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3:13 PM
[Tensor visualisation] Hmm... not very insightful...
of course they are ellipsoids, but so what
 
Is there a name for the theorem that "sum of all exterior angles of a convex polygon is 360 degrees"?
 
I don't think it has a name despite its widespread use in euclidean geometry
 
"Exterior angle sum theorem" maybe?
Doesn't sound "official", but it also doesn't sound "wrong" and people will probably know what you mean
Thought: if the lengths of $a$ and $b$ are measured in cm, then the length of $a\times b$ is measured in cm^2
(vectors)
 
Is anyone here familiar with convex hulls?
 
You know, during my study of Riemannian geometry, it never occurred to me to think about the units of the quantities. What are the units of $\langle R(X,Y)Z,W\rangle$? @BalarkaSen
@nbro I'm familiar
 
3:27 PM
Are you familiar with Chan's algorithm? Maybe being familiar only with the orientation test is enough to help me determine the correctness of an algorithm I have just written down
 
@nbro I'm not. I could look it up
 
Essentially, in Chan's algorithm, once mini-hulls are computed, we need to "merge" them
 
This is not terribly insightful. But perhaps, it gives us a way to map matrix polynomials to complex functions of one variable...
 
In the process of merging them, we need to find the "right tangent" from a point outside one of this mini-hulls to the mini-hull itself
So, basically, a right tangent leaves all points of the mini-hull the the left of itself
 
3:30 PM
How do you compute mini-hulls? Is it recursive- do you use Chan's algorithm on the pieces?
 
Initially, Chan's algorithm partitions the set of input points into sub-sets of similar size.
For each of these subsets, it computes the convex hull (using e.g. Graham's scan)
Again, once we have these mini-hulls, we need somehow to merge them, right?
 
One possible way to approach this might be: Since every matrix can be mapped into ellipsoids of colors in the complex plane, we can perhaps do the same trick as in 3B1B's video, in that see what happens when we nudge the entries of the matrix in the matrix polynomial incrementally, how the ellipsoid changes shape and the quantifying this morphing somehow using a color map of sorts...
The investigation will continue tomorrow after the last batch of chemistry data is taken cared of
 
@AkivaWeinberger It's not recursive. You can think of Chan's as a two-step algorithm.
1. Compute mini-hulls
2. Merge them
 
@nbro ok so what's your question?
 
Why is this faster than using Graham on the whole thing? Or using Chan on the subsets?
 
3:33 PM
As I was explaining, during the process of merging the mini-hulls, we need to find these "right tangents"
 
@nbro Arright
 
So, I wrote an algorithm to find them, but I am not 100% sure it's correct
You just need to be familiar with orientation tests and binary-search to understand my algorithm.
 
is it on git? can i take a look?
 
I know binary search, I'm not familiar with orientation tests, sorry
 
I will post a picture here
 
3:35 PM
Did you test it on examples? Assuming it's not just pseudocode
 
It's just pseudo-code. I don't have much time to test it on examples...
 
Oh, I see, given three points you can ask whether or not they're oriented clockwise or counterclockwise
 
So, $C$, in the algorithm, is a list/vector of vertices ordered in counter-clockwise representing one mini-hull.
Whatever. I had written it in the description.
@AkivaWeinberger Yes, exactly.
Now, I am not sure about the "else if" and "else" parts. I am not sure about those conditions.
But I am pretty sure that we find the right tangent once the condition in the "if" part is true
 
4:24 PM
@MatheinBoulomenos mathein :D I kinda need your help:D please text me when you see this =p
 
4:42 PM
@KasmirKhaan what's up?
 
Inb4 "the sky"
 
Hi, could anyone take a look at my question, please? There's no responses so far.
0
Q: Relationship between the product & sum of digits

DarkRunnerI was collecting data on 3 particular functions, to find a relationship between them: Let $f(n)$ be the product of all the nonzero digits of a number so that $P(n)$ is the sum of all $f(n)\le n$ $S(n)$ is simply the sum of all consecutive integers $\le n$, basically $\cfrac { n(n+1) }{ 2 } $ $...

 
What is meant by an integral that is "generalized"?
More to the point, my book says $\Gamma(z) = \int_{0}^{\infty} t^{z-1}e^{-t} dt$ (i.e the gamma function) is generalized in its endpoints
Does it mean $|\lim_{t \to 0 \lor \infty} t^{z-1}e^{-t}|<\infty$
For $\Re z>0$ ^^
 
5:18 PM
@MatheinBoulomenos Ill send ya email :D
sorry was not here =p
 
@KasmirKhaan apology NOT accepted. >:(
 
@Daminark -.- dont make me angry dami , you wont like me when am angry :D
 
Zee
@Daminark how did you find algebraic topology compared to algebraic geometry?
 
