Mathematics - 2020-04-13 (page 1 of 2)

nbro

00:48

0

Q: How can neural networks approximate any continuous function but have $\mathcal{VC}$ dimension only proportional to their number of parameters?

Neural networks typically have $\mathcal{VC}$ dimension that is proportional to their number of parameters and inputs. For example, see the papers Vapnik-Chervonenkis dimension of recurrent neural networks (1998) by Pascal Koirana and Eduardo D. Sontag and VC Dimension of Neural Networks (1998) b...

DarkRunner

03:13

I don't understand the point of this example:

Ted Shifrin

03:25

It is illustrating the fundamental equation $Ax\cdot y = x\cdot A^\top y$.

Balarka Sen

Hi @Ted

Ted Shifrin

rehi, a @Balarka

DarkRunner

@TedShifrin Ah, I realize it now;

But on the last line, it states that

Ted Shifrin

I told my classes that that is "Ted's favorite formula."

DarkRunner

Ted Shifrin

03:29

Yes. Multiply a mystery matrix by $y$ and get that answer.

DarkRunner

Ah so we are defining the transpose from the dot product in a way

Indeed, that's exactly what the book states

Ted Shifrin

That's why the transpose is so important — it moves the matrix across the dot product.

DarkRunner

I analytically understand the statement -- that the transpose and the dot product are linked via that equation. But my question is how can I understand it intuitively (if that makes any sense)?

Ted Shifrin

I have no answer to that ... each person's "intuition" is individual. I think that formula is beautiful and important, and that's how one should understand why the transpose is important.

DarkRunner

So we have three things: Two vectors $x$, $y$, and a matrix $A$. Matrix $A$ acts on vector $x$ and so we have the resulting vector $Ax$. Likewise, the transpose of Matrix $A$ acts on vector $y$ and we have the resulting vector ${ A }^{ T }y$. Now, what I say is this: the angle between vector $Ax$ and $y$ is the same as the angle between vector $x$ and ${ A }^{ T }y$.

Balarka Sen

03:42

No, $v \cdot w$ does not compute the angle between $v$ and $w$. The length of $v$ and $w$ also contribute to $v \cdot w$.

Ted Shifrin

Yup, not angles.

Balarka Sen

@TedShifrin I like a Thurston quote, which goes something like "one man's intuition is the other man's intimidation"

DarkRunner

Oh right

Whoops

Ted Shifrin

Yeah, intuition is quite personal. But I don't know how to give anything other than what I said for Ted's favorite formula unless we assume something like $A$ an isometry.

@DarkRunner: One of my favorite examples of applying this formula (which I have in my books) is this. You can find this is one of my early YouTube videos. Let $A$ be the ingredient-recipe matrix for producing different kinds of cookies. $x$ is the vector that tells you how many of each kind of cookie you want to product. $Ax$ tells you the amounts of ingredients needed. $y$ is the price vector for the ingredients. Then $Ax\cdot y$ tells you the cost to product $x$ of each kind of cookie.

Then the equation $Ax\cdot y = x\cdot A^\top y$ can be interpreted this way: $A^\top y$ tells you the cost of making each kind of cookie, and dotting with $x$ tells you the total cost.

DarkRunner

04:06

OK, Thanks for the idea

Jack Ohara

05:17

@LeakyNun Hello ,you know some logic_

EnjoysMath

06:05

0

Q: The smallest conceivable yet still interesting diagram chase. "Mini snake lemma".

Let objects be $R$-modules for a ring $R$ and morphisms $R$-module homomorphisms. I think the below diagram proves well-definedness of a map $N \to M'$ given a commuting square and exact rows (one surjection on top, one injection on the bottom side of the square). Since I didn't do anything o...

category-theory modules group-homomorphism exact-sequence diagram-chasing

I invented that

probably a re-invention of something well-known

hchar

06:21

I see an exercise to "Prove that the set of all polynomials of degree $\le n$ is a subspace of $\mathscr{P}\mathit{l}$". Any idea what $\mathscr{P}\mathit{l}$ is?

