6:03 PM
@Clash figured it out yet?

@JM I've just noticed that that answer was downvoted. Why was that, I wonder?

Lenna is a beauty.

@robjohn Yeah, I don't understand why it was downvoted...

@QED not really, no... I've been staring at it and looking at the answer below, which says you should get 4 equations, which I do not get... I've been tryting to solve it like this, we have for example a(1, ...) + b(0, ...) -x(0, ...) - y(1, ...), and then I write something like a -y= 0 => a=y. Is this one of the four equations I should be trying to get?

@QED ...and now the image processing community adores her. :)

6:05 PM
@Clash, try to write it as a matrix multiplied by a vector [a,b,x,y]

@JM They tracked her down in Sweden and flew her to an image processing conference (I think to commemorate the 25th anniversary or something)

hah
I'd not heard that one Yoda

@QED done, I guess I do gauß now? let me see what I get

@yoda Oh yeah, I read that; she became a fundraiser or something. She had no idea how famous she became...

Well don't just eliminate without know what to do once you've done it
but yes, that's the idea

6:07 PM

I think this has just become one of my favorite titles of a math paper: A separable somewhat reflexive Banach space with nonseparable dual

@yoda BTW: good thing you decided to write an addendum. Congrats on your first "Nice Answer" here!
@tb "somewhat"? :)

@JM hah, thanks :) I should spend more time here... Math.se was the first site I joined (much before I went to SO).
Plenty to learn :)

weird, I got a=-y, b=0, x=y and y could be any value. I must have done something wrong. Here is the matrix that I have
1 0 0 -1
1 1 1 -2
0 3 2 -2
-1 1 -1 2

@Clash, when you do row operations you've got to also change the variables around
i.e. "a" might be replaced with "a-b"
"augmented matrix"

6:14 PM
Can I transpose the matrix and do row operations then?

$\left( \begin{array}{rrrr} 1 & 1 & 0 & -1 \\ 0 & 1 & 3 & 1 \\ 0 & -1 & -2 & 1 \\ 1 & 2 & 2 & -2 \end{array} \right)$
@Clash Transpose of what you got :-)

thanks @robjohn, im trying it right now
i pray for my life

If you have Mathematica, I put that matrix through NullSpace[Transpose[a]]

oh well, that did not work out either... i hate doing stuff without understand why it should work

@Clash what are you trying to do?
You want to do column ops on your matrix or row ops on mine

6:21 PM
I have just done row operations on your matrix and got y can be anything, x=-2y, b=5y and a=4y
and I should get x=1, y=1, a=1, b=0

Anyone interested in discussing some problems regarding extrema of three multivariable functions and lagrange multipliers? :)

$\left[ \begin{array}{rrrr} -1&0&1&1 \end{array} \right] \left[ \begin{array}{rrrr} 1 & 1 & 0 & -1 \\ 0 & 1 & 3 & 1 \\ 0 & -1 & -2 & 1 \\ 1 & 2 & 2 & -2 \end{array} \right] = \left[ \begin{array}{rrrr} 0&0&0&0 \end{array} \right]$
The null space of the matrix has 1 dimension
and $\{\left[ \begin{array}{rrrr} -1&0&1&1 \end{array} \right]\}$ is its basis

@robjohn. thanks! but why did row operations on that matrix fail me to get the desired results?

what row ops did you do? what were you trying to do with the row ops?

I got the same result as wolframalpha
http://www.wolframalpha.com/input/?i=NullSpace%5B%7B%7B1%2C1%2C0%2C-1%7D%2C%7B0%2C1%2C3%2C1%7D%2C%7B0%2C-1%2C-2%2C1%7D%2C%7B1%2C2%2C2%2C-2%7D%7D%5D

@Clash You are getting the right side null space there...
@Clash Try the Transpose of the matrix

transpose the matrix again? won't i have again
1 0 0 -1
1 1 1 -2
0 3 2 -2
-1 1 -1 2
I do get your results with the matrix above. (-a, 0, a, a)
however he has a different answer here: math.stackexchange.com/questions/25371/…: 1, 0, 1, 1

Can someone help me? I want to plot some graph and I don't have access to Mathematica. [Wolfram|Alpha times out.]

what graph do you want to plot?