I haven't seen any algebraic geometry so I can't say
@KasmirKhaan :O sorry
 
@Daminark forgiven and forgotten :D
 
Zee
5:25 PM
@Daminark I never thought I would say this but am starting to get seduced by the algebraic parts of math
 
is A*A hermitian?
 
They can have an appeal if done right for sure.
@JoeShmo do you know the adjoint of a product?
 
no, thats why im asking
 
So, $(AB)^* = B^*A^*$. Try to prove this
And then see how it gives your desired answer. I've gotta go on airplane mode but good luck!
 
If $A,B$ are complex matrices then this basically just goes to how the transpose works.
 
5:35 PM
yes, complex matrices.
 
Right. Then it's useful to ask: how is matrix multiplication defined?
There's a 'visual' way we usually do it, but that's not the definition.
 
how is it different than matrix multiplication over the reals?
or any other field for that matter
 
It's not.
 
then why is it useful to ask
 
The only difference with complex matrices is really that the Hermitian conjugate involves both transposition and complex conjugation.
Hmm. Well, let me ask a different question then: Do you know why $(AB)^\top = B^\top A^\top$?
 
5:38 PM
top H?
 
What?
 
what is (AB)^\top
 
transpose.
 
oh. yeah sure
 
I prefer $(AB)^T$ myself but Ted's been after me on that.
 
5:39 PM
follows from definition of matrix multiplication
stand your ground
 
lol
Right. Then you can just take the complex conjugate of both sides to get $(AB)^*=B^* A^*$
 
yeah, the last paragraph here -- mathworld.wolfram.com/ConjugateTranspose.html
spells out the details
 
(i'm assuming $A^*=\overline{A^T}$ is your definition)
In physics we tend to write it as $A^\dagger$
 
are you a physicist?
 
probably because it looks like T still but not quite
 
5:42 PM
i've seen the dagger notation on wikipedia
 
physics grad student
so close enough
 
what do you study
 
@Semiclassical They're both $\sum_ka_{jk}b_{ki}$ essentially. Is there a way to write that so that it works with the Einstein summation convention?
 
Or $\langle (AB)^\ast v,w\rangle=\langle v,ABw\rangle=\langle A^\ast v,Bw\rangle=\langle B^\ast A^\ast v,w\rangle$ implies $(AB)^\ast=B^\ast A^\ast$ if you're allergic to coordinates.
 
You've just expanded my mind @anon
 
5:45 PM
@AkivaWeinberger eh, I'd say that in Einstein summation convention the statement is already embedded into the notation.
 
Do I need to write $a_i^j$ for $A$? And then $AB$ is $a_i^kb_k^j$ or something
That looks wrong
 
Depends what you're doing.
Usually when doing matrix stuff you don't worry about contra/covariant index stuff
So you might as well make them both subscripts
 
$(AB)^i_k=A^i_jB^j_k$
 
yeah, that's probably the more proper way to do it
 
It's weird
 
5:53 PM
hello
anyone here
 
@Semiclassical did we conclude last night that the dimension of the space of hermitian matrices is n^2 over the complex numbers?
 
yeah
1 from each of the n (real) diagonal elements, 2 from each of the (n^2-n)/2 (complex) subdiagonal elements
 
yes
 
@Semiclassical gdi,do you know how to cite arxiv in bibtex
It seems impossible
 
6:08 PM
yeah, idfk
 
@0celo7 What seems hard about it?
 
I can't get the arxiv number to appear in the citation
 
Where are you putting it?
 
the bloody bibliography is one of the reasons (read: excuses) I've been procrastinating on my thesis
 
@Semiclassical the bibliography is the only thing I have left, basically
@TobiasKildetoft I don't know what you mean
 
6:09 PM
Just put arxiv:XXXXXX as the Journal
 
dealing with bibtex is miserable
 
@TobiasKildetoft Ugh.
Why isn't this native?
 
native to what?
 
bibtex
 
as in, why isn't there a specific entry in bibtex for it
 
6:10 PM
You mean bibtex should have a specific form for arXiv papers?
 
why isn't there an applet that conntects bibtex to latex, mathscinet, inspirehep, and harvard
and you just do everything there
instead I need 4 fucking screens to have all of the shit on at once
 
I think there are some programs that help with that
 
There is a site that gives you bibtex entries based on arxiv identifier
 
I am using it, but they are wrong
it's formatted incorrectly
 
yeah, the one I found used some weird type of entry
 
6:12 PM
 
right, for some reason those people decided it should be a misc, which is silly
 
do you have a reference format that I can use?
 
