Mathematics - 2012-10-10 (page 1 of 4)

Peter Tamaroff

00:07

Anyone around?

Link

00:21

hey

Peter Tamaroff

@Link Any Linear Algebra knowledge there?

Link

ugh, no, still in calc at intergerals

Peter Tamaroff

@Link OK.

Link

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?

Peter Tamaroff

@Link There are nice eqns for that

Link

00:27

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

Peter Tamaroff

@Link Oh, the angle is just the slope at that moment.

Link

but how would you get it?

Peter Tamaroff

@Link Say $dy/dx=p$. Then $\tan \alpha =p$

@Link Make a drawing.

Link

okay

Peter Tamaroff

@Link You should get a concave parabola. Now make a tangent to it at $t=$ something.

Then slope of that tangent $=p=dy/dx=\tan\alpha$

Link

00:29

ah

okay

Peter Tamaroff

Where $\alpha$ is the angle it makes with the $x$ axis.

Link

so one i get p

i just do arctan?

Peter Tamaroff

@Link Yes, just be careful about the quadrants.

Link

and get angle?

thanks

:D

teacher's explanition was a bit difficult.

Peter Tamaroff

@Link Sometimes that happens.

Thelonius

00:39

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?

Peter Tamaroff

@anon Man, I have a Lin Algebra mid term tomorrow, could you help me with something?

Thelonius

@PeterTamaroff surely you jest

anon

@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.

Peter Tamaroff

@anon Well, I have done almost all this exercise, I'm almost done.

@Thelonius ?¿

@Thelonius Why?

@anon I have three subspaces.

Thelonius

@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?

anon

00:43

How is it not already "for any submodules A,B"?

Peter Tamaroff

$$\eqalign{
& S = \left\{ {x:{x_2} - {x_3} - {x_4} = 0} \right\} \cr
& T = \left\{ {x:{x_1} + {x_2} + {x_4} = 0} \right\} \cr
& H = \left\{ {x:{x_1} + 2{x_2} - {x_3} + 2{x_4} = 0} \right\} \cr} $$

All in $\mathbb R^4$.

Thelonius

@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

Peter Tamaroff

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.

anon

@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...

Peter Tamaroff

$$\eqalign{
& S = \left\langle {\left( {1,0,0,0} \right)\left( {0,1,1,0} \right)\left( {0,1,0,1} \right)} \right\rangle \cr
& T = \left\langle {\left( {0,0,1,0} \right)\left( {1,0,0, - 1} \right)\left( {0,1,0, - 1} \right)} \right\rangle \cr
& H = \left\langle {\left( {0, - 1,0,1} \right)\left( {1,0,1,0} \right)\left( {0,1,2,0} \right)} \right\rangle \cr} $$

And

Thelonius

00:47

@anon what if $A$ and $B$ are finitely generated?

anon

forgetting some \perps I believe

Peter Tamaroff

$$\eqalign{
& S \cap T = \left\langle {\left( { - 1,1,1,0} \right)\left( { - 1,0, - 1,1} \right)} \right\rangle \cr
& S^\perp = \left\langle {\left( {0,1, - 1, - 1} \right)} \right\rangle \cr
& T^\perp= \left\langle {\left( {1,1,0,1} \right)} \right\rangle \cr
& S^\perp+ T^\perp = \left\langle {\left( {1,1,0,1} \right)\left( {0,1, - 1, - 1} \right)} \right\rangle \cr} $$

anon

@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...

Thelonius

@anon your sight is better than mine

anon

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.

Thelonius

00:50

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

anon

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.

Thelonius

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

anon

In general, $$\sum_{i=1}^k A_i\cong \underbrace{\left(\bigoplus_{i=1}^k A_i\right)}_{E}~\big/~\mathrm{ker}\left(E\to M:(a_i)_i\mapsto \sum_{i=1}^k a_i\right).$$

Thelonius

ahh, and this kernel must be a submodule!

M is the internal direct sum?

anon

no, $M$ is the overmodule that everything takes place in

the $A_i$ are each submodules of $M$

Thelonius

00:59

@anon weird. he makes no reference to an overmodule...

i guess it doesn't matter which one...

still holonomic

anon

@Thelonius How do you add submodules together internally if they are not all submodules of something? That doesn't even make sense!

Thelonius

i see what you mean

anon

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 :-)

Peter Tamaroff

@anon Yeah. No problem.I think I got it.

Thelonius

thanks, @anon

Peter Tamaroff

01:15

@anon If you have any keen observation, just tell me

Thelonius

01:29

@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

Peter Tamaroff

02:22

@JayeshBadwaik Could you help me with some linear algebra?

Nevermind.

Peter Tamaroff

02:57

@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.

Jayesh Badwaik

03:15

@PeterTamaroff Sorry, was AFK. if you still are having the problem, maybe I can help.

