \right\rangle \cr} $$
 
@Thelonius What difference does it make? The intersection of two groups (e.g. $\Bbb Z$-modules) is (first of all empty unless they are both subgroups of a bigger group, but also) not determined at all by the isomorphism class of said subgroups. I don't see how f.g. has anything to do with the issue...
 
@anon your sight is better than mine
 
You might be vaguely interested in something called amalgams of groups, after all. (A research topic of MSE's Arturo Magidin, as it happens.) It has to do with how different groups may be "joined at the hip" in an overgroup in different ways.
 
12:50 AM
i want to justify something in Coutinho, which says basically that the finite sum of holonomic modules is holonomic because submodules and quotients of holonomic modules are holonomic and the direct sum of holonomic modules are holonomic
i don't get the because
 
What does holonomic mean, and is the sum internal and not necessarily direct?
But the implication would follow since, as you note, internal summations of submodules are isomorphic to direct sums quotiented by intersections.
 
he just says "finite sum". the definition of holonomic is not so important, his argument is. holonomic means a module over the Weyl algebra A_n which is of dimension n
if the sum were direct my question would be trivial
@anon just trying to understand what a reasonable expression of $\sum_1^N \text{holonomic}$ would be in terms of direct sums and quotients
 
In general, $$\sum_{i=1}^k A_i\cong \underbrace{\left(\bigoplus_{i=1}^k A_i\right)}_{E}~\big/~\mathrm{ker}\left(E\to M:(a_i)_i\mapsto \sum_{i=1}^k a_i\right).$$
 
ahh, and this kernel must be a submodule!
M is the internal direct sum?
 
no, $M$ is the overmodule that everything takes place in
the $A_i$ are each submodules of $M$
 
12:59 AM
@anon weird. he makes no reference to an overmodule...
i guess it doesn't matter which one...
still holonomic
 
@Thelonius How do you add submodules together internally if they are not all submodules of something? That doesn't even make sense!
 
i see what you mean
 
as an example in action: think of 1d subspaces (lines through the origin) of R^2. the external direct sum of n lines will be iso to R^n, but the internal sum of distinct subspaces will just be R^2.
sorry @Peter, I like this algebra more and my time is limited :-)
 
@anon Yeah. No problem.I think I got it.
 
thanks, @anon
 
1:29 AM
@anon given $I\subset J$ two left ideals of some ring $R$ (Noetherian if you like), can you express $I/J$ as a combo of sums, submodules or quotients of $R/K$, where $K$ is any left ideal you like
if you know this, even a hint would be fine
still working through same chapter
 
2:22 AM
@JayeshBadwaik Could you help me with some linear algebra?
Nevermind.
 
2:57 AM
@Thelonius I have a silly question
(ormaybe not)
Suppose $(5;-1;7)$ is the unique solution to $$Ax = \left( {3;4;1} \right)$$ where $A$ is $3\times 3$ and (the $;$ denote column vectors)
Let $B$ be the matrix
1 ,2,-3,-1
-1 ,0, 4, 2
1, 3, -1, 1
What are all the solutions to $(AB)x=(3;4;1)$?
Now since $Ax=b$, $b\neq \bf 0$, has unique solution, $A$ is inversible.
 
3:15 AM
@PeterTamaroff Sorry, was AFK. if you still are having the problem, maybe I can help.
 
@JayeshBadwaik That above
@JayeshBadwaik On the matrix eqns
 
Okay.
@PeterTamaroff B is a 4x4 matrix. How can it be multiplied with a 3x3 matrix?
 
@JayeshBadwaik $3\times 4$!
=)
 
Sorry Sorry
Hmm.
You can find $A$.
 
@JayeshBadwaik What I can think of is considering $Bx=A^{-1}(3;4;1)$
@JayeshBadwaik I can?
 
3:19 AM
@PeterTamaroff Yes that is what I was thinking.
 
@JayeshBadwaik Since $A$ is inversible.
 
You can determine $A$ upto similarity I guess.
But another thing is check the rank of $B$, it might come out to be $4$, and probably no solution exists.
 
@JayeshBadwaik Oh! That is it.
@JayeshBadwaik I think that does it.
 
