« first day (4875 days earlier)      last day (79 days later) » 

12:05 AM
miktex comes with a simple tex editor (texworks) that has a click-and-compile-and-display feature. full disclosure, i don't know how it compares with overleaf, and i personally use command line. but i have at least experimented with texworks, and it does not require input at the level of the command line.
miktex may be windows-specific, but i think texworks is cross platform
another tried and true strategy is to see if there is an older version of something that you like that runs better on an older machine. even if whoever maintains the package that you like doesn't keep old versions available, there are sites dedicated to archiving old versions of stuff.
 
12:20 AM
BaKoma TeX was nice 10 years ago when I installed / bought it
now it's probably nicer
WYSIWYG type
Overleaf.com is another option
 
 
1 hour later…
1:28 AM
@DanielDonnelly LyX, too.
 
Many moons ago,,I bought Textures for the Mac. At some point, they went out of business. I switched to TeXShop, which is free. It is excellent. Occasionally when I switch computers, I have issues with reinstalling fonts that II’ve bought, but that’s mostly my fault.
 
2:16 AM
@TedShifrin I have a friend who uses TeXShop. He swears by it.
 
3:00 AM
Let $f(x)\in\mathbb{Z}[x]$ and let $a_1$ be integer satisfies $f(a_1)= 0\mod{p}$ and $f'(a_1)\neq 0\mod{p}$. Prove there exists $f(a_n)=0\mod p^n$ and $a_n = a_{n-1} \mod {p^{n-1}}$, for $n\geq 2$. How to start constructing $a_2$? I need some ideas.
 
is there something other than command line?
 
I read the proof somehow. But didn't get that why it only takes the first-order approximation.
nvm I found this on MSE
 
3:17 AM
make the ansatz $a_2=a_1+mp$ (you know $a_2$ has to have this form for some $m$)
 
@XanderHenderson It’s produced three of my four books :)
 
Yes but I don't get that why $f(a_2)=f(a_1)+f'(a_1)mp$ using taylor expansion
 
mod $p^2$ only!
 
I see.
I knew how to do know I think. Thanks!
 
3:34 AM
nice
 
3:52 AM
1
Q: Find the change of basis matrix $P$ (say) for the change from the basis $B=\{(1,0,0),(0,1,0),(0,0,1)\}$ to $B_1=\{ (1,1,0),(0,0,1),(0,1,0)\}.$

Thomas FinleyFind the change of basis matrix $P$ (say) for the change from the basis $B=\{(1,0,0),(0,1,0),(0,0,1)\}$ to $B_1=\{ (1,1,0),(0,0,1),(0,1,0)\}.$ I am familiar with change of coordinate matrix. My definition of change of coordinate matrix is: Let $V$ be a finite dimensional vector space such that, $...

Need help with this one, please.
 
4:33 AM
@Thorgott emmm, what does it mean when for example $f'(a_2)=0 \mod p$ and I obtained in this case $f(a_3)=f(a_2) \mod p^3$
How do I interpret this? It may or may not have root $a_3$ for $f(a_3)=0 \mod p^3$?
 
5:23 AM
Let $X$ be a set and denote $\Delta:=\{(x,x) \ | \ x \in X\}$ the diagonal. Is this proof by contrapositive of $[A \cap B =\emptyset] \iff [(A \times B) \cap \Delta =\emptyset]$ correct?

Proof. Assume $A \cap B \ne \emptyset$, so there exists $x \in A \cap B$ and hence then $x \in A$ and $x \in B$. Thus, by definition of cartesian product, $(x,x) \in A \times B$. By definition of diagonal, $(x,x) \in \Delta$ and so $(x,x) \in (A \times B) \cap \Delta$. Thus, $(A \times B) \cap \Delta \ne \emptyset$.
 
