Mathematics - 2019-06-14

Q: Proof logical statement with interference rules

Existentialism probably contains the most irreflexive binary relation I have ever seen:

That how the self interprets itself does not necessarily defines itself

Tuki

Maybe someone could check this?

1

Proof following statement with interference rules ( without truth table) that $$ (\neg C \wedge B \wedge (A \rightarrow C) \wedge (B \rightarrow D ) )\implies (\neg A \wedge D ) $$ Attempt to proof $B$ (premise) $B \rightarrow D$ (premise) $D$ (Modus ponens 1,2) $A \rightarrow C $ (premise) ...

I'm always quite uncertain when trying to write any kind of proofs

The guy saying "You really have only one premise", I don't think I quite get what he means

12:44 AM

@AlessandroCodenotti algebraically closed fields are classified by transcendence degree and characteristic

but for uncountable, transcendence degree = cardinality

N. Maneesh

12:57 AM

@akhilkrishnan I don't understand your question.

5:29 AM

Gallian says: In PID, any strictly increasing chain of ideals $I_1\subset I_2\subset\cdots$ must be finite in length. I don't see why we need strictly increasing, proof does not seem to use strictly anywhere.

5:40 AM

@Silent surely $(1) \subseteq (1) \subseteq \cdots$ is infinite in length.

substitute this example to the proof.

huh :) Thank you!

5:55 AM

@MatheinBoulomenos if I have f.p. groups G1 and G2 such that for any finite group H, |Hom(G1,H)| = |Hom(G2,H)|... then is it true that G1 = G2?

6:41 AM

Just to confirm, if $\omega\in \Lambda^k(V)$ for $V$ an $n$-dimensional vector space,

Just to confirm, if $\omega\in \Lambda^k(V)$ for $V$ an $n$-dimensional vector space, $$\omega = \sum_{1=i_1<\dots<i_k=n}a_{i_1,\dots,i_k}dx_{i_1}\wedge\dots\wedge dx_{i_k}$$

where $$\omega(v_1,\dots,v_k)=\sum_{1=i_1<\dots<i_k=n}a_{i_1,\dots,i_k}\begin{vmatrix}dx_{i_1}(v_1)&\dots& dx_{i_1}(v_k)\\\dots\\dx_{i_k}(v_1)&\dots&dx_{i_k}(v_k)\end{vmatrix}$$

That's how we actually apply this form to a collection of $k$ vectors of $n$-dimensional $V$ right?

Where $dx_{i_m}$ is the dual vector to $e_{i_m}$ so in particular the above is:
$$\omega(v_1,\dots,v_k)=\sum_{1=i_1<\dots<i_k=n}a_{i_1,\dots,i_k}\begin{vmatrix}v_1^{i_1}&\dots& v_k^{i_1}\\\dots\\v_1^{i_k}&\dots&v_k^{i_k}\end{vmatrix}$$

where $v_i^j$ means the $j^{th}$ component of the vector $v_i$

@F.White Yeah that's right

I think

@TedShifrin Please confirm

Thanks for partial confirmation :)

7:14 AM

Higman group

In mathematics, the Higman group, introduced by Graham Higman (1951), was the first example of an infinite finitely presented group with no non-trivial finite quotients. The quotient by the maximal proper normal subgroup is a finitely generated infinite simple group. Higman (1974) later found some finitely presented infinite groups Gn,r that are simple if n is even and have a simple subgroup of index 2 if n is odd, one of which is one of the Thompson groups. Higman's group is generated by 4 elements a, b, c, d with the relations a −...

7:37 AM

##

woops, dropped ketchup on my keyboard and wiped it off lol

I came off quite well considering

it's the meta hash tag

lol

Andrews

11:19 AM

compact Riemann surface $X$, $p\in X$ is divisor, $\mathcal O_p$ is sheaf.

How can I prove $\dim H^0(X, \mathcal O_{p})=1$ ?

$p\in X$ is a point in $X$, treated as a divisor.

$\mathcal O_{p}$ is set of meromorphism s.t. $(f)+p >=0$

genus $g>0$

Ok, I figured it out

11:43 AM

@Ryan are you here by any chance?