Leaky Nun

@hchar polynomials

Edward Evans

Morning

hchar

Thanks; just about to say found the definition. It's just all the polynomials of real coefficeints.

First time using this chat platform. Would be nicer if it supports Latex.

Leaky Nun

see room description

Secret

0^i is NOT 0

►

hchar

06:49

@LeakyNun Thanks. "start ChatJax" bookmark works in Firebox.

user76284

@Secret $0^0 = 1$.

hchar

@user76284 per wikipedia there is no agreed value on $0^0$.

user76284

@hchar There are good reasons why $0^0 = 1$ and no good reasons to claim it's undefined.

$0^0$ is an empty product, and therefore equal to the identity element of the naturals under multiplication, namely 1.

$0^0$ is the number of functions from a 0-element set to a 0-element set, and there is exactly 1 such function (the empty function), so again we have $0^0 = 1$.

$0^0 = 1$ yields the expected result for many combinatorial identities such as the binomial theorem, which would otherwise have to be mutilated with ugly special cases.

Leaky Nun

the limit $x^y$ as $(x,y) \to (0,0)$ is undefined

user76284

$x^0$ is the identity element of the polynomial ring whose evaluation homomorphism yields 1, so $x^0 = 1$ regardless of the value of $x$.

@LeakyNun Sure, that doesn't mean $0^0$ is undefined. Just because a function is discontinuous at a point doesn't mean it has to be undefined at that point.

Leaky Nun

06:57

but it's a good reason to claim that it's undefined

user76284

That function is undefined in the negative region anyway.

Leaky Nun

all your reasons rely on the fact that the exponent is supposed to take on integer values

but this is not true in analysis

user76284

@LeakyNun Combinatorial arguments tell us $0^0 = 1$. Exponentiation on the reals should extend exponentiation on the naturals. Therefore $0^0 = 1$ "ought" to be the case in the context of the reals as well.

It's a straightforward argument.

If you agree that real exponentiation should extend natural exponentiation where the latter is defined, then you should agree that $0^0 = 1$.

I love arguing with people why $0^0 = 1$ :) Knuth is right

$0^0 = 1$ in combinatorics. $0^0 = 1$ in the binomial theorem. $0^0 = 1$ in the cardinals. $0^0 = 1$ in the ordinals. $0^0 = 1$ in the nonzero analytic functions. $0^0 = 1$ in the polynomial ring.

Leaky Nun

you define $x^y$ for irrational $y$ by demanding it be continuous in $y$ (e.g. $10^\pi = $ limit of $10^3$, $10^{3.1}$, $10^{3.14}$, ...) so $0^0 = \lim_{x \to 0^+} 0^x = 0$

again, all your arguments rely on the fact that we never use anything non integer for the exponent

user76284

@LeakyNun Again, do you agree that real exponentiation should extend natural exponentiation where the latter is defined?

There are many good reasons to assign the value 1 to $0^0$, and no good reasons not to.

Leaky Nun

07:02

do you agree that real exponentiation should be continuous in the exponent?

I just gave you a good reason not to

user76284

@LeakyNun It will never be continuous in the exponent no matter how you define it.

Leaky Nun

if you define $0^0=0$ then it will be continuous in the exponent

user76284

No, it will not.

Leaky Nun

yes it is

user76284

Oh, so you're dropping one of the arguments? :)

I see now you said "in the exponent".

Leaky Nun

07:04

how about recognize the fact that the complex function $z^w$ has an essential singularity at $(0,0)$ and leave it undefined

user76284

Then the same argument applies for $x^0$ which yields the value 1 and your argument cancels out.

Leaky Nun

so that's why it should be undefined :)

user76284

Not to mention that $x^0$ is actually defined for $x < 0$ while $0^x$ is not.

Leaky Nun

you can't define $x^\pi$ by having it continuous in $x$

user76284

@LeakyNun Why do you say it should be undefined?

