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12:18 AM
I iterated it some more. It looks very wacky
 
 
2 hours later…
2:45 AM
what demonry is this
 
Alessandro is the demon!
How was Munchkin’s reaction to the vaccine?
 
@leslietownes Q: What? A: Demonry is this.
 
3:02 AM
ted: no side effects whatsoever. apparently, not too uncommon with kids her age.
 
Interesting.
 
she didn't even feel the shot, apparently. while she complained to no end of her flu shot. we're still hearing about it months later.
 
Oh no I messed up
 
shake your phone a few times until it settles the right way
 
There we go
In any case: neat fact that those intersection points that look like they lie on the circle, do, in fact, lie on the circle
$(0,\frac12)$, $(\frac15,\frac25)$, $(\frac13,\frac13)$, and $(\frac12,\frac14)$ all lie on a straight line
(namely, $x+2y=1$)
 
3:36 AM
Hello :-)
 
4:01 AM
some dan brown novel thing going on???
 
Let $\{x\}$ refer to the fractional part of $x$. Let $A$ be the operator defined by $Af(x)=\int_0^1\{\frac xt\}f(t)dt$. Is $A$ injective on $L^2$?
PSYCH that's equivalent to the Riemann hypothesis
 
Given a SVD $A = U \Sigma V^T$, for the column vectors in $U$ - does it matter what order they are in? Does the order affect the order of the columns in $V$ in any way?
 
They change the order of the diagonal elements of $\Sigma$
You have to reorder the columns of $V$ the same way
 
$A= \sum_k \sigma_k u_k v_k^T$.
 
You're missing an index surely
 
4:08 AM
starting square
 
$A_{ij}=\sum_k\sigma_k u_{ik}v_{jk}$ perhaps?
(missing two indices)
 
no. look at the bottom of the first paragraph in en.wikipedia.org/wiki/Singular_value_decomposition
 
copper, did you watch the movies KGF and KGF2 ?
 
sorry I mean the columns in $V$ that's in the kernel of $A$ - does their ordering affect the ordering of columns in $U$?
 
Oh, I see
How is a matrix the sum of scalars...?
Oh duh
$uv^T$, not $u^Tv$
Right, yes, OK
And indeed $(u_kv_k^T)_{ij}=u_{ik}v_{jk}$
 
4:19 AM
If $A$ does not have full rank, then there will be columns in $V$ that's in $\ker A$ and some columns in $U$ maybe be in kernel of $A^T$, right? Does the ordering of these vectors in $U$ affect the ones in $V$? I don't think so, right?
 
It's still weird to me that $\frac1p+\frac1q=1$ is equivalent to $(p-1)(q-1)=1$
@Steve Sounds plausible to me
 
i was just gonna tell u no way, that's wrong, but its right lol @AkivaWeinberger
 
It's probably cleanest to think of a generic example
like $p=\frac{11}8$, $q=\frac{11}3$ or something
 
@Koro not yet :-) just finished lost in space last night.
 
And sure enough, $\frac8{11}+\frac3{11}=1$ and $\frac38\frac83=1$
 
4:26 AM
well the p-1,q-1 one says one of them, say p-1≤1 and the other ≥1.
the 1/p and 1/q one is saying, 1/p≤1/2, and 1/q≥1/2 or the other way
 
Indeed
Fun fact: an $m\times n$ square has area less than perimeter iff $(m-2)(n-2)<4$
 
lol
 
Akiva, you need to get out more
 
he's out in the field gathering lyrics for his next song
 
iff $\frac1m+\frac1n>\frac12$
 
4:28 AM
"if and only if"
 
I didn't actually make that last connection until just now
The fact that the area equals perimeter iff $(m-2)(n-2)=4$ actually has a really clean geometric interpretation
I wonder if the fact that that's equivalent to $\frac1m+\frac1n=\frac12$ also has a clean geometric interpretation
 
next up is akiva and the corollaries...
 
