Mathematics

Conjecture. The following arithmetic function is never zero, for any $t \in \Bbb{R}$, and $t \geq 5$: $$ g(t) := \sum_{d \mid p_n\#}(-1)^{\omega(d)}\lfloor\frac{t}{d}\rfloor|G_d| $$ where $G_d = \{ x \in \Bbb{Z}/d : x^2 = 1\pmod d\}$, and where $n(x) := \pi(\sqrt{x + 1})$ varies with $x$. What...

Koro

Hi DLeftadjoint

Daniel Donnelly

7:38 AM

@Koro HI, I invented that formula

it's got groups in it

It's weird, right?

Thx, bra

Sourav Ghosh

@Koro Yes. Choose three points discrete space.

Koro

@SouravGhosh no, we want a connected space example.

@SouravGhosh i.e. a counterexample to this.

counterexample if any will show up only if we have more than 2 path components.

Daniel Donnelly

@Koro, want to learn more about my formula?

It has groups!

Subgroups of the group of units modulo square-free $d$

Square-free just means not divisible by any $q^2$ prime squared

Koro

not right now

:(

Daniel Donnelly

:'(

All ops needed to know formula: floor, +, -, /, *, |.| (group order) and # (primorial)

Koro

7:49 AM

If $A$ is measurable subset of R of measure $>0$, then there is a $d>0$ such that $A\cap (A+x)$ is non empty whenever $|x|<d$.
how to prove this?

If I show that f(x)= m(A\cap (A+x)) is continuous at 0 then I am done.
But I'm having difficulty in showing that: $f(x)- f(0)= m(A\cap (A+x))- m(A)\le |A|-|A|=0$

how to show $f(0)-f(x)$ is also small when x is near 0?

Sourav Ghosh

@Koro Discrete space is an example where union of two path component is always disconnected.

So we need an example of a topological space where union of two path components is always connected.

Koro

no, we need an example of a connected space in which union of two path components is NOT connected.

@SouravGhosh this is topologist's sine curve.

12 mins ago, by Koro

counterexample if any will show up only if we have more than 2 path components.

Sourav Ghosh

@Koro Trivial for connected space with exactly two path compoents.

Koro

indeed

Sourav Ghosh

@Koro Trivial for connected space with exactly two path components.

ZaWarudo

7:58 AM

Assume $0<a_n \le a_{n+1}$ for each $n\in\mathbb{N}$, show that $\sum_{n=1}^\infty a_n$ diverges. My attempt: since $(a_n)_n$ is increasing, its limit exists with $\lim_{n \to +\infty} a_n= \sup_{n\in\mathbb{N}} a_n$. But $\sup_{n\in\mathbb{N}} a_n \ge a_n >0$, hence $\lim_{n \to +\infty} a_n>0$. This implies that the necessary condition for convergence is not satisfied. Does this work?

Koro

@ZaWarudo you are trying to prove a wrong statement.

@SouravGhosh more trivial for a connected space with exactly one path component. 😅

oh, my bad @ZaWarudo. I didn't see $\sum_{n=1}^\infty$ there.

I misunderstood that as $\lim a_n$.

ZaWarudo

@Koro: No worries, thank you for the reply :) do you think my proof works?

Koro

Let me take a look at it. (Earlier, I didn't see it after what I thought to be $\lim$ which was actually a sum.)

@ZaWarudo it looks fine to me.

although for definiteness, you could have said: $\lim a_n=\sup a_n>= a_1>0$ or add ' for all n' after your 'But $\sup a_n \ge a_n\gt 0$'.

ZaWarudo

Yes, indeed I was just thinking that maybe it would be better to fix the estimation from below with the first term. Thanks again :)

Is this necessary because from $a_n>0$ I can only deduce $\lim_{n \to +\infty} a_n \ge 0$, right?