 
1 hour later…
6:33 AM
Consider a countably long sequence of real numbers $\{a_n\}$ which exhibits divergence due to oscillatory phenomena of its terms (that is, there is a upper and lower bound on where the $a_n$ s are). Can the set of all such sequences along with the set of all convergent sequences, be combined to form a large set that preserves as much of the field properties of real numbers?
 
 
4 hours later…
10:28 AM
@ZaWarudo you could write $x = (x_A, x_B) = (x_\Delta, x_\Delta)$ hence from properties of ordered pair $x_A = x_B = x_\Delta$
But sure its correct
 
10:52 AM
@Secret This is the vector space of all bounded sequences $\ell^\infty$
 
11:12 AM
@ThomasFinley I've added an answer
 
X4J
Can one's say that if $D \in \mathbb{M}_n$ is Jordan matrix with Jordan blocks $J(\lambda_1), \dots J(\lambda_k)$ at the diagonal then $\chi_D(X) = \chi_{J(\lambda_1)}(X)\cdots\chi_{J(\lambda_k)}(X)$ where $\chi$ is the characteristic polymomial of the corresponding matrix?
Since I use to get confused when things get more abstract, I am trying to understand if my intuition is correct now
 
12:06 PM
@X4J yes, and more generally that holds for any block diagonal matrix
it should follow from the result saying the determinant of such a matrix is the product of the determinants of the blocks
 
12:28 PM
@oscarmetalbreak that cannot happen
 
12:48 PM
What is the largest value of $n$ for which there exist a set $\{A_1,A_2,...,A_n\}$ of distinct nonzero matrices in $M_k(\Bbb{C})$ such that $A_i^\ast A_j$ has zero trace for all $1\leq i\leq j\leq n$? .
Any idea on this one?
 
1:04 PM
are you sure you don't want $i<j$
 
oh right, yes $i<j$
 
@Thorgott Do you mean $f'(x)=0$ won't happen?
 
1:22 PM
yes, you lift simple roots to simple roots
 
Can I calculate this?
for example $f'(a_2)=$?
I was trying to do the reverse by expanding $f(a_1)=f(a_2-mp)$
 
the point is that under the map $\mathbb{Z}/p^2\mathbb{Z}\rightarrow\mathbb{Z}/p\mathbb{Z}$, the preimage of a unit is a unit
@SoumikMukherjee try showing that $\langle A_i^{\ast},A_j\rangle=A_i^{\ast}A_j$ defines an inner product on matrices
 
@Thorgott Got it, so the answer is $n=k^2$. Thank you
 
Sorry. Can you elaborate a bit? How does the map you said relate to $f'\neq 0$
 
1:40 PM
@SoumikMukherjee Actually, you don't need the full strength of what I said. Show that $A_i^{\ast}A_i\neq0$ if $A_i\neq0$ and then you can argue directly that such a set is linearly independent, hence has size at most $k^2$. On the other hand, the stanadrd basis $M_k(\mathbb{C})$ gives an explicit example example of such a set of size $k^2$.
@oscarmetalbreak $a_2\mod p^2$ maps to $a_1\mod p$ by construction, so $f^{\prime}(a_2)\mod p^2$ maps to $f^{\prime}(a_1)\mod p$. the latter is a unit, hence the former is a unit by the fact I stated.
 
Thorgott Okay
 
2:14 PM
@SineoftheTime I have accepted your answer(+1 from me). But I don't understand your comment below my OP, i.e
your third vector in $B_1$ is different from $e_3$ — Sine of the Time 6 hours ago
 
2:26 PM
@ThomasFinley in $B_1$ there is $(0,1,0)$ whereas $e_3=(0,1,1)$
 
2:55 PM
interesting that 'martingale' is in fact a gambling name
 
Let $f$ be continuous BV and $f'$ exists except for a countable set. Then $f'$ is AC: we know $f'$ is gauge integrable and $f$ is its indefinite integral, so $f$ maps null sets to null sets. From Banach-Zarecki, $f$ is AC
Let $f$ be continuous BV on $[a, b]$ and $f$ is AC on $[c, b]$ for $c \in (a, b)$. Then $f$ is AC: if $Z\subseteq [a, b]$ is a null set then so is $f(Z\cap [a-1/n, b])$ for all $n$, so $f(Z)$ is a null set. From Banach-Zarecki $f$ is AC
 
