Mathematics - 2020-08-07

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S.D.

4:01 AM

Is there any easy way to see that any arbitrary union of right open intervals of the form $[a, b)$ in $\mathbb R$ has a minimal element?

user76284

@S.D. $\bigcup_{\varepsilon > 0} [\varepsilon, 1)$ ?

S.D.

4:24 AM

@user76284 Interesting. So that is not actually true. Thanks!

2 hours later…

Balarka Sen

6:34 AM

@TedShifrin @MikeMiller What I want has been proved by Forster and Rampsott, see first paragraph in the introduction here.

4 hours later…

Threnody

10:19 AM

Am I right in saying: (Complex Analysis)
If our f is holomorphic in some region D, but we have some closed contour C that goes around some hole somewhere inside (but not included in) D, then we cannot use Cauchy's result to conclude the integral is 0, since f is hence not holomorphic on and inside C. (Since we're only given that f is holomorphic in D and not the 'hole')

Thorgott

10:36 AM

correct

Threnody

So in this question:

Q: Cauchy's Theorem, Deformation Theorem, similarities and differences - Complex Variable

1) Let $f$ have an antiderivative $F$ in a region $D$ containing a $\underline{closed \,\,\,contour}$ $C$. Then \begin{equation} \int_Cf(z)dz = 0 \end{equation} 2) Also, Cauchy's theorem says: If f is $\underline{holomorphic}$ inside and on a $\underline{closed \,\,\,contour}$ $C$ then \begin{eq...

The question says 'Is it now obvious? Of course they are, they are zero!'

The error in this reasoning is that there's an assumption that the contour's interiors are in D

Am I correct?

Thorgott

yeah, the integrals aren't necessarily $0$

Threnody

I get that, but it's because of my argument or something else?

Thorgott

well, the reason why they aren't necessarily $0$ is that counter-examples exist; your argument explains why this doesn't contradict Cauchy's theorem

Threnody

Ok, thank you :)

1010011010

10:47 AM

Hi everyone

I actually hail from the physics side of things, but its MP so that could be a compromise

My object of interest is the case where I have a product of many many rational expressions in some variable

So I already covered the case where I have more zeroes in the denominator, I found something like this:

Anyway, I'd also like to find something like this variant of the Heaviside rule but for polynomial long division

Finding reading material on this question seems to be a rather difficult task, I was hoping maybe someone here knew how to divide a general polynomial by another general polynomial, ofc assuming these polynomials don't have the same coefficients

And assuming that the polynomial in the denom P_D has a smaller degree than the one in the one in the numerator

P_N say

ng.newbie

11:27 AM

Does anyone know what "two symmetric edges" means in graph theory ?

Alessandro Codenotti

Two edges between the same two vertices I think (with the same orientation if the graph is oriented maybe)

Knight

11:59 AM

@robjohn Hello sir! Can you please give me a proof that $$f(x)= \begin{cases} 0 & if~x~is~irrational \\ 1/q & if ~x~is~of~the ~form~p/q~(it ~in ~simplest~form~and~p,q \in \mathbb Z , q\neq 0 \end{cases}$$ ($0\lt x\lt 1$) have a limit zero at any $0\lt a \lt 1$.

Here is the proof given by Spivak

But I understand your proofs quite easily, so if you have time can you please write up an elegant proof (like you do). It’s a request not a demand.

Lelouch

@Knight The idea is simple: If you fix the denominator, there a discrete amount of rationals so the smallest distance from $a$ to the nearest such one can't be too small