12:07 PM

I wish Ted or Balarka were here

12:20 PM

@AlessandroCodenotti I am now

I asked on MSE but I'm not quite sure what the guy in the comments is suggesting

he means linear isometry

Wait so $L^p([0,1])$, $p\neq 2$ is isometric to a Hilbert space? That's crazy

uhhh

well no, it doesn't satisfy the pythagorean theorem

so what does he mean haha

did you look up the result?

The paper only say "isomorphic to an Hilbert space"

10.1007/BF02771592 it's the DOI

12:30 PM

idk how to use doi, whats the title

Lindenstrauss, J., & Tzafriri, L. (1971). On the complemented subspaces problem. Israel Journal of Mathematics, 9(2), 263–269. doi:10.1007/bf02771592

ok I found that too

they definitely mean that the norm satisfies the parallelogram law

complemented means what exactly here

just in the linear algebra sense?

the guy says bounded projections, do they say that here?

@AlessandroCodenotti their assertion that this is trivial if X is finite dimensional is completely crazy

maybe they just mean there's a bounded invertible map

matwbn.icm.edu.pl/ksiazki/or/or2/or214.pdf

@RyanUnger I assumed "$N$ is complemented in $X$" means that there is another subspace $M$ with $N+M=X$ and trivial intersection

in this survey they do distinguish between isom and isometric isom

and for this result they say only isom

so they just mean bounded invertible linear map

Alright, but if it's not an isometry I don't see how it answers the question on MSE

12:36 PM

are P1 and P2 preserved under isomorphism?

Hmmm I'm not sure

Isomorphism here means linear bijection which is also an homeomorphism

So a continuous bijective linear map is an isomorphism

This is the terminology Conway uses to state this result by Lindenstrauss and Tzafriri

@AlessandroCodenotti that's what I said

well, corrected myself to say

12:55 PM

Lindenstrauss and Tzafriri wrote a book called "classical Banach spaces", but as far as I can see this result is not in the book

The ultimate trolls

what is the question now?

@ÉricoMeloSilva one of the papers Rod told me to read is 160 pages

the second is 60 pages

the third 93 pages

Whether this Lindenstrauss-Tzariri result is indeed enough to solve the question on MSE

it's not clear to me

does a linear algebraic splitting $V\oplus U$ give you bounded projections onto the summands?

what he says seems slightly stronger but it might be very basic functional analysis

1:27 PM

How to prove or disprove that continuous increasing function is surjective?

Obviously, it is injective.

I assume you mean a function $\Bbb R\to\Bbb R$?

yes, $f: \mathbb{R} \to \mathbb{R}$

Do you think this is true or false?

I think it is true

Why?

1:34 PM

Suppose that we have $y_j$ such that for every $x_i$ $f(x_i) \neq y_i$.
$f$ is injective. That means there $is$ $x_j$ such that $f(x_j)=y_j$. Otherwise $y_j$ is not in the range of $f$.

It is flawed, but I thought in this direction.

@BalarkaSen I’ve spotted the guy

Mats Granvik

Why is discrete optimization not used in number theory? After all, we can get a square root bound on a problem involving mainly negative numbers.

1:49 PM

let $f: D \to R$ be continuous increasing function. To prove surjectivity:

Suppose that $\exists y_j: \; \forall x_i f(x_i) \neq y_j$.
$f$ is inj. $\implies \; [\forall \;x_1,x_2 \in D f(x_1) < f(x_2) \implies x_1 < x_2]$.
Since $f$ is continuous then $y_j $ is out of range out of $f$.

Does this prove surjectivity of $f$ ?

You assume $y_j$ is not in the range of $f$ and conclude $y_j$ is not in the range of $f$.

Also what are you even indexing over?

Trying to prove surjectivity of fn.

I just found out that increasing function is not surjective. But still, how to prove it ?

I understand that's what you were trying to do. The question was for which purpose you are indexing the variables and what set they and the indices range over.

Mats Granvik

The partial sums of the Möbius inverse of the Harmonic numbers fit within this linear programming problem:
https://pastebin.com/WJT75cSE
Setting the upper bound on the variables to zero gives us a square root bound, but then it is no longer an
arithmetic sequence.