:-)
 
akiva conjecture
:)
 
Unrelatedly, if you place $m$ $n$-gons around a vertex, that forms a polyhedron iff... $(m-2)(n-2)<4$ iff $\frac1m+\frac1n<\frac12$
If they're equalities, it's a Euclidean tiling, and if they're $>$, it's a hyperbolic tiling
As far as I know it's a coincidence that the equations end up the same
 
4:31 AM
surely that's got to be related
 
but this tells us that, because the Euclidean tilings are square (4,4), triangle (6,3), and hexagon (3,6), those are the three integer-sidelength rectangles with area equal to perimeter
and the five (integer sidelength) rectangles with area less than perimeter correspond to the Platonic solids
Wait
 
there's no time
 
They don't correspond to polyhedra if $m$ or $n$ is 2
7 mins ago, by Akiva Weinberger
Unrelatedly, if you place $m$ $n$-gons around a vertex, that forms a polyhedron iff... $(m-2)(n-2)<4$ iff $\frac1m+\frac1n<\frac12$
*$\frac1m+\frac1n>\frac12$, I meant
 
4:51 AM
@copper.hat :)
 
@Koro $\overset{👀}{\smile}$
 
haha, this is very nice. :)
 
$\overset{\Large👀}{O}$
 
ü
(the umlaut is eyes)
My face when someone says they don't like umlauts: Ö
 
5:03 AM
What's the reaon when you ask a question but receive no answer?
 
@LibertarianFeudalistBot there could be many. maybe your question is messy and no one wants to read it. Or if your question is written well, maybe your problem is just hard. Or maybe its just your luck that no one clicked on your question. It happens.
There could be other reasons I'm sure
you can try to see if this list helps. You can also post your question here and someone might click
 
I didn't receive downvote. I posted my question last night but only 13 views so far.
https://math.stackexchange.com/questions/4481654/connecting-morphism-of-exact-triangle-of-cohomology

This was my question. If anybody is interested please
 
'only 13 views' could be a big part of it. give it time
 
Hi leslie good morning
So i am going to write the question from last night
 
you mean the question from this morning :)
don't tell me you were lazy and slept all day
 
5:12 AM
Ah yes. iterate the reason why intergalactic human civilization will never work
anyways
 
what clock would the iss use if it flew around the earth once every two seconds
 
I would check if thats even possible..
 
same clock everyone uses, calvin. pacific time
which is pacific daylight time, if you mean right now
 
if its really daylight time right now, then do tell, where the heck is the daylight?
 
It's actually kinda odd that the columns of a square matrix are orthonormal iff the rows are
Like, it's not clear that $a^2+b^2=1$, $c^2+d^2=1$, $ac+bd=0$ should imply $a^2+c^2=1$, $b^2+d^2=1$, $ab+cd=0$
 
5:18 AM
that's AA^T = I iff A^T A = I right. proof is clean with matrix multiplication but would be weird at the level of rows/columns
 
Is there a pure algebraic (but not linear algebraic / matrix algebraic) way to see this?
Well in this case you can work out $a=d$ and $b=-c$ or $a=-d$ and $b=c$ I guess
but with a 3x3 it becomes much worse
 
Given the gram matrix
$ \begin{bmatrix}
1&0\\
0&-1
\end{bmatrix}$
which is nothing but the one dimensional minkowski metrik.
the question states, find bases in $\mathbb{R^2} \times \mathbb{R^2}$
And considering the "unknown" bilinearform $b$ which belongs to this matrix. The question states, find basis $(v_1,v_2): \forall \: i \; b(v_i,v_i) > 0 ,b(v_i,v_i)<0, b(v_i,v_i)=0 $
Thus we conclude, that the definitness of a bilinear can not be examined n a basis.

so this is what i meant last night, especially, the last comment.
 
mad: i guess the lesson i would take from this is that you can't detect definiteness from diagonal entries of a matrix
i.e. perhaps you need the whole matrix in general and can't just look down the diagonal
 
its been so long since i thought about why dct is true that it seems like magic again...
 
it is the baire category theorem that is always a bit of magic to me
found and answered a convex psq. expecting lots of singular serial voter reversals and downvotes.
 