Leaky Nun

07:05

1 min ago, by Leaky Nun

how about recognize the fact that the complex function $z^w$ has an essential singularity at $(0,0)$ and leave it undefined

because if you take different paths you end up with different limits

just as you demonstrated

user76284

Like I said earlier, just because a function is discontinuous at a point doesn't mean it's undefined at that point.

Leaky Nun

but it should be

user76284

Why?

Leaky Nun

because taking different paths gives you different results

user76284

Fun fact: If $f$ and $g$ are analytic and $f$ is positive in a neighborhood of $0$, then $\lim_{t \rightarrow 0^+} f(t), g(t) = 0$ implies $\lim_{t \rightarrow 0^+} f(t)^{g(t)} = 1$.

Always.

@LeakyNun I know that.

That's not a good reason.

Leaky Nun

07:09

that's an odd restriction. why should we only care about analytic functions?

user76284

It's just math winking its eye at us.

And what do you have to say about the point I made regarding extension? I notice you haven't addressed it :)

Oh, and how would you like to mutilate the binomial theorem? Hehe.

Secret

$$0^{0^{0}}$$

Leaky Nun

here's the story of the "extending natural numbers". you define $x^y$ by first defining it for all natural $y$ (include $0^0$ if you want), and then extending to all (nonnegative/positive) rational $y$ by taking roots, and then extending to all (nonnegative/positive) real $y$ by continuity; but then you realize $0^y$ is not continuous at $0$ anymore

@user76284 in the context of combinatorics the convention is $0^0=1$

so the binomial theorem is fine

Balarka Sen

Lol $0^y$

user76284

@LeakyNun The base can be continuous in the binomial theorem.

Leaky Nun

07:12

in the binomial theorem $0^0=1$

user76284

@LeakyNun Right. Your story shows why $0^0 = 1$ ought to be true in our extension.

Leaky Nun

no, my story shows that $0^0=1$ is inconsistent with extending it continuously to all real numbers

user76284

@LeakyNun Your extension is not continuous either (unless you restrict the base a la $0^x$, but that's cheating because the same argument applies to $x^0$).

If I read your point correctly.

Leaky Nun

how do you define $e^\pi$?

@BalarkaSen pick a side

user76284

@LeakyNun You first use the theorem that each positive real has a unique $n$th root.

Leaky Nun

07:16

ok then?

user76284

That gives you denominators.

Then use the standard repeated-multiplication definition for numerators.

Then take a sequence of rationals that converges to pi.

Leaky Nun

you see

user76284

And take the corresponding sequence $e^x$.

Leaky Nun

you define it by taking a limit on the exponent

user76284

Yes that's right :) For positive arguments.

Leaky Nun

07:17

so you're making $0$ an exception

user76284

Well, $0^0 = 1$ in the naturals like you said.

:)

Are you asking about the $0^x$ for $x > 0$ case?

Leaky Nun

I said $0^0=1$ in discrete/combinatorics contexts

loch

maybe you guys should compromise and define 0^0 = 1/2

Balarka Sen

@LeakyNun which side? the diff eq and the series are the same thing because ODEs are elliptic so by elliptic regularity has an analytic solution so the Taylor series is the function

user76284

@loch Ha!

Balarka Sen

07:18

:3

user76284

@LeakyNun Right, but we're defining real-exponentiation in terms of rational-exponentiation.

Leaky Nun

@loch if 0^0 could be both 1/2 and 1 then it is best to leave it undefined :)

user76284

natural-exponentiation*

Since natural-exponentiation dictates that $0^0 = 1$ (through the various combinatorial, etc. arguments), real-exponentiation morally ought to as well.

Leaky Nun

but insisting $0^0=1$ for reals makes the story inconsistent

because fun fact: if you take a sequence of rationals $q_i$ that converge to another rational $q$, then $x^{q_i} \to x^q$

this is consistency

user76284

How does $0^0 = 0$ preserve consistency? You'd have the same problem in the base.

See what I mean?

That argument cancels out :)

Leaky Nun

07:20

no, I didn't say $0^0=0$ preserve consistency

I say that you can't assign a value to $0^0$ that makes the story consistent

and that is why you should leave it undefined

user76284

Well, then we go back to what I was saying: There's no other good candidate for $0^0$. But 1 is a good candidate for $0^0$.

