$f:[a,b]\to\mathbb{R}$ is continuous, is it obvious that for every fixed $n$ there exist a unique polynomial $p$ of degree $n$ minimizing $||f-p||_\infty$?
My numerical analysis book defines the best approximating polynomial as the one with the property above but neither existence nor uniqueness are discussed
@Alessandro you can think of $\|f-p\|_\infty$ as a function on the coefficients of the polynomials, then you need to think why this function realises its minimal value
@becko Hmm. If there's an obvious way to identify binary numbers containing $m$ digits, you could have a mapping which converts the bare binary representation to that.
BTW, Balarka, DogAteMy gave a nice proof for something I asked him. Did you see my question about the maximum breadth of a (smooth closed) convex curve (bounding in terms of curvature of the curve)?
I actually really like the proof of Ramsey theorem, I found it very ingenious to prove the existence by induction using smaller Ramsey numbers to bound a bigger one
@TedShifrin The last time this fellow compiled an exam, we had a problem which asked to find the conditions for making a box of given dimensions pass through a door of given area.
It's that, hey, they're power series, so the generating function is holomorphic at zero. Hence you can extract the coefficients by contour integration.
@Semiclassical And even some analysis of what sort of singularity happens on the edge, which can tell you about subexponential corrections to your estimates
Hmm ... it's parametrizing the line joining $\xi$ and $\eta$, so you can think of it as a projective transformation from $\Bbb P^1$ to that particular $\Bbb P^1$, I guess.
You need $n$ everywhere linearly independent ones, Karim.
@TedShifrin and, the solutions of $(s\xi+t\eta) \cdot \mathbf{x} = 0$ would be the plane whose normal is the projective line $[s,t]$ after that projective transformation is applied (which would cause it to become $[s\xi + t\eta]$, as mentioned above)
Choose a chart $(U,\psi)$ be a chart based at $e \in G$. We have the map $\theta : U \rightarrow TG$. We will extend this to global section. Consider $\sigma : G \rightarrow TG$ defined by sending $y \mapsto \phi_{g_{*}}(\phi(g^{-1}(x))$ This is well defined and extends $\sigma$
Yeah. Isn't that amazing? Make sure you work out some concrete examples. I actually had a student do a masters thesis on trying to understand this more deeply.
"if we apply two possibly different linear transformations to two possibly different lines, then as we vary the lines, their intersections draw a conic in $\mathbb{R}^2$"
Choose any projective transformation mapping $\lambda_P$ to $\lambda_Q$. (These are both $\Bbb P^1$'s.) So $t\in\lambda_P$ maps to $f(t)\in\lambda_Q$. The conic is the locus of the intersection points of those two lines as $t$ varies.
So an interesting question is: Can you tell from the projective transformation (i.e., LFT) what kind of conic you're going to get in terms of Euclidean geometry? (Projectively, parabola, ellipse, hyperbola are all the same, of course.)
I took my linear algebra final today, @Balarka. After having taken the exam, I think I now get the basic idea with inner products and the notion of length/distance.
@TedShifrin The professor had "no calculator" and "calculator allowed" problems mixed in the same exam packet, and she wasn't really watching what we were doing. Of course, some students ignored those advisories.
I made up problems so that arithmetic came out very easily ... and repeatedly reminded students that if they were getting yucky arithmetic, they were making mistakes.
@TedShifrin Yeah, it's pretty silly. I've seen people type facts/formulaes in letters into various places on their device, as tedious and unnecessary as that is.
Well, I wanted students to have basic computational skills and made up the numbers so that it came out very easily. Plus, on some problems, essentially the matrix was already in reduced echelon form.
I kill time doing constructive things, like I did in my language test writing down Lewis Carroll quotes in a footnote on an analysis of Shakespeare's Macbeth.
@TedShifrin For more constructive examples, I cut off the controversial China and Pakistan-owned territory in my map of India in the geography test the previous year, if you like.
If $TM$ is isom to $M \times \Bbb R^n$, you can just pick an orientated basis on $\Bbb R^n$. That gives an oriented basis on $T_pM$ for each $p \in M$.
@G.Bergeron I am not sure what you mean. I am just getting a bundle-orientation on $TM$ from a trivialization. A bundle orientation on $TM$ is exactly an orientation on $M$.
@BalarkaSen Isn't the parallelizability of Lie groups coming from the action of the group on the algebra and itself which leads to diffeomorphism groups generating the vector fields? By dualizing this you can relate the parallelizability to the invariant 1-form and the non-vanishing n-form