I want to plot the region $6x^3y^2 - 4y^3 + 3x^3y - 4x^6 - x^3y^3 \geqslant 0$, where $0 \leqslant x, y \leqslant 1$.

6:39 PM
Eep, inequalities...

@JM This one is a beauty. [If I have done everything correctly...]

@QED Gnuplot can do such complicated things?

we'll see..

Hm, interesting.

6:41 PM
The "eep" wasn't for the inequality itself; I was thinking that W|A is the only web-based thing I know that does inequalities...

@JM Aw, that's right. I am overly dependent on WA.

"non-integer passed to boolean operator"
lets fix that

@QED - what does that mean? Which is the boolean operator?

@Clash are you still here?

yes @robjohn

6:44 PM
To record our row ops, we start with my matrix and an identity matrix:
$\left[ \begin{array}{rrrr} 1 & 1 & 0 & -1 \\ 0 & 1 & 3 & 1 \\ 0 & -1 & -2 & 1 \\ 1 & 2 & 2 & -2 \end{array} \right] \left| \left[ \begin{array}{rrrr} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array} \right] \right.$
Then we subtract the top row from the bottom row on both

wel..
it displays it but it closes immediately afterwards

$\left[ \begin{array}{rrrr} 1 & 1 & 0 & -1 \\ 0 & 1 & 3 & 1 \\ 0 & -1 & -2 & 1 \\ 0 & 1 & 2 & -1 \end{array} \right] \left| \left[ \begin{array}{rrrr} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ -1 & 0 & 0 & 1 \end{array} \right] \right.$

@Srivatsan Tsk, I tried out RegionPlot[6*x^3*y^2 - 4*y^3 + 3*x^3*y - 4*x^6 - x^3*y^3 > 0, {x, 0, 1}, {y, 0, 1}] on W|A and it chokes. Unfortunately I'm not on my Mathematica box...

@QED Um, =).
@JM Yes, it did choke =)

@Clash can you see my matrices?

6:46 PM
@Srivatsan that's the best I can manage at the moment:

@robjohn yes i can

is it supposed to look like that?

@tb Wow, that's nice.
At least for now. :-)

6:46 PM
...and that's what's called independent confirmation, people. :)

Thanks, both.
QED used gnuplot. What about you, @tb?

I used grapher.app That's the only tool I can handle a bit.

okay, so then we add the second row to the third, and subtract it from the fourth:
$\left[ \begin{array}{rrrr} 1 & 1 & 0 & -1 \\ 0 & 1 & 3 & 1 \\ 0 & 0 & 1 & 2 \\ 0 & 0 & -1 & -2 \end{array} \right] \left| \left[ \begin{array}{rrrr} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 1 & 1 & 0 \\ -1 & -1 & 0 & 1 \end{array} \right] \right.$

@tb Oh, ok. Thanks.
@QED As to whether it's supposed to look like that, perhaps. I do not have much of an idea =)
I am basically experimenting.

Then we add the third row to the fourth
$\left[ \begin{array}{rrrr} 1 & 1 & 0 & -1 \\ 0 & 1 & 3 & 1 \\ 0 & 0 & 1 & 2 \\ 0 & 0 & 0 & 0 \end{array} \right] \left| \left[ \begin{array}{rrrr} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 1 & 1 & 0 \\ -1 & 0 & 1 & 1 \end{array} \right] \right.$

6:51 PM
@robjohn In case you are interested, the shaded region in that plot is the feasible region for the (AM, GM, HM) triple for $n=3$, where I have normalised the AM to be $1$. The x-axis represents the GM, and the y-axis is the HM. @tb @JM

@Clash: The matrix on the right is what was multiplied on the left of the original matrix to get the matrix on the left
@Srivatsan Now to $n=4$ :-)

@robjohn I don't know how to do it. My approach does not extend.

