3:12 AM
In this paper, the author gives three different examples of fractals which show the concept of fractal compression. Are there any other fractals which are easy to illustrate the concept like the three in the paper?
I know of the von Koch curve
What are some others?

5 hours later…
8:01 AM
@leslietownes mm pizza

8 hours later…
3:55 PM
@robjohn I understood it, thank you

Is @leslie baking us fresh pizza?

I’ll get the grill from Texas

4:14 PM
I am suspicious of pizza made by americans

And rightly so. Some of it is beyond abominable.
But I make my own dough and don't make it goopy with sauce and cheese.

I've a question regarding complex numbers. Our professor said: consider $\bar z$ as a variable and compute $\frac{\partial z}{\partial \bar z}$ where $\bar z$ is the conjugate. What does that mean? Is the derivative simply $0$?

@TedShifrin I grow my own wheat

@Sine This is always a confusing issue. It makes sense when you get to a fancier context (with complexified tangent and cotangent bundles). In the meantime, you have the definition $$\frac{\partial}{\partial\bar z} = \frac12\left(\frac{\partial}{\partial x}+i\frac{\partial}{\partial y}\right).$$ So what do you get when you apply it to $z=x+iy$?
You want to think about expanding a (real) analytic function of $x$ and $y$ as a function of $z$ and $\bar z$. Then holomorphicity is equivalent to having no $\bar z$ appear.

4:23 PM
Can anyone help me on how to determine the error term in the Numerical integration of the Numerical integration?

@ted I can't find this definition in my notes, I'll try to obtain it. Give me some time :)

Oh, your professor must have defined something. Partial derivative with respect to a variable doesn't even make sense. You need to specify what the independent variables are.

we started this discussion from the differential and we compute the increment along the axis

That makes sense if you're talking about $f(u,v)$, say, and write $df=\frac{\partial f}{\partial u}du + \frac{\partial f}{\partial v}dv$. As I said, you have a complete set of independent variables.
So you have to make sense out of saying the $z,\bar z$ form a coordinate system with independent variables. I'm not sure your professor has done so.

in general we are talking about $\Bbb R$ and $\Bbb C$- linear function

4:32 PM
So that is the explanation of the Cauchy-Riemann equations. When is a linear map $\Bbb R^2\to\Bbb R^2$ actually a $\Bbb C$-linear map. Separate issue.

we wrote an $\Bbb R$-linear function as $f(z)=az+b\bar z$ and compute $\frac{\partial f}{\partial z}$ and $\frac{\partial f}{\partial \bar z}$ and he left the exercise I wrote before

It makes no sense to write $f(z)$. You have to write $f(z,\bar z)$.

I know, but I must stick with that notation because it's used in class

Ask him how you differentiate $f(x) = e^x + \sin y$ with respect to $x$ and $y$.

but maybe I understood the point: we found that $b=\frac 12 (f_x+if_y)$ and $f_{\bar z}=b$, so using these info I can see that I find the definition you sent before
makes more sense now
so basically $\frac{\partial z}{\partial \bar z}=0$ right?

4:40 PM
Yes.
Indeed, the Cauchy-Riemann equations will tell you that a $C^1$ function $f$ is holomorphic (complex-differentiable) if and only if $\partial f/\partial\bar z = 0$.

thank you. I like complex analysis but the professor is always in a rush and doesn't want to teach
C-R equations are the topic of the next lesson

Yeah, it sounds like he's not exactly explaining things carefully. I used to be a complex differential geometer, so I taught this material a number of times. :)
Ah, OK.

moreover, he's an algebraist :(

So much for all the beautiful geometry and topology in complex analysis :D

4:45 PM
I'm sure my former colleagues were not happy when I taught abstract algebra (and even wrote a book) in a way that did not fit their formal taste in algebra.

4:59 PM
Is there a definition for when a sequence of functions diverges to infinity?

What do you mean by that, @sunny?

Hmm, well, that $f_n(x)$ diverges to infinity for some $x$ as $n\to\infty$, does that make sense?

Yes, but you have a fixed $x$. So the functions are only going to $\infty$ at that particular $x$.

Yes.

g'morning, @robjohn
@sunny When you say "the sequence of functions diverges to infinity," I expect that to happen at every $x$, not necessarily in a uniform way.

5:05 PM
Ok.

There is a section titled as "Determination of the Error Term" in the first pic
But I thought they just determined the error term in the last line of that section preceeding it as they wrote something like $R_n=...$
This seems really very weird
@SoumikMukherjee Can you please take a look at this ?