Peter Tamaroff

@JayeshBadwaik That above

@JayeshBadwaik On the matrix eqns

Jayesh Badwaik

Okay.

@PeterTamaroff B is a 4x4 matrix. How can it be multiplied with a 3x3 matrix?

Peter Tamaroff

@JayeshBadwaik $3\times 4$!

=)

Jayesh Badwaik

Sorry Sorry

Hmm.

You can find $A$.

Peter Tamaroff

@JayeshBadwaik What I can think of is considering $Bx=A^{-1}(3;4;1)$

@JayeshBadwaik I can?

Jayesh Badwaik

03:19

@PeterTamaroff Yes that is what I was thinking.

Peter Tamaroff

@JayeshBadwaik Since $A$ is inversible.

Jayesh Badwaik

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.

Peter Tamaroff

@JayeshBadwaik Oh! That is it.

@JayeshBadwaik I think that does it.

Jayesh Badwaik

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 $Ax = (3;4;1)$

Peter Tamaroff

@JayeshBadwaik Hmmm.... I'll sleep now. I'll ask my prof tomorrow.

Jayesh Badwaik

03:28

@PeterTamaroff Good night.

Peter Tamaroff

@JayeshBadwaik Night

Jayesh Badwaik

Now, I will go away, and this room will be an empty room in JaX Shell.

user19161

@JayeshBadwaik Hi!

Jayesh Badwaik

@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

03:32

I see that the past few starred messages only have one star each mostly.

Jayesh Badwaik

@JasperLoy Yup. As I said, no real line punchers currently.

user19161

@JayeshBadwaik That definitely looks like KDE.

Jayesh Badwaik

@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?

Jayesh Badwaik

@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

03:45

@JayeshBadwaik Hmm, I agree with your above paragraph.

Jayesh Badwaik

@JasperLoy okay.

user19161

@JayeshBadwaik Hmm, maybe I will tell you more about me in email some time. I feel very bad these days.

Jayesh Badwaik

@JasperLoy You are welcome. I would do my best to help.

Joe

04:50

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.

Jayesh Badwaik

05:05

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.)

robjohn

@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 $\int_{-1}^2\frac{1}{x}dx$

I 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...

robjohn

@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.

Jayesh Badwaik

@robjohn Ohh! Nice. :-) No stinking special functions it is.

Joe

05:21

I'm just not sure how you can immediately evaluate the first integral since it still isn't defined at -1, that's all.

robjohn

@Joe That is the idea of the Principal Value, you don't split it into two completely separate integrals, they are always in sync.

Jayesh Badwaik

@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)?

robjohn

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 :-)

Jayesh Badwaik

@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

05:27

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. :)

robjohn

@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.

Jayesh Badwaik

@robjohn Yes.

robjohn

@Joe: did you understand about the integral of an odd function being $0$?

Joe

Yes, that's trivial. Perhaps not with Cauchy though.

robjohn

@Joe Well, not so trivial... consider the odd function $\dfrac{x}{x^2+1}$

Joe

05:33

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?

robjohn

@Joe I didn't disregard them, they cancelled.

Joe

Ah, with the second integral, yes.

Looking at $\frac{x}{x^2+1}$ now.

robjohn

@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 $$\frac{1}{2} \ln \left(\frac{4N^2+1} {N^2+1}\right)$$

robjohn

06:04

@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 $0$?

robjohn

@Joe Not so obvious any more :-)

Joe

How can you justify the symmetry about the origin from $-N$ to $2N$?

robjohn

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

robjohn

06:11

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.

robjohn

@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.

robjohn

So too, in the $\int\frac{x}{x^2+1}\mathrm{d}x$

N3buchadnezzar

@Joe Hi cotton eyed joe =)

robjohn

06:15

@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.

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

:)

N3buchadnezzar

@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.

robjohn

06:23

@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

N3buchadnezzar

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? =)

robjohn

@N3buchadnezzar $|z|^2=|\mathrm{Re}(z)|^2+|\mathrm{Im}(z)|^2$

N3buchadnezzar

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

06:52

@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

07:07

@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!

N3buchadnezzar

@JasperLoy Loy!

user19161

@N3buchadnezzar Hey hey, hope you slept well.

N3buchadnezzar

@JasperLoy I slept for about 5 hours =)

@JasperLoy You ?

N3buchadnezzar

07:28

Does anyone know where to find a purely geometric proof of the parallelogram law ? eg
$$ \| x + y\|^2 + \| x - y \|^2 = 2 \|x\|^2 + 2 \|y\|^2$$

Jayesh Badwaik

You can use the cosine rule on the sides of the triangle $x,y,x+y$ and $x,y,x-y$.

N3buchadnezzar

That is a trigonometric proof

Jayesh Badwaik

@N3buchadnezzar Hmm, simple pythagoras will do with some constructions then (which is essentially the same thing).