@Srivatsan Looks like a banana. Or a slug. Or a bit of both.

Btw, thanks to @tb and @QED. =)

@Srivatsan Do you know what the bounding curves are?

6:54 PM
@robjohn thanks! alright, I also have this result, what does this mean? Why did you the last line of the right matrix as solution?

@robjohn No clue. It'll just be that huge expression set to zero, right?

Use the gradient to determine a unit vector normal to the graph of 4ycosx-y^2=3 at the point (pi,1)

@Clash because that row vector times the original matrix gives the $0$ vector

the gradient here would be <-4ysinx,4cosx-2y>, no?

robjohn, JM: I can tell you what I did. In case you want to play with it. [Also independent confirmation will be nice.]

6:55 PM
@Clash That is given by the row on the bottom of the left matrix
@Srivatsan sure :-)

@robjohn why? why don't the other lines matter?

@JM I personally thought it was a hammock at an angle, but I didn't want to follow your comment because I've learned to not use the words "banana" and "hammock" in close proximity =)

@Clash The other lines show that the rank of the original matrix is 3, but it is the last row that is in the null space of the original matrix

@yoda I can see why. :)

@robjohn ok so we have -1,0,1,1, why is this still different from the original answer? or are both valid? (ps: many thanks)

6:58 PM

For $n=3$, the idea is simple. So we have three numbers: $(A, G, H)$ and three unknowns $(a_1, a_2, a_3)$ such that $A$ is the AM of these numbers, $G$ is the GM, and $H$ is the HM. So we can write a cubic equation such that $a_1, a_2, a_3$ are the roots. This cubic turns out to be
$$x^3 - 3A x^2 + \frac{3G^3}{H} x - G^3 = 0.$$
Now write out the condition that this cubic has 3 real, nonnegative roots...

@yoda: ooh, gardening.SE. My wife would love that...
@Clash You still need to process that to get the intersection...

@robjohn ohhhhhh

@Matt Cool. It timed out for me. Thanks =)

@robjohn Hey, you should get her to join our site! We certainly could use more folks :)

6:59 PM
Well, I've capped today, and I'm sleepy. See you all later!

@Clash What we have is a linear combinations of the basis vectors of the two subspaces that equal the same thing (i.e. their difference is 0)

@JM Bye, JM.

@JM Darn you! I couldn't cap yesterday, and that was the closest I've come recently!
@JM and have a good night :-)

Bye J.M.

@robjohn Boy, yesterday, I should've got like at least 350 reps from upvotes alone. That plus, 60 from accepts. =)
Well, I just thought you might be interested in this fact... =)
Today I am back to normal though

7:03 PM
@Clash: So from that last line, we see that $\left[ \begin{array}{rrrr} -1 & 0 & 1 & 1 \end{array} \right] \left[ \begin{array}{rrrr} 1 & 1 & 0 & -1 \\ 0 & 1 & 3 & 1 \\ 0 & -1 & -2 & 1 \\ 1 & 2 & 2 & -2 \end{array} \right] = \left[ \begin{array}{rrrr} 0 & 0 & 0 & 0 \end{array} \right]$
@Srivatsan :-p :-p :-p

ok, what do we have to do to get the intersection?

@Clash: the top two rows and the bottom two rows are the bases for the subspaces. Can you see how to get a vector that belongs to both from that matrix equation?

By the way, @robjohn, the most interesting feature of the plot for me was that the HM cannot be too large or too small compared to the GM. Of course, there is the $HM \leqslant GM$ bound, but there's some reverse relationship between them as well... That is, $HM \geqslant$ some function of $GM$.

@yoda I will definitely suggest it to her.

@robjohn no i cant :(

7:08 PM
I need to draw a minimal diagram. Some V shaped thing.
Any way to do that without packages?

@robjohn maybe the upper two bases * [-1 0 1 1] (transposed)?

@Clash: do you see that the negative of the top row plus the sum of the bottom two rows = 0?

@robjohn Yes it is the OU operator.

