Mathematics - 2012-05-02 (page 1 of 6)

Antonio Vargas

00:02

@Chris I think assuming that the space in question, solSet, is a solution to a system of equations is a red herring.

anon

@Chris: A homogeneous system is when the b vector is 0. Every subspace of a vector space must contain 0, which means if a solution set is a subspace it contains 0, which means 0 is a solution ie A0=b, but A0=0 so b=0.

Antonio Vargas

You are essentially starting with an arbitrary subspace. You assume that it has dimension 1, which is fine, but unnecessary. Since projections are linear transformations, any projection onto your subspace gives you a linear transformation whose image is your subspace.

anon

@Chris: Another possibility is that you're actually thinking about "the set of all vectors b such that Ax=b has a solution in x." This is a subspace, and it is the image of the linear transformation A.

Chris

@AntonioVargas: We are being asked if the solution set ( a specific one, we 've found it) is the image of a linear transformation. So the space in question is the solution set.

Projection in a 1-dimensional subspace is something like the identity function?
So the answer would be something like: Assuming we are in $ R^{1} $ the projection's solution set is the image of the projection, itself. In example, given the number x, the projection of x in $ R^{1} $ is the x itself. Would that be ok?

@anon: I fully understand your 1st answer. The 2nd one, not that much :/

Antonio Vargas

00:19

@Chris I don't understand what you wrote. Maybe we should start from the beginning. Could you write out the question in full?

Chris

We are given a system of linear equations: $ 6*x + 5*y + 7*z + 4*w = 0, 6*x + 5*y + 3*z + 4*w = 0, 14*x + 1*y + 2*z + 4*w = 0 $, the solution set is all the x,y,z in R that (x,y,z,w) = w(-1/4, -1/2, 0, 1), where w belongs in R.
Is that correct up to here? @AntonioVargas

Antonio Vargas

Uh, well I'd have to solve it myself to know, and I'm too lazy, so let's assume that's correct.

Chris

@AntonioVargas: Here: http://tinyurl.com/d2a65cm

So we are given this statement: "The solution set of the system is the image of a linear transformation", true or false?

@AntonioVargas: A little change on the coefficient of x coming up

Antonio Vargas

It's true. The solution set $S = \{w(-1/4,-1/2,0,1) \colon w \in \mathbb{R}\}$ is a subspace of $\mathbb{R}^4$, so we can construct a map which projects any vector in $\mathbb{R}^4$ onto $S$. This map is a linear transformation whose range is $S$. Are you familiar with projecting onto a subspace?

This is exactly what Arturo does in #4 of his answer.

Chris

00:35

@AntonioVargas: The change of the coefficient:

$ 9*x + 5*y + 7*z + 4*w = 0,
9*x + 5*y + 3*z + 4*w = 0,
14*x + 1*y + 2*z + 4*w = 0 $, the solution set is all the x,y,z in R that (x,y,z,w) = w(-16/61, -20/61, 0, 1), where w belongs in R.

Ok, I got it better now! ("so we can construct a map which projects any vector in R4 onto S" :-) )
Range is something like the rank of a matrix?
No, that's the bad thing about it, our teacher hasn't talked about projections :-/

Antonio Vargas

The range of a transformation $T$ is the set of all vectors $v$ such that you can find a vector $x$ satisfying $Tx = v$.

You might speak with your teacher about projections, they are interesting and useful.

Chris

I will surely do!
So it's something like the opposite of the kernel ? ( I am trying to correlate with the terms that I am already familiar with)

Thank you for your help!

Antonio Vargas

It's kind of the opposite of the kernel, I guess. I think of it as the opposite of the domain. The domain of a transformation is the set of vectors which you can put into it, and the range is the set of vectors you can get out of it.

Chris

Great! Thank you for the clarification. :-)

Antonio Vargas

No problem! Glad to help.

robjohn

00:58

@tb it appears as if I did, too. Maybe next time ;-)

Jordan

Is there a better video series than khan academy for calculus?

robjohn

@Jordan I really don't know. Have you tried to Google "calculus video"?

