3, last section of 4, and then 6 (which is very heavy going). Definitely need inverse/implicit function theorems (but not proofs) and then the introduction to manifolds.
You can consider differential operators $P(D)$ where $D$ is a differential operator and $P$ is a polynomial, but the differential operators are not polynomials.
Let $\Omega \subset \Bbb R^n$ be an open domain. You can let a polynomial ring $\Bbb R[\partial_1, \cdots, \partial_n]$ on $n$ formal indeterminates (with suggestive notation) act on $C^\infty(\Omega)$ by defining $\partial_i \cdot f = \partial f/\partial x_i$. This is the connection you're looking for, and what Ted mentioned is a special case.
It's true, in general, that if you have a linear transformation $T\colon V\to V$, then you can consider its powers (under composition), and then for any polynomial $P$ (with coefficients in the field over which $V$ is a vector space) $P(T)$ will be another linear transformation. But we never think of this as a polynomial. We think of it as a linear transformation. That's the context of your differential operator thing.
Hello, in a paper there is this inequality that is used without any explanation $\frac{1-\frac{\lambda}{\mu}}{(1+\sqrt{\frac{\lambda}{\mu}})^2}$$ $$\leq\tanh{\frac{\Delta}{4}}$ assuming that $\Delta \geq 0$ and $\ln (\frac{\mu}{\lambda})\leq\Delta$ Do you know where it comes from?
@TedShifrin Okay, I get what you're saying, although I'm fairly sure I've been told $P(T)$ can be thought of as a polynomial. Perhaps this is in the same category of simplification as "multiplying by $dx$" to solve a separable ODE?
the length of $R^n$ is $n\cdot\mathrm{length}(R)$ by induction, since we have a SES $0\rightarrow R\rightarrow R^n\rightarrow R^{n-1}\rightarrow 0$ and length is an isomorphism invariant, so $R^n\cong R^m$ implies $n=m$
that's much nicer than tensoring with a residue field, is it not
I imagine (x, y) as a C[x, y]-module as a stick over the origin in C^2. (x) and (y) are continuum of sticks over the x-axis and the y-axis respectively.
If you intersect the pictures for (x) and (y) you get the picture for (x, y)
So this seems like a manifestation of the two sequence (0) in (x) in (x, y) and (0) in (y) in (x, y)
in barycentric subdivision lemma ($C_n(A+B) \to C_n(A\cup B)$ is a quasi-iso), why do we need interiors to cover the space, why isn't covering by the sets themselves good enough?
so vick says, $S^1 = x \cup (S^1 -x)$ should give a problem
Even with finitely many wiggles, how do you subdivide a 2-simplex with a parabola running over the middle so that each piece is a simplex and belongs to either the top or bottom of the parabola? There is not a linear way to do it.
When I mean a 2-simplex I really mean a geometric 2-simplex
You can't say "oh we include the wiggly curve as an edge in my subdivision" because that's not an invariant construction; barycentric subdivision is a linear thing
I see. Suppose you included your parabola in the upper half. Then the lower half of the simplex cannot be a singular simplex because its not compact, right?
This is also why people use Borel-Moore homology by the way. You can subdivide D infinitely such that each piece lies in A or B
Borel-Moore chains have nice restriction maps $C(X) \to C(U)$ for any open subset $U$ of $X$. Just intersect a simplex in $X$ with $U$ and subdivide like mad
hi i neede help on a question:For the set of whole numbers from 1 to 20 inclusive,Tammy knows that some numbers are odd .She is going to write each number on a different ball and place the balls in a box. If one ball is randomly selected from the box,what is the probability,to the nearest tenth,that the number written on it is divisible by 5 or is an odd number?
The point is that the original sentence was a negation of a sentence with qualifiers in it. Negating that whole thing just removes the negation. This is not like negating a positive sentence with quantifiers, where you think about what it means for "for every blah, something happens" to fail.
in my first year we had a weird non-credit class called writing of mathematics and the professor assigned some ridiculous negation exercises, like youd have to negative "$f$ is uniformly continuous" or something but all the $\varepsilon$'s and $\delta$'s were switched
Yes. One of my least favorite problems in Spivak's Calculus is a multi-part problem with the $\delta$-$\epsilon$ definition jumbled every possible way.
g(grad f, grad f) is the directional derivative of f along the direction of steepest ascent of f, so i guess it makes sense that this guy is constant given the directions of steepest ascent are acceleration zero?
There must be some nice way. ||grad f||^2 is like energy of the function f : M -> R at a point, and if g : I -> M are integral curves of grad f, g' = grad f, so ||g'||^2 is also energy of the flowlines at a point. Being a geodesic means energy minimizing...
find a maximum $m$ of ||grad f||^2 on M and draw a tiny flowline of length $\varepsilon$ of grad f passing through that point. it's energy is like $m \varepsilon$
sanity check: if $R$ is a commutative ring, $M$ a $R$-module and $\mathfrak{m}\subset R$ an ideal, then a generating set for $M/\mathfrak{m}M$ as $R$-module is the same thing as a generating set for $M/\mathfrak{m}M$ as $R/\mathfrak{m}$-module, right? since $\mathfrak{m}$ annihilates $M/\mathfrak{m}M$
@TedShifrin Nevermind, figured it out. The map $X \mapsto \nabla_X \text{grad} f$ is self-adjoint, so $0 = \langle \nabla_{\text{grad} f} \text{grad} f, X \rangle = \langle \text{grad} f, \nabla_X \text{grad} f \rangle = \frac12 X \langle \text{grad} f, \text{grad} f \rangle$
Which does imply |grad f| is constant as this holds for all X
Weird but alright
The self-adjoincy is just <N_X grad f, Y> = X<grad f, Y> - <grad f, N_X Y> = XY(f) - (N_X Y)(f), and then observe that this is the same as YX(f) - (N_Y X)(f) by symmetry of the connection