But again, $Bx$ should not be possible, since $B$ is a $3 \times 4$ while $x$ is $3 \times 1$.
Unless this is a different $x$ from the one in $Ax = (3;4;1)$
 
@JayeshBadwaik Hmmm.... I'll sleep now. I'll ask my prof tomorrow.
 
3:28 AM
@PeterTamaroff Good night.
 
@JayeshBadwaik Night
 
Now, I will go away, and this room will be an empty room in JaX Shell.
 
user19161
@JayeshBadwaik Hi!
 
@JasperLoy Good morning!! I guess you woke up, or not yet slept?
 
user19161
@JayeshBadwaik I woke up. I am trying to regulate my sleep to normal hours but keep failing.
 
user19161
3:32 AM
I see that the past few starred messages only have one star each mostly.
 
@JasperLoy Yup. As I said, no real line punchers currently.
 
user19161
@JayeshBadwaik That definitely looks like KDE.
 
@JasperLoy It is. It is.
Wait a min. Will repost that image.
 
user19161
@JayeshBadwaik Hey, you are not considering other countries for your grad school application?
 
@JasperLoy I am considering, but most european countries demand a maths bachelor, and US countries demand a math GRE. There is very less chance otherwise (applying without a math subject GRE score for a non-math major). I am giving math GRE in November (that is the only time it is offered here.)
@JasperLoy what do you think?
 
user19161
3:45 AM
@JayeshBadwaik Hmm, I agree with your above paragraph.
 
@JasperLoy okay.
 
user19161
@JayeshBadwaik Hmm, maybe I will tell you more about me in email some time. I feel very bad these days.
 
@JasperLoy You are welcome. I would do my best to help.
 
 
1 hour later…
Joe
4:50 AM
Any help on this integral please?
$$\int \frac {\sqrt{\tan \theta}} {\sin 2\theta} \ d \theta$$

I tried rewriting it as $$\int {\sqrt{\tan \theta}} \cdot \csc(2\theta) \ d\theta$$

Supposedly letting $u = \sqrt{\tan \theta}$ cleans up the integral to just $1$, but I don't see how. $$du = \frac{\sec^2 \theta}{2\sqrt{\tan(\theta)}} d\theta \implies 2u \ du = \sec^2 \theta \ d\theta$$

Here's where I'm not sure how expressing $\csc(2\theta)$ as a double angle and in terms of $u$ cleans it up so nicely.
 
user19161
@Joe Try posting on the main site.
 
Joe
I was considering it, but I figured it was a subtle thing I was missing that would be easier here.
 
5:05 AM
Chat went wrong
@Joe Try putting $\sin 2\theta = 2 \sin \theta \cos \theta$ and then use the beta functions. Else substitute $(\sin \theta)^{-1/2} = t$ and try to solve the resulting integral (if possible in closed form.)
 
@JayeshBadwaik It was answered on main :-) using $u=\tan(\theta)$
@JayeshBadwaik we don't need no stinkin' special functions!
@Joe I was about to post that answer here, but since I saw it on main, I posted there... It gets you some points, too :-)
 
Joe
Ha, it's fine rob.
I'm looking back at one of your answers to an old improper integrals question that I might want you to elaborate on.
If you don't mind. I'm not sure why it's justifiable to apply Cauchy's PV.
4
Q: Evaluating $\int_{-1}^2\frac{1}{x}dx$

JoeI understand that when evaluating $$ \int_{-1}^{2} \frac{1}{x} \mathrm dx = \ln 2$$ It's simple integration, I understand. I'm more focused on the theory behind if it even exists. I had a question from Larson, Edward's Calculus 9th edition that was a true or false relating to this earlier toda...

 
@Joe I saw, the one with the Cauchy Principal Value
 
Joe
Since once you split it into the two integrals, the first one still isn't defined at -1.
Despite letting epsilon be the $0$ bound and taking the limit from the right.
 
@robjohn Ohh! Nice. :-) No stinking special functions it is.
 
Joe
5:21 AM
I'm just not sure how you can immediately evaluate the first integral since it still isn't defined at -1, that's all.
 
@Joe That is the idea of the Principal Value, you don't split it into two completely separate integrals, they are always in sync.
 
@robjohn Cauchy principle value reminds me of renormalization.
 