5:25 AM
Leslie, do you have good things to say to remove my confusion, i am considering opening a question about this. Since i am still deeply bothered by this, and i have no grasp over it. and i am not sure i understand what you mean.. why is a diagonal matrix anydifferent from other matrix to another base, they are all isomorph.
IE i cant write the matrix back in none diagonal form.. i dont understand......
 
the matrix of a bilinear form is generally not a diagonal matrix
that's one bit of confusion that needs to be cleared up
 
so. Is this the concept of definitness now related to basis choice?!
Is it generally not and there are exceptions? (like that)
 
nooooooooooooooooooooooooooooooooo
 
the other bit is, knowing what b(v_i,v_i) is for a basis {v_1, v_2} isn't enough to tell you the bilinear form
these questions are so ill-formed i can't answer them
 
Ill formed due to ill understanding lol
 
5:28 AM
what's your definition of definiteness of a form, it might help to start there
 
well you know is it not defined the same everywhere. Say bigger than zero for all smaller yadda yadda?
 
well that's not going to work at all, you need to pick a definition
if there's no starting point, there's no explanation except, go around and read all the various sources together... that's time consuming and we can't really help
 
Also quick interruption, the determenant is none basis dependent right? but what happens if i choose a base with reversed orientation, will i get a minus sign in the determinant?
 
maybe an analogy would help. copper's analogy was about the determinant. it's a function of matrix entries. but it's also invariant under a change of basis.
be careful there, "orientation" itself is often defined in terms of determinants.
 
-.-
 
5:32 AM
if {v_1, v_2} is an ordered basis of R^2, and M is the matrix of a linear transformation on R^2 with respect to that basis, and N is the matrix of the same linear transformation with respect to the basis {v_2, v_1}, then det(M) = det(N)
 
Yea i know because for me, detirminant means VOLUME meassurement, it matters not how i look at it! its the same!
But i dont understand this shitty term of definitness
 
well stepping away from 'volume' measurement, the determinant it's also a function of matrix entries, right? and matrix entries generally do depend on the choice of basis
but the determinant doesn't
it's maybe surprising that it doesn't, but it doesn't
 
i never looked at math as just numbers, i awlays tried to understand whats behind. is there meaning to definitness
People just didnt make this up for the lols
 
well, it relates to the eigenvalues of the matrix of the form (which again, the matrix may depend on the basis, but the eigenvalues do not)
what that 'means' for whatever your application is may vary depending on what you're interested in
but in terms of just what the concept is mathematically, it can be characterized in terms of eigenvalues
and eigenvalues are again something that can be calculated in a way that depends on matrix entries, but also don't change under a change of basis
 
Alright so if i understand you, returning to our example
Are you saying, that the bilinear form to the matrix i given, has an intrinsic properity of definitness (or none for that matter) that is given the actual naked explicit form of b, can be calculated, and this is for all basis the same?
 
5:39 AM
yes. what might not be the same for all bases are individual entries of the matrix used to represent the form in that basis
those can vary, and perhaps pointing in that direction is even the purpose of the exercise
 
Okay so the REVERSE engineering the defeitness from the GRAM matrix is not possible, however, if we have explicit form, then the definitness is unique and stays so nomatter change of vision (basis )
Pls say yes
:D
 
slow down
the numbers b(v_1,v_1) and b(v_2,v_2) are the diagonal entries of the matrix of b with respect to {v_1,v_2}. you might call the full matrix the gram matrix with respect to {v_1, v_2}
those numbers are not the entire gram matrix
i might regret this, but stepping one level of generality away, the point of the exercise is that the diagonal entries of the gram matrix of b are not, in general, enough to tell you whether b is positive definite, or negative definite, or some other mix
 
@TedShifrin I lost 170.
 
Leslie, is it not that the digonalized matrix is equal to the undiagonalized * some transformation matrices, you mean this operation messes with our result?
 
you absolutely can tell what is going on with b's definiteness if you have b(v_1,v_1), b(v_2,v_1), b(v_1,v_2), and b(v_2,v_2) - i.e. the full gram matrix. but that's not what this exercise is focused on. it's looking only at the diagonal entries
mad: you generally can't tell how a matrix M will diagonalize if you only know the diagonal entries of M
if i said, i have some matrix, and it's diagonalizable, and it has all zeros down the diagonal, how does it diagonalize, the answer is "who knows?" there's not enough information. it's partial information about the eigenvalues but not enough to say what they are
similarly if all i tell you about a bilinear form is the diagonal entries of its matrix with respect to some basis, you can't determine the eigenvalues of that matrix from that, and so you can't determine if the form is positive definite or singular or what its signature is
the exercise above seems to be oriented toward leading you to see it in a concrete example
 
