@LeakyNun I think for étale cohomology you just take a pullback along the inclusion $Z \to X$ of a codimension $i$ irreducible subvariety and mess around with Poincaré duality
Is this a known change of independent variable for solving ODE's with variable coefficients? I saw it on page 93 of this textbook. Does it have a name? it is very nice. the book is by Markarets. differential equations.
und sie sagen, dass $E^1_{p-1,q} \to E^1_{p,q} \to E^1_{p+1,q}$ hat Kohomologie 0
wo $p > 1$
aber dass verstehe ich nicht
and even if I accept this, ich kann noch nicht sehen wie wir können unsere gewünschte Funktion haben
because I don't know how to make sure that the map $H^{2j}(X \times \Delta^{2j-i}, \Bbb F_\ell(j)) \to H^i(X,\Bbb F_\ell(j))$ that I end up with isn't just $0$
this whole spectral sequence thing sounds like they're taking diagonals but then the summands don't match up so you just end up getting 0 maps
and they need to intersect properly with all the subsimplices
so that you form a complex $\cdots \to z^j(X,2) \to z^j(X,1) \to z^j(X,0)$
where $z^j(X,i)$ are codimension-$j$ cycles in $X \times \Delta^i$ whose intersection with all the $X \times F$ has codimension $j$, across $F$ subsimplices
Let $d$ be the dimension of $X$. Then if $Z$ has codimension $i$ in $X \times \Delta^{2j-i}$, $Z$ has dimension $d+2j-2i$
Note that we have if we work with cohomology with compact support $H_c^{2(2j-i)}(\Delta^{2j-i})=\Bbb F_\ell$
Now use Künneth for cohomology with compact support to get a map $H_c^{2(d-i)}(X)= H_c^{2(d-i)}(X) \otimes H_c^{2(2j-i)}(\Delta^{2j-i}) \to H_c^{2(d+2j-2i)}(X \times \Delta^{2j-i})$
now consider the inclusion $\iota:Z \hookrightarrow X \times \Delta^{2j-i}$, this induces a pullback map on cohomology with compact support $H_c^{2(d+2j-2i)}(X \times \Delta^{2j-i}) \to H_c^{2(d+2j-2i)}(Z)$
True or false. There exists a function $f:{\Bbb R^*}^n\to\Bbb R$ such that the Hausdorff dimension of a set $X\in\Bbb R^n$ equals $\sup_{x\in X^*}f(x)$, where $\Bbb R^*$ are the hyperreals and $X^*$ is the hyper extension of $X$
$T\in L(V)$ and suppose that there is a basis w.r.t. which T has an upper triangular matrix M(T) and let dim V=n. Then, if all diagonal entries of the matrix are non zero, then T is invertible. Proof: Showing range T =V suffices to prove the result. Let $v_1,...,v_n$ be the basis (let's call it B) w.r.t. which T has M(T). range T=span ($Tv_1,...,Tv_n$). RHS is isomorphic to $span([Tv_1]_B,...,[Tv_n]_B)$, which is column space of the matrix $([Tv_1]_B,...,[Tv_n]_B)$,
whose column space won't change by doing column operations and since diagonal entries are non-zero, we can convert the matrix to an identity matrix using column operations. Hence the matrix is of dimension $n$. It follows that span $([Tv_1]_B,...,[Tv_n]_B)$ is of dimension $n$ that is, $range T$ is of dimension $n$. This proves the claim. :)
this is where LADR might diverge from a 'normal' linear algebra book. most people think of invertibility of A as a question about solvabilityo f Ax = b, row and column spaces, gaussian elimination. which is one view.
i gave the axler-ish view above.
the nonzeroness lets you do 'backsubstitution' to solve Ax = b for x.
a refinement of the result is that under those hypotheses, dim T {v_1, ..., v_k} = k for all 1 <= k <= n. the k = n case tells you that T is invertible.
Let $f(z) = e^{\frac{z+1}{z-1}}$. I am trying to argue that $|f(z)| < 1$ for all $z \in \Bbb{D}$. It's not difficult to argue that $|f(z)| = 1$ if $z$ is any point on the unit circle other than $1$. I would like to just simply apply the maximum modulus principle at this point; however, there is a singularity at $z=1$ which I don't believe is removable, so $f$ isn't continuous in the boundary of $\Bbb{D}$.