I've only heard the word 'gauge' in Gauge theory
 
Not that gauge
@Jakobian meant to write $f$ is AC
 
3:28 PM
I have a basic question. Consider a square, invertible matrix $A(t)$ depending on some real variable $t$. If I invert $A(t)$ and then evaluate at $t_0$, is this the same as evaluating $A(t)$ at $t_0$ and then inverting the matrix?
The reason for the question is I'm studying two different textbooks, same subject, same equation. In one, they use the notation $A^{-1}(t_0)$ and in the other the notation $(A(t_0))^{-1}$.
 
4:12 PM
Never mind my question, I actually overlooked that the first notation is not $A^{-1}(t_0)$ but $A(t_0)^{-1}$. This makes more sense.
 
X4J
0
Q: $\ker f^k=\operatorname{Im}f^k$ for $k \in \mathbb N$ for nilpotent $f$ $\Rightarrow$ lengths of chains in basis are equal

atefsawaed If $f:\mathbb R^n \rightarrow \mathbb R ^n$ is a nilpotent linear operator with $\ker f^k=\operatorname{Im}f^k$ for some $k \in \mathbb N$, then all the chains that are in a chain basis of $f$ have the same length? If a look at the Jordan Matrix representing $f$, then I am looking for the si...

@TedShifrin I've found this solution, however I don't understand the explanation (maybe it skips some trivial reasoning which I still havent grasped)
 
5:03 PM
@Jakobian of course there is more direct way to do this, showing this for monotone continuous functions (hence also for continuous functions of bounded variation). But here's a variation on this exercise: Suppose $f$ is continuous, BV and $f$ is AC on intervals $I_1, I_2, ...$ where $[a, b]\setminus \bigcup_{i=1}^\infty I_n$ is countable. Then $f$ is AC
 
@XanderHenderson thought you might have good answers for this question: what sort of entities or commitees exist that coordinate (1) the difficulty of the course such that all teachers give a class that has a similar difficulty for the same subject, and (2) prerequisites, such that the teachers for who your class is a prerequisite will see their desired prerequisites adequately satisfied?
if you have time, of course
 
5:32 PM
@X4J Have you already studied Jordan form?
 
5:50 PM
@shin Aside from uniform final exams (which some universities use for calculus-level classes), it's very difficult to ensure. I often encountered students with Cs or better in prerequisite classes who really hadn't learned much of anything. It was well known to students (and, presumably, to faculty) that a student who got a C from me often was more competent than a student who got an A from some of my colleagues.
Indeed, one metric to measure someone's teaching effectiveness is to study how his/her students performed in the next class. I actually did this sometimes when I was writing the teaching dossier for someone's promotion or a teaching award.
 
Hello @TedShifrin I have a question. How can I come up with research ideas?
 
No clue. Start by working super-hard questions on the material you're studying. The notion of "research" doesn't make sense for a mid-level undergraduate, unless you just mean "I'm going to play around with a hard question I don't know how to do." Maybe some computer experimentation to try to find evidence.
 
@TedShifrin yeah! I thought it because I can never run out of math books which teach me new stuff. So I thought why not to think about research.
 
What does "research" mean?
People love to throw the word around.
 
Lemme try, Research is something we do to find something new or to get to know about more which is already known.
For example people knew about some series which converge and some which diverge. Then they start doing research on what type or condition should a series should follow or obey to converge.
 