5:48 AM
@robjohn Thanks a lot for guiding me professor Rob. Thank you so much.
170 may include some of fan voting from me.
 
i had a string of serial upvotes of length one apparently
 
"oh wow, with the same simple bilinear form, if i change what basis i use, i can make the diagonal of its gram matrix look like all sorts of things." -> leading to the message "i guess the diagonal of the gram matrix with respect to some random basis won't be enough to tell me much about the form"
copper: i only do that because i want attention
 
:-)
gn folks. have to get up in 4 hrs to drive my wife & son to the airport so they can fly to athens and meet a friend of mine.
someone needs to water the flowers and generate $$$
i am waiting for the deluge in response to my convex psq answer
 
@copper.hat night!
 
Alright leslie, i guess thats somewhat satisfying, thanks!. i will look further into it, i seem to not find alot of search results in english, am i naming this wrongly? how do you call it?
 
5:52 AM
Thanks Calvin!
 
mad: i dunno. "bilinear forms," "quadratic forms," "classification" of them, the "signature" of a quadratic form,
who knows
 
Alright, well i got to go work. You have a lovely evening.
Thank you!
 
cheers
 
 
1 hour later…
7:02 AM
Is there a name for the following equation? I want to be able to look it up to be able to characterise it correctly:
$n(x,y,z)=d\cdot(1-x^2/a^2-y^2/b^2-z^2/c^2)$, where $a,b,c,d$ are constants.
At first I thought it was an inverted parabola, since if it was missing a dimension it would be one, then I thought maybe it's a type of ellipsoid, but n varies. So now I am at a loss of what to look up in order to understand it properly.
 
well, it looks like "level sets" of n are going to be ellipsoids, or a point, or empty.
hard to visualize the graph of n itself (which would exist in R^4)
 
I just got to a reference calling it a 4-elliptic paraboloid. So, a 4D paraboloid.
Ah, I sent that a little too late.
 
In the Taylor series $x^0=1$ for all x is not 0 but why do we say $0^0=1$? I know it makes sense for $x^x$ when x approaches 0. It still bothers me if it is okay to do this.
 
it's just a useful convention. there are reasons unrelated to that limit to adopt the convention. it makes it easy to state the binomial theorem without qualification, for example
and other expressions involving polynomials. i would include taylor series in that
 
@leslietownes Thanks for mentioning this. So far it's been better than the webpage I found (which appears to be a forum post anyway) at visualising the equation.
 
7:18 AM
user: the level set n(x,y,z) = k seems to be classifiable in terms of the sign of k/d - 1. if it's 0, you get a point. if it's negative, you get an ellipsoid that can be rewritten in 'standard form' by rescaling a, b, and c. and if it's positive, no solutions.
 
In the context I am using the equation, d is always equal to, or less than n.
 
as for visualizing those ellipsoids, i dunno. they all appear to be similar to one another, with the relative scale of the axes established by a, b, c. and as k goes to d from below, they get smaller and go to a point.
 
why is there notation for the resolvent R(x, T) when $(x-T)^{-1}$ is perfectly fine, and as a bonus no one needs to guess what your convention for the resolvent is ((T-x) vs (x-T))
 
@leslietownes I think I don't understand what you mean by useful convention.
 
calvin: you're welcome to use that. some people find that stuff like the resolvent identity and other identities become easier to grok if you have a name for that.
 