@AlessandroCodenotti Yes. Most American pizza is a far cry from Italian pizza, much like most Chinese restaurants. But there are a handful good ones. @冥王Hades When I used to live in North Texas I drove quite far to Cavalli Pizza, the first to earn certification from the Associazione Verace Pizza Napoletana.

5:20 PM
@F.White I finally have a solution (to my regret I didn't come up with the idea myself, even though it was a rather sensible thing to try), and I've written it out as an answer under my post: math.stackexchange.com/questions/4778846/…

@ThomasFinley They are using first mean value theorem for the integrals to write things in a better way, what is weird in that?

Just a small reminder: if we set $R=\mathbb Z[\sqrt{-3}]$ and $S=\mathbb Z[1/2+1/2\sqrt{-3}]$ then I wanted to prove that for any $R$-ideal $I$ we have $I=Ry_0$ or $I=Sy_0$ for some $y_0\in R$. Now, a way to guess this generator was to consider $y_0\neq 0$ with minimal norm, which is possible, since $R$ is discrete
That was such a sensible idea!
I have to get used to these kinds of arguments...

@Sha Yes, that would be ideal. :D

loooool Ted x'D

@TedShifrin hey, there. Been a busy morning. Then I have PT this afternoon.

5:26 PM
Ah, mine is tomorrow, but I have to do my home exercises later.

Yeah, I was not able to do all of my exercises while on vacation. I did a lot of hiking, but I don't think that is a replacement.

Well, good to get healthful exercise, too :)

Yes. It's great hiking with my dog.

@SoumikMukherjee Did that make things better? I mean well...ok, but where did the $\eta$ come from? Sorry, if I sound stupid...

$\eta$ is a point in $(a,b)$

5:39 PM
@SoumikMukherjee wasn't it $zhi$ before?
I thought $\zhi$ was a constant.

@ThomasFinley yes, in many occasions, it gives a bound
@ThomasFinley \xi

@SoumikMukherjee agh...I am sorry :?)

@SoumikMukherjee but it was $xi$ before and then they changed to $\eta$. But why? Is $\xi=\eta$

5:56 PM
@ThomasFinley I think so, not sure why they are using different notations there
How to show that the long line is not bihomogeneous?
Intuitively, it is clear to me, but how to prove it?

Why is it intuitively clear to you?

6:10 PM
The part that it is homogeneous is clear to me, we are sliding the point $a$ to the place of the point $b$. But to make it bihomogeneous, we need to slide in the opposite direction simultaneously, which we can't do

How is that sliding given by a homeomorphism of the space? How would you show $\Bbb R$ is bihomogeneous? Certainly not by sliding. What goes wrong if you try to adapt that to the long line?

6:53 PM
@TedShifrin For $\Bbb{R}$ we can flip it by the middle point of $a$ and $b$
for the long line, we can't do that as one side is paracompact and other side is not

Well, the long line typically has a smallest element, too.

So that would mess you up with trying to "flip" :)
I still am not sure what the homeomorphism is that takes $a$ to $b$ in the long line.

Sorry, I forgot to mention that we are considering the long line without the initial point

Right, sure.

7:12 PM
@TedShifrin I think the long line as a train with $\omega_1$ number of compartments and each compartment is $[0,1)$, so first we slide the the compartment of $a$ to the compartment of $b$, then shrink/stretch to match the place. Is this wrong?

7:26 PM
If I take $M\in M_d(\Bbb{C})$ and define $\|M\|^2=\sum_{i,j}|X_{ij}|^2$, then how can I show that for every orthonormal basis $(u_i)_{1\leq i\leq d}$ of $\Bbb{C}^d$ we have $\|M\|^2=\sum_{i,j}|\langle Xu_i,u_j\rangle|^2$?

We get the hint to first prove that for all unitary matrices $A,B$ $\|AXB\|=\|X\|$ which was pretty easy. But now I don't see how to deduce the first part from this hint.
I know that $\langle u_i,u_j\rangle=1_{i=j}$
I would be done if I could show $|\langle Xu_i,u_j\rangle|=|X_{ij}|$

7:46 PM
@SoumikMukherjee You have to say what to do with the compartments to the left and right, too, but yes, I think this will work.
@user123234 That's clearly not correct.
Try thinking about the matrix $P$ whose column vectors are the $u_i$ (written in terms of the standard basis).

7:58 PM
ah and we want to show $P$ is unitary?
to use the hint?

Sounds promising :)

@TedShifrin Do you know if there's any way of proving that a finitely generated ideal $I$ of a commutative ring such that $I^2 = I$ is generated by an idempotent, without using determinants?