5:59 PM
Since most of the math you've learned (probably all) was known by the 18th or 19th century, it's going to take more learning to get to the frontiers. This is one reason elementary number theory (which I dislike) is so popular; everyone can understand the basic statements about integers and modular arithmetic.
I'll be surprised if you can find a result on series to prove that is not already known. There are very esoteric results and tests that most of us don't know exist.
I still believe that you should start by working on hard challenge problems in subjects you've been studying. I always assigned some of those to my students at every level.
 
What is the English name for this theorem? $D$ is a bounded domain such that $f(z)\neq 0$ on $\partial D$, then $\frac1{2\pi i} \oint_{\partial D} \frac{f'}f dz= \text{number of zeros of f - numbers of poles of f}$
 
Argument principle
 
faster that light :)
 
There are also nice generalizations to smooth functions (not analytic) in $\Bbb R^n$ — indeed, on manifolds. Differential topology rules.
 
@TedShifrin oh yeah. Elementary Number Theory is literally elementary. Thank you Sir! I would work on hard challenge problems from now.
Is it okay to see proofs or solutions? Or I should always try to prove it myself
 
6:03 PM
Of course you should do it yourself.
How are you going to solve something unsolved if all you know how to do is read books with answers and internet posts with solutions?
I gave an elementary proof to do by mathematical induction a few days ago. You can warm up with that. Do not look for other people's proofs. Prove by induction: If you choose any $n+1$ numbers from $1$ to $2n$ inclusive, then there is a pair $a,b$ among your choice with $a$ dividing $b$.
 
@TedShifrin yeah, I should have some weapons in my arsenal. I was asking because sometimes I get stuck on problems despite trying too hard. In that case....
 
I didn't believe this when I was first told it. So trying examples of course is where one should start, but it's hard to turn examples into a proof.
 
@TedShifrin hm I see how uniform exams could help with this stuff. I guess there's some sort of statistical adjustment after the fact to level scores according to teachers?
 
elementary number theory is actually the opposite of elementary
it's a trap
 
@TedShifrin yeah sure. I am going to tell you my attempt here after some time :)
 
6:08 PM
@shin That's the political problem. Most math profs are very independent and insist on autonomy. So if they can give any grades they want, despite the uniform final results, it is pretty much meaningless.
When I experienced this as a TA in grad school, we were basically told that we had to assign grades according to the uniform final results as a whole. That is, if my students earned 7 A's, 12 B's, 20 C's, 5 D's, and 2 F's on the final, that had to be (within epsilon) the net grades I assigned my class.
 
hm difficult problem to navigate
 
@Thor But at least the statements are all super elementary. That's not true for research questions in geometry, topology, analysis, algebra, etc.
 
that's why it's so dangerous. they look simpler, but are harder
 
@shin My first year at UGA we had uniform exams in all three quarters of calculus. It was publicized what our average results were for each class, but no one checked to see what grades we gave. I was second in every class (the first was an experienced instructor who was famous for being a superb teacher). Uniform exams went away the next year. Too many faculty disliked it.
And, in fairness, some of the exams were not well written and caused problems.
 
@TedShifrin I can do it by modular arithmetic. But from induction.......
 
6:13 PM
By modular arithmetic? Really? I don't know that.
How?
I like it for induction because it's a meaningful proof by induction, not a formal manipulation of algebra.
 
@TedShifrin I was thinking that: we have $2n$ numbers so we have $a_1,a_2,....a_n$ basically $n$ numbers which are congruent to $s={0,1,2,...n-1}$ $\pmod(n)$ and we are choosing $n+1$ numbers so any one number will be equal to a number which is in $s$.
But I am sorry It doesn't give a proof of a problem. Your problem is very interesting...
 
@LuckyChouhan take n=6, 5 is congruent to 11 but does not divide it
 
Write out your proofs carefully and critique them thoroughly before presenting them to us.
As I said, there is (at least one) proof by other means. I hadn't thought of it, but some of my students quickly gave it to me more than twenty years ago. But I insist on induction here.
 
