Does anybody know how to smooth out the Ricci scalar R (for a 2D surface) that diverges when it approaches a hypersurface H (i.e. I am thinking of decreasing the Ricci scalar when it gets closer to H, then passes H smoothly, and then I could increase R again)? The metric tensor has the following form: g=E(u,v)(du)^2+G(u,v)(dv)^2.
Yeah, that's pretty rough, Leaky. I just remembered that you're a first year student there, right? There have been rumors that my university is going to close down after spring break for a week, too. But it's pretty strongly indicated that, if we do shut down, we'll be expected to continue working online.
Hello. There's a bounty of mine running out of time. It's not something I think I can answer myself without help but I reckon some of the right people could if they wanted. Here it is:
This is Exercise II.6 of Mac Lane and Moerdijk's, "Sheaves in Geometry and Logic [. . .]". According to the first few pages of this Approach0 search, it is new to MSE.
The Details:
The functors $\Lambda$ and $\Gamma$ are discussed in the following:
Just what is Mac Lane & Moerdijk's $\Lambda$...
I think I have to find the terminal object (binary products, equaliser, resp.) of the domains of each of $\Gamma$, $\Lambda$, and the associated sheaf functor then show that they each map them to that (terminal object, binary products, equaliser, resp.) of their codomains.
Consider a probability space $(\Omega, \mathcal{F}, P)$ and a random variable $X \in\mathcal{F}$. Let $\mathcal{G}\subset\mathcal{F}$ be a smaller $\sigma$-algebra. Show that if $X\in\mathcal{G}$, $Y (ω) \in\,L^1(\Omega, P)$, and $X(\omega)Y (\omega)\in\,L^1(\Omega, P)$, then, $$E[XY\,|\,\mathcal{G}]=XE[Y\,|\,\mathcal{G}]$$
I had tried using indicator and condition on $G$, still no luck
The following are the definitions of conditional expectation and conditional probability, could you fit your case in?
Definition 1: If $\mathcal F \subseteq \mathcal G$ are two $\sigma$-fields, and $X$ a $\mathcal G$-measurable integrable random variable, then $\mathbb E[X | \mathcal F]$ is d...
I have got a conceptual problem. $\rho$ is the mass density and $\mathbf v$ is the velocity (although we don’t need to specify any meaning to them) then an equation is written like this $$ \frac{d \rho }{dt} = \frac{\partial \rho}{\partial t} + \mathbf v \cdot ~grad~ \rho$$
Now, the book writes “if $\rho$ changes with space and if $\frac{d\rhp}{dt}$ is zero, then” $$ \frac{\partial \rho}{\partial t} = - \mathbf v \cdot ~grad~\rho$$
“Which describes the temporal fluctuations of $\rho$ at a fixed point of obersvation, while the material rate of change ($\frac{d\rho}{dt}$) vanishes”.
My problems is how if the derivative of $\rho$ w.r.t. time is zero then how it’s partial derivative w.r.t. to time is non zero. I think it’s more a question of Calculus.
for instance if $\rho(t, x(t)) = t + x(t)$, then ${d\over dt}\rho = 1+x'(t)$. If $x'(t) = -1$ then the total derivative is 0 and but the partial derivative wrt t is always 1
for the other one, you definitely need to allow (x,y,z) to depend on time. conceptually, that means you're not considering the density rho at a fixed point (x0,y0,z0). rather, you track the trajectory of some particular point (x(t),y(t),z(t)) over time and consider what density you'd observe as that point changes position
@Semiclassical The point (ii) of fact 4.2 is wrong. Consider $$ 2 \lt 3 \\ -2 \lt -1 \\ -4 \lt -3$$ there is no flipping of inequalities but accorinding to it the inequality should flip
that relies on the following: x^6 >y^6>0 iff x>y>0
(it would not work without the positive condition. -1>-2, but (-1)^6 < (-2)^6. you're only asking about positive numbers, however, so this doesn't matter)
First, an easy case: Suppose you picked a certain observation point (x0,y0,z0) within your fluid and let that observation point remain unchanged for all time
then the velocity of that observation point is zero, so the statement collapses to $d\rho/dt=\partial\rho/\partial t$
that is: if the density at a fixed observation point changes, it's because the density has actually changed over time
on the other hand, suppose you have a fluid for which 1) the density doesn't change over time, but 2) you move your observation point over time
in that case, you'll still see changes in density if your observation point moves between points of different density
if the little sensor stays fixed at that point for all time, then the only way that I'll see changes in density is if the density at that point really is changing over time
By contrast, suppose I consider a column of atmosphere. Then it's well-known that the density of air is higher at the bottom of the column than at the top.