Jordan

yup

robjohn

a number of decent looking options come up

However, I don't have any experience with any of them.

Jordan

Ok I was just wondering if there was an agreed upon "best" video series, khan academy skips some of what we cover in class

t.b.

01:15

@robjohn let's hope so :)

robjohn

@tb I seem to be missing you, unless you are a phantom :-)

watching chat from the transcript

@Benjamin: hello!

gotta run to the park. bbl

Benjamin Lim

01:34

hi guys

@robjohn Hi

t.b.

01:45

@robjohn Yes, we seem to play hide-and-seek :) I just saw that you decided run for moderator, after all. No big surprise... Thanks for doing this, good luck!

Bedtime for me... Good night!

anilorap

02:02

hi

Eric Gregor

any hint for the following. Show that a complex square matrix can be uniquely decomposed into the sum $A+iB$, were $A$ and $B$ are Hermitian?

t.b.

@EricGregor $A = \frac{1}{2}(M + M^\ast)$ and $B = \cdots$. If you have two decompositions $A - A' = i(B' - B)$ is hermitian and skew-hermitian, so...

Eric Gregor

doh

thanks @tb

anilorap

Hi, I am doing differential equations. And I was asked to find the general solution and as an eigenvalue I had a complex number and my system look like:

Eric Gregor

wait, @tb, is if $B$ is hermitian is $iB$ skew hermitian?

anilorap

02:12

-2x+3y=i(squ.root of 2)x

Eric Gregor

is that by definition?

anilorap

-2x+2y=i(squ.rootof 2)y

t.b.

@EricGregor $(iB)^\ast = -i B^\ast = -(iB)$

Eric Gregor

thanks!

robjohn

@tb still up I see :-)

t.b.

02:15

@robjohn yeah, so much for my healthy intentions... :)

robjohn

@tb who needs sleep?

anilorap

:(.. so I would like to know, how do I check if they are redundant?

t.b.

@robjohn that darn body of mine...

robjohn

@tb such an impediment to the operation of the mind :-D

J. M.

@robjohn "normal" people (whatever the hell "normal" means.) :D

t.b.

02:17

@JM I'd be surprised to find them hereabouts :)

(and hi, n'all)

J. M.

@tb Indeed. (And hi!)

@anilorap You didn't say what your original DE system was, so it's hard to say anything about correctness.

anilorap

my original system is dx/dt=(-2,3) dy/dt=(-2,2)

J. M.

Hmm, I see rob is now a candidate, and there are now four people with the "Tenacious" badge. One of these two things is a little bit troubling... ;)

@anilorap That's a bit hard to make sense of; you have a derivative on the left and ordered pairs on the right?

anilorap

am sorry Its my first time using this.. but is a matrix

top [-2 3] bottom [-2 2]

J. M.

@anilorap I see, so you have a $2\times 2$ matrix multiplied with the vector $\begin{pmatrix}x&y\end{pmatrix}^\top$.

anilorap

02:23

:)

so i found my eigenvalue

J. M.

Well then you do have a pair of conjugate eigenvalues...

anilorap

Lambda = +- i (sq.rootof 2)

is a complex eigenvalue.. so i will be working with +i(sq.rootof 2)

so i need to find my eigenvector by solving Ay=(lambda)y

robjohn

@anilorap it would seem so.

anilorap

thats how i got my two equations... -2x+3y=i(sq.rootof2)x and -2x+2y=i(sq.rootof2)y

t.b.

@JM oh, at least we still don't have an unsung one...

J. M.

02:27

@tb Well, that's one of the few comforts...

t.b.

(this is not implying that there are robjohn songs that I'm aware of :))

anilorap

so how do i know this two eq. are redundant?

robjohn

@tb however, there are a lot of Robert Johnson songs

@anilorap start to solve them.

J. M.

@anilorap What happens if you try row reduction?

anilorap

ive been doing it for hours... i dont know what am doing wrong

@JM can i do that?

t.b.

02:31

@robjohn Yes, but you wouldn't sing sweet home Chicago, for example, would you?

