when we do substitution of variables in a double integral, we introduce mod of Jacobian determinant. Suppose that we are changing coordinates from (x,y) to (u,v) then what determines the order of integration in (u,v)?
Suppose that we do a substitution $xy=u$ and $y/x=v$. The region in x-y was given as bounded by the hyperbolas $xy=1$ and $xy=9$ and the lines $y=x$ and $y=9x$. Then I believe that $\int \int dx dy= \int_{v=1}^{v=9}\int_{u=1}^{u=9} |J|du dv$
But why are u-limits on RHS not like $\int _{u=9}^{u=1}$?
2. There is an arbitrary function of 4 variables $f(u,v,w,x)$ and:
$u∈[-2\pi,2\pi]$
$v∈[-2\pi,2\pi]$
$w∈[-2\pi,2\pi]$
From these ranges, an arbitrary combination of $uvw$ is taken, and this combination is then substituted into the function $f$, and then $x$ is searched for such that $f$ has a real solution. Thus, we study the space of real solutions of the function $f$. How to generate triples in such a way that the sample of these triples has a minimum size, but at the same time covers the specified ranges of $f$ as widely as possible?
But result for constant function 1 may not hold in general for $\int\int f(x,y) dx dy$. :(
Suppose we change it to u, v: $\int\int F(u,v) |J| dudv $, then we won’t know limits of integration on the latter on the basis of knowing them in case of constant function.
Ah, I think, I got what you’re saying.
We are only finding limits of integration which is same as saying that we are doing something on ‘region of integration’.
And the region of integration has nothing to do with the integrand.
So if some sign convention holds for the constant function, it does for any function f as well.
@Govind75 Ansatz is a German word meaning "beginning" and is cognate with English "onset"; but mathematically it means an initial hypothesis/assumption
If I have a (probability) measure space $(X, \mu)$ (I guess the measure space comes from a topology on $X$), and $Y \subseteq X$, what are Borel subsets of $Y$? Are they simply sets of the form $Y \cap B$, where $B$ is a Borel subset of $X$?
That's what I suspected. I guess one way of thinking about this is to take subspace topology of $Y$ and form the Borel $\sigma$-algebra associated to the subspace topology. This should give the same exact thing.
@leslietownes But does the notion of a Borel set make sense without an underlying topology?
not as far as i know. if you think of borel sets as just random elements of a sigma algebra then the generalization would be a sigma-algebra that might not arise from a topology.
Lol okay...Yeah, that concept shouldn't really come up. I'm only thinking about probability measure spaces in connection with (abelian) von Neumann algebras, because I am trying to understand the crossed product construction.
This is because picard's theorem guarantees existence of a solution in an interval around $t_0$ (initial point). But for an order 1 linear ode, we get uniqueness and existence for the entire interval where all coefficients are continuous.
@leslietownes thanks. I feel much better. Everything salty is tasting too much salty. And hot water and ginger tea taste just fine.
picard's theorem is one of those things that is so variously stated that it might as well be "gauss's theorem," in terms of vagueness. although leaky did include a specific result.
my sense of taste appears to be fine, i just don't want to eat.
my daughter was funny last night. when going to bed she said "do i have covid?" we haven't talked about this at home. i said, maybe. she said "i don't have covid, i have the coronavirus in my mouth." ... OK.
the lateral flow tests are not very reliable, assuming you get a good swab, which usually you don't. false positives are rarer than false negatives, though, i think.
we had a client who wanted to sue someone with a patent that they were interpreting broadly enough to cover lateral flow tests. i pointed out, home pregnancy tests were doing exactly this for about two decades before your patent, which could be a problem. a bit of silence after that.
they didn't. they really wanted to. it was awkward. ordinarily people at my level of seniority don't say these things directly to the client, but i was being pushed.
so i said, explain to me how what you just said is any different from a home pregnancy test.
you can find a law firm to say almost anything, although i think you would be laughed out of any court for asserting a patent supposedly covering all lateral flow antibody/hormone tests.
koro that was exactly it. the patent was owned by an investment entity and they needed to do something to hit some target.
their investors were patient, but getting nervous. they wanted to see some result.
in the US, how and when it appears depends on the agreement between the client and the firm. things are very flexible.
we've done some work where we aren't paid until maybe a year or two after a 'win' in court. and other work we are paid on a monthly basis irrespective of outcome.
some of the agreements we do in the US would be illegal in other countries, or would have been illegal in the US in the past. it is regarded as unseemly for lawyers to own a stake in the outcome of a case.
because I think that if such firm is listed on Nasdaq (in your case), then looking at their balance sheet and seeing that their income from non-operational sources is increasing would discourage people from investing in their stocks.
oh, unless we're talking about the client and not the law firm. this client was not publicly traded. their investors were a handful of people personally known by the founder.