Joe
By in sync, I presume you mean the upper bound is the same epsilon that the lower bound of the next is (or the reverse)?
 
The definition of the Principal Value is $\int_{-1}^{-\epsilon}\frac1x\mathrm{d}x+\int_{\epsilon}^{2}\frac1x\mathrm{d}x$
@Joe By in sync, I mean that the hole is symmetric about the singularity at $0$. the upper limit of the lower integral is the same distance as the lower limit of the upper integral.
The reason for doing this becomes clearer as you learn more, especially when you learn some contour integration (complex analysis)
But it also just seems a natural thing to do, if you're going to do such a thing :-)
 
@Joe In this example, you are basically getting around the infinity part of the integral by using the fact that $ \frac{1}{x} = - \frac{1}{-x} $ and then canceling them from the integral. You don't need to evaluate the terms, you just state they are equal in magnitude and oppoiste in sign and hence, their integral will cancel.
@robjohn is this correct (in the essence, if not the details) ?
 
Joe
5:27 AM
Okay. I've read a bit about complex analysis and have seen several answers on here over time involving contour integration and singularities, but it's still above my level as you know. :)
 
@JayeshBadwaik yes. For instance you'd like to say that the integral of an odd function is $0$. Think of the odd function $\frac1x$ when $|x|<1$...
@JayeshBadwaik The Cauchy Principal Value gives the result of $0$ as you might expect.
 
@robjohn Yes.
 
@Joe: did you understand about the integral of an odd function being $0$?
 
Joe
Yes, that's trivial. Perhaps not with Cauchy though.
 
@Joe Well, not so trivial... consider the odd function $\dfrac{x}{x^2+1}$
 
Joe
5:33 AM
When you evaluated the splitted integrals, why did you disregard the $\ln|t|$ term for the first integral (because of Cauchy?) because shouldn't $t \rightarrow$ negative infinity?
 
@Joe I didn't disregard them, they cancelled.
 
Joe
Ah, with the second integral, yes.
Looking at $\frac{x}{x^2+1}$ now.
 
@Joe Don't think of it as two integrals, think of it as one integral over a domain with a hole.
@Joe Now that integral you can actually compute... $\frac12\log(x^2+1)+C$
You might think, "oh, it's odd, its integral should obviously be $0$"
But say we integrat from $-N$ to $2N$.
what do we get?
 
Joe
Sorry, friend came in room. One minute.
Just looking at it now.
Where would the singularity on the reals be though since the bottom is an irreducible quadratic?
Looks like we would get $$\frac{1}{2} \ln \left(\frac{4N^2+1} {N^2+1}\right)$$
 
6:04 AM
@Joe and as $N\to\infty$
 
Joe
@robjohn Apply L'Hopital to get $$\frac{1}{2} \ln (4)$$ as $N \rightarrow \infty$. But that's just $\ln (2)$ - why not $0$?
 
@Joe Not so obvious any more :-)
 
Joe
How can you justify the symmetry about the origin from $-N$ to $2N$?
 
This is an example of conditional convergence.
 
Joe
I saw a similar example with Cauchy on Wiki, but the limits were reversed.
proceeds to look up conditional convergence
 
6:11 AM
The sum $\displaystyle\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n}$
 
Joe
Ah, I think I'll be getting to that next week. I'm on sequences and series right now in Calc 2.
 
@Joe :-)
@Joe Normally, summing as we normally do, the sum is $\log(2)$
@Joe But by rearranging terms, it can sum to anything
 
Joe
Interesting.
 
So too, in the $\int\frac{x}{x^2+1}\mathrm{d}x$
 
@Joe Hi cotton eyed joe =)
 
6:15 AM
@Joe by changing how the limits go to $\infty$ we can make it come to anything we want.
 
Joe
Right now we've just introduced a few theorems and got into the formal basics of sequences and series.
Hi.
That's kind of a neat concept @robjohn.
 
@Joe The integral on $(0,\infty)$ diverges to $+\infty$, so to on $(-\infty,0)$ it diverges to $-\infty$
@Joe So we can choose how much of each to give to the integral by changing the limits
@Joe Of course, the standard way we do things gives the integral of $0$
And that leads back t the Cauchy Principal Value
 
Joe
:)
 
@robjohn Very nice post on stack excahnge about the integrl that evaluated to $\Omega(1)$
 
Joe
I may have to start coming back to the chat more @robjohn. I was in here late most nights over summer with you and Peter learning away. :)
It's fun learning new things.
 