6:31 PM
@TedShifrin I have a question. Are $2n$ numbers should be consecutive or just random numbers?
 
Did you read the problem I typed?
 
@LuckyChouhan If they were randoms, can't you just take primes?
 
@SoumikMukherjee that's what I just thought while writing.
Yeah it was from $1$ to $2n$.
 
You've got to learn to pay attention to details and not ask questions due to sloppiness.
 
@TedShifrin yeah sure.
 
6:41 PM
I'm so depressed that it's saturday night and I'm studying complex analysis
 
Well, go out for a while with some friends ;)
 
I don't have friends
 
acquire friends
I don't know, for me friends aren't important, I never was a person like that
 
6:57 PM
Well, I'm not here to tell you how to live your life. Go do something that you enjoy.
 
7:07 PM
Can I have some feedback on this, please?
 
7:18 PM
$$2\sqrt{s}\dfrac{\partial}{\partial s} \sqrt{\mp \Omega_s(x)}=\sqrt{x}\sqrt{\pm\dfrac{\partial}{\partial x}\Omega_s(x)}$$

Anyone know how to take the Mellin transform of this equation wrt the $x$ variable? All I can figure out is that the LHS remains the same and the RHS changes in some way, but the square root of the partial derivative is tripping me up.
It's a degenerate equation for $s=x=0$ which is why I'm thinking the mellin transform may help. Also my initial condition/ Cauchy data is $\Omega_s(x)|_{s=0}$ if that helps.
 
8:16 PM
@smth I don't understand your (removed) question.
 
8:50 PM
i need people
 
I think this lemma is false
I think there is a function which maps every neighbourhood of a point to a set of measure $\geq c > 0$ for some fixed $c$
Like $f(x) = \sin(1/x)$ and $x = 0$
 
I do not understand the statement.
 
clearly you haven't fried your brain enough on henstock integrals
 
9:12 PM
I'm not confident what $|f|(\mathcal{P})$ means but I think it should have been $|f(\mathcal{P})|$
 
ah I think I misunderstood
 
@mick prime numbers, mon
There's already Martin Hopf's prime numbers, so let's populate that room instead of starting our own
Just remember though, how long / complex Wiles' FLT proof was. It's gonna take some advanced study to crack any of these prime number open problems
 
I was sleepy the entire day
winter is just so perfect for sleep when the thermo is turned on
 
9:33 PM
@leslietownes Bartle is pretty nice introduction so far, but I wouldn't call it the most general or anything
 
 
1 hour later…
11:00 PM
I have a basic question. There is this corollary in my book which has me pondering.
> Let $A$ be a linear map on a vector space $V$ over $\mathbb C$. Then there are subspaces $V_-, V_+, V_0$, invariant under $A$ such that (a) $V=V_-\oplus V_+\oplus V_0$ (b) all eigenvalues of $A: V_-\to V_-$ have a negative real part (c) all eigenvalues of $A: V_+\to V_+$ have a postive real part (d) all eigenvalues of $A: V_0\to V_0$ have a real part equal to zero.
Now, I'm solving a linear system $x'=Ax$ ($A$ is $3\times 3$) and to compute $e^{tA}$ I have to decompose the space into invariant subspaces. Since the eigenvalues are $1$ (with multiplicity $2$) and $-3$, I get the (generalized) eigenspaces $\ker(A+3I)$ and $\ker((A-I)^2)$.
Now a TA has pointed out that $V_-$ is in fact $\ker(A+3I)$. Maybe this is trivial, but I would like to verify this. The restriction of $A$ to $V_-$ should have only eigenvalues with a negative real part. What is the restriction of a linear map to its eigenspace?
 
What is the restriction of a function (map) in general?
 
It is the same function defined on some subset of the domain, right?
 
0
Q: Hausdorff topological space and graph of a continuos function.

SigmaAlgebraLet $X$ an Hausdorff topological space and $A$ a subspace of $X$. Show that if exists a continuos application $f:X\to A$ such that $f_{|A}=id_A$, then $A$ is a close set in $X$. My second attempt is this: Because $X$ is Hausdorff, the subspace $\Delta_X=\{(x,x)\in X\times X, \quad x\in X\}$ is cl...

pretty sure they've asked this question twice now
must have deleted it
@psie If $f:A\to B$ then restriction of $f$ to $C\subseteq A$ is the function $f\restriction_C :C\to B$ defined by $f\restriction_C(x) = f(x)$ for $x\in C$
 
right
 
a linear map restricted to an eigenspace just becomes a multiplication operator.
since the eigenspace is $A$ invariant, if you start in the eigenspace then the solutions of $x'=Ax$ also remain in that eigenspace and so have a particularly simple form.
hence the preoccupation of control engineers with eigenvalues
 
11:17 PM
hmm ok, could you define what a multiplication operator is more exactly? Multiplication by some fixed number, right?
 
$x \mapsto c x$ for some constant $c$.
 
ok 👍
 
if $x \in \ker (A+3I)$, what is $Ax$?
 
$Ax=-3x$?
 
yup
 
11:19 PM
ok, makes sense
 
does the eigenspace corresponding to $1$ have one or two dimensions?
 
it has two dimensions
sorry
the generalized eigenspace has two dimensions
 
i know that. does the eigenspace itself have one or two dimensions.
 
@shintuku By and large, such systems are, in my experience, rare.
 
@copper.hat it must have one dimension, because the direct sum of the (generalized) eigenspaces makes up $\mathbb R^3$ in this case
 
11:25 PM
i am not sure why you say must, as $1$ has multiplicity $2$, but you have the $A$, not me.
 
@DanielDonnelly ok
 
For intro level classes, I've seen common finals. And I've seen coordinated courses where one faculty administers several other instructors.
 
@psie i don't like repeating myself. i am not asking about the generalised eigenspace.
you are trying to compute $e^{At}$, correct?
 
yes
 
But most US institutions have a philosophy of academic freedom which is subject in conflict with too much coordination.
 
11:28 PM
the eigenspace associated with $1$ (multiplicity $2$) has dimension $1$ and that of $-3$ (multiplicity $1$) has also dimension $1$
 
We do also track student performance across classes and instructors as part of our accreditation.
 
@psie can you solve $x'=Ax$ restricted to the generalised eigenspace?
 
I'm restrict YOUR eigenspace!
 
i live in an unstable manifold
 
@copper.hat I probably could if you gave me some directions :)
my book has other terminology and it's a bit confusing
 
11:31 PM
don't listen to what strangers tell you on internet_ my mom
 
@psie have you found a basis for $\ker (A-I)$?
 
@XanderHenderson Academic freedom is typically misinterpreted/misunderstood.
 
it means tenure
 
It is meant to protect controversial research, not to give the right not to teach the syllabus with appropriate standards.
Academia will take further hits under the dictatorship.
 
@copper.hat yes I have, this is the single eigenvector associated with the eigenvalue $1$ of multiplicity $2$
 
11:36 PM
Sorry, I meant $\ker (A-I)^2$.
 
yes, I also have
 
the solutions will be of the form $e^t v + t e^t w $ for some vectors $v,w$.
 
ok, but one needs to add the solution from the eigenspace $\ker (A+3I)$ to that, right?
alright, we were talking about restricting to the generalized eigenspace, so no solution from $\ker (A+3I)$
 
you have $A v_1 = v_1, A v_2 = v_2 + v_1, A v_2 = - v_3$ for some li. $v_k$. in the basis $v_k$ the matrix looks like [1 1 0 ; 0 1 0 ; 0 0 -3] and solving that for initial conditions [1 0 0]', [0 1 0]', [0 0 1]' is straightforward.
 

« first day (4875 days earlier)      last day (79 days later) »