@robjohn yes

@JonasTeuwen Am I correct in thinking that there is some boundary where it ceases to be elliptical?
@Clash: So if I take the sum of the bottom 2 rows (which is in one of the subspaces) and compare it to the top row (which is in the other subspace) what do I see?

7:12 PM
@Srivatsan Funny. For me it's quite fast.

@robjohn I didn't check that out.

@Matt My internet connection isn't too great, but I don't see why that would affect the computation time.

@robjohn yay! \o/

Help with the diagram?
Is it possible without external files, and or packages?

@robjohn it is the first vector inverted and thus also in the first subspace?

7:15 PM
The first row = the sum of the bottom two rows, so that vector is in the intersection of the subspaces.

oh I see thats why the first coefficient is 1, the 2nd 0, the 3rd 1 and the 4th 1
oh my god, finally it clicked
jesus, sorry for my slowness robjohn, many thanks for your help!

@Clash: $\left[ \begin{array}{rrrr} x & y & 0 & 0 \end{array} \right] \left[ \begin{array}{rrrr} 1 & 1 & 0 & -1 \\ 0 & 1 & 3 & 1 \\ 0 & -1 & -2 & 1 \\ 1 & 2 & 2 & -2 \end{array} \right]$
is a general vector in the first subspace
@Clash: and $\left[ \begin{array}{rrrr} 0 & 0 & u & v \end{array} \right] \left[ \begin{array}{rrrr} 1 & 1 & 0 & -1 \\ 0 & 1 & 3 & 1 \\ 0 & -1 & -2 & 1 \\ 1 & 2 & 2 & -2 \end{array} \right]$
is a general vector in the other subspace

x=1, y=0, u=1, v=1

@Clash: so if we can find $\left[ \begin{array}{rrrr} x & y & u & v \end{array} \right] \left[ \begin{array}{rrrr} 1 & 1 & 0 & -1 \\ 0 & 1 & 3 & 1 \\ 0 & -1 & -2 & 1 \\ 1 & 2 & 2 & -2 \end{array} \right] = \left[ \begin{array}{rrrr} 0 & 0 & 0 & 0 \end{array} \right]$
@Clash: we can break that into a sum of a vector in one subspace plus a vector in the other subspace
@Clash: negate one and they are the same.
@Clash That is the choice to get thet two vectors that are equal :-)

alright, many thanks!

7:21 PM
Does that clear it up?

In a latex document, is there any way to prevent it from indenting the first line of a new paragraph?

@Matt \noindent at the beginning of the line

@robjohn Thanks robjohn, and if I want to turn it off document-wide?

@Matt I think you can change the \parindent variable

@robjohn I'll try that. Ta!

7:25 PM
@Matt \setlength{\parindent}{0cm}

@Matt search for parindent on this page

@tb Nice, that did the trick, thanks!

@tb Hmm, I seem to recall that that messes up other stuff as well.
Or was it the line separation that is often used in conjunction with that?

@JonasTeuwen many people set simultaneously a greater value for \parskip and this messes up lists and enumerations and the like.
(after all, one shouldn't fiddle around with these things anyway)

Yes.
That was it.

7:29 PM
@tb That is why I just use \noindent :-)
Off to feed the dog. bbl

See you.

later, robjohn

Bye robjohn.

@robjohn You can feed the dog with Matt's dog.

@tb, cab we discuss the probability question now?

7:43 PM
Sure.

What is the question by the way? The way you understand it...

I'm not sure but I believe that the question might be asking why some people impose that the probability function is defined on a $\sigma$-algebra and $\sigma$-additive (Kolmogorov's axioms) and some don't require the $\sigma$.

@tb And I proved that if $P$ is countably subadditive, then $\sigma$-additivity comes for free. Yes?

Provided that you have finite additivity, yes.

[Just making sure I am not making some implicit assumption that I don't see.]

7:47 PM
Dog fed...

So there are probability functions that are not countably subadditive?

Wooo! Time to go out for burgers! See you guys later.

Enjoy!