J. M.

@anilorap Sure, you can row reduce. Arithmetic still works on complex numbers...

t.b.

Does the announced $\color{red}{\text{(status-completed)}}$ here work for you?

(I presume we have to wait for the next build to roll out)

robjohn

$(-2-i\sqrt{2})x+3y=0$ and $-2x+(2-i\sqrt{2})y=0$ do they look any more dependent?

J. M.

@tb Well, it says "rev 2012.5.1.2423" at the bottom, and it's May second already...

anilorap

:(

robjohn

02:34

@tb effectuated? why not simply effected?

J. M.

If the May 2 build rolls out and the bug's still there, then we have cause for a riot... >:)

t.b.

@JM oh, yes. I'm still living in the past

robjohn

@tb Tull is a great band. I've seen them twice in concert.

J. M.

@robjohn You preach truth, brother. ;)

t.b.

@robjohn because my English builds on my French, I assume. effectué sounds like the reason for this, but effectuated is not wrong, is it?

robjohn

02:37

@tb I don't know; I've never heard the word before :-)

t.b.

@robjohn Oh, yeah. I only saw them once, but I saw two concerts of Ian solo

J. M.

@tb It's not wrong, but not very common either.

robjohn

@JM I agree, the word is correct, but not common.

t.b.

@robjohn I see. Thanks for pointing it out!

robjohn

@tb otherwise, it works for me :-)

Jeff

02:56

@robjohn Hi Rob. I cannot duplicate your partial fractions decomp (I don't know the Heavside method). I need to finish this problem tonight, though. If I accept your decomposition, what is my next step on that problem? I am supposed to be using residue theorem.

robjohn

@Jeff I will write up the decomposition.

Jeff

@robjohn wow. cool, thx.

Benjamin Lim

@DavidWheeler Such an extension does not exist

robjohn

03:18

$$
\begin{align}
\frac{1}{x^5+1}
&=\frac{A_0}{x+1}+\frac{A_1}{x+\zeta}+\frac{A_2}{x+\zeta^2}+\frac{A_3}{x+\zeta^3}+\frac{A_4}{x+\zeta^4}\\
1&=A_0\frac{x^5+1}{x+1}+A_1\frac{x^5+1}{x+\zeta}+A_2\frac{x^5+1}{x+\zeta^2}+A_3\frac{x^5+1}{x+\zeta^3}+A_4\frac{x^5+1}{x+\zeta^4}
\end{align}
$$
To compute $A_k$, we need to compute $\dfrac{x^5+1}{x+\zeta^k}$ and evaluate at $x=-\zeta^k$. This is because for $j\not=k$, $\dfrac{x^5+1}{x+\zeta^j}=0$ when $x=-\zeta^k$. Using L'Hospital, we get $\dfrac{x^5+1}{x+\zeta^k}=5x^4=5\zeta^{4k}$. That is,

t.b.

Good news for Asaf :)

robjohn

@tb at least interesting news.

Jeff

@robjohn this is all with $\zeta=e^{\frac{2\pi}{5}}$, right?

anon

I changed my gravatar nearly 24hr ago. Should I have to wait this long? Is there a way to force-propogate it?

robjohn

@Jeff $\zeta=e^{i2\pi/5}$.

@anon It usually comes over as soon as I log out from gravatar.com

t.b.

03:25

@anon It did change for me a few hours back (I see a round rainbow now).

anon

Y U NO WRITE $i$ AFTER $2\pi$??

Jeff

i forgot the i (i meant it like you have it). last time you had a coefficient of 1/5 in front of the whole thing.

@anon if that's me, it's cuz I forgot.

robjohn

@anon are you talking to me or Jeff?

anon

You.

robjohn

is it not the case that $i2\pi/5=2\pi i/5$?

anon

03:27

@t.b. By round rainbow, do you mean this?

kahen

Is it unreasonable to ask for extra macros on math.sx? In particular it'd be really nice to have a \midrel defined by \def\midrel#1{\mathrel{}\middle#1\mathrel{}}. This would make writing sets much easier: $\left\{a \in \mathbb R \midrel| a > 0\right\}$.

anon

@robjohn: Aesthetically, no!!

kahen

as per tex.stackexchange.com/questions/5502/…

t.b.