2:06 PM

@Thorgott Oh, $x_i \in D$, $i,j \in \Bbb N$,
$y_j \in R$

The idea was to show that there exists $y$ for every $x \in D$ such that $f(x) \neq y$

Maybe seems like a mess.

Why are you enumerating these elements? is $D$ countable? Is $y_j$ supposed to be an arbitrary element of $R$? (If yes, you will run into problems with enumeration) Clearly there exists a $y$ for every $x\in D$ such that $f(x)\neq y$, because there are more than one real number (did you mean to put the quantifiers in a different order there?),

in The h Bar, 1 hour ago, by bolbteppa

'Patterns That Eventually Fail' [link](https://johncarlosbaez.wordpress.com/2018/09/20/patterns-that-eventually-fail)

"$$\displaystyle{ \int_0^\infty \frac{\sin(x)}{x} \, dx= \frac{\pi}{2} }$$

$$\displaystyle{ \int_0^\infty \frac{\sin(x)}{x}\frac{\sin(x/3)}{x/3} \, dx = \frac{\pi}{2} }$$

$$\displaystyle{ \int_0^\infty \frac{\sin(x)}{x}\, \frac{\sin(x/3)}{x/3} \, \frac{\sin(x/5)}{x/5} \, dx = \frac{\pi}{2} }$$

where the pattern continues until

$$\displaystyle{ \int_0^\infty \frac{\sin(x)}{x} \, \frac{\sin(x/3)}{x/3}\cdots\frac{\sin(x/13)}{x/13} \, dx = \frac{\pi}{2} }$$

Eventually diverging sequence, the opposite of a net

@Secret you know some physics right

depends on what topic

@Ultradark ask

Yes, quantifies are in wrong order too : $\forall x \exists y : f(x)\neq y$

2:15 PM

That statement is obviously true though. For each $x$, $f(x)$ is just a real number, so you can just pick $y$ to be any other real number (and this is clearly possible because there are more than one real number).

thanks

It has nothing to do with whether $f$ is surjective, continuous, increasing, anything else or not though.

wait, what would $\exists y \forall x: f(x) \neq y$ mean ?

@secret I want to attach a natural geometry to a discretized space (a network)

what is a "natural geometry"?

2:17 PM

That means there is a $y$ that is not contained in the range of $f$, i.e. $f$ is not surjective.

What I mean by that is, a discrete realization of say, hyperbolic space.

I'd like to find a geometry that supports such a discrete space

like the network might be a discrete "slice" of hyperbolic space

the network should be a discrete realization of some geometry (not sure which geometry matches up correctly)

sciencedirect.com/science/article/pii/B9780121894207500360

Discrete hyperbolic geometry appears to be a thing

@F.White The only thing that's wrong is that you should have $1\le i_1<\dots<i_k\le n$. :)

yeah I am not making it up lol

Well I found some reference, however I don't have enough knowledge to comprehend so many things (root systems, combinitorics etc.), so good luck reading it

2:25 PM

I think I'm going to move on to other things because I don't have the knowledge yet either

Does anyone know a good reference that contains a proof of the analytic implicit function theorem without complex analysis?

anakhro

2:41 PM

@Thorgott I have found it listed as an exercise with a reference to this theorem......

"We leave it to the reader to apply the inductive method of Dini (Section 3.2) to derive a general real analytic implicit function theorem from our Theorem 6.1.2."

- Krantz & Park (The Implicit Function Theorem: History, Theory, and Applications)

Flajolet and Sedgewick's analytic combinatorics book has 3 proofs in the appendix for a bivariate one.

With a reference to "Hille’s book [334]" --> [334] HILLE , E. Analytic Function Theory. Blaisdell Publishing Company, Waltham, 1962. 2 Volumes.

@Thorgott hopefully one of these is sufficient...otherwise...*shruggles.*

creative-mathematics.cunbm.utcluj.ro/wp-content/uploads/…

dma.unina.it/hamiltonianPDE/mate/tolandCapri.pdf

y a

3:04 PM

@TedShifrin Thank you :)

I've known Krantz & Parks, but the others are new to me. Thanks a lot!