You mistake me for an algebraist :) I don't even see how determinants are arising.

16

What follows is partially excerpted from my Historia Matematica post of 2005.01.04 on this topic, in reply to questions by Colin McLarty and Martin Davis. The Lemma below is from Gilmer's 1970 Monthly Classroom Note[1]. Gilmer mentions that it is also presented on p. 58 of his textbook[2].$\:$ LE...

this lemma

8:07 PM
My version of Nakayama's lemma is for the Jacobson radical
unless I can use it somehow... hmm
alright, I'll try using Nakayama's lemma somehow. I wouldn't think of it myself

8:26 PM
I've a question on the complexification of a real vector space. If we define the norm on the complexified vector space, to prove the homogeneity property, the scalar should be in $\Bbb R$ or in $\Bbb C$?

The latter.
It’s supposed to be a complex vector space, right?

yes
but I can't prove the property

What is the correct statement?

the norm is $\|u+iv\|=\max_{x\in [-\pi,\pi]}\|u\cos x -v\sin x\|_{\Bbb R}$
on the right there's basically the norm on the "original" vector space

OK, interesting. So what are you trying to prove?

8:33 PM
@TedShifrin I showed $P$ is unitary so I know $\|PMP\|=\|M\|$. But is there a fast way to show that $(PMP)_{ij}=|\langle Xu_i,u_j\rangle|$? Or do we need to compute it by hand

that's actuallly a norm. I've proven all properties except the homogeneity. If the scalar is real, then it's trivial, but I'm finding problems if the scalar is complex

No, if you understand how to write the inner product with the matrices, it’s easy. The absolute value should not be there.
Check for $e^{i\theta}$, Sine.

Sorry, what do you mean?

That’s a complex scalar.

ok, let me try

8:42 PM
@TedShifrin So you mean rewrite $\|PMP|=\langle PMP, PMP\rangle$

No, no. What is the $i$th column of $MP$?

I get something like $\max \|\cos (x+\theta)u-\sin(x+\theta)v \|$

And so?

it does the job

So you’re done.

8:57 PM
yes, thank you

Yup.

@TedShifrin it is $c_{ij}=\sum_{k=1}^d m_{ik}u_{kj}$ for all $1\leq j\leq d$

Agh.
Be conceptual.

sorry I edited it

If you multiply $AB$, how do you tell a beginning student what its $i$th column is — as a vector?
Go back to how matrix multiplication is defined.

9:03 PM
@Jakobian Prove this version of Nakayama: If $I$ is an ideal and $M$ a f.g. module s.t. $IM=M$ there is an $i\in I$ such that $im=m$ for all $m\in M$

Oh, of course I should have said to ask Thor and Lukas.

Sorry I don't get it, I mean we have defied the i-th column as the vector $(c_{i1},...,c_{id})$

Draw two matrices on your paper and use your fingers to find the $i$th column of their product.

so one could also write it as $m_i\cdot P$ where $m_i$ denotes the i-th row.

It's not the geometric version of Nakayama (which is arguably the coolest in terms of concept), but it is the one I find easiest to remember and apply

9:10 PM
Sorry if $i$ is confusing. I chose it instead of $j$ because of the formula you want to prove. I asked for column, not row.

Ah so you take $M\cdot u_j=\langle M, u_j\rangle$

No, no, don’t jump yet!
What’s the $i$th column of $MP$? What vector?

Wouldn't it be $M\cdot u_j$?

Why the $\cdot$? And why $j$?

because you wrote I should use $j$ instead of $i$ but we can also replace it. So I mean I multiply each row of M with the j'th column

9:16 PM
No, I didn’t write that! The answer is that we take $Mu_i$. Look familiar?

aha sure, I mean I only wanted to write a product but it is not the dot product but yes I agree

Now, how do you take the inner product of that with $u_j$ using $P$ again?

would it be $u_jMu_i$

Not quite. How do you make sense of that as a product?

You mean if $u_j$ is written as a column I should need the transpose on $u_j$

9:49 PM
Yes, and probably a conjugate, too.

@TedShifrin i've seen this advice (and/or the need for it) pop up so many times recently that i wonder if anybody does this anymore.

10:03 PM
I sure did every time I taught. In my books, there are no fingers, but there is shading :)
In grad school, when I was teaching MATH 41 my 4th year, I messed up my arm and shoulder. I needed to ask for a guest gesturer from among my students.

I’ve messed up mine without teaching MATH 41

Teaching didn’t mess it up. An iron chair did.

2 hours later…
11:49 PM
@Thorgott can I prove this using the version with Jacobson radical?