Bye, Asaf

@AsafKaragila I'll be doing the same soon!

7:48 PM
@Srivatsan Yes. A standard example would be a Banach limit which can be interpreted as a finitely additive measure on $\mathbb{N}$.

@tb I see.

In fact, the dual space of $\ell^\infty$ can be interpreted as the space of (signed) finitely additive measures on $\mathbb{N}$. Under this identification the $\sigma$-additive measures are precisely the elements of the dual space that come from $\ell^1$ and the probabilities are the non-negative sequences summing to $1$.

Bye Asaf.

Wait, I am still hung up on Banach limits.
This is clearly a generalisation of limits, yes. But is this some specific example of some more general concept?
Like, is there a Banach limit for general topological spaces?

Well, yes, there is: Tao can explain this better than me. Also, Martin has some nice related answers.

8:00 PM
@tb Ok. =) I have seen this post, but I didn't understand much of it.
Anyway...
@tb How exactly do I interpret the Banach limit as a finitely additive measure on $\mathbb N$?

If you have disjoint sets $A, B \subset \mathbb{N}$ and $L$ is our Banach limit, then by linearity $L([A \cup B]) = L([A]+[B]) = L([A]) + L([B])$ and by normalization $L([\mathbb{N}]) = 1$.
..and furthermore $L \geq 0$, of course.

Oh, my problem was a bit more basic, but I figured out based on your comment. See if I make sense: Given a set $A \subseteq \mathbb N$, the measure of $A$ is $L(\text{characteristic sequence of } A)$.

exactly.
Now if $A$ is a one point set $L([A]) = 0$, and this shows that $L$ isn't $\sigma$-additive.

Right. Ok, makes sense.
One tangential question: if $A$ has a natural density, then $L(A)$ will be that natural density, right?
I think I can verify this easily for one simple case. When $A = k \mathbb N$, then I can verify that $L(A) = \frac{1}{k}$.
By natural density, I mean $\lim_{n \to \infty} \frac{|A \cap \{1, \ldots, n\}|}{n}$. I guess this is standard terminology, but not sure how standard.

I'm not quite sure if it is true in general, but you can certainly arrange that it holds.
(natural density certainly is very standard terminology)

8:14 PM
@tb Ok, that is certainly good.
Ok, back to your original comment.
@tb This makes a lot of sense now.

What was my original comment? :)

@tb The one my next sentence (the line below) points to... =)
The one about the dual space of $\ell^{\infty}$ and $\ell^1$...

I see.
I don't know what exactly the reasons are, but there are people saying that it is more natural to use finitely additive probabilities instead of $\sigma$-additive ones. Probably some sort of Ockham's razor vs. utilitarian argument. I don't know, I never really thought about it. Anyway, my point was that I think the question might have been about this.

I thought that as well. A friend of mine posted some answer using Exts and stuff I'm not so comfortable with and then took it down. Not sure why

@tb Yes, I see that now. I just wish the OP was clearer though =).

8:23 PM
@DylanMoreland Oh, he accidentally proved the wrong direction.

It looked good on the face of it.
Aha.
That's a problem.

And, thanks for the patient explanation, @tb.

The argument he gave was perfectly fine. Although I wouldn't have mentioned Ext.
@Srivatsan You're welcome.
(Always glad when someone listens :))

@tb Not sure what that means. Who doesn't listen? =)

That's a long list.
@tb I wish the OP for that question had shown a few more attempts. There's certainly stuff you can try.

8:34 PM
Aw, I see. Natural density of a set is just the "Cesàro limit" of its characteristic sequence. Interesting.

@Srivatsan Exactly. By the way at the end of my answer here I give an outline for the easiest construction of a Banach limit I know of.

It does seem like something homological would be much cleaner.

It must be a well known property in homological algebra. What I'm really curious about is: Which injective modules $I$ over a ring have the property that $\operatorname{Hom}{({-},I)}$ reflects exactness?

Hm, it's an exercise in Lang. I don't know what that tells me.
Another reason why I didn't think about the problem earlier is that I figured Matt would come along and kill it, but that hasn't happened yet.