6:23 AM
@Joe There are often discussions about things that I don't know, like algebraic geometry.
 
Joe
Ah, grad math courses. That's a ways away for me :P
 
I need to show that $|z|\geq |\text{Re}(z)|$ and $|z|\geq |\text{Im}(z)|$.
I used that $|z| = |\text{Re}(z) + \text{Im}(z)|$, but not quite sure how to proceed from here, any tips? =)
 
@N3buchadnezzar $|z|^2=|\mathrm{Re}(z)|^2+|\mathrm{Im}(z)|^2$
 
Thanks! Would something like this suffice as a proof?
By the pythagorean theorem we have that
$$ |z|^2=|\mathrm{Re}(z)|^2+|\mathrm{Im}(z)|^2 $$
and now since both $|\mathrm{Re}(z)|$ and $|\mathrm{Im}(z)|$ are
positive we conclude that $|z|\geq|\mathrm{Re}(z)|$ and $|z|\geq|\mathrm{Im}(z)|$, with equality only at respectively the real and imaginary axis.
 
Joe
6:52 AM
@robjohn Random thought: suppose we have a function that were are integrating that looks like $$\int \left(\sqrt{f(x)}\right)^2 \ dx$$. Why can we justify only taking the principal branch when integrating? Why do we disregard the other branch? Is it because I am working in $\mathbb{R}$ most likely?
 
user19161
7:07 AM
@rob The chatjax message has become unpinned! But I guess it is not necessary as the link is in blue in the subtitle of the room.
 
user19161
In other news, Firefox 16 is out, fresh and hot!
 
@JasperLoy Loy!
 
user19161
@N3buchadnezzar Hey hey, hope you slept well.
 
@JasperLoy I slept for about 5 hours =)
@JasperLoy You ?
 
7:28 AM
Does anyone know where to find a purely geometric proof of the parallelogram law ? eg
$$ \| x + y\|^2 + \| x - y \|^2 = 2 \|x\|^2 + 2 \|y\|^2$$
 
You can use the cosine rule on the sides of the triangle $x,y,x+y$ and $x,y,x-y$.
 
That is a trigonometric proof
 
@N3buchadnezzar Hmm, simple pythagoras will do with some constructions then (which is essentially the same thing).
 
8:12 AM
@JayeshBadwaik Okay thanks =)
 
 
1 hour later…
9:32 AM
Good day! :)
What a hell with MathType :( There are no circ symbol for composition of functions?! It looks like a degree symbol here
 
9:45 AM
XXX


I'm still not sure what it means for a machine to be turing complete vs. a language to be turing complete. The first part seems me intuitive, but regarding the second one: If you have a language which is complicated enought to encode all programs, don't you still need some sort of compiler (specific to the language you consider), which translates it to machine code, for some particular machine. What is the totality of things the turing complete language has to express? The rules+the input?
 
Turing completeness is defined in terms of turing machine. If a machine can simulate any arbitrary single tape turing machine, then it is turing complete and the language it uses to do so is a turing complete language.
@NickKidman Also you might be a little mistaken in thinking that compiler is somehow outside the language. However, compiler is just a program at the end. For example, haskell compiler is written in haskell, so glasgow haskell compiler is just a haskell program at the end of the day.
I hope it makes things clear.
 
10:14 AM
@Joe You need to say which branch you are taking and justify your answer. You can take any branch you wish, as long as your answer worrks
 
10:59 AM
Good morning people!
Could someone tell me the name of a "good" book about Mathematical Analisys II (from the series ahead).. Something easy to understand..
 
 
2 hours later…
12:51 PM
@JayeshBadwaik: Thanks for the response. I don't understand it yet. The grammar is part of the language right? Then to make sense of turing completeness, the machine has to match/understand the grammar of said language. So in the end, a language alone can't be enought to be turing complete. After all, you need to specify in which way the language encodes the single tape turing machine. How this is done, the reading of the input, is part of the machine. This is why I spoke of compilers.
 
