and no, that if you have a matrix with polynomial entries, first evaluating the matrix entrywise and then taking the determinant or first taking the determinant and then evaluate the resulting polynomial yielding the same result is not the homomorphy of the determinant, it's naturality
if R is a commutative unital ring, A is a matrix with det(A) a unit, can't you localize, pushforward A by SL_n(R) -> SL_n(R_m), use elementary generation of SL_n(R_m), pullback?
today,I see a amazing math problem:
show that
$$\sum_{n=1}^{\infty}\dfrac{\binom{4n-4}{n-1}}{2^{4n-3}(3n-2)}=\dfrac{\sqrt[3]{17+3\sqrt{33}}}{3}-\dfrac{2}{3\sqrt[3]{17+3\sqrt{33}}}-\dfrac{1}{3}$$
This problem is from here.
But I consider sometimes,and I think it maybe use Taylor therom
...
#4 I actually have used in the Spivak course. One of my former students (who got his PhD at Princeton and quit academia) stumped me on that for five minutes originally.
How come $y'=y^2 / 2$ blows up in finite time but $y'=y^2-1$ doesn't? I mean if we start with a value like $y(x_0)=10$ then the derivative of the second is much bigger
@Emolga @Leaky: I don't follow. If I take $c=\frac12\log(9/11)$, this is what I have for $y_0=10$. There are solutions with $|y|<1$ but also different solutions with $|y|>1$.
Emolga, your solution is only valid for part of it. Remember that there's an absolute value when you integrate to get log.
But, yes, this shows that antiderivatives can be very sensitive to a little wiggle in the function.
you're wiggling the vector field, not y; $f \mapsto \int f$ is of course a continuous function on $C^0$, so whatever the symbols look like in the end, the antiderivative of a function only changes by (a constant multiple of) as much as the function changed
I'm just objecting to calling this an antiderivative instead of an integral curve
I'm talking about perturbing $\arctan(x/c)$ to $-1/x$ to $\log((x+c)/(x-c))$ even though the derivatives are "close" in some sense (obviously not uniformly) as functions.
Actually, that was always one of my favorite puzzles to Calc II students. How is it that the antiderivatives of $1/(x^2-1)$ and $1/(x^2+1)$ look so different. Of course, they don't if we use $\Bbb C$.
It's some sort of lack of stability in some sense. Of course I'm intentionally being vague. It can probably be phrased in terms of non-transversality to some Thom-Boardman stratum in function space.
Well, certainly, over $\Bbb R$, anyhow, something is explained by looking at $1/(x^2-\varepsilon)$ versus $1/(x^2+\varepsilon)$. But it's surprising that $1/x^2$ is so different from $1/(x^2+\varepsilon)$.
Well, but again that's just because x^2 has a zero. But I will admit that I probably only am saying "oh, it's clear" because I've seen this.
I do think that what we're seeing here is not obvious and interesting, even if these examples show it immediately --- that the function "vector field $X \mapsto$ integral curve of $X$ through, say, $(0,0)$" is only continuous in a very local sense --- is surprising
And now I've chased off anyone interested in this to begin with
Fall is Topology and Calc IV (multivar integration, with a bit at the end of intro to complex variables). Should be fun.
I mean, here in Chicago Lori opened up the bars and restaurants all by herself. Reckless abandon is widespread; we all worship that Aztec god, the economy
it's essentially zombified and has been for a decade, kept alive through increasingly intense quantitative easing efforts (pouring money into it to keep it high)
The market may not be crashing, but the people are. Yet another case study in the fact that the market does not reflect the reality as felt by the people
Well, just "The plane which your curve most nearly sits in" sounds more convoluted than "The circle which is the best approximation to your curve", no?
Approximating a 1D thing by a 2D thing throws people off
I agree. I made that comment to you when you were teaching it last time. I never have taught this stuff in multivariable calc. There are more important things.
I do remember finding it weird that my calc book in high school developed the Frenet frame and then never used it again, instead forging on toward surfaces and (the divergence/Green's/Stokes') theorem.
Those times I do cover it (and the way we split up the calc courses here, Calc III is starved for content to begin with) I try to focus on how it allows us to better visualize velocity & acceleration
Yeah, this is a historical relic ... I mean in undergrad diff geo it's a good excuse to make people learn the single-variable chain rule. I can see talking a little about the physics of motion and understanding centripetal acceleration as it relates to banking roads.
It makes some sense, but to me the sophistication required for physics/math/engineering is just different, unless the econ folks are intending to go to grad school, in which case they need more (like linear algebra and multivariable analysis).
I try to make III novel by including a discussion of the basics of linear algebra and matrices so I can give an honest explanation of the chain rule, instead of the awful one I'm used to seeing
anyone know a hint for evaluating this? $\frac13\frac{\partial^{-1/3}\beta(x, y)}{\partial^{-2/3}x\partial^{1/3}y}\bigg|_{x \rightarrow 0, y\rightarrow 1}$
Hello everyone! I'm having trouble with some differential geometry topics... I'm trying to prove that a Riemannian mfld $(M,g)$ is conformally flat iff $Ric=\lambda g$ (for some function $\lambda$); does anybody have a hint?
Next is to ask whether conformally flat manifolds are Einstein, which again I think the answer is no but I don't remember Besse well enough to say off the top of my head
My fault, $\lambda$ isn't a very nice name for a function. The fact is, the definition of Einstein mfld "given in class" to us was the weaker "non-constant $\lambda$" version, so I'm struggling a bit
@TedShifrin Suppose I have a piece of paper and I bend it so as to create positive Gaussian curvature. Then I take the top of the paper and, while holding the bottom fixed, I twist it. Is there a concept that would describe such a twist?
I think of it as analogous to the exercise in curves/surfaces that if you have an umbilic point at every point, then in fact the principal curvatures have to be constant.
You can't, @Stan. Papers can't be distorted.
This is part of the consequence of Gauss's Theorema Egregium.
j'essaie de réécrire la definition de la continuité <.> est continue si pour tout epsilon >0 il exist delta >0, ||(x,y)-(x_0,y_0)||<\delta implique |<x,y>-<x_0,y_0>|<epsilon
@Stan: The question is — what is "it"? If that's a band of paper, it's still flat no matter what.
Anonymous
10:38 PM
The notation $\{0, 1\}^{\mathbb N}$ means the set of all maps $\mathbb N \to \{0, 1\}$. Does $\{0, 1\}^{\mathbb R}$ mean the set of all maps $\mathbb R \to \{0, 1\}$?
I'm sorry to bother you again, but since we established the existence of non-conformally flat Einstein manifolds, I'm starting to wonder what the link between Einstein manifolds and the Weyl tensor would be.
Is it just something like, if $W=0$ then $R=f g\wedge g$ ($\wedge$ being the Kulkarni-Nomizu product)? Or there's something else?