@0celo7, question
T_A M(n) = M(n), and T_f(A) S(n) = S(n) because M(n), S(n) are linear?
 
T_A ?
 
Tangent space of the matrix A
 
6:16 PM
@0celo7 I use @article {amrw17,
AUTHOR = {Pramod Achar and Shotaro Makisumi and Simon Riche and Geordie Williamson},
TITLE = {Koszul duality for Kac-Moody groups and characters of tilting modules},
JOURNAL = {arXiv:1706.00183 [math.RT]},
YEAR = {2017},
}
 
@TobiasKildetoft thanks.
 
@MatheinBoulomenos email sent! :D
 
Got the email
 
6:19 PM
as allways mathein thanks alot ! :D and i understand if you have other things to do :D
I am gonna use those as preparation for the exam:D
 
@TobiasKildetoft It puts the arxiv thing in italics. Is that normal?
 
You are very nice to me and I wish I could return the favor one day :D @MatheinBoulomenos
@TobiasKildetoft hey Tobias :D
 
6:32 PM
can a hermitian matrix be decomposed into two parts the same way a symmetric matrix can?
 
well, how would you do it?
 
well i can picture a variety of ways. i can break into upper and lower triangular.
buti don't think i need to
 
If I have $\pi(d)x = \pi(r)$ does "precomposing with $\Psi$" mean $\Psi \circ \pi(d) \Psi(x) = \Psi \circ \pi(r)$?
 
Think by comparison with symmetric matrices. How do you write the symmetric and antisymmetric parts of a matrix $A$?
 
well for symmetric matrices C = (1/2)C + (1/2) C^T, but that's trivial since C^T = C
the decomposition is trivial
 
6:45 PM
yeah, I should've by comparison with real matrices
But then I'm not sure what decomposition you had in mind for symmetric matrices
 
this guy ^^ well for symmetric matrices C = (1/2)C + (1/2) C^T, but that's trivial since C^T = C
 
yeah
for real matrices, you decompose it into symmetric and antisymmetric parts
for complex matrices, you decompose it into hermitian and anti-hermitian parts.
 
yeah i see it on wiki
its not as straight forward
 
so for hermitian matrices in particular it's not very interesting.
 
right
im trying to show f_A(B) = BA + AB is onto H(n)
they it's done for S(n) and O(n) is decomposing symmetric matrices as above and yada yada
what to do in the case of unitary matrices?
excuse me,
f_A(B) = B*A + A*B
^* being the conjugate transpose
 
6:51 PM
what branch of math they study manifolds? or they dont anymore?
 
differential manifolds = differential geometry + differential topology
 
H(n) is what?
 
and the connections with algebraic geomtry?
 
hermitians
 
6:52 PM
Anyone ever used taylors series to derive four point numerical differeniation formula
 
not sure. most probably varieties
 
So you're wanting to show that $A^* B+B^*A$ is hermitian if $A,B$ are hermitian?
 
no. that's trivial
given hermitian C, I would like to find A,B satisfying A^* B+B^*A = C
 
Well, $A,B$ are both assumed to be hermitian, right?
 
for the first claim, (A^* B+B^*A)^* = B^*A+A^* B is hermitian
nope. arbitrary matrices
 
6:54 PM
huh.
Well, if you're only looking for some A,B
then one simple route is to consider the case of B=I.
 
arbitrary in M_n(C)
 
In which case it collapses to finding some matrix A such that A^*+A = C
And I can think of one particularly simple example of such A...
 
the function is a function of B
i.e. we must find a B that satisfies the equation, for a fixed A
 
Ah, point.
 
do we know anything about algebraic conjugation in the hermitians?
 
6:59 PM
Not sure what you mean by that.
 
what does A* B A* = ?
so that's why my *'s disappeared up there
 
yeah, it's annoying
 
i suspect the decomposition of the hermitian matrix has something to do with it
 
can some1 point me an algebraic geometric result that can be seen geometrically?
 
7:00 PM
maybe start by figuring out how it's proven in the case of A^* B+B^* A = I.
 
since in the case of symmetric matrices, that's exactly what's being done -- imgur.com/85t3tDf
 
Which I guess would be B^* = A^{-1}/2
So in general what you want is to find a matrix B such that B^* A has hermitian part C/2.
What if you choose B^* = C A^{-1}/2?
 
whats the decomposition of the hermitian that you are using?
 