@NickKidman This might help.
 
 
1 hour later…
hhh
2:06 PM
(confused)[math.stackexchange.com/questions/210465/… ... could someone help with this simple differential prob?
(this thing does not work well in iPad...I cannot scroll up to fix typos)
 
@hhh What has differentiation got to do with it?
 
hhh
differential of velocity is acceleration, in that sense.
My friend agrees that my teacher was not careful in his statement, the equations should be correct.
I claimed that $|\bar v|^2=|v_x|^2+|v_y|^2$ but my teacher of course 2.4133 says that it is not correct in some situations, he was unable to provide rigorous mathematical explanation. I even claim $|\dot{\bar v}|^2=|\dot{v_x}|^2+|\dot{v_y}|^2$ or $|\bar a|^2=|a_x|^2+|a_y|^2$ and also for distances $|\bar s|^2=|s_x|^2+|s_y|^2$, directly from Pythagoras but I would like to see some sort of differential way to prove or disprove the situations, is my teacher right and why is it so?
Lenght of velocity -vector and acceleration -vector, the differential -forms correct?
I moved this to chat, I think my statements must be correct.
 
the equation is correct, I do not see why it should be wrong. It is possible you did not properly formulate the problem and it is quite possible your teacher was thinking in 3 dimensions.
 
I believe the more important question to ask here is this purely Newtonian mechanics?
 
hhh
yes
This is the course page, nothing to do with velocities near speed of light -- problem with slow moving glider.
 
user19161
2:23 PM
@unNaturhal Just read Rudin.
 
hi @robjohn
 
@skullpatrol anything exciting on your calendar today?
 
user19161
@robjohn What about yours?
 
@Thelonius $I/J\cong (R/J)/(R/I)$, right? (I don't feel like checking.)
 
@robjohn Not much...
 
2:25 PM
@JasperLoy Work and then dogsit for a friend in the hospital with cancer.
 
hhh
@robjohn any idea how to have LaTex shown nicely on iPad? I need to somehow execute the JS -script.
 
@hhh If you are talking about the browsers, chatjax should do well.
 
@hhh is there a problem with LaTeX on the iPad?
 
Hi guys
 
@RajeshD Hi guy :-)
 
hhh
2:27 PM
@robjohn I can see the LaTex on the main site but I dont know how to create a clickable button for the script in iPad...works differenty to desktop.
so I cannot see any LaTex here well.
 
user19161
@hhh By the way, the X should be capitalised.
 
user19161
@skullpatrol Hi SKULL.
 
hhh
@JasperLoy what is X?
 
is there a compact way to write that a module M is generated by two elements a and b?
 
@JasperLoy Yo, whatz up?
 
user19161
2:28 PM
@hhh LaTeX.
 
user19161
@skullpatrol My XXX.
 
M=<a,b>? M=Ra+Rb? What is more standard
 
hhh
@robjohn things such as Tab etc does not work in iPad, the mobile version of this site is limited?
(up down scrolling does not work also)
 
user19161
@hhh Mobile chat is different.
 
@JasperLoy Is "X" an imaginary variable?
 
user19161
2:30 PM
There are mobile versions of other sites too like blogger.
 
@skullpatrol ROTFL.
 
@JayeshBadwaik :-D
 
user19161
@skullpatrol It is a real variable, the last I checked.
 
hhh
@robjohn I changed now to Desktop version and it does not work well, Tab is still partly broken.
 
@JasperLoy Are you sure it doesn't have an imaginary component?
 
user19161
2:31 PM
@skullpatrol Hmm, the imaginary part just takes it to greater heights.
 
hhh
@robjohn Tab is not auto-completion but moving away from writing box >(
 
user19161
@hhh Have you typed @ plus the first letter before tabbing?
 
hhh
...and this page goes up down up in iPad, they have not tested on iPad, not working well.
 
@JasperLoy Only in your imagination, not mine :-)
 
hhh
@JasperLoy Yes
 
user19161
2:33 PM
@skullpatrol You are in my imagination.
 