Let $R[x]$ be a UFD. Is it true that n-th degree polynomial in $R[x]$ has at most $n$ roots?

anakhro

No

Oh wait

at most

Yes, that's true.

Thank you!!

anakhro

@Silent this only needs integral domain.

Well, that $R$ is an integral domain to be more specific.

Érico Melo Silva

3:16 PM

@RyanUnger rip

This raises the question: does $R[X]$ being a UFD imply that $R$ is a UFD?

Probably should because $R$ embeds into $R[X]$

yes

Is there an integral domain $R$ such that $R[x]$ not UFD?

@Thorgott Wow!

@Silent Sure, pick $R$ not a UFD

right! thanks.

3:24 PM

However if $R$ is an UFD, then $R[X]$ is a UFD as well

wow! so much to think about

An example of such a ring is $\mathbb{Z}[\sqrt{-6}]$, iirc.

On the importance of the difference between "good" and "great"

@Thorgott if you pick your $D$ small enough then $\Bbb Z[\sqrt{D}]$ is a UFD by the Minkowski bound

Say the cystic fibrosis patients at Hospital 1 have a 0.5% chance of getting sick on any given day

and the CF patients at Hospital 2 have a 0.05% chance of getting sick on any given day

What are the odds that they get through the whole year without getting sick?

This is the difference between good and great

Excellence comes from seeing, on a daily basis, the difference between being 99.5% successful and 99.95% successful

(The answer to the question is 16% and 83% respectively!)

3:35 PM

That's interesting. $-6$ seems big enough though.

@AlessandroCodenotti Is $X$ a transcendental variable?

i.e. is $R[X]$ the ring of polynomials?

Yes, I'm talking about the polynomial ring

Right, so $\Bbb Z[\sqrt{-5}]$ (which is not a UFD) doesn't count

What's a transcendental variable?

If any nonzero polynomial of $X$ is nonzero

$\sqrt{-5}$ is not transcendental because if you apply the polynomial $x^2+5$ to it you get zero

3:38 PM

@Thorgott well the minkowski bound depends on the discriminant of your number field so if you take it small enough the Minkowski bound won't exceed $2$, in which case you have no prime ideals to generate the ideal class group

(but obviously these are not the only cases of rings of integers of number fields that are UFDs because the bound also depends on the degree of the extension and the number of complex embeddings)

(Note that $\Bbb Z[\pi]$ is isomorphic to the ring of polynomials over $\Bbb Z$)

@AkivaWeinberger You mean $\Bbb (Z[\sqrt{-5}])[x]$ not a UFD by that comment?

$\Bbb Z[\sqrt{-5}]$ is not.

That's $R[X]$ where $R=\Bbb Z$, $~X=\sqrt{-5}$

But if $X$ represents a formal variable then it works

That's beyond me, but I'll take it.

Oh, I misunderstood Thorgott's commet

@Thorgott I thought you were claiming that was a UFD

rather than the opposite

(6 rather than 5 but whatever)

3:42 PM

$\Bbb Z[\pi + e]$

Both aren't UFDs, albeit integral domains.

$6=2\cdot3=(1+\sqrt{-5})(1-\sqrt{-5})$ and $10=2\cdot5=(2+\sqrt{-6})(2-\sqrt{-6})$

So both are integral domains, whose polynomial ring is not a UFD, which is what Silent was asking for.

yes.

@Thorgott Oh

I stand corrected

4:55 PM

how to maximize the value of line integral by choosing a appropriate curve within a domain in the given vector field ?

Ah, that's one of my favorite questions, @Ajay. You have to be in $\Bbb R^2$, of course?

And I assume you're trying to find a closed curve $\gamma$ that maximizes the line integral.

No, open curve. Thanks for replying.

We don't call them open curves.

Give me the explicit question, please.

Q: Finding path for maximum value of line integral

Yeah, I am in $ \mathbf R^2$

1

I need help in the following question: I have a field: $$F=\left(y^3-3y+xy^2,3x-x^3+x^2y\right)$$ bounded in region $D$ defined by $x^2+y^2\leq 2.$ I need to find a path $C$ that goes from $(1,1)$ to $(-1,-1)$ inside $D$ such that the value of the line integral/work of $F$ on the path $C$ is max...

line-integrals

I ain't the person who asked it, just curious to learn.