@anon exactly that.

robjohn

@anon I don't see that :-(

@anon have you logged out from gravatar.com?

t.b.

03:29

@robjohn I guess you have to reload the chat window.

robjohn

@tb I've done that several times and I still see the old rainbow

anon

I've been logging in/out and refreshing everything, nothing changes.

robjohn

@anon you're sure the email addresses match?

is the computer plugged in?

anon

lol

I've never had any other address associated to mse or gravatar.

robjohn

then I am at a loss.

anon

03:31

now, gravatar tells me I have the new icon. MSE doesn't.

t.b.

Caching problems, maybe?

anon

I'll check.

Cache cleared, nothing changed. Except it wouldn't let me log out of the main for a few moments.

lewist

@anon By changing the size parameter of your icon in chat, I can see either the new one or the old one.

t.b.

@kahen People don't tend to react very favorably to such proposals.

(there were a few other, similar meta threads but I can't seem to find them right now, maybe they were deleted by the OPs due to the huge number of downvotes)

lewist

@anon When the size is a power of two, I seem to get the older gravatar, but any other value displays the new one (in a different size, of course).

anon

03:39

O.o

robjohn

Still no joy...

Mariano Suárez-Alvarez

@kahen, shouldn't that be \mathrel{\middle#1} ?

robjohn

let me try a non-power of two

lewist

... and now the pattern's breaking. For $s = 256$, we get the new gravatar.

Mariano Suárez-Alvarez

using \mathrel{} for spacing is weird

robjohn

03:42

yes! when I look at 300 pixels, I get the new one.

old gravatar and new gravatar

kahen

mariano: you'd THINK it should be that, but that won't work because \middle will fail to see \left and \right in that case. Try it out yourself

Jeff

@robjohn ``This is because for $\displaystyle j \ne k,\ \frac{x^5+1}{x+\zeta^j}=0$ when $x=-\zeta^k$''.

Let $k=2,\ j=3$, then $x=-\zeta^2$ and $\displaystyle\frac{(-\zeta^2)^5+1}{(-\zeta^2)+(\zeta)^3} = \frac{1-\zeta^{10}}{\zeta^2+\zeta^3}=\frac{1-1}{\text{not } 0}=0$.

Mariano Suárez-Alvarez

oh

just say \big#1 then

robjohn

@Jeff yes, that is simply the distributive property.

@Jeff indeed

t.b.

03:57

@MarianoSuárezAlvarez But that wouldn't scale, would it?

Mariano Suárez-Alvarez

\newcommand\midrel[2][]{\mathrel{#1#2}}
allows you to say \midrel{|}, \midrel[\big]{|}, or \midrel[\bigg]{|}

J. M.

@anon: nice mandala

Mariano Suárez-Alvarez

@tb, I don't think having it scale looks good at all :)

robjohn

@anon: I still don't see it on chat :-(

Mariano Suárez-Alvarez

in fact, I personally hate a bar there, and exclusively use a colon myself

anon

03:59

Me either. I give up for now.

robjohn

@MarianoSuárezAlvarez I do, too :-)

t.b.

@MarianoSuárezAlvarez That's why I always use the colon there... But scaling was the goal if I understood correctly.

@MarianoSuárezAlvarez oh :)

robjohn

@anon did you try the links here?

lewist

@anon Well, there's something screwy. Removing the rating parameter (r=PG) causes it to display correctly, regardless of size.

anon

@rob Yes, they correspond to your descriptions.

robjohn

04:01

Do people see what is described in these links?

anon

What about on the mainsite?

Antonio Vargas

@robjohn I see the old one.

t.b.

@anon It's funny, on FF it works just fine for me, while on other browsers (Safari and Chrome) I get your old Gravatar...

Mariano Suárez-Alvarez

*The* correct way of implementing a resizable bar nowadays would be, I think, using \DeclarePairedDelimiterX from mathtools
This allows you to say \mathrel{\delimsize\mid} to get the corrent size

robjohn

@AntonioVargas but not the new one?

anon

04:02

@tb Madness!