A^* B+B^* A = C = C/2+C^*/2
 
how do we know the latter?
 
7:05 PM
C^*=C
it's hermitian, after all.
 
oh
dur
for the same reason as above
wait a minute
the argument follows in the exact same way
oh bloody hell
 
Illuminate me. Is the sum of $i$ up to $\log(n)$ something special?
 
Hi, can anyone take a look at my Math SE question?
It's been up for hours, but there's been no response at all
0
Q: Relationship between the product & sum of digits

DarkRunnerI was collecting data on 3 particular functions, to find a relationship between them: Let $f(n)$ be the product of all the nonzero digits of a number so that $P(n)$ is the sum of all $f(n)\le n$ $S(n)$ is simply the sum of all consecutive integers $\le n$, basically $\cfrac { n(n+1) }{ 2 } $ $...

 
@nbro are you an algorithmist?
 
7:08 PM
What do you mean by "algorithmist"?
Are you asking it because of the algorithm I shared with you before?
 
yes
are you a grad student and if so, what do you study
 
@DarkRunner one thing you could add to those plots is the fit equation for each of them. (if you're plotting in log-log scale, then it being a straight line should translate into an exponential fit)
so presumably for S(n) you'd end up with S(n) = n(n+1)/2 ~ (1/2)n^2
 
algorithmist is well defined -- networks, combinatorial optimisation, approximation, probabilistic, distributed
 
with the others you'll end up with some other exponent and prefactor
So the relation between them would just be that they all behave like F(n) ~ a*n^k for some a and some k.
 
@Semiclassical As in the slope/best fit line?
 
7:27 PM
@BalarkaSen So say you have two geometric circles in $\Bbb R^3$. Even if they don't link, it's possible for the disks they bound to intersect.
Now say you have two geometric circles in the surface of the unit ball in $\Bbb R^4$ (so they're in $S^3$). Apparently, if I'm reading this right, they link if and only if the disks they bound intersect.
 
Let $G$ be a group with exactly two elements $a,b$ that have order 3. Show that for each $x\in G$, we have $ax=xa$ or $ax=xb$. Since $a$ has order 3, we know that $a^2\neq e$ has order 3, so it follows that $b=a^2$. Now, if $G$ is abelian, then $ax=xa$ for each $x\in G$, so assume $G$ is non-Abelian. I don't know how to proceed from here on..
 
("The disks they bound" being subsets of $\Bbb R^4$, not $S^3$.)
 
what's the property called that a function is almost linear except that it conjugates $\ell(\lambda v + w) = \bar{\lambda}\ell(v) +\ell(w)$ where $\lambda \in \mathbb C$?
can I call it "conjugate linear"?
 
@ShaVuklia What's the order of $x^{-1}ax$?
 
found it on wikipedia, conjugate-linear or antilinear
 
7:31 PM
@AkivaWeinberger let me see
(o I have to go for a sec, will be back)
 
@DarkRunner yes. But you should pick the fit that’s appropriate for what you’re plotting.
Ah, drat. I suggested exponential above when I meant power-law
That was pretty silly.
 
@DarkRunner it seems that all these functions can be close-formed
 
What is Closed formed? "An equation is said to be a closed-form solution if it solves a given problem in terms of functions and mathematical operations from a given generally-accepted set"
 
@AkivaWeinberger ohh so clever! thanks, I got it!
 
the aggregator function D, or even maybe P, can be studied following the incremental behavior of unit digits, D is simpler to get closed form.
 
7:46 PM
how did you get the idea for this tho? @Akiva is it standard to just take the conjugation, or what's kind of the reasoning behind this?
 
@Semiclassical is what I get after power law
 
I guess it just makes sense to try it out
 
8:08 PM
@DarkRunner Hrm, that looks wrong. I was expecting it to do the power law for in the underling variables eg n, S(n)
But apparently not. but what I really was after was: what are the equations that go with those plots?
Or are the lines on the original plot just connecting the dots?
 
@Semiclassical I have the whole document here
 
I’m here on mobile atm so I can’t help there
 
Hm, let me find the equations
 
@Semiclassical Have you read Griffith's Introduction to QM?
 