N3buchadnezzar

08:12

@JayeshBadwaik Okay thanks =)

Nimza

09:32

Good day! :)

What a hell with MathType :( There are no circ symbol for composition of functions?! It looks like a degree symbol here

Nick Kidman

09:45

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?

Jayesh Badwaik

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.

robjohn

10:14

@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

unNaturhal

10:59

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..

Nick Kidman

12:51

@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.

Jayesh Badwaik

@NickKidman This might help.

hhh

14:06

(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)

Jayesh Badwaik

@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.

Jayesh Badwaik

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.

infinitesimal

I believe the more important question to ask here is this purely Newtonian mechanics?

hhh

yes

noppa.aalto.fi/noppa/kurssi/mat-2.4133/etusivu

This is the course page, nothing to do with velocities near speed of light -- problem with slow moving glider.

user19161

14:23

@unNaturhal Just read Rudin.

skullpatrol

hi @robjohn

robjohn

@skullpatrol anything exciting on your calendar today?

user19161

@robjohn What about yours?

anon

@Thelonius $I/J\cong (R/J)/(R/I)$, right? (I don't feel like checking.)

skullpatrol

@robjohn Not much...

robjohn

14:25

@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.

Jayesh Badwaik

@hhh If you are talking about the browsers, chatjax should do well.

robjohn

@hhh is there a problem with LaTeX on the iPad?

Rajesh D

Hi guys

skullpatrol

@RajeshD Hi guy :-)

hhh

14:27

@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?

Helmut

is there a compact way to write that a module M is generated by two elements a and b?

skullpatrol

@JasperLoy Yo, whatz up?

user19161

14:28

@hhh LaTeX.

user19161

@skullpatrol My XXX.

Helmut

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.

skullpatrol

@JasperLoy Is "X" an imaginary variable?

user19161

14:30

There are mobile versions of other sites too like blogger.

Jayesh Badwaik

@skullpatrol ROTFL.

skullpatrol

@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.

skullpatrol

@JasperLoy Are you sure it doesn't have an imaginary component?

user19161

14:31

@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.

skullpatrol

@JasperLoy Only in your imagination, not mine :-)

hhh

@JasperLoy Yes

user19161

14:33

@skullpatrol You are in my imagination.

skullpatrol

@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 :)

robjohn

@hhh what browser are you using?

skullpatrol

@robjohn I came up with a corny psuedo-proverb about math; wanna hear it?

robjohn

@skullpatrol would it matter? ;-)

skullpatrol

14:44

no

robjohn

swhut I thought :-)

skullpatrol

just wanted your learned opinion...

Rajesh D

@skull its ok if it is "never say a pseudo-proverb about math".

robjohn

@skullpatrol well...?

hhh

@robjohn I am testing now in Safari, simiar bugs. The report in Meta was tested with iPad Google Chrome.

skullpatrol

14:46

If you find mathematics easy to understand you are not doing it right.

robjohn

@hhh Okay, I am misunderstanding something. What do you mean the "Desktop" version?

Helmut

@anon can i ask you two dumb ring/module theory questions?

anon

only if I can give you two dumb answers

Helmut

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?

anon

cyclic?

Helmut

14:50

a cyclic module

anon

hmm, wasn't aware of that term

@Helmut well, wait: what if $I$ is an ideal of $R$, and we consider $Iu$?

Helmut

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

Rajesh D

You don't have to ask if something is trivial, I guess

Helmut

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

Rajesh D

@Helmut no worries

Helmut

14:55

can $Ru$ have nontrivial submodules?

anon

sure. Z has submodules nZ for every n in Z.

Helmut

right...

i told you they were dumb questions

the point is this...

Rajesh D

Hi @robjohn

Helmut

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

Helmut

14:59

$M$ is artinian, so i think this makes sense

i retract my questions @anon. i think...

anon

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.

robjohn

@hhh You mean a non-mobile browser

skullpatrol

@robjohn I warned you that it was corny :-D

unNaturhal

15:48

@JasperLoy Rudin?

user19161

@unNaturhal Principles of mathematical analysis. Probably the most famous classical analysis book in the world.

unNaturhal

@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.

unNaturhal

@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

15:53

@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.

skullpatrol

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..

unNaturhal

@JasperLoy Is this one:amazon.com/Principles-Mathematical-Analysis-Third-Walter/dp/… ? O_O

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.

skullpatrol

This is the comment that sparked my corny psuedo-proverb

unNaturhal

15:55

97 dollars? It's a bit expansive...

user19161

@unNaturhal OK, I recommend Browder's Mathematical Analysis.

user19161

@unNaturhal expensive, not expansive

Jayesh Badwaik

@unNaturhal 97 Dollars! Where do you live? Hmm, I see even paperback is at 54 dollars. hmm.

unNaturhal

@JasperLoy Ops..

Jayesh Badwaik

Sorry for pinging you, I did not read the above comments. Now I know you live in Italy.

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$Mathematics$ Mathematics