Antonio Vargas

No, just the old one, here in chat and in the list on the right. Chrome here.

robjohn

@tb I am using Firefox and I see the old gravatar at power of two sizes.

J. M.

@anon, could you maybe check what rating you assigned to your rainbow mandala?

Jeff

@robjohn can you explain why computing $\displaystyle\frac{x^5+1}{x+\zeta^k}$ at $\displaystyle x=\zeta^k$ yields $A_k$?

anon

Checked; I assigned G.

robjohn

04:04

@anon the one shown here is PG...

anon

@Jeff: Do you mean $x=-\zeta^k$? l'Hospital

t.b.

@robjohn From main: FF, Safari

Antonio Vargas

I also see the "old" one on anon's math.SE profile page. And I doubt it's a caching thing since I don't believe I've seen that particular gravatar before.

robjohn

@Jeff $\displaystyle1=A_0\frac{x^5+1}{x+1}+A_1\frac{x^5+1}{x+\zeta}+A_2\frac{x^5+1}{x+\‌zeta^2}+ A_3\frac{x^5+1}{x+\zeta^3}+A_4\frac{x^5+1}{x+\zeta^4}$

J. M.

How about removing and reuploading, anon?

robjohn

04:06

@Jeff all but one of the coefficients of $A_k$ is $0$

anon

lewist

@tb Out of curiosity, what is the URL for both of those images? I understand if you're busy.

anon

I'll try that JM

WOAH nelly, it just clicked

Jeff

@anon yes, forgot the negative sign :(

robjohn

@anon I see it now :-)

lewist

04:08

@tb Nevermind, it appears to have resolved.

Jeff

@robjohn Aaahhh. OK. Is the l'Hosp you're using the same one in calculus used to calculate limits? Why does it work here?

t.b.

@lewist Oh, it seems the https did the trick here (I have https everywhere installed on FF)

(and on Safari I just had the http link)

kahen

Jeff: FYI there is no 's' in l'Hôpital.

anon

@kahen My whole life has been a lie!

lewist

Ahh neat, thanks t.b.

Antonio Vargas

04:10

@anon Works for me now.

robjohn

@Jeff whether we cancel the polynomials or we limit their quotient as we approach $\zeta^k$, we get the same thing.

anon

Actually, Mathworld claims the two spellings are equivalent in French

And both used in English.

robjohn

@kahen both seem to be used in many places.

t.b.

@kahen his book was written under the spelling l'Hospital

kahen

I see. l'Hôpital must be a later spelling then

t.b.

04:13

It might have to do with capitalization, though (like 'SZ' vs. ß in German)

Jeff

@robjohn are you saying factoring (again!) the top and bottom and cancelling yields the same thing as using l'hosp? Also, did you mean "as we approach $-\zeta^k$?

robjohn

@Jeff yes as $x\to-\zeta^k$

anon

You could also use a change of variables and the geometric sum formula

Mariano Suárez-Alvarez

both ô and os are correct; the spelling of that was changing at the time

robjohn

@MarianoSuárezAlvarez and ôs shall be right out :-)

Jeff

04:16

@robjohn that makes sense. it's sort of like the the second example everyone uses when teaching limits: $\displaystyle \lim_{x \rightarrow 1}\frac{x^2-1}{x-1}$

t.b.

@robjohn Three, Sir!

robjohn

@Jeff indeed

@tb Three.

t.b.

:)

Jeff

@robjohn which i just showed to a student today! :D

robjohn

that rabbit's dynamite!

Jeff

04:19

@robjohn now the rest of your explanation falls into place.... so now, to integrate, i need to use residue theorem. i think just a straightforward applicatoin of that theorem solves the integral.

kahen

It is a bit sad that you are (mostly) limited to examples using rational functions where the denominator divides the numerator. at least until you have some result on sin, cos and tan.

robjohn

@Jeff let me know if you have any trouble.

@kahen eh?