as such, if I dropped my sensor from the top of the column and let it fall to the bottom, then my sensor will report that the density is changing
that's not a matter of the density -at a given point- changing: it's because I'm changing which point I'm looking at
So in that case I'd have $\partial \rho/\partial t=0$ (the local density is not actually changing) but still $d\rho/dt\neq 0$ (because my observation point is changing)
mathematically, the idea is merely that I'm not considering a fixed point (x0,y0,z0) but rather a moving point $(x(t),y(t),z(t))$
And thus the 'observed density' is $\rho(x(t),y(t),z(t),t)$
Yes, $$\frac{\partial \rho}{\partial t}$$ is zero because it means “keep everything constant and let the time vary” well in that case we know density will be same at a point
And $$\frac{d\rho}{dt}$$ is not zero because density is changing with time (no restriction is there about the space all we want is if it is changing with time)
but you could also imagine cases where the density is not changing with time, but the observed density is still changing because you change what point you're looking at
so those are two special cases: 1) observed density is changing because the local density at a fixed point is changing, and 2) observed density is changing because the local density is fixed but I'm changing what point I'm observing at
the first gives you $d\rho/dt=\partial \rho/\partial t$. the second gives you $d\rho/dt = \vec{v}\cdot \nabla \rho$. together, you get $d\rho/dt=\partial \rho/\partial t+\vec{v}\cdot \nabla \rho$
in terms of the chain rule, we have $$\frac{d\rho}{dt} = \frac{\partial \rho}{\partial t}+\frac{dx}{dt}\frac{\partial \rho}{\partial x}+\frac{dy}{dt}\frac{\partial \rho}{\partial y}+\frac{dz}{dt}\frac{\partial \rho}{\partial z}$$
which is the same as the previous formula
(again, this is using the fact that $\rho$ is a function of $(t,x(t),y(t),z(t))$)
@nbro assuming you mean by function a map $\Bbb R \to \Bbb R$, yes. A continuous function is determined by its restriction to the rationals which are countable, so there are at most $(2^{\aleph_0})^{\aleph_0}=2^{\aleph_0}$ continuous functions. In contrast, there are $(2^{\aleph_0})^{2^{\aleph_0}}$ functions $\Bbb R \to \Bbb R$ which is bigger than $2^{\aleph_0}$ since it is at least $2^{2^{\aleph_0}}$ which is bigger than $2^{\aleph_0}$ by Cantor's theorem
and if you allowed $y$ to depend directly on $x$ i.e. $y=f(x,u(x),v(x))$, then that becomes $dy/dx=\partial y/\partial x +(\partial y/\partial u)(du/dx)+(\partial y/\partial v)(dv/dx)$
which is directly analogous to your case of interest (albeit with different labels)
but physically the interpretation is just: fluctuations in observed density can be either due to local temporal fluctuations or due to your observation point changing over time
(actually would you be interested in reading some "set theoretic algebra", I don't know how to call it, I'm reading the book "almost free modules: set theoretic methods")
"Spring break will be extended on the Duluth, Rochester, and Twin Cities campuses until Wednesday, March 18. Students will resume classes, through online or alternative instruction, on March 18."