Leaky Nun

fun fact: "$0^0=1$ is inconsistent" doesn't mean "$0^0=0$ is consistent"

user76284

It is perfectly consistent, though.

Leaky Nun

2 mins ago, by Leaky Nun

because fun fact: if you take a sequence of rationals $q_i$ that converge to another rational $q$, then $x^{q_i} \to x^q$

user76284

But that's just not true.

I was saying that if it were true (that we had to demand this property), the same "problem" would arise in the base.

Leaky Nun

07:22

so make it undefined

user76284

Nah, I'll stick with $0^0 = 1$ and not make a bunch of special cases for my identities :)

Leaky Nun

such as $x^y/x^z = x^{y-z}$?

user76284

It's just more concise, more convenient, more clean, more natural!

@LeakyNun That identity can never be satisfied for all possible values of $x, y, z$.

At best it only holds when you restrict them in certain ways.

Leaky Nun

how about $0^x=0$

user76284

That's true for $x > 0$.

By the way, as a parenthetical, I'm sure we agree about the technical details. I'm merely highlighting the "aesthetic" or "moral" nature of this kind of issue.

Leaky Nun

07:28

whereas I'm highlighting the stupidity of this issue

user76284

What I mean by that is something like the following: Consider how to define $0!$. I could say $0! = 7$ and modify all the other formulas of my system (e.g. by inserting special cases) to be consistent with that choice.

Surely you agree that would be stupid, right?

Well, claiming it's not true that $0^0 = 1$ is an equally stupid convention :)

Leaky Nun

what is 0/0?

user76284

In what system?

A wheel?

Leaky Nun

what?

user76284

Wheel theory

Wheels are a type of algebra, in the sense of universal algebra, where division is always defined. In particular, division by zero is meaningful. The real numbers can be extended to a wheel, as can any commutative ring. The Riemann sphere can also be extended to a wheel by adjoining an element ⊥ {\displaystyle \bot } , where 0 / 0 = ⊥ {\displaystyle 0/0=\bot } . The Riemann sphere is an extension of the complex plane by an element ∞...

Leaky Nun

07:31

looks like they have a lot of special cases for their identites

someone wouldn't like that

user76284

I'm not sure what you mean by that.

Leaky Nun

should $0/0=1$?

user76284

I don't see how any of the reasons I gave for why $0^0=1$ apply here :)

I don't see a combinatorial interpretation, for instance.

Leaky Nun

I don't like special cases for my identity $x/x = 1$

Secret

What happens when we actually left $0^0$ as it is in binomial theorem

I will figure the $0^0$ will absorb all the powers

when you e.g. expand $(a+b)^n$

user76284

07:35

True! But I don't like any special cases for my identity $x^2/x = x$ either :)

Leaky Nun

so let's leave $0/0$ undefined

user76284

Well (takes a look around) are there any arguments for assigning $0/0=1$ any particular value (that aren't cancelled out by similar arguments for other values)?

Non-rhetorical question.

I actually want to think about that one for a bit.

Secret

$0/0$ is actually the worst of the indeterminates

user76284

Of course, defining $0/0$ is actually much more dangerous than defining $0^0$.

Secret

Even in many algebraic systems where it is defined, it is mostly an additive absorber

like in wheels for example

user76284

07:38

"Dangerous" in the sense that the resulting aberrations spill over the rest of your system, so to speak.

Secret

it is even more problematic than $n/0$ for $n \neq 0$

user76284

@Secret Yeah that's along the lines I was thinking, good point.

You probably have to end up loosing all sorts of algebraic properties.

Secret

At minimum, defining division by zero will lose all except power associativity

It's something that live in a highly nonassociative corner of mathematics

user76284

Yeah I haven't seen much connection between wheels and other math topics, which might lead one to believe they're not a very "natural" algebraic structure.

I'm reading the paper to see.

Knight

@Secret Hi!

user76284

07:49

Hah, turns out Wikipedia has a typo.