While saying $a_n \ge a_1>0$ implies $\lim_{n \to +\infty} a_n \ge a_1>0$ and so $\lim_{n \to +\infty} a_n>0$

leslie townes

8:14 AM

if you want to 'optimize' this proof by making it shorter or involving less things, it's maybe worth thinking about your 'necessary condition for convergence.' to deduce divergence, it likely doesn't require that lim a_n exist. it likely only requires you to know that it is not the case that lim a_n = 0. you could write a proof with fewer inferences from this, if you want to take that test as a given.

all of this stuff is only a few lines removed from the definitions, so you might also try a proof from the definitions (i.e., not routing through the test for divergence)

ZaWarudo

@leslietownes You mean something like this: $\sum_{n=1}^\infty =\lim_{N\to+\infty} \sum_{n=1}^N a_n \ge \lim_{N\to+\infty} \sum_{n=1}^N a_1=\lim_{n\to+\infty} Na_1=+\infty$? Here, I only used that $0<a_n \le a_{n+1}$ for each $n\in\mathbb{N}$ (to say that $a_1>0$ and so the latter limit is $+\infty$) and I am not using the existence of $\lim_{n \to +\infty} a_n$.

Sorry, I edited because I forgot a sum :D

leslie townes

i was thinking of a proof that began "let e > 0 be given." just for fun. but yes, any number of arguments like that will work too.

its helpful to be able use basic results (like the 'test for divergence') as black boxes in arguments, because that tracks how we often make use of results that we 'know' after we know them. it's also helpful to be able to write proofs of this kind of stuff from the definitions, because when it's possible to do so, as it is here. the arguments used to establish all of those basic results about the algebra of limits are just as useful as the results themselves.

but yeah, your formulation above is interesting because it proves a more general theorem (i.e., that if a_n is a sequence satisfying a_n >= a_1 > 0 for all n, then sum a_n diverges) and without making use of the existence of lim a_n in the case where a_n happens to be increasing.

you're still kinda using the extended real numbers to assign a "value" +infty to lim N a_1 there, and some theorems about what happens to inequalities when you pass to a limit there, and if you wanted to you could remove those things too.

its just fun to look at these kinds of things from multiple perspectives.

Mad

8:34 AM

hello, i am having issues with notation of cyclic sums.
Suppose i have two numbers a_i b_i that are cyclic in i =0...n such that n+m = m for m = 0...n
I want to show that the sum from j = 1 to n of a_j-1 b_j is the same as the sum of j=1 to n of a_j b_j-1

more over , consider them not to be numbers, but a as a vector and b as a linear map

CroCo

@TedShifrin I usually do that, but I want to know why using the rank's notion doesn't lead to satisfactory results. May be I'm wrong after all.

leslie townes

8:57 AM

mad: by 'a is cyclic' do you mean that a_{n+m} = a_m for all m and similarly for b? (you've written just "n+m = m" above, which can't be what you mean.) what happens if n = 3 and (a_0, a_1, a_2) = (1,0,0) and (b_0, b_1, b_2) = (0,1,0)? isn't a_0 b_1 + a_1 b_2 + a_2 b_3 = 1*1 + 0*0 + 0*0 = 1 while a_1 b_0 + a_2 b_1 + a_3 b_2 = 0*0 + 0*1 + 1*0 = 0?

Mad

nevermind

the statemnt is not true anyway

Mad

9:22 AM

let me correct.

$v_{l \pm n} = v_{l }$ where as $v_l$ is a base vector. for $l = 0...n$
i want ot show that the operator give throught $\sum_{l=1}^n v_l \phi_{l-1}$
$\phi_l= <v_l, ->$ is hermetian.

the heremtian of $v_l$ is $\phi_l$ and that of $\phi_{l-1} $ is $v_{l-1}$

This is physics so dont ask about details, none are provided

some hilbert space of quantum mechanical mumbo jumbo

so the hermetian is given through $ \sum_{l=1}^n v_{l-1} \phi_{l}$

Now i need to show that thi is aquivalent. i thought, i show hte sums are the same, which i dont think they are... then, lets try to apply a random base vector $v_j$

but its so confusing with the cyclic indices

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$Mathematics$ Mathematics