@JasperLoy But you are not in mine...
 
hhh
@robjohn I moved the bug report to meta: meta.math.stackexchange.com/questions/6333/…
...I have filed millions of bug reports to Apple :P I did not expect the same here, anyway have to go now to do a paper for friday and do my old homeworks, taking extra time to move from a laptop to iPad. I try only use iPad for doing assignments :)
 
@hhh what browser are you using?
 
@robjohn I came up with a corny psuedo-proverb about math; wanna hear it?
 
@skullpatrol would it matter? ;-)
 
2:44 PM
no
 
swhut I thought :-)
 
just wanted your learned opinion...
 
@skull its ok if it is "never say a pseudo-proverb about math".
 
@skullpatrol well...?
 
hhh
@robjohn I am testing now in Safari, simiar bugs. The report in Meta was tested with iPad Google Chrome.
 
2:46 PM
If you find mathematics easy to understand you are not doing it right.
 
@hhh Okay, I am misunderstanding something. What do you mean the "Desktop" version?
 
@anon can i ask you two dumb ring/module theory questions?
 
only if I can give you two dumb answers
 
Let $M$ a module over $R$ generated by two elements, $u$ and $v$, such that $M$ has a finite length, say $r$. It is trivial to say that $Ru$ has length 1, no?
Isn't $Ru$ cyclic by definition and thus can have no submodules?
 
cyclic?
 
2:50 PM
a cyclic module
 
hmm, wasn't aware of that term
@Helmut well, wait: what if $I$ is an ideal of $R$, and we consider $Iu$?
 
i was looking at an argument which inducts on the length of a module. one supposes $M$ is generated by two elements, so that $M=Ru+Rv$. We want to show that $M$ is actually cyclic, generated by one guy. But the argument goes that $Ru$ has merely finite length
 
You don't have to ask if something is trivial, I guess
 
implying it's not obviously cyclic, but saying that you can at least say it has a cyclic submodule
cyclic means generated by one guy
 
@Helmut no worries
 
2:55 PM
can $Ru$ have nontrivial submodules?
 
sure. Z has submodules nZ for every n in Z.
 
right...
i told you they were dumb questions
the point is this...
 
Hi @robjohn
 
i want to say that $M/something$ has smaller length than $M$. So $M$ is generated by two guys, $u$ and $v$. The idea I'm told to consider is that you want to find a submodule of $Ru$ which is simple, call it $Ruy$. Then considering $M/Ruy$ has length less than $M$
 
hhh
@robjohn the picture shows you the Desktop
 
2:59 PM
$M$ is artinian, so i think this makes sense
i retract my questions @anon. i think...
 
Isn't there an isomorphism/correspondence theorem for modules? Then you can take a series for $M/S$ and lift up to a series for $M$, which must be $\le $ the length of $M$ in size.
 
hhh
iPad Google Chrome has two versions: Mobile and Desktop.
 
@hhh You mean a non-mobile browser
 
@robjohn I warned you that it was corny :-D
 
3:48 PM
@JasperLoy Rudin?
 
user19161
@unNaturhal Principles of mathematical analysis. Probably the most famous classical analysis book in the world.
 
@JasperLoy Could you give me a link?
 
user19161
@unNaturhal Wait are you looking for a book or online notes?
 
user19161
You can just check it out from your library. Every decent library should have a copy.
 
@JasperLoy I'm in Italy :D I'm not really sure that they could have an italian version :p So I just have to take an english one..
 
user19161
3:53 PM
@unNaturhal Ah I don't know about Italian books. And I only have seen the German and French and Russian ones translated into English so far.
 
5 hours ago, by unNaturhal
Could someone tell me the name of a "good" book about Mathematical Analisys II (from the series ahead).. Something easy to understand..
 
user19161
Hmm, I think I should list my favourite nine books in my profile. That way more people will know about them.
 
user19161
@unNaturhal Si, si.
 
This is the comment that sparked my corny psuedo-proverb
 
3:55 PM
97 dollars? It's a bit expansive...
 
user19161
@unNaturhal OK, I recommend Browder's Mathematical Analysis.
 
user19161
@unNaturhal expensive, not expansive
 
@unNaturhal 97 Dollars! Where do you live? Hmm, I see even paperback is at 54 dollars. hmm.
 
@JasperLoy Ops..
 
Sorry for pinging you, I did not read the above comments. Now I know you live in Italy.
 

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