In the "Wheel of fractions" section. I was wondering why the "denominator" for the multiplication rule was non-symmetric (and therefore ugly).

Oh wait nevermind, they're using a different notation from the paper.

I guess it is symmetric then. Not as bad as I thought.

It looks more natural when presented with the wheel of fractions definition than with the conditions stated at the beginning of the Wikipedia article.

Reminiscent of the Grothendieck group construction.

(e.g. how you define the integers as pairs of naturals, and the new versions of the usual arithmetic operations between them)

But yeah you still don't have the algebraic properties we think of as "nice".

Compare these constructions:
https://en.wikipedia.org/wiki/Integer#Construction
https://en.wikipedia.org/wiki/Rational_number#Formal_construction
https://en.wikipedia.org/wiki/Complex_number#Construction_as_ordered_pairs
https://en.wikipedia.org/wiki/Wheel_theory#Wheel_of_fractions
Some of these rules lead to "nice" algebraic structures, and others don't.

@Secret I'm also of the opinion that $\sum_{n=0}^\infty 2^n = -1$ ($p$-adically convergent, or more generally, convergent under any stable and homogeneous generalized-limit) is just as "true" as $\sum_{n=1}^\infty 2^{-n} = 1$ (Cauchy-convergent). But don't tell anyone that or I'll get yelled at with some talk about "the real world" and such :)

Secret

08:06

The central philosophy of how wheel multiplication is defined, seemed to be trying to collect as many zero terms together as possible

user76284

So the addition and multiplication rules are the same as those of the rational construction, but there's those extra rules.

$A$ is assumed to be a commutative ring with identity, and $S$ a multiplicative submonoid of it.

Secret

08:52

I am suspecting the rules for the zero terms in wheels are in a way, treating them as invariant to involution and multiplication

For example:

$xz+yz = xz+(y+0)z = xz+yz+0z = (x+y)z+0z$

$(x+yz)/y = (x+yz+0)/y =x/y + yz/y + 0/y = x/y + z +/(y0) = x/y + z + 0y$

Though this guess does not seemed to work for
$(x+0y)z = xz+0y$

Math geek

09:32

Any physics students here? May I get help to clear this basics. here is my question.

in Let’s do some Physics , 1 min ago, by Math geek

-1

Q: What is the difference between $F_2$ and $W_A$?

Underlined statements are definitions of $F_1$ and $F_2.$ I am learning the basics of free body diagram. I am not able to judge the distinction between $F_2$ and $W_A$. $W_A$ is the weight of $A$ acting on $B$. $F_1$ is the reaction force of $B$ to $A$ as the consequence of Newton's th...

classical-mechanics

Secret

09:47

Math geek

10:37

What is the difference between $F_2$ and $W_A$?

Edward Evans

Hmm, what will be the Lebesgue volume of the set $\lbrace (x_i) \in \prod_i \Bbb R : \sum_{i} \lvert x_i \rvert < t \rbrace$ for some fixed $t$

with $i \in \lbrace 1, \dots, r \rbrace$

I mean, $\operatorname{vol}\left( \bigcup_i (-x_i, x_i) \right)$ is gonna be $\frac{1}{2}\sum_i \lvert x_i \rvert$ if the intervals are disjoint, I think, in which case I'll get something like $2^rt^r$

Leaky Nun

@EdwardEvans high school maths time?

Edward Evans

yep

Leaky Nun

a triangle is half the area of its rectangle; a pyramid is a sixth of the volume of its cuboid

so by pattern matching, if the number of dimensions is $r$, then you divide by $r!$

Edward Evans

actually I'm computing the volume of $X = \lbrace (z_\tau) \in K_\Bbb R : \sum_\tau \lvert z_\tau \rvert < t\rbrace$

