Mathematics - 2025-02-08

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Elsa F.

00:04

@AkivaWeinberger Been thinking about some similar graph theory questions recently, (a combinatorial approach to finding the number of distinct unlabeled spanning trees on $n$ vertices, plus a variation involving minimal-weight spanning trees). Cayley's formula gives us that the number of labeled spanning trees on $n$ vertices is $n^{n-2}$, so I wonder if you could do an approach from the per

**perspective of directions as edge-labeling? You'd have in/out directions as your options, depending on the directed tree you're hoping to construct. Choose a root, and if your graph is connected, you'll always have |V|-1 edges that need to be labeled, and choice of roots shouldn't change that/impact how you reach edges further from the root

If anyone knows more graph theory, please edit/correct me -- just an intuition dump... but not near a proof haha. It's an interesting question :)

leslie townes

00:28

i ran into rooted forests when proving the positivity of some matrix elements last year math.stackexchange.com/questions/4873471/…

bringing matrices into it actually does lend itself to some ideas about some bijections, but the matrix has to be 'choice'

hbghlyj

01:25

A Course in the Theory of Groups p.15 remarked that the distributive law doesn't hold for subgroup lattices in general. Can you give an counterexample?

user image

Thorgott

think about lines in $\mathbb{R}^2$

hbghlyj

Let $H,K,L$ be three distinct lines through the origin in $\mathbb{R}^{2}$. Then $H \cap L=K \cap L=\{0\}$, so that $(H \cap L)(K \cap L)=\{0\}$, while $(H K) \cap L=L$.

Ok

I noticed that subgroup lattice of $\mathbb Z$ is distributive. Is there a name for the groups whose subgroup lattice is distributive?

Gwyn

02:13

Reading about modus ponens, I read that since $[p \wedge (p \implies q)] \implies q$ is a tautology, we can deduce that $q$ is true when $p$ is true and $p \implies q$ is true because in the truth table of modus ponens the only case where $p$ is true and $p \implies q$ is when $q$ is true. Since $[p \wedge \neg (p \implies q)] \implies \neg q$ is also a tautology, can I deduce $\neg q$ is true (that is, $q$ is false) by showing that $p$ is true and $\neg (p \implies q)$ is true?

In the second line, I meant "the only case where $p$ is true and $p \implies q$ is true is when $q$ is true"

Thorgott

03:19

TFAE:
(1) the lattice of subgroups is distributive
(2) every finitely generated subgroup is cyclic
(3) the group is a subquotient of $\mathbb{Q}$

3 hours later…

Anacardium

06:33

Consider the set $S$ of all sequences of $0$'s and $1$'s with no consecutive $1$'s. Does there exist any one to one correspondence from $S$ onto the Cantor set?

3 hours later…

Mats Granvik

09:46

Are the solutions to all integrals actually obfuscations of root sums of some function, in disguise?

3 hours later…

Thorgott

12:33

@Anacardium this is clearly a subset of the Cantor set. conversely, take a sequence in the Cantor set and insert a $0$ between any two digits, this yields an injection of the Cantor set into $S$. conclude by CSB.

3 hours later…

Anacardium

15:14

@Thorgott: Hmm got it. Thanks.

It may seem a bit silly question to ask but I having hard time getting this. By double derivative test, it is easy to see that both $e^x$ and $e^{-x}$ are convex, as in both the cases the derivatives are strictly increasing function of $x.$ But then I found the following link : math.stackexchange.com/a/532690/512080

This tells us that both $e^{x}$ and $e^{-x}$ are simultaneously convex as well as concave which is possible if and only if they are linear functions in $x.$ This is completely absurd.

Where did I make mistake?

ZaWarudo

15:31

In the commonly accepted definition of field $\mathbb{K}$, do we assume $1_{\mathbb{K}} \ne 0_{\mathbb{K}}$? I was wondering about this because if we assume $1_{\mathbb{K}} = 0_{\mathbb{K}}$ then for each $k \in \mathbb{K}$ from the definition of unity follows that $k = k1_{\mathbb{K}} = k 0_{\mathbb{K}} = 0_{\mathbb{K}}$ and so $\mathbb{K}=\{0_{\mathbb{K}}\}$, not very interesting.
Moreover, if $1_{\mathbb{K}} \ne 0_{\mathbb{K}}$, we can deduce from the definition of inverse for product that $k^{-1} \ne 0_{\mathbb{K}}$ because assuming that leads to the contradiction $k^{-1} = 0_\mathbb{K}…

Thorgott

yes, $1\neq0$ is one of the field axioms

2 hours later…

leslie townes

17:10

@Anacardium how does that link show that those two functions are concave? or how would you otherwise deduce that they are "simultaneously convex and concave"? the question there starts with a function that is assumed to be concave (which e^x and e^(-x) are not)

it's a one-way implication being proved there, not an if and only if. if that helps

Ben Steffan

18:04

@Thorgott To get back to an older discussion: $\Omega^\infty\colon \mathrm{Sp} \to \mathrm{An}$ is an equivalence when restricted to connective spectra on the domain and grouplike commutative monoids on its codomain

Thorgott

18:28

yeah, that's essentially the recognition principle

1 hour later…

Anacardium

19:30

@leslietownes: If $e^x$ is convex what is $-e^x$?