Jeff

@robjohn i will (let you know, that is, not have trouble). and, btw, thank you.

kahen

well. examples like x^2-1/x-1, where x-1 | x^2-1.

robjohn

@Jeff no problem

kahen

04:21

i guess exponentials could come into play as well

robjohn

@kahen The case at hand was a polynomial, so that's all we needed.

t.b.

Why are we talking l'Hôpital here anyway, I call that the definition of the derivative...

anon

You could also say $$x=-\zeta w\implies\frac{x^n+1}{x+\zeta}=\frac{1}{\zeta}\frac{w^n-1}{w-1}=\frac{w^{n-1}+\cdots+w^1+w^0}{\zeta}\xrightarrow{w\to1}\frac{n}{\zeta}$$ for $n$ odd and $\zeta$ any $n$th root of unity

robjohn

@tb well, in the case of a linear denominator, they are the same :-)

@anon that was the point of my comment that whether dividing and evaluating or limiting, we get the same thing.

anon

Ah, I wasn't paying that much attention.

t.b.

04:24

@robjohn but l'Hôpital isn't exactly exciting for analytic functions

robjohn

@tb it still applies if you approach along the same path.

J. M.

@tb chicken-egg... :D

t.b.

@JM the whole point and only difficulty of l'Hôpital's rule is that you can infer the existence of a limit that doesn't obviously exist by looking at the derivatives.

Jeff

right now, i'm just applying what you showed me to $\frac{1}{x^5+a^5}$. i don't think the $a$ changes much. I'm still using $\zeta=e^{\frac{2 \pi i}{5}}$. Should I use $\zeta=ae^{\frac{2 \pi i}{5}}$ instead? (Oh, also, I'm using $\omega$, cuz I like it better).

robjohn

going from $x^5+1$ to $x^5+a^5$ is a simple change of variables.

Benjamin Lim

04:39

Hey

does anyone know what it means to say

in a commutative ring that an ideal $I$ is a direct summand of $R$?

anon

R can be written as a direct sum, with one of the summands being I?

Mariano Suárez-Alvarez

that it is a direct summand of R when you view the latter as an R module

Benjamin Lim

@MarianoSuárezAlvarez What do you mean?

Mariano Suárez-Alvarez

what I wrote :)

a submodule $N\subseteq M$ of a module is a direct summand

when there is another $N'\subseteq M$ such that $N\oplus N'=M$

kahen

Direct sum of modules

In abstract algebra, the direct sum is a construction which combines several modules into a new, larger module. The result of the direct summation of modules is the "smallest general" module which contains the given modules as submodules. This is an example of a coproduct. Contrast with the direct product, which is the dual notion. The most familiar examples of this construction occur when considering vector spaces (modules over a field) and abelian groups (modules over the ring Z of integers). The construction may also be extended to cover Banach spaces and Hilbert spaces. Constructi...

Benjamin Lim

04:41

Ah ok@MarianoSuárezAlvarez That is clear

Mariano Suárez-Alvarez

now $R$ is a module over itseld, and $I$ being an ideal means precisely that it is a submodule of $R$

Benjamin Lim

yes

kahen

wow. chat's clever. it pastes from wikipedia

Benjamin Lim

so this means that there exists an ideal $J$ such that $I \oplus J = R$

Mariano Suárez-Alvarez

yes

as modules

for example, the ideal $\mathbb R\times0$ of the ring $\mathbb R\times\mathbb R$ is a summand

kahen

04:43

where I \oplus J = IJ = set of finite sums of terms of the form ij where i \in I and j\in J

Mariano Suárez-Alvarez

no

I \oplus J is the direct sum of modules

kahen

oh. directy product, right.

-y

Mariano Suárez-Alvarez

(what you wrote is the product as ideals)

(which has nothing to do with the direct product :) )

Benjamin Lim

exactly

@kahen

@MarianoSuárezAlvarez

@MarianoSuárezAlvarez It's ok I got it now

Jeff

@robjohn wait a minute, when $j=k$ the fractions reduces to $0/0$.