Leaky Nun

10:44

how about no

Edward Evans

but via an isomorphism $f : K_\Bbb R\to \prod_\tau \Bbb R$ one can move from the Minkowski volume to the Lebesgue volume, if you multiply by $2^s$

lol

Leaky Nun

oh, I thought $K_\Bbb R$ means $K^{\Bbb R}$ for some reason

Edward Evans

hahaha nah

Leaky Nun

you mean $\prod_{r+2s}\Bbb R$

Edward Evans

$\tau \in \operatorname{Hom}_\Bbb Q(K, \Bbb C)$, but yeah there are $r + 2s$ of those

Leaky Nun

10:48

so $\dfrac{(2t)^{r+s}}{(r+s)!}$ multiplied by or divided by $2^s$

I'm too lazy to figure out which one

Edward Evans

From the answer it's gonna be divided by lol

Leaky Nun

cool

I can sense some class number formula

Edward Evans

I'm attempting to show that the volume of $X$ is $2^r\pi^s\frac{t^n}{n!}$

Leaky Nun

I can almost smell it

where tf does the $\pi$ come from

yeah my $r+s$ should be $r+2s$

and then maybe divide by $2^{2s}$

Edward Evans

If you split $\operatorname{Hom}_\Bbb Q(K, \Bbb C)$ into real and complex embeddings and then write $X$ as a union of the factors of $\prod_\tau \Bbb R$ indexed by real and complex embeddings resp. then you get for the first set some union of intervals in $\Bbb R$ and for the complex embeddings some discs in $\Bbb R^2$

so that's where the $\pi$ is coming from

Leaky Nun

10:53

oh right

something is wrong wi' me

Edward Evans

Lol in Neukirch he's just like "Yeah so obviously the volume is this" and then I spent an hour trying to work out why it's so obvious

Leaky Nun

just use induction

Edward Evans

I'll use proof by "oh gosh look at the time I better move on to the next section if I wanna get finished in time for exams"

Leaky Nun

lmfao

Alessandro Codenotti

11:19

@EdwardEvans "obviously" means "with trivial but too painful to write computations"

Royi

11:30

Hi, I have interesting case about the solution of @greg to the following question:

1

Q: Gradient of the spectral norm of a matrix

Let $X \in \mathbb{R}^{a \times b}$ and $$\|X\|_2 = \sigma_{\max}(X) = \sqrt{\lambda_{\max} \left( X^T X \right)}$$ How can I compute $\nabla_X \|AX\|_2$, where $A \in \mathbb{R}^{c \times a}$ is some known matrix?

linear-algebra derivatives norm matrix-calculus singular-values

In case we have few singular values with the same value (Maximum) @greg says to sum over all corresponding vector of $ U $ and $ V $.

Yet, when I compare it to Numerical Differentiation I get factor of 2 in the answer.

It works perfectly (His derivation matches the numerical differentiation for the case there is single maximum value. But in case of multiple something doesn't work.

Sebastiano

11:44

Good morning everybody

excuse me for this post....

But I can not close this question. I think that is a duplicate because this question has been asked yesterday.

math.stackexchange.com/questions/3623306/…

Sebastiano

12:14

In the previous closed question there is an answer.

$\sqrt{2-2\cos\theta}=\sqrt{2(1-\cos\theta)}=\sqrt{2\cdot 2\sin^2{\frac{\theta}{2}}}=2\sin\frac{\theta}{2}$ and the absolute value where is?

Knight

12:51

@Sebastiano Hi! You remember me?

@Sebastiano for any $x$ we always have $\sqrt {x^2}= |x|$

Mojibake

I'd like to protest a duplicate.

math.stackexchange.com/questions/3622932/… Isn't the same as the question linked in the duplicate.

The question linked is form natural n,ms and it's different when they can be negative values

Mike Miller

13:07

It is true that they are different, but not that different. The case of the set {1/m - 1/n | m,n positive integers} is the most work, but guy can show this only accumulates to the set {1/m | m nonzero integer} cup {0}.

Mojibake

@MikeMiller There's also the point that the question I asked was if my solution was correct, and that hasn't been answered.

Mike Miller

I don't get involved in site mechanics, I'm just commenting on your problem.

Mojibake

And while that is possible, it isn't explicity stated in the answer, so I don't think my question should be declared a duplicate.

OK, I was just slightly annoyed, and I know you were just commenting.