I see. I usually ask the question for a closed curve passing through a given point. You want to make a closed curve out of this question (so, close it up by following the curve and then the line segment from $(-1,-1)$ to $(1,1)$). Now use Green's Theorem and look at what double integral you're getting.

4:59 PM

I think, it is like tangent of the curve at every point should be collinear with the vector field direction at that point. But I don't know how to express this in equation.

It's highly unlikely you can follow what you said and get from one point to the other.

If you could, I agree that it would be nice.

@TedShifrin I am going to try that. Thanks! :)

Have fun!

5:38 PM

I think it's useful for young people who have an interest in pursuing an academic career to think of the eduation system less as a hierarchy of degree attainment than a continuous progression from layperson to novice to apprentice and that a chain of degrees is more a progression through your "training" for your eventual job

at least it sounds like a perspective I wish I'd had as a younger person

5:49 PM

What's a good notation for redefining the value of a function $f$ at a point $x$? Best I can think of is $f - \{(x,f(x))\} \cup \{(x,y)\}$ but it's hardly concise.

You could try something like $\tilde{f}=1_{\{x\}^C}f+1_{\{x\}}y$, though I'd personally just go for doing it by cases.

I've seen $f|^x_y$ used before in set theory

@AlessandroCodenotti $f(x)-f(y)$?

Pfff that's something only an analyst would thing

well literally anyone who uses calculus

6:01 PM

$f[x/y]$ could work too I guess then

what on earth are you trying to write

Maybe $f|_{\{x\}^\complement} \cup \{(x,y)\}$.

Q: Traces of manifold-valued Sobolev maps

0

Let $(M^m,g)$ be a compact Riemannian manifold with smooth nonempty boundary, and $N^n\subseteq \Bbb R^d$ a boundaryless isometrically embedded Riemannian manifold. For $1\le p<\infty$ we define as usual $$W^{1,p}(M,N):=\{u\in W^{1,p}(M,\Bbb R^d):u(x)\in N\text{ a.e.}\}.$$ Using the Euclidean tr...

dg.differential-geometry ap.analysis-of-pdes riemannian-geometry

6:20 PM

@AlessandroCodenotti I asked this question on meta math.meta.stackexchange.com/questions/30375/… and it seems like it was discussed before.

Something else occurred to me though: Why not just add the bookmarklet link directly to the room description, with a name like "Render LaTeX"?

@user76284: The standard thing (let's not use $x$ for a specific point, please) is to use cases notation: $f(x) = \begin{cases} F(x), & x\ne a \\ b, & x=a\end{cases}$.

You're working with the graph as the definition, but rarely do we really do that.

@Ted Yeah that's what I'd usually do, the problem is I'm doing this multiple times in a single expression.

I don't understand what you mean.

I meant that this "function redefinition" occurs multiple times in a single expression.

I think I might just define a new "operator" $R(f,x,y)$ that does this.

I still don't understand. You mean you're redefining it at several points?

By the way, it is rarely informative to repeat exactly what someone said he did not understand.

6:27 PM

I'm not sure how else to explain it. Picture something like $R(f,x,y) + R(g,a,b) \leq h(w) + R(m,c,d)^{R(n,e,f)}$ where each occurrence of $R$ requires case notation. It quickly gets unwieldy.

Multiple occurrences of case notation in a single expression I suppose.

Well, what you've written makes absolutely no sense to me. If $R(f,x,y)$ is the redefinition of a function at $x$ only, you need its value at $w$ for what you've written to make sense, so you would need $R(f,x,y)(w)$, etc. That difficulty propagates in what you've written.

Yeah I forgot to add the evaluations.

But you get the idea :-)

Yeah, I can't think of this as anything that would naturally show up in any mathematics I've ever done. But I can see that from a programmer's perspective, it might be useful.

I would just define the functions ahead of time and then write the inequality.

user246622

8:38 PM

Hello\

user246622

is any one there

user246622

doubt?

What's up

user193319

8:50 PM

If I have a linear functional defined a proper subspace of a Banach space, how does one use Hahn-Banach to extend it to the entire vector space? Rudin simply says this can be done, but he doesn't show how it is done.