P(n)=$\frac{12167}{25}x-389344$
$S(n)=\frac{11001}{2}x-5000000$
$D(n)=\frac{37}{2}x-4999$
@Semiclassical those are the equations of best fit(If I haven't made any mistakes)
 
8:20 PM
I'm not seeing how the author's "easily implies" easily implies here,
 
@Semiclassical how does one write the acknowledgements section
 
for a linear map $A$ on a Hilbert space, there is a real positive scalar $\beta$ , he proves
 
@DarkRunner oops, wrong person
 
$\beta \Vert u \Vert \le \Vert Au \Vert$, this "easily implies' $A$ is injective and the range of $A$ is closed in $H$
 
@0celo7 shrug
@Lozansky yeah
 
8:21 PM
I'm not seeing why :(
 
@0celo7 You think about the people you want to acknowledge. Then you put them in that section :P
 
@GFauxPas it easily implies injective, agreed?
@ACuriousMind I have a list of people.
 
no, because it's a weak inequality
?
 
I don't know what else to write
 
@DarkRunner I’ll re-examine that once I’m able ti
 
8:22 PM
@GFauxPas so?
 
@Semiclassical OK, thanks!
 
@0celo7 I'd add a brief indication of what you're thanking them for to each.
 
@ACuriousMind Yes, clearly
 
I'm missing something obvious
 
But how do I do that
 
8:23 PM
Is this about how to format it or how to come up with the reasons to thank people? :P
 
Well, I can write down bullet points for what to thank them for
Except for Balarka. I'll probably take him out.
Why is he on this list anyway
 
For an undergrad senior thesis it probably matters less
 
Ok, I've eliminated the weak link
 
@0celo7 I don't see it :(
 
@Semiclassical Probably, yes, but this is like the last thing before I start editing it
@GFauxPas do you know what injective means
so I want to get this done
 
8:24 PM
Fair.
 
$x \ne y \implies Ax \ne Ay$
 
for a linear map, I mean
 
$\ker(A) = \{0\}$?
 
wot
take $x\in \ker A$ and let me know what happens
 
that the only solution to $Ax = 0$ is $x = 0$
oh
there we go
okay, great :) so its injective
 
8:26 PM
is A bounded?
 
@DarkRunner did you get these curves following an algebraic way or just statistically connecting them togather ?
 
yes
 
Consider $(y_n)\subset \ran A$ converging to some $y\in H$
Show that $y\in \ran A$
Hint: it's a Cauchy sequence
@ACuriousMind Well, really, how does one write it
"I would like to thank blah for X. I would like to thank blah2 for Y. etc"?
This sounds awkward
 
Sure
 
@Abra001 I used google sheets to choose power series.
 
8:29 PM
Well, don't just put these sentences next to each other, vary the structure a bit, add connectors like "also", etc., but that's basically it
 
thanks 0celot )
:)
appreciate the help
 
did you get it?
 
@DarkRunner ok so it is an embedded feature in google docs if i understand it right.
 
I can do $||y_n - \beta I y_n ||$, right?
 
uh
what?
 
8:32 PM
It's easy to come up with a relation between S and D, P poses a problem.
 
@Abra001 Yup
 
oh I got my variables confused
man I need more coffee
been working for hours
okay, let me try again
 
@Semiclassical Does it require any fancy mathematical background? Or is PDE/Fourier enough?
 
doesn't it just follow from $A$ being continuous, as its bounded?
 
@Lozansky yeah, that should be plenty. Might be good to review linear algebra as well
 
8:39 PM
what an absolute pain in the ass
 
L
Isnt it enough that A is continuous?
 
@Semiclassical Alright cool, thanks!
 
Last night dream I asked a surprisingly not weird question to Leaky in the math chat, on whether it is possible to find a complex function such that there's a bounded region where there are countably many roots and hence a closed curve surrounding that region will have an infinite winding number. I was initially suspecting that it is always true in the neighbourhood of any essential singularity. The dream never resolved that question. Reality check, however, gave the following definite answer:
5
Q: Can the winding number be infinite?

mezLet $z$ be a point in the complex plane, and $\gamma$ be a closed curve. Is it possible that $$n(\gamma,z) = \frac{1}{2\pi i}\int_\gamma \frac{dw}{w-z}$$ becomes unbounded? In other words, is it possible to find a curve $\gamma$ such that it winds around a fixed point infinitely many times?

 
@GFauxPas no
 
32
Q: Is There A Polynomial That Has Infinitely Many Roots?

КарпатськийIs there a polynomial function $P(x)$ with real coefficients that has an infinite number of roots? What about if $P(x)$ is the null polynomial, $P(x)=0$ for all x?