Mike Miller

I don't think you've actually linked your post though, just the dupe target

Mojibake

Hmm..it work fine on my computer? I was saying that this shouldn't be a dupe target?

Mike Miller

13:10

Yeah, but I don't know what got closed as a duplicate.

Oh I see.

Mojibake

OK, I don't get why this isn't working for you.

Mike Miller

Without an account it just redirects me to the target, I don't get to see the closed question

Sorry about that, my bad.

Mojibake

Yeah, that's fine.

Mike Miller

If you reproduce your argument here I can read it

Mojibake

Yeah, so the argument I used was first that $X_n=\{ \frac{1}{n} +\frac{1}{m}\ \mid |m|\ge |n|, m \in \mathbb{Z}\}$ for fixed, non-zero $n$ has only $\frac{1}{n}$ as it's limit point.

Then, I supposed $x$ was a limit point of the given set and let $Y_n$ denote set of natural $m$'s for which $X_m$ and $B_{\frac{1}{n}}(x) \setminus \{x\}$ have non-empty intersections.

$Y_n$ is not empty for any $n$ since $x$ is positive and is a limit point. By the well-ordering property, the sequence $y_n=\min Y_n$ is well defined. Note that $Y_{n+1} \subset Y_n$ so we have $y_n \le y_{n+1}$. If $y_n$ is bounded, it thus converges to a natural number $y$ as all $y_n$ are natural numbers. We may say that $x$ is a limit point of $X_y$, and by Step 1 we have that $x=\frac{1}{y}$.

If $y_n$ is not convergent and unbounded, $y_n \to \infty$. But for $k > \frac{4}{x}$, note that $X_{k}$ lies between $[0, \frac{2}{k})$, while for $\epsilon<\frac{x}{2}$, $\epsilon$ neighbourhoods lie in $(\frac{x}{2}, \infty)$. Thus a contradiction, as $Y_k$ is empty if $k > \frac{4}{x}$. We repeat a similar logic for $x<0$. Thus, every limit points of $X$ are of the form $\{\frac{1}{n} \mid n \in \mathbb{Z} \setminus {0} \} \cup \{0\}$

(I copied and pasted so the wording might be a little awkward)

Mary Star

13:23

Hello!! We have that $\lim_{x\rightarrow +\infty}f(x)=L\in \mathbb{R}$ and I want to show that $\lim_{x\rightarrow +\infty}(f(x+1)-f(x))=L$.

I have done the following using the mean value theorem:
For some $\xi\in (x, x+1)$ we have that $f'(\xi )=f(x+1)-f(x)$.
If we take the limit as $x\rightarrow \infty$ then it is also $\xi \rightarrow \infty$ and so we get $$L=\lim_{\xi\rightarrow \infty}f'(\xi)=\lim_{x\rightarrow \infty}[f(x+1)-f(x)]$$
Is this correct?

Leaky Nun

@MaryStar yes

Mojibake

Could you possibly verify this? @MikeMiller(If it isn't that much trouble)

Mike Miller

Sorry, @Mojibake, I got distracted by something at home. Reading now.

Mojibake

Thank you. I appreciate this.

Mary Star

Great!! And when $L=\pm\infty$ ? Then from $\lim_{x\rightarrow +\infty}f'(x)=L$ we get that $\forall x>M$ for some $M>0$ we have that $f'(x)>M$. D we use here again the mean value theorem to get the same result, i.e. that \lim_{x\rightarrow +\infty}(f(x+1)-f(x))=L$ ? Or is this not possible in this case? @LeakyNun

anakhro

13:30

@MikeMiller where does the Tor appear with Z/2 coefficients in the Kunneth formula for homology, as opposed to using Z/3 coefficients

Leaky Nun

@MaryStar looks good to me

Mary Star

It is exactly the same proof as previously, right? We didn't use anywhere the fact that $L$ is real, right? @LeakyNun

Mike Miller

@Mojibake It works (you never explicitly assumed $x>0$ at the start, but it was clear you meant to when you found a contradiction). I would probably have written that the epsilon-neighborhoods lie in (x/2, 3x/2); the infty threw me off.