Mariano Suárez-Alvarez

04:56

hey

be careful!

if you divide by zero this chat thing crashes

Jeff

@MarianoSuárezAlvarez haha... fortunately i wrapped it in dollar signs

Mariano Suárez-Alvarez

money fixes everything

3

The Chaz

$everything$

anon

Unfortunately, you can't fix a lack of money with money, because you have none.

Mariano Suárez-Alvarez

well, you can fix lack of money with money, no?

you could, rather

t.b.

04:59

@TheChaz try \$everything\$ :)

The Chaz

Oh you :D

kahen

Is $..$ supposed to be go through mathjax here?

Or do you just put it in chat out of habit?

t.b.

@kahen yes

kahen

weird. it doesn't work for me. i just see LaTeX source and it doesn't get mathjax'ed

anon

Did you read the link sufficiently? :P

ie are you using the bookmark?

mixedmath

05:04

oh, that's really funny

I've been using greasemonkey to see the latex on chat for a long time

I didn't realize that we could do that

that's a really slick trick

J. M.

@mixedmath We have rob to thank...

mixedmath

that's really extraordinary

we should make and give him a hacker badge, indeed

robjohn

@Jeff yes, and that is why we either cancel the fractions or take the limit.

Jeff

@robjohn, you're awake! if $\displaystyle\zeta=e^{\frac{2 \pi i}{5}}$, how did you get $\displaystyle 5x^4=5 \zeta ^{4k}$.

anon

plugging in $x=-\zeta^k$?

robjohn

05:14

@Jeff if $x=-\zeta^k$... yeah, what anon said :-)

Jeff

oh. duh! :D

kahen

Hm... I took the link and converted it to a greasemonkey script, but it doesn't seem to work. $x$...

nope. no worky

mixedmath

oh, I would recommending using the bookmark directly

robjohn

@kahen The bookmark used to work automatically, but something happened to the Ajax "complete" event and now we have to invoke the bookmark periodically.

the LaTeX renders for a fraction of a second and then reverts to plain text

Jeff

@robjohn when i type latex in here, the interpretation (the math format, whatever you call it) shows up for a moment, then it un-translates back into text

robjohn

05:24

1 min ago, by robjohn

the LaTeX renders for a fraction of a second and then reverts to plain text

Jeff

i think i finally completed the partial frac decomp. the $a^5$ doesn't change too much. I get that $\displaystyle A_k=\frac{1}{5a}\omega^{k}$

@robjohn oops. i didn't read all of the script :D (i've been working on a complex partial fraction decomposition). sorry

Benjamin Lim

@mixedmath Interested in some commutative algebra?

mixedmath

ooh, commutative algebra

this should be good

Benjamin Lim

Now

I am trying to show that if the $A/R$, where $R$ is the nilradical is absolutely flat

then every prime ideal is maximal

my idea is this:

If $p$ is a prime ideal

Then we consider $A/p$

If I can show that this is absolutely flat and a local ring, then it is a field and hence $p$ is a maximal ideal.

Now I am guessing I have a surjective map $f : A/R \rightarrow A/p$

Mariano Suárez-Alvarez

don't guess

Benjamin Lim

05:34

hahahahahaha

Mariano Suárez-Alvarez

A/p is not going to be local

Benjamin Lim

hmmmmmmm

there goes my idea....

Mariano Suárez-Alvarez

suppose $A$ is a domain

Benjamin Lim

yes

Mariano Suárez-Alvarez

so that $p=0$ is a prime ideal

you are saying that all abs. flat domains are local

Benjamin Lim

05:36

I don't think that is true

Mariano Suárez-Alvarez

hm, my counter example is not commutative :)

you commutative people!

2

Benjamin Lim

@MarianoSuárezAlvarez Hmmm, there goes my approach now I have to think of another one.

mixedmath

hmm

Mariano Suárez-Alvarez

notice that you can suppose $R=0$

otherwise, you'd just mod out by $R$

Benjamin Lim

you mean I can suppose the nilradical is zero

Mariano Suárez-Alvarez

05:38

(because modding R out does not change the conclusion: R is contained in all primes)

Jonas M.

Hi!