'closely held' companies are more likely to be dumb in court than publicly traded ones. because their decisionmaking boils down to whatever pleases a small group of people, with little repercussions if there is mismanagement.
something you see a lot in the US is investment groups populated by high-income people such as medical doctors or dentists, making very bad decisions, because dentists don't know much of anything other than dentistry. then they sue each other and lose.
mostly because the vast majority of corporate law cases concern publicly traded companies where it is assumed that investors are essentially gambling and nothing short of gross negligence will be a problem.
even if your dentist friend used the wrong contractor.
my dad's cousin is a medical doctor. i gave him free legal advice: do not invest with your doctor friends in whatever it is they are planning to do. he thanked me later.
I have an assignment due on Thursday. I was stuck for hours yesterday because I was asked to solve for a term that had not been defined in class. I tried Google, but kept getting inconclusive answers. It turns out the definition was in the readings assigned for today...
i think it comes at a weird point in the curriculum. you can't assume much about what people have already seen, other than calculus. so the connections are hard to make.
and if you do have linear algebra it's finite dimensional linear algebra which differential equations are mostly not about.
Yep, I asked my professor about this as well. He was describing a solution set to an equation and said it's basically an infinite-dimensional span. Way beyond what we did in intro LA.
(That description may or may not be accurate. He used some fancy terminology.)
@LearningCHelpMeV2 Here's a hint: Note that the roots of $x^3-3x-2\cos 3\alpha=0$ are twice of roots of $4x^3-3x-\cos 3\alpha=0$. Using this hint, you can even generalise $(iii)$.
I want to integrate $\int\int_S e^{\frac x{x+y}}dx dy$, where $S=\{(x,y): x\ge 0, y\ge 0, x+y\le 1\}$.
The solution given involves substituting $u=x+y, y=uv$. But I think that's not correct because the transformation mapping is not one one.
Quick question. Does this pattern look familiar to anyone? I could swear I've seen it somewhere. A type of tree structure. desmos.com/calculator/dquroyjauz
According to OEIS, searching the sequence 1,3,7,15,7,3,1 gives one result: symmetrical triangle sequence oeis.org/A143086
@TedShifrin Thank you for reply. I don't know what text will be used but I think it will be again my lecturer notes(in Armenian). And what I need from geometry(euclidean,3d,analytic)?
Not much specifically about geometry, other than things that show up in linear algebra (lots with dot products). You can look at my differential geometry text (linked on my profile) and see what's in there.
Does anyone know this result: the Gauss-Lucas theorem also holds for functions of the form $p(x)e^{ax+bx^2}$ where $p$ is a polynomial? More generally for functions of the form $p(x)e^{g(x)}$ where $g$ is a function whose derivative preserves the arguments of the input in a certain way
@Koro First of all, you can just see which function solves the ODE. But it looks like the OP didn't do the computation correctly. But I think the guy you're arguing is right that this has nothing to do with reduction of order.
@Koro If you restrict the domain appropriately, the mapping is certainly one-to-one.
Ah, OK, @Derivative. No, I haven't seen generalizations of it.
Just look at the usual proof and see how it applies to these other functions. The exponential will factor out, so it's only going to depend on the derivative of the argument of the exponential.
@Koro Oh, you're right. Because the region is symmetric about $y=x$ you cannot. Stupid change of variables. Just use a linear one instead. $u=x+y$, $v=x-y$. Much easier anyhow.
Stupid "solution."
@copper.hat You're thinking of curds and whey, of course.
Alright, let's do that: Jacobian determinant J should be $1/2$ so $|J|=1/2$. $x=0$ gets mapped onto $u=-v$, $y=0$ gets mapped to $u=v$ and $x+y=1$ gets mapped to $u=1$. We have a triangular region in $uv$ plane
Since on $x-y$ plane $x+y=k$ represents a line parallel to the line $y=-x$ we have that
$u=x+y \implies 0\le u\le 1$
then for any $u$ we have
$0\le y\le u\implies 0\le v\le 1$
I got a very basic linear algebra question. This is regarding transversality condition in Guillieum/Pollack (p28). It says in my book that "But dg: Ty(Y) -> R is surjective linear map whose kernel is Ty(Z). Thus dg carries a subspace of Ty(Y) onto R^l precisely if that subspace and T_y(Z) span all of Ty(Y)". Why does it and T_y(Z) have anything to do with dg already being a surjective map? If dg is already a surjective map, isn't it already going to carry that subspace onto R^l?
So you have a surjective map $L:E\to V$. You know its kernel $U\subset E$. Let $V\subset E$ be a subspace. On what condition is the restriction of $L$ to $V$ surjective?
So you have a surjective map $L:E\to F$. You know its kernel $U\subset E$. Let $V\subset E$ be a subspace. On what condition is the restriction of $L$ to $V$ surjective?
Exercise for Hawk when everything's done: Suppose $S\colon U\to V$, $T\colon V\to W$ are linear maps. Then $T\circ S$ is surjective $\iff \text{image}(S)+\ker(T) = V$.
I'm trying to solve the general DE $x' = a(t)x +g(t)$ w/ initial condition $x(t_0) = x_0$, the question recommended using ansatz $\lambda (t) = c(t) e^{\int_{t_0}^{t}a(s) ds}$. I've obtained my solution (very general), but I am trying to prove uniqueness. The lecturer suggested to examine $\frac{d}{dt}[\frac{\mu (t)}{\lambda (t)}]$ for another solution $\mu (t)$ to show we get $\lambda (t) = \mu (t)$ but I am struggling so heavily with it
@TedShifrin thanks a lot professor Ted. I completed my solution and it also matches with the answer in the book. I think I should write to the author of the book about justifying the weird substitution (espacially its one-oneness). I have taken some classes from the author so I can contact him. :)
@TedShifrin and @leslietownes I passed my complex prelim. Thank you two for help you've provided over the past couple months I've been studying for it!
One first step is solving that specific example: on which subspaces of $\mathbb R^2$ is the restriction of the identity still surjective? Second step is to understand the problem when the map is any isomorphism. Last step is the general case.
Problem:
Let $M\subseteq \mathbb{R}^3$ be a smooth surface. If $p\in M$ and $(U,x^1,x^2,x^3)$ is a chart in $\mathbb{R}^3$ about $p$ for which $x^{3}(p)=0\iff p \in U\cap S$ then $(U,x,y,z)$ is a chart for which $p\in U\cap S$ iff $z(p)=0$
My attempt:
Let $p\in M$. Choose a chart $(U,\phi)=(U,x^...
my daughter was yelling about how she has the coronavirus this morning. she means she has this rubber ball with protrusions that resemble how the virus is depicted. i regret calling her toy that and told her not to say that school.
First of all, @monoidal, you really should have not just $p$ but all of $M\cap U$ satisfy $x^3=0$. Second of all, are you telling me that $x,y,z$ is the standard coordinate system on $\Bbb R^3$? Of course not. So I continue not to understand the point.
@TedShifrin I've been told (in a paper by Brian Conrad) that every complex abelian variety is isogenous to a principally polarized one and that this can be proved "complex multi-linear algebraically" using Pfaffians; would you know about those?
Pfaffians are exterior algebra or old name for differential forms ... but I guess the whole point is that it's an abelian variety, not just any old torus. I haven't thought about this stuff in many years.
@TedShifrin Let Λ be a full lattice in V = C^g with a Hermitian form H on V, such that for any λ1 and λ2 in Λ, H(λ1, λ2) has integral imaginary part. Let Λ' = { v in V | im H(v, Λ) ⊆ Z }, so naturally Λ ⊆ Λ', and the statement is that [Λ' : Λ] is a square (I suspect there is a stronger statement)
Say I have a probability space $(\Omega, \mathcal{A}, \mathbb{P})$ and a random variable $X \colon \Omega \to \mathbb{R}$. Can anyone recommend some consistent notation for the induced probability distribution $\colon X(\Omega) \to [0,1]$, pmf of $X$, pdf of $X$ and cumulative distribution function of $X$?
the DE you're trying to solve is not linear because there is a constant term, $g(t)$. The difference between solutions of the DE with a constant term, however, satisfies the homogeneized system (the same DE without the constant term)
Since on $x-y$ plane $x+y=k$ represents a line parallel to the line $y=-x$ we have that
$u=x+y \implies 0\le u\le 1$
then for any $u$ we have
$0\le y\le u\implies 0\le v\le 1$