Mary Star

One last question... Knowing that $\lim_{x\rightarrow +\infty}(f(x+1)-f(x))=L$ and knowing that $\lim_{x\rightarrow +\infty}f'(x)$ exists in $\mathbb{R}\cup \{\pm \}$ does it hold that $\lim_{x\rightarrow +\infty}f'(x)=L$ ? Do we use again the same procedure?

For some $\xi\in (x, x+1)$ we have that $f'(\xi )=f(x+1)-f(x)$.
If we take the limit as $x\rightarrow \infty$ then it is also $\xi \rightarrow \infty$ and so we get $$\lim_{\xi\rightarrow \infty}f'(\xi)=\lim_{x\rightarrow \infty}[f(x+1)-f(x)]=L$$

Mojibake

Ah that's good. Thanks. @MikeMiller

Leaky Nun

13:37

you have exhausted the cases in the previous two cases

so you can case on the value of lim f'(x)

this is reverse analysis

Mary Star

So the proof I wrote for the reverse direction of the previous one is correct, or not? I got stuck now. @LeakyNun

Balarka Sen

14:18

Strange. If $M$ is a surface of constant Gaussian curvature and $\mathbf{x} : U \subset \Bbb R^2 \to M$ is a parametrization such that the coordinate directions of $\mathbf{x}$ are asymptotic curves, then the surface normal is parallel $\mathbf{x}_{uv}$.

Sayan Chattopadhyay

Is there an example of a non-constant holomorphic function whose integral over two non homotopic paths is the same?

Balarka Sen

Sure, $f : \Bbb C^* \to \Bbb C^*$, $f(z) = z$, let the paths be two arcs of the unit circle with endpoints $\pm 1$.

Sayan Chattopadhyay

So the converse of the theorem that the integral over two homotopic paths is the same should also hold true right?

Balarka Sen

State the converse.

Sayan Chattopadhyay

Nvm, forget it. I messed up, you just gave an example

What quarantine meth are you doing these days @BalarkaSen

Balarka Sen

14:31

Quarantine meth? Haha.

I suppose some differential geometry. Mostly Riemannian geometry.

Sayan Chattopadhyay

Oh cool.

Balarka Sen

15:01

What's $(v_1 \times v_2) \times v_3$ again? lol

$(v_1 \cdot v_3) v_2 - (v_2 \cdot v_3) v_1$?

Looks right

Alessandro Codenotti

@BalarkaSen Do we look like physicists to you?

Balarka Sen

No, differential geometers

Alessandro Codenotti

I'm not sure whether that's better or worse

Balarka Sen

Lol

Tobias Kildetoft

@BalarkaSen Well, which cross product algebra is it in? :)

Balarka Sen

15:05

Ah so you're saying I need to ask a Clifford algebraist

ms._VerkhovtsevaKatya

Hello! I apologize for interrupting the discussion. While searching the site, I found a post that actually answers my question, but on a more advanced level than mine, so I thought it would be better to ask on chat than making a duplicate question:

49

Q: How to prove and interpret $\operatorname{rank}(AB) \leq \operatorname{min}(\operatorname{rank}(A), \operatorname{rank}(B))$?

Let $A$ and $B$ be two matrices which can be multiplied. Then $$\operatorname{rank}(AB) \leq \operatorname{min}(\operatorname{rank}(A), \operatorname{rank}(B)).$$ I proved $\operatorname{rank}(AB) \leq \operatorname{rank}(B)$ interpreting $AB$ as a composition of linear maps, observing that ...

linear-algebra matrices inequality matrix-rank

Wrong link

6

Q: The rank and eigenvalues of the operator $T(M) = AM - MA$ on the space of matrices

This problem is from Artin Algebra Second edition, 5.2.3. Let $A$ be an $n\times n$ complex matrix. (a) Consider the linear operator $T$ defined on the space $\mathbb{C}^{n\times n}$ of all complex $n\times n$ matrices by the rule $T(M) