Mathematics - 2021-09-08 (page 1 of 2)

user193319

00:32

Let $f : \Bbb{C} \to \Bbb{C}$ be an entire function such that $|f(z_1+z_2)| \le |f(z_1)| + |f(z_2)|$ for every $z_1,z_2 \in \Bbb{C}$. Is it possible to prove that $|f(az)| \le a |f(z)|$ for all $a > 0$ and $z \in \Bbb{C}$?

Obviously it works for $a \in \Bbb{N}_0$

Paulo

do you guys know a similar result for functions of several variables?

robjohn

Taylor's theorem

In calculus, Taylor's theorem gives an approximation of a k-times differentiable function around a given point by a polynomial of degree k, called the kth-order Taylor polynomial. For a smooth function, the Taylor polynomial is the truncation at the order k of the Taylor series of the function. The first-order Taylor polynomial is the linear approximation of the function, and the second-order Taylor polynomial is often referred to as the quadratic approximation. There are several versions of Taylor's theorem, some giving explicit estimates of the approximation error of the function by its Taylor...

That link is for multivariable

Ted Shifrin

@user193319 Then don’t the usual methods prove it for $a\in\Bbb Q$?

gian

Is there a way to show that the structure group G for some vector bundle is not further reducible? Many references I've seen just ascribe the inability to further reduce G to the existence of some topological obstruction, but don't elaborate any further.

For instance, what stops me from assigning an $\{e\}$-structure to the sphere $S^2$?

Paulo

00:48

@robjohn its not exactly that. I want to know that if a function is represented by its taylor series at some neigboorhood B of 0, i can also represent it by its taylor series at different points of B (in some smaller neighboorhood). Or is it the same thing and I'm missing something?

robjohn

If your neighborhood of the new point is inside the neighborhood of the point for which you know convergence, then it converges.

if you know that the radius of convergence about $(0,0,0)$ is $14$, then the radius of convergence about $(3,4,12)$ will be at least $1$.

user193319

@TedShifrin yeah, i think so; and then just use a density argument. thanks!

Paulo

I understand the same series converges, but if say, $F(x) = \sum_{m=0}^\infty \frac{1}{m!} \sum_{|\alpha|=m} \partial^\alpha F(0)x^\alpha$ around $\overline{B}(0,R)$, then $F(x) = \sum_{m=0}^\infty \frac{1}{m!} \sum_{|\alpha|=m} \partial^\alpha F(a)(x-a)^\alpha$ on some $B(a,R_a)$ ($R_a$ taken small enough) ? its that simple?

$a\inB(0,R)$

leslie townes

01:07

yeah, the series at a will converge if a is inside (0,R).

leslie townes

01:17

ballot received and counted. blegh.

Paulo

how would i go about to prove that? i tried to replicate the 1 dimensional case and it didnt workout nicely?

Ted Shifrin

If you’re doing a disk within the original disk of convergence, it’s the identity principle.

Paulo

that would be for 1 complex variable right? i want for $\mathbb{R^n}$

leslie townes

01:32

mm, may need some hypotheses on F. i was assuming F was complex analytic or perhaps at least solved a nice enough PDE.

if F is just real analytic, i dunno.

Paulo

I wouldn't mind adding hypothesis, but i thought it would end up working with the same ones given on the 1 dimensional case.

Thorgott