It should be concave. Right?

So according to the link $-e^{-x}$ is convex. Hence $e^{-x}$ is concave. Is there any mistake here?

leslie townes

@Anacardium look at the definition of the set S that the OP's function "f" is proved to be convex on

Anacardium

@leslietownes$:$ Could you please tell me the error that I am making?

@leslietownes: I see. But can't we similarly prove reciprocal of a positive convex function is concave?

leslie townes

well, no, as your example shows. but there's no conflict between your example and that result

e^x and 1/e^x are reciprocals of one another, and they're both convex, and neither is concave. we can be sure of that. right?

Anacardium

This is given as an assignment problem of a course in which I have been assigned as a TA. The problem states "Show that reciprocal of positive convex function is concave."

@leslietownes That's what I came up with. That's why the confusion arose.

leslie townes

well, in the sense of that exercise, is a "positive convex function" a function that is assumed to be positive and convex on all of R? note that this too is not the setup of the question linked above

Anacardium

19:40

So it's a wrong problem as stated.

@leslietownes The question in the link is more general.

leslie townes

i would talk with whoever set the problem and ask if they meant to ask something else. maybe there are some hypotheses missing that make it doable

Anacardium

You can prove the same thing if the setup of the linked question is changed to positive everywhere.

I think the same argument goes through.

leslie townes

but that would be about a function that is positive and concave everywhere. different setup

3

Q: Question About Concave Functions

It easy to prove that no non-constant positive concave function exists (for example by integrating: $ u'' \leq 0 \to u' \leq c \to u \leq cx+c_2 $ and since $u>0$ , we obviously get a contradiction. Can this result be generalized to $ \Bbb R^2 $ and the Laplacian? Is there an easy way to see thi...

real-analysis calculus functions

there aren't a lot of functions that are positive and concave on all of R. only the constant functions (see robjohns answer there)

and it's actually true that constant functions are both convex and concave and their reciprocals are too

if you just think geometrically about concavity, if you fix some point on the graph at which the function is positive, and the tangent line has a nonzero slope, the graph, being "stuck" under that tangent line, will have to cross the x-axis and can't be positive forever

Anacardium

Hmm. That is really nice to know. So the question would not be interesting if one replaces convex by concave.

leslie townes

so that thing about intervals and only looking at the set on which "g" is positive in the post linked above is i guess what you could call an essential part of the setup

i love how there's a robjohn answer for almost everything :)

Anacardium

19:49

Nice observation. But positive convex functions do exist. Such as $x^2, e^x, e^{-x}$ etc.

Because the tangent line is now passing below the curve at any point.

So even though it cuts the x-axis, the curve might not.

I get your point. Thanks a lot.

leslie townes

yeah it almost feels like there should be a complete symmetry between concave/convex, but there kinda isn't

or if there is i guess it would be between positive concave functions and negative convex functions :)

Anacardium

Hmm right.

By the same analogy negative convex functions do not exist which is non-constant; although plenty of negative concave function do exist.

imbAF

20:28

What is the difference between an axiom and a tautology?

In both cases, we have a statement which is taken to be true, without the need of a proof

Am I wrong?

Xander Henderson

20:50

An axiom is a statement which is assumed to be true. It is true by fiat.

A tautology is a statement which is true by its logical structure, e.g. for any statement $P$, $P$ implies $P$.

robjohn

it's raining or it isn't.

imbAF

21:07

So what is the difference between assuming it to be true and being true?

Do you need to undertake some action for one, that is different from the other?

Xander Henderson

Again, a tautology is about the structure of a statement. An axiom is simply a statement which is assumed to be true.

The real numbers may be defined axiomatically. One axiom is that addition is commutative. The axiom is the statement "If $a$ and $b$ are real numbers, then $a+b = b+a$."

A tautological statement would be "Addition is commutative because addition is commutative."

imbAF

Ok I see

psie

21:43

I'm working the following exercise in Rudin's book (see below). To prove that $g$ is differentiable, we put $f(x)=y$ and $f(x+h)=y+k$ and $$\frac{g(y+k)-g(y)}{k}-\frac{1}{f'(x)}=\frac{1}{\frac{f(x+h)-f(x)}{h}}-\frac1{f'(x)}.$$ Now, for all $\epsilon>0$, there exists $\eta>0$ such that $$\left|\frac{1}{\frac{f(x+h)-f(x)}{h}}-\frac1{f'(x)}\right|<\epsilon\tag1$$ if $0<|h|<\eta$. But this is not what I want. I want $0<|k|<\delta$ to imply $(1)$.

I feel like I need to establish that $g$ is continuous. Rudin has this theorem that if $f$ maps from a compact metric space and is 1-1 and onto its codomain, the inverse is continuous. But $(a,b)$ is not compact :(

user image

psie

21:54

being differentiable is a local property, right? like continuity? maybe it suffices to restrict myself to a compact subset of $(a,b)$...hmm.

1 hour later…

imbAF

23:18

Yes, I have a question

In this statement

A fundamental principle of modern mathematics is that the way to understand
a space M, given as some set of points, is to look at F(M), the set of
functions on this space. This “linearizes” the problem, since the function space
is a vector space, no matter what the geometrical structure of the original set
is. If the set has a finite number of elements, the function space will be a finite
dimensional vector space.

If you take the set with the element 1

What prevents me from taking an infinite nr. of functions

?

Alessandro Codenotti

23:33

@imbAF What kind of functions are there on a set with one element?

imbAF

A function can be anything can it not?

x+1, x^2 , \sqrt(1)

etc

Alessandro Codenotti

Sure, but the space you're starting from has only point

imbAF

yes

Alessandro Codenotti

So the function is entirely determined by its value on the only available point

imbAF

which i use as an input for different functions

Yes

Alessandro Codenotti

23:35

Or in other words there is a 1-1 correspondence between real-valued functions on the space with one point and real numbers

imbAF

x+1 and 2x both can be evaluated at that point

Alessandro Codenotti

But the real numbers are a one dimensional vector space, so in particular a finite dimensional one, as claimed

imbAF

I am sorry, are you in your statements limiting yourself to an arbitrary function

to one*

Alessandro Codenotti

I'm taking arbitrary functions $\{x\}\to\Bbb R$

imbAF

arbitrary function

ILikeMathematics

23:37

@BenSteffan Do you know of any lecture notes to "Einführung in die Geometrie und Topologie" at Bonn?

imbAF

Ok, so where is the issue here?

Alessandro Codenotti

There is no issue, functions on this space are exactly the same as real numbers, hence a one dimensional vector space

imbAF

Because the statement above implies that you have a set of functions that has as many elements as elements of the set that is used as the definition region

Maybe I missunderstand the meaning of 1D

Alessandro Codenotti

No, it only says that the set of functions will be a finite dimensional vector space

imbAF

One sec

Alessandro Codenotti

23:38

Not that there will be finitely many functions

imbAF

But isn't it an obvious thing and unrelated to the nr. of elements of the set, the nr. of elements of the vector space?

I will given an example

Alessandro Codenotti

The dimension of the vector space will be the same as the number of elements in the set

imbAF

if you have S={x} a set of one element, and you have G={x+1,x+2,....x^2...}, G is clearly infinite, or can be as such

As dimension of the vector space

I consider the nr. of different elements

is that not correct ?

Alessandro Codenotti

Any (real) vector space will have infinitely many elements, regardless of its dimension

imbAF

But that is what I consider as dimensions, the nr, of elements, at least that is what's the case in physics

For example the group of rotations, is an infinite group

because it contains infinite elements

And, in physics at least, the dimensions of a vector space are tied with the nr. of basis elements

from which all the elements of the vector space, can be generated

Idk, maybe I am misunderstanding something simple

@AlessandroCodenotti Ok I see that you relate the dimensionality with the basis of the vector space, which in case of real numbers, the basis is 1D and contains the nr.1

I get what you mean with dimensionality of the vector space

Ben Steffan

23:54

@ILikeMathematics No, but Munkres should serve you well

Actually, no, I do know of a set of notes: uni-bonn.sciebo.de/s/…

but they are a bit atypical: sections 5 & 6 are not usually taught as part of the course

Alessandro Codenotti

Of course even the most basic topology course in Bonn is heavily algebraic topology oriented :P

Ben Steffan

it also has other, uh, idiosyncracies, such as not discussing separation properties apart from Hausdorffness

@AlessandroCodenotti It's not usually like that I swear :)))

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