Well, Matt wasn't around for four hours, and the question was posted about an hour ago. Tetrapharmakon claims in a comment that this article settles the problem. I don't understand exactly why, yet.

8:47 PM
Ah, right, one can check. I always thought that was a funny feature.
I'm surprised that you can't turn it off.

Yeah, I'd prefer to be able to turn that off.

I originally started visiting this site just so I could figure out if he was near his computer so I could bug him about math or a grant letter that was due in an hour.
I didn't think I'd use the site proper for anything. But questions that you know the answers to are sort of addictive in a way.

That's too true...

Shoot. I was so confused reading the conversation because I didn't realise Matt refers to ME. :/
This is not the first time, that too =)

Ah, sorry.

8:53 PM
As a rule of thumb: Dylan is around and Matt's hat is dimmed, so it must be Matt E we're talking about.

@DylanMoreland Of course, no problem. It's just so amusing and obvious once I realise.
@tb - what's his MO handle? I cannot find him there...

Aw, that's a catch. I don't think I could've guessed that anyway...
If it matters: He is really close to posting 400 answers in MO.

@DylanMoreland I don't get it.

Yeah, strange.

9:06 PM
Given $x \in \mathbb \ell^\infty$ and $y \in \ell^1$ (seen as the dual space of $\ell^\infty$), can you show how $y$ acts on $x$?

$x \mapsto \langle x, y \rangle = \sum_{n=1}^{\infty} x_n y_n$.

@tb Ok, checks out. Thanks.
I'm trying to verify your claim that no $y \in \ell^1$ will satisfy the OP's requirements. [Check.]

Tell me if you want a hint :)

It's clear, thanks :)

9:35 PM
I have a clarification about the proof: a positive linear functional maps positive elements to a nonnegative real number, yes?
What are the positive elements of $\ell^{\infty}$?

Yes // sequences with non-negative entries.

Alright. Thanks.
Neat proof. =)
But it seems kind of coincidental that the all-ones sequences is at a positive distance from $U$ [where $U$ is the space of sequences of the form $Tx-x$ with $x \in \ell^\infty$].

I assume QiL is Qing Liu
Guy is an algebraic geometry machine
his book is great too

@Srivatsan I like it, too. I think it's pretty close to Banach's original argument, but I'd need to check. Two more related threads: here and here
@DylanMoreland Zhen has the same suspicion. See also here and here for some confirmation.
@Srivatsan How would you get close to that sequence by sequences of the form $Tx-x$?

@tb Well, I could "proof" it, but still it feels a bit weird. [I don't have any issue with the other steps.] I guess digesting this thing will take some time, that's all...

9:49 PM
@Srivatsan My intuition is very silly: let's assume $x_1 = 1$ then in order for $Tx - x$ to be close to $(1,1,1,\ldots)$ we must have $x_2 \approx -2$ thus $x_3 \approx 3$ and so on, so the sequence must be unbounded.

@tb Yes, is it the case that $U$ is precisely the space of sequences with bounded partial sums?

Ick, how did this get burped back up? I know that there was some tag editing and a title change a couple of days ago, but why is it still in the current main page?

@robjohn Mike Jones

@tb Oh boy.

Ah, I missed his edit 18 min ago.

9:52 PM
@Srivatsan I think so, yes.
(but I'm not sure)

conundrum, paradox, enigma, thought experiement, fallaciously
too many big words. =)
Well, $Tx-x = (x_2 - x_1, x_3 - x_2, \ldots, x_{n+1} - x_{n}, \ldots)$. So given a sequence $y$, we can "solve" for the right $x$ such that $Tx -x = y$. The solution is $x = x_1(1, 1, 1, \ldots) + (0, y_1, y_1+y_2, y_1+y_2+y_3, \ldots)$. We want this $x$ to be in $\ell^{\infty}$.

@Srivatsan Which is the dual to sequences of bounded variation (I think)

@robjohn In what sense? Are you using "dual" in a technical sense?