Am I mistaken about $z=1$ being a non-removable singularity? If not, how do I argue $|f(z)| < 1$ for all $z \in \Bbb{D}?
semi: determinants are basically unavoidable in multivar changes of variables. other than that, they are objectively bad, even if axler is ok with them (and therefore wrong)
user, would the same methods you use to conclude |f|=1 on the boundary not apply inside the disc? since |e^z| = e^(Re z), isn't this just about where the fractional linear transformation (z+1)/(z-1) sends the unit disc (not just its boundary, but its interior)?
shouldn't that thing map a disc to either another disc or a half plane, and then it's just, what disc or what half plane
Wait, I misspoke. I wasn't able to show that $|f|<1$---that's precisely what I am trying argue! I am being an idiot. Give me a second and I'll type up what I have (max. mod. isn't needed/helpful).
i was wondering about that. i'd figure out what that fractional linear transformation does. enough to figure out what happens to three points, like i, -i, and -1.
I was able to show that if $z \in \Bbb{D}$, then $Re(\frac{z+1}{z-1}) < 0$. Hence, since $x \mapsto e^x$ is a strictly increasing function, we have that $$|f(z)| = |e^{\frac{z+1}{z-1}}| = e^{Re(\frac{z+1}{z-1})} < e^0 = 1$$ for every $z \in \Bbb{D}$.
it looks like i, -i, -1 all get mapped to the imaginary axis. because 0 maps to -1 this means the image of the unit disc under that transformation is the left half-plane.
yeah.
i guess it's enough to figure out what happens to four points. you need to know not just where the boundary goes but what side of it you end up on under the transformation.
i remember mentally completely skipping the section on fractional linear transformations when i took my first complex analysis class.
it seemed boring. it still kind of does. don't tell anyone.
don't get me started on contour integration. use a semicircle! use a keyhole contour! use some adrenaline, injected right into my heart, or i will fall asleep from how boring all of this is.
the word on the street was "complex analysis is easy, it's just calculus again," unless you had the ill fortune to actually take it from someone who bothered to teach it.
not enough is said about the mediocrity of postdoc teaching. i guess because it's unkind, and although highly general it feels like you are singling particular people out.
the game at my undergrad was to try to take every one of the major requirements from a postdoc, because they were easy. not hard to accomplish that goal, either.
houshou what is unintuitive about this? i'm not contesting the point, just wondering. which sequences having small probabilities, or being potentially more likely than other sequences, seem unusual to you.
Okay, so now that I got the first part figured out, how about the second part. The second part asks me to show that $|f(z)| = 1$ for almost every $z$ on the unit circle. But this is easy, right, because $Re(\frac{z+1}{z-1}) = 0$ for every $z \neq 1$ on the unit circle, and certainly $\{1\}$ is a set of measure $0$. Does this seem right?
negative effects of the time change: daughter wakes up before 6 am. positive effects: she is usually ready for day care by 7:45. (this never happened before)
Question: If you were trying to show something using a table in Latex, would you rather build the table using the \begin{tabular} or insert a screen shot image?
i'd build in some form in native latex. easier to modify later and more accessible for users.
unless there's something specific about the graphics in the presentation that seems hard to capture in latex. then i'd just render wherever i wanted and drop it in.
someone probably knows this better than i do but i wonder how latex integrates with spreadsheets or csv. e.g. for a table of numbers that might be recalculated in a spreadsheet when new or different data arrives.
absolutely nobody uses this functionality in word/excel. i showed my wife how to do it, and i still see her manually typing in three-digit numbers with decimal points in them, and redoing it.
there are sometimes reasons for this. my job has a document management system that lets you open multiple documents from the system at once, but won't let them 'talk' to each other because it creates disjoint copies in weird places that destroy any links between them.
i would hope so but don't know. something in this ballpark might be a good question for the latex SE. in particular, if anyone has implemented linking latex tables to documents in other places, or something resembling conditional formatting.
i could see something like that interacting with an excel-generated .CSV where you do the conditional formatting on the excel side. but there would be a need for some kind of script in the middle to generate the tex.
ah. that's handy. i was thinking of stuff like, highlight one color if p < .001 and another if .001 <= p < .01. or some kind of continuous shading of that. maybe nobody is doing this.