7:22 AM
the only way i can remember the resolvent identity (at the moment at least) is by working it out for rational functions on C
 
e.g. if A and B are not going to commute you still might want to recognize R(z,A) and R(z,B) when you see them even if you are generally not comfortable with formally inverting more complicated expressions in A and B.
yeah, that's how i do it too.
 
ive run out of things to rant about
 
not: if you would like to be able to write a_0 + a_1 x + a_2 x^2 as sum_{k=0..2} a_k x^k then you need to adopt the convention that 0^0 is 1. or you can treat the constant term separately and not adopt the convention. it doesn't really matter.
same with the binomial formula (a + b)^n = sum k=0..n C(n,k) a^k b^{n-k} which looks great although if you are uncomfy with interpreting a^0 and b^0 in general then maybe you would have to write both the k = 0 and k = n terms separately.
but if you adopt the convention then you don't have to treat those things separately.
just pointing out that you might want to do that even if you have never heard of limits, or don't care about what the limit of x^x is as x goes to 0. in adopting the convention you can just be adopting a notational convention, not deciding what limits are.
"0^0" in calculus class is what people call an 'indeterminate form,' i.e. generally knowing that f goes to 0 and g goes to 0 as x goes to c isn't enough to tell you what f^g goes to as x goes to c. so even the fact that x^x goes to 0 as x goes to 0 is something of a happy coincidence.
 
7:37 AM
I understand now thanks for pointing out $a_0 + a_1 x + a_2 x^2+...$.
 
i think the first calculus book i learned from deliberately wrote the first term separately to avoid raising this issue
 
does someone know about marty's theorem in complex analysis?
We have the following form. Let F be a family of meromorphic functions on \Omega where \Omega is open and connected. then F is spherically normal iff F^#=\{f^#: f\in F\} is locally bounded. We denote by f^# the sperical derivative of f.
But now I asked myself don't we need some more information to speak about local boundedness. So I mean where is ist locally bounded? Would't it make more sense if we say locally bounded in \Omega?
 
 
2 hours later…
9:48 AM
Can maybe someone help me here:
0
Q: How can I show that the set of complex functions $F=\{z^n\}$ is normal on $\Omega$?

Wave Let $F=\{z^n\}$ be the set of complex functions and I want to show the following 3 claims: $F$ is normal on $\Bbb{D}$ $F$ is normal on $\Bbb{C}\setminus \Bbb{\bar D}$ $F$ is not normal on $|z|<2$ In the lecture our prof gave us the hint to use Marty's theorem. So let me quickly introduce our ...

 
10:39 AM
So about this polar topology, I switched to absolute polars instead because I've noticed that they only really consider balanced sets anyway. So that's that. I think they just had absolute polars in mind from the start.
If there is some counter-example where using real polars doesn't generate a Hausdorff locally compact TVS then feel free to share
(I believe the greatest issue here is proving whetever the multiplication is continuous)
 
 
2 hours later…
1:05 PM
@Koro In the article linked in my comment, the extra votes came from a glitch in the software that, in most cases, created accidental double votes.
 
1:19 PM
Ouch! oh well, lucky they're only fake internet points :-)
social media has a nasty habit of displaying the voters
 
 
1 hour later…
2:47 PM
why do people write $\frac{\delta F}{\delta \rho}(\rho)$ for the first variation of a functional $F$ at $\rho$, like why not just $\frac{\delta F}{\delta \rho}$
 
3:34 PM
Let $Y \subseteq X$ be (discrete?) spaces, and let $i_X : X \to \beta X$ be the map associated to the Stone-Cech compactification of $X$. Is it true that $\beta Y$ can be identified with $\overline{i_X(Y)}$?
It seems like it should and it's just about chasing definitions down and the universal property, but I just wanted to cech (pun intended).
 
Hello if we have a function $f$ defined on some space $V$, and $V$ is a direct sum of others, how would you express f on the direct sum? $V= \bigoplus M_i$
Say something like $f_{\big| \bigoplus M_i}= f_{\big|M_1} + ... ?$
 
$f_{\big| \bigoplus M_i} = f_{\big| V} = f$
 
Wow !! i never thought of that!! what a great reply
 
@MadSpaces $f=\sum_if|_{M_i}$ I suppose
 
Dude obviously i want to get seperate components..
 
3:48 PM
Basically what you wrote
 
Hmm yea akiva but it feels wrong
then you need to define the restriction to be zero on points else where.
 