J. M.

@MarianoSuárezAlvarez I don't believe people are commutative... you put them in a long line, and if you swap any two of them, at least one becomes upset.

Benjamin Lim

ok so you're claiming that if $A$ is absolutely flat then every prime idea is maximal, is that even true?

robjohn

@Jeff I get $\displaystyle\frac{1}{x^5+a^5}=\frac{1}{5a^4}\left( \frac{1}{x+a}+\frac{\zeta}{x+a\zeta}+\frac{\zeta^2} {x+a\zeta^2}+\frac{\zeta^3}{x+a\zeta^3}+\frac{\zeta^4}{x+a\zeta^4}\right)$

Mariano Suárez-Alvarez

I am claiming that if $A$ is abs flat and has trivial nilradical, then every prime is maximal

Jeff

05:40

I don't have $\displaystyle \frac 1{5a}$.

Benjamin Lim

Hmmm a boolean ring has every prime ideal maximal :D :D

Jeff

...no exponent on the $a$

robjohn

@Jeff plug in $x=0$, I think yours will fail.

hhh

How can you integrate?

$\int\sqrt{1-\cos(t)} dt$

Benjamin Lim

@MarianoSuárezAlvarez We have a surjective map from $A/R$ to $A/p$ because

consider the canonical projection $\pi : A \rightarrow A/p$

robjohn

05:44

@hhh $1-\cos(t)=2\sin^2(t/2)$

Benjamin Lim

then because $R \subset \ker \pi$ we have an $A$ - linear map $f : A/R \rightarrow A/p$

Mariano Suárez-Alvarez

yes

Benjamin Lim

where $f$ is defined by $f(\bar{a})$ = $\pi(a)$

But then because $\pi$ is surjective

the map $f$ is surjective too

hhh

@robjohn double angle -identity, well I wish I could remember those easily...yes now it is easy with that tip

Benjamin Lim

this establishes why we have a surjective map $f: A/R \rightarrow A/p$

hhh

05:48

@robjohn I could find something here to help me remember things like that...

Jeff

when $x=0$ i get $\displaystyle\frac 1{a^5}=\frac{1}{5a} \cdot \frac 5a$. Pretty clearly not right.

robjohn

@hhh another to be familiar with is $\cos(x)=2\cos^2(x/2)-1$

J. M.

Today's XKCD is a bit nice...

robjohn

@JM A friend sent me this link which I thought of as we were having gravatar problems with anon.

Jonas M.

I really hate to interrupt like this, but does maybe anyone have access to this paper? A friend of mine needs it for his bachelor thesis, and we're both not allowed to view it.

(I promise, it's the last time I'll ask ;)

Jeff

05:54

@robjohn what did you plug in for $x$ when trying to find the $A_k$s? I used $x=-a\zeta^k$. But that's not working out so that the numerators are $0$.

Benjamin Lim

@MarianoSuárezAlvarez

One thing I don't get is this flatness stuff

hhh

$\cos(x+y)=\cos x\cos y - \sin x\sin y$ and $\sin(x+y)=\sin x\cos y +\sin y\cos x$, how to deduce them?

Jeff

i think i need $x=-\zeta^k$ so that it becomes $-1$ and the numerator adds to $0$.

robjohn

@hhh look here: i.sstatic.net/L371s.png

hhh

(I have probably seen them earlier but forgot)

kahen

05:56

or you use $e^{i\theta} = \cos \theta + i \sin \theta$

robjohn

@kahen type up-arrow to get back a previous comment for editing.

@kahen :-)

hhh

@robjohn thanks, I saw that in the past in some book but never learnt to remember that perhaps I shuold start with assupmtion: two angles a and b, deduce the angles...and needequalities from thta...perhaps unit circle as a help?

robjohn

that works up to two minutes after the comment is entered.

kahen

$\cos(x+y) + i\sin(x+y) = e^{i(x+y)} = e^{ix}e^{iy} = (\cos x + i\sin x)(\cos y + i\sin y)$. The real and imaginary parts of each expression have to be equal, so you just need to expand the RHS

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