Yeah. I'm using to show a pattern on a table. so... it looks messy, code-wise as I have to add it into each cell of the tables. But it is accomplishing what I want it to do. Making the pattern easier to see
Now it's a matter of formatting the table and introducing my predictions.... Gods. I wish I could just submit my excel journal
i've been telling my wife for over 10 years that she needs some way of automating the process of putting things from spreadsheets into papers. word does allow for this but she's never taken the plunge. it would be simpler if people could just submit excel.
Personally, for me. It really would. I've already done all this work with formulas in Excel. Showing a pattern. A pattern of growth. How that Pattern trends to 33% / 67%.
Which then translates into a Probability Trend of 0% and Non-0%
When applying number theory and infinity on a certain Non-Linear Recurrence Relation.
Transcribing it to a Tex Document for submission is a pain
Studying a bit of math a question came to my mind. We use mathematical analysis in physics frequently, which assumes quantities infinitely small. Hence, we are assuming that matter behaves like that, we can "think" in infinitely small quantities of matter. Right?
mm, i dunno. analysis as normally done does not involve infinitely small anything, only arbitrarily small things. in physics the models are chosen to mimic reality in some aspects but not others. e.g. the fact that mass, matter, angular momentum, etc. might be quantized in some models and continuous in others is just a choice of modeling.
Like, we are asumming things infinitely small in physics but to solve that paradox we can't assume that the paint pot is made of infinitely small particles
the revolution stuff is a subtlety but not really the point. for any positive a_n with sum a_n convergent it is possible to imagine unit boxes with heights a_n that fulfill the same role.
The horn has finite volume but infinite surface area, which means we would require a finite amount of pain to fill the volume but infinite amount of paint for the surface
But wait, yeah, you have to assume that paint to be infinitely divisible which doesnt happen in physics
The issue is that we would require an infinite amount of paint to paint the sides but we would require a finite amount of paint to fill the volume generated by rotating the surface
But filling the volume should also paint the sides, but the correspondent integral is divergent
Which is contradictory
By the way i found this paradox in my calculus book of Robert Adams but just came to my mind
I think i am assuming a horn with a layer of finite width in my head
Gonna give an example of what is confusing me about "infinitely small quantities": When computing the mass of a wire of variable composition streched along the x-axis, (assume from $x=0$ to $x=L$) with density $\rho=kx$, we solve the integral $m = \int_0^L kx dx$. This is a riemann integral.
@Odestheory12 You still haven't stated it precisely. Why is it contradictory? Your last sentence seems to be the relevant thing. Paint molecules have a positive diameter. So what happens eventually?
Oh. Suppose $V$ is an irreducible representation of $C_n$ choose $v\in V\backslash \{0\}$. Then $V'=span(gv: g\in C_n)$ is finite dimensional and is a subrepresentation of V$.
Ah, OK, and so you're saying the volume has to be infinite. But I'm looking for the physical explanation of the paradox. You haven't quite given it to me.
That's the physical explanation. We can not pain the surface of the horn. There is no painting molecules with 0 diameter. Also it would require an infinite amount of time
@TedShifrin Uhm yeah, that's confusing me. I am associating the integral with the correspondent Riemann sum, which yields me to think we are "partitioning" infinitely that wire
But the mass can't be divided infinitely, which forces my brain to go in 404 mode
Any Riemann sum is an approximation, just as a secant line is an approximation to the tangent line. You don't have to think about "infinity" or "infinitesimals" ever.
2
Mathematics is about approximating things, hopefully better and better.
Is it always relatively simple to solve quasilinear first order PDEs using the Lagrange method, or does the method only work for semilinear problems and the occasional, nice strictly quasilinear problem?
I went to office hours and my CS professor asked me to excuse his slow speaking, as he had been at the University all-night working on a project. I couldn't help but be slightly inspired.
Actually, answering to my last question, yes, the matter is assumed to be infinitely divisible. I didnt study continuum mechanics yet and it seems that concept of infinitely small in physics is studied there