Oh yeah you're right
@MadSpaces That's not linear
 
Can we just use the unity of sets, edit no we can not
 
Since $V=\bigoplus M_i$, we can write any $v$ as $v=\sum v_i$ with $v_i\in M_i$. Let $f_i(v)=f(v_i)$. Then $f=\sum f_i$
@MadSpaces You want $f_i$ to be $f$ on $M_i$, $0$ on $\bigoplus_{j\ne i}M_i$, and linear on the rest
 
That's just taking the dual
 
3:53 PM
@Astyx Sure. $f_i=f\circ e^i$ where $e^i$ is the function picking out the $i$th component
and the $e^i$ are a basis of the dual
Well, no
 
if we defined the restriction to be the identity outside.. and then use composition.. maybe it works., problem would be when you reach the function where the point is not ouside.. then you just get a new element, this new element might be in one of the sets. if our function goes into the same set, so we have a problem.
 
I want $e^i$ to be $V\to M_i$, not $V\to\Bbb R$
I should probably write $p_i$ for "projection", then (instead of $e^i$ or whatever)
Note that if $V=M_1\oplus M_2$, then $p_1,p_2$ each depend on both components
 
0
Q: Formal usage of direct sum in a set of functions

High GPADefine $f_i$ as functions indexed by $i\in I$. Is it ok to say that $\bigoplus_{i\in I}f_i = f=(f_1,f_2,f_3,......)$? I understand that the direct sum is often used in vector spaces or algebra structures. Is the usage here in functions common? I have never seen this kind of usage before, though.

Second comment.
He basically writes f as f_1 + f_2 ... so on.
 
Hi @shin!
 
i cant find wikipedia article on the term he is using "direct sum of maps"
 
4:01 PM
I mean, we don't know the category you're working in
 
Category of vector spaces surely
 
yea
 
hello @Koro
 
A direct sum is defined on matrices, we can say matrices are linear functions and define something similiar?
 
Well if you have $f:A\to C$ and $g:B\to C$ you can define $h:A\oplus B\to C$ by $h=(f,g)$, but if $f:A\to B$ and $g:A\to C$ you can define $h:A\to B\oplus C$ by $h=f+g$
I think the latter is technically the direct sum?
And the former is the direct product?
 
4:06 PM
yes
 
The weirdness is that, in the category of vector spaces, finite products and finite coproducts (finite direct sums) are the same thing
Wait no I think I have it backwards lol
 
Man i swear to god i just wanted to prove this statement and i stopped at this question
i want to calculate the gram matrix of a bilin form $f$ inductively on sub spaces, whose direct sum produce the vector space. so i am trying to get to the result where i can say, the left matrix is equal to the direct sum of matrices on the right
Because, wikipedia is telling me thats the answer lol!!
@TedShifrin uwu
 
no uwu in chat please
 
@TedShifrin help me step prof, i am stuck!
well, i think we all stuck.
 
So here's a random realization
This is a double spiral. (Notice how it has two separate branches.) This is another double spiral. If I lay them on top of each other, I get this image.
Now, let's do something weird: let's flip the colors in the top-right quadrant. We get this image. If we focus on each color individually, we see that black looks like this and red looks like this. They're each wacky-looking single spirals now!
So the union of two double spirals equals the union of two wacky single spirals!
...What do we do with this information? [shrugs]
 
4:15 PM
Mad, what is the concrete, precise problem? All I see is vagueness.
 
Akiva its optical illusion, nothing more
 
"Illusion"?
 
He alludes to illusion.
 
ted its about writing a function on a direct sum in some kind of singular functions defined on each summand
 
Still too vague.
If this is linear stuff, then use linearity and add what the functions do.
 
4:21 PM
the main problem was i want to show the matrix of a bilin form, has a spesific construction, for that, the proof i am doing right now, which lead me to this point, is i can write the vector space, as a direct sum of smaller ones, and i know how the matrix of the form looks on those smaller ones. so according to the formula that if weh ave matrices then the direct sum of those, are hte matrices on a diagonal of a bigger one, which is actually the final form of the matrix of te function.
However in order to reach the last result, i need to show that the matrix on the left, is truly the one on the right, so this means, the matrix of the function on the direct sum, will equal the direct sum of the matrices.
Get it ?
So if we denot A as the matrix for f, i want to show that $A = A_1 \oplus A_2 ....$, where as the $A_i$ are the restrictions on the smaller subspaces whose addition result the main vector space
 
No. Use the definition of the actual mapping to check whether this holds.
 