3
Q: The set of zeros of a holomorphic function is finite in compact sets

user100106Statement Let $f:\mathbb \Omega \to \mathbb C$ be a holomorphic function, $f \neq 0$ ($\Omega$ is a region, i.e., an open, nonempty, connected set). Prove that in every compact subset $K$ of $\Omega$, the set of zeros of $f$ is finite. I've read in Stein's textbook a proof of the statement "Sup...

 
8:56 PM
$y_n \to y \in H$ because it's a Hilbert space, so we have $\beta ||u_m - u_n || \le ||A(u_m - u_n)|| = ||A u_m - A u_n || \to 0$ for $m, n$ suff. large...
with $y_n = Au_n$
so $u_n$ is also Cauchy
so we have both that $Au_n$ and $u_n$ are Cauchy
 
For a more concrete answer to the dream's question:
3
Q: Holomorphic function having finitely many zeros in the open unit disc

nji1Suppose $f$ is continuous on the closed unit disc $\overline{\mathbb{D}}$ and is holomorphic on the open unit disc $\mathbb{D}$. Must $f$ have finitely many zeros in $\mathbb{D}$? I know that this is true if $f$ is holomorphic in $\overline{\mathbb{D}}$ (by compactness of the closed unit disc), b...

 
and then, uh
well $u_n \to u \in H$ because Hilbert
 
Holomorphic functions are very nice. There are lots of non-holomorphic functions that are less nice.
 
and then $\lim_n A u_n = A \lim_n u_n = Au$, and that's all I got
@0celo7, because $A$ bdd means $A$ continuous
 
@GFauxPas you don't know that $\lim u_n$ exists.
 
9:06 PM
https://math.stackexchange.com/questions/2707892/show-that-h-x-in-gax-xa-has-index-2-g-contains-2-elements-of-order-3
anyone an idea for this question?
 
@GFauxPas yes, this is the correct approach.
 
yes I do, because $H$ is Hilbert and $u_n$ is Cauchy
?
 
As for what inspires the dream. This paper outlined something very similar to the root finding algorithm in the 3B1B video seen yesterday, which mentioned about winding number encloses the same number of roots in polynomials in some bounded region. Later on when experimenting with complex plots just before going to sleep, the wikpsdia article about essential singularity is read and the phrase
> ...in every neighborhood of an essential singularity a, the function f takes on every complex value, except possibly one, infinitely many times...
caught my attention. Somehow I suspect a that question popped up in my mind after that, might be too fast for me to be conscious of thus it only showed itself in the dream owing to its byzantine dream constructing rules that I don't fully understood in my mind happened to select that question (and probably distorted it somehow
 
alright well I certainly have a lot more insight than "easily implies" first gave me
thanks guys
 
also typo ")"
 
9:11 PM
@Semiclassical In essence, I created the graphs to solve the following question: Let P(n) be the product of all the nonzero digits of a number n. Find P(1)+P(2)+...+P(10^2018-1)
I thought that creating said graph would result in some form of a nice equation, or formula, but I really don't know what the end result is showing
I mean clearly, one way to group such numbers is palindromes, since they will have the same product, or replacing any numbers containining 0 (i.e., 016) with 1s (to 116), but otherwise, I have no idea how to begin the problem
 
9:32 PM
If $A$ is an adjacency matrix of a graph $G$ with vertex set $\lbrace 1,\dots,n\rbrace$, and $x$ is a vector real entries in positions corresponding to the vertex set of $G$, what is the best way to interpret $\langle Ax,x\rangle$? I feel like $\langle Ax,x\rangle=\sum_{ij\in E}x_ix_j$.
 
good question
 
0
Q: Algebraic of degree n numbers and two equivalent? theorems

FaustMy book says that the following two theorems are equivalent but i can't seem to show one direction, please help. Liouvilles theorem: if a number $\xi $ is a real algebraic number of degree $n\geq 2 $ then there exists a constant $c>0$ s.t $$| \xi-\frac {p}{q}| < \frac {c}{q^n }$$ has no solution...

\can anyone help?
 
9:51 PM
[Random]
40
A: Is learning (theoretical) physics useful/important for a mathematician?

Tim van BeekIt is certainly possible to study all kinds of topics in pure mathematics without any knowledge of physics, because you will always find literature/researchers who are used to explain the key concepts to fellow mathematicians without any knowledge in physics. But here are some examples of useful...

There is no set theory on that list, but pretty much everything else is there
 
everyone and there dog has posted on that thing eh?
 
guys, if you find an easier proof for your own question (which has received answers), is it then lame to post it as a separate answer?
 

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