"We are suspending in-person instruction, including field experiences and clinicals, across our five campuses and are moving to online, or alternative, instruction." So I can't see how that wouldn't include lab
I just don't know how they do lab on-line, @Semiclassic. That discussion came up with some of my math friends on Facebook who are trying to figure out various ways of delivering lectures on-line.
I don't know how much help I can be, but I've created a chat room where people who suddenly find themselves in a situation where they have to entirely change their teaching toolset in the next few weeks can post resources, or at the very least, collectively whine.
I suppose this meta-thread can ...
I wonder if the Math Educators site has discussion about on-line delivery. I already got an email from DogAteMy's instructor at Yale asking if he can use my videos for his course. :P
for the second course, though, the lab grade is based off of 1) an online pre-lab quiz on the relevant material, 2) an in-lab portion guided by online content, 3) a TA check for participation, and 4) a post-lab online quiz
that one is more readily adapted to an online-only format
uhm, I'm not sure. They're dual to projective modules which are a straightforward generalization of the free modules (they basically share the homological nice properties of free modules). Injective modules are acyclic for _every_ left exact functor (that's an iff as you can take Hom functors), so if you want to understand the "homological properties" of a left exact functor, then it behaves simply for injectives in some sense. So resolving an object by injectives is a natural thing to do. Injectives resolutions are even nicer than any old acyclic resolution, too: any map on the base object…
@Semiclassic: In the days that UGA had a calculus lab accompanying the first calc, Newton's method was covered only in the lab (which makes sense). Of course, the one time I volunteered to teach it, I quickly learned that they (i.e., some of they) didn't know how to find the tangent line to a graph :(
@loch: In principle, it was supposed to improve their understanding of the basics, although so many faculty teaching calculus did their best to sabotage the lab experience :(
1101: algebra-based physics. (i think it's only a 1-semester course) 122x: two semesters of physics for pre-med/biology majors 130x: two (maybe 3?) semesters of physics for scientists and engineers
i watched that on the bus and couldn't stop chuckling near the end because I could tell what was going to happen
(to sum up: rubber band starts out grey, aka about midway between the two temperature ranges. they pull it apart quickly, it becomes warmer and so turns white on the thermograph; when they immediately relax it, it returns to being grey. then they pull it apart suddenly, wait for the band to cool off, then release tension)
(it's already at room temperature at that point, so now releasing tension means that the rubber band becomes -cooler- than room temp. hence, black :) )
Hi guys, I need to do Taylor expansion to order $3$ centred to $0$ of $e^{(\frac{z}{z-1})^2}$
I'm following these steps: 1) I'm searching Taylor expansion of $(\frac{z}{z-1})^2$ in 0 order 3, I take it fast thanks geometric series. Result is y=$-z-z^2-z^3+o(z^3)$
2)I substitute y in $e^y=1+y+y^2+y^3+o(y^3)$
The problem is in the substitution: too much arithmetic, a lot of calculus. I'm following a slow way? There is a faster way than this?
The only way that those can't be poles is if you have a 0/0 expression — that is, if either $(z-1)^3$ or $e^{(z/(z-1))^2}$ is 0. In those cases, you have to run the Taylor series to see whether it's a pole or whether it has a valid limit.
This is Exercise II.6 of Mac Lane and Moerdijk's, "Sheaves in Geometry and Logic [. . .]". According to the first few pages of this Approach0 search, it is new to MSE.
The Details:
The functors $\Lambda$ and $\Gamma$ are discussed in the following:
Just what is Mac Lane & Moerdijk's $\Lambda$...
The following are the definitions of conditional expectation and conditional probability, could you fit your case in?
Definition 1: If $\mathcal F \subseteq \mathcal G$ are two $\sigma$-fields, and $X$ a $\mathcal G$-measurable integrable random variable, then $\mathbb E[X | \mathcal F]$ is d...