Now i have seen the final result, and it is actually TRUE, that $A$ will have that form, however i am stuck at the part where i write the function as constraint on the subspaces, and dont know how to procceed..
 
I'm telling you you have to look at what the actual mapping is and see if breaking it into pieces is valid. Certainly if $V=V_1\oplus V_2$ and $f\colon V\to W$ is linear, then $f(v_1+v_2) = f(v_1)+f(v_2)$. There is no more to say.
 
Suppose it is true as you wrote it.
nevermind.. this is too annoying. you are not getting me, and i am not explaining myself clearly. so let it die
 
Maybe it would help you to say what basis (bases) you are using to give your matrix representation. Otherwise, a matrix is meaningless.
 
4:47 PM
Turns out linear algebra was too complicated for the superior mathematical mind of this particular german
 
Indeed. Superiority is vastly overrated.
Haven't seen you in ages, @Astyx.
 
I haven't been here in ages
Well, at least I haven't sent a message in ages
how have you been?
 
Hanging in there. Covid is going nuts again. And the US is back in the 1800s.
 
yeah, so I've heard
 
someone pls help the hegemon back on its feet
i need it to hold out until i get my diploma
 
4:50 PM
Are you in the US, @shin?
 
well, the us is certainly in me and around
there's some geometry joke here somewhere
 
You can hate it all you want, you're still trapped here
That is to say,
You can take the America out of the Shintuku, but you can't take the Shintuku out of America
 
Maybe this will be a new brand of automobile.
 
5:51 PM
Is it okay to rearrange the Fourier series? After learning a bit about how scary it is to rearrange infinite series in calculus I find it disturbing to even think about it. I don't understand how should I prove if it is absolutely convergent.
 
@Ted Shifrin Hi, in this answer math.stackexchange.com/questions/3894028/… you state that $\frac{\partial\phi}{\partial x} = F(\tfrac yx) -\frac yx F(\tfrac yx) \quad\text{and}\quad \frac{\partial\phi}{\partial y} = F(\tfrac yx)$ but shouldn't it be $\frac{\partial\phi}{\partial x}=F(\frac{y}{x})-\frac{y}{x}\frac{\partial F}{\partial x}$ and $\frac{\partial\phi}{\partial y}=\frac{\partial F}{\partial y}$? What am I missing? Thanks
 
Maybe it is because we have to check the series is absolute convergent beforehand.
 
6:09 PM
@NotTfue This depends on how you define $\sum\limits_{-\infty}^\infty$
 
@robjohn By "depends " you mean whether it can be rearranged?
 
there is no standard definition for the double ended sum in non-absolutely convergent series
 
@NotTfue I wouldn't call this rearranging at all!
 
Or do you mean the equation?
 
that equation could be considered how this book/paper is defining the double ended sum
 
6:15 PM
@lorenzo There certainly were primes missing. I don't know how that happened. But, no, $\partial F/\partial x$ and $\partial F/\partial y$ make no sense. $F$ is a function of a single variable.
 
@TedShifrin Isn't the meaning of rearranging rearranging the order? Why wouldn't you call it rearranging?
 
what is the original arrangement?
this is what I meant about there being no "standard" definition of the double ended sum
 
Anyhow, @lorenzo, thanks for calling my attention to that flaw. I can't believe the primes were missing, but it doesn't look like anyone edited it ...
@NotTfue If I have a usual series $\sum_1^\infty a_n$ and I write $\sum_1^\infty a_{2n-1}+a_{2n}$, is that a rearrangement?
 
@robjohn original arrangement was not shown I assumed there is a standard definition.
 
As robjohn said, the usual definition is to split the doubly infinite sum into two infinite sums, one with negative terms, one with positive terms. So you "assumed" that.
 
6:21 PM
and that is what you've written above
 
@TedShifrin No.
 
what you've written above could be considered the definition of the double ended sum
 
OK. So this is what I would consider the analogy to the case of the doubly-infinite sum.
 
Ok, I see what you mean.
Thanks
 
6:39 PM
@TedShifrin I see, so $F'(\frac{y}{x})=\frac{dF(\frac{y}{x})}{dx_1}$ and, for example, $\frac{\partial \phi}{\partial x}=\frac{\partial }{\partial x}\left(xF(\frac{y}{x})\right)=F(\frac{y}{x})+\frac{\partial F}{\partial x}=F(\frac{y}{x})+x\frac{d F}{dx_1}\cdot\frac{\partial x_1}{\partial x}=F(\frac{y}{x})+x\frac{d F}{dx_1}\cdot (-\frac{y}{x^2})=F(\frac{y}{x})-\frac{y}{x}\frac{dF}{dx_1}$ right?
(You're welcome, and thank you for the help in understanding the exercise.)
 
Careful. What you have written makes no sense. Writing $\dfrac{dF}{dx}\big|_{x=a}$ is very different from $\dfrac{dF(a)}{dx}$. The prime notation is better.
You’re still writing $\partial F/\partial x$, which makes mo sense. STOP!
 
I don't see why: doesn't $F$ depend on $x$ via $x_1$?
 
There’s a composition of functions. Break your terrible calculus habits of using the same letter for a function and for its composition with another function. This is horrible single-variable habits!
 
6:59 PM
so if I understand correctly I cannot write $\frac{\partial F}{\partial x}$ because $F$ doesn't depend directly on $x$, so would a notation like $\frac{\partial (F\circ x_1)}{x}$ make more sense?
 
$x_1$ is a variable, not a function. I don’t like it.
Define functions as explicitly as you can.
 
7:20 PM
Do someone have a nice example which we can solve using runges theorem? I haven't found some when I google.
 
@robjohn Is it not $\lim_{M,N\to\infty}\sum_{-M}^N$?
You could also do $\lim_{N\to\infty}\sum_{-N}^N$, but there are instances where this converges and the above does not
 
What would be the "pinch map" of $S^1\to S^1\vee S^1$?
My notes refer to the pinch map without much context
I also can't find it in standard literature (Hatcher, Dieck)
 
I seem to be forgetting some basic general topology and I am hoping someone can correct me. Let $X$ be a Hausdorff space and $A$ a subspace. Is it true that $X/A$ is Hausdorff if and only if $A$ is closed?
 
Like you have a circle, and you pinch is and it becomes two circles
 
ohhhh
haha thanks
sometimes I forget to use common sense
 
7:36 PM
Oops, my wedge is the wrong way around
 
@user193319 Hausdorff is definitely not enough, but $T_3$ looks good
 
hahah yea I got it xD
 
The point being that if $\in X$ cannot be separated from $A$, then $\pi(x)$ and $\pi(A)$ are two points that cannot be separated in the quotient
 
Write $1\Bbb Q$ for the indicator function of the rationals. Then $1\Bbb Q$ has a symmetric derivative at every rational but not at any irrational
(the symmetric derivative is $\lim (f(x+h)-f(x-h))/(2h)$)
@ShaVuklia It defines it in the next sentence: "The latter is the quotient map identifying the basepoint of $S^1$ with its antipodal point"
(this is the same as Astyx's drawing)
I suppose this is $\theta\to\begin{cases}(2\theta,0),&0<\theta<\pi\\ *,&\theta=0,\pi\\ (2\theta,1),&\pi<\theta<2\pi\end{cases}$, or something like that
 
@AkivaWeinberger Hm, right, because any choice of quotient map would be equivalent
so we can speak of "the" quotient map
 
7:49 PM
There's "the" quotient map $S^1\to S^1/(* \sim-* )$, and then "a" homeomorphism $S^1/(* \sim-*)$ to $S^1\vee S^1$, I think
where $*$ is the base point (and also ruining ChatJax… I should have used $p$)
 
@AkivaWeinberger Ah, right
I initially thought we were looking at the quotient map $S^1\sqcup S^1\to S^1\vee S^1$, so I was confused, but what you say makes sense
 
@AkivaWeinberger That one is the "principal value" sum, but the first one you wrote is the same one as Not Thue gave.
hence I was wondering what the original ordering they were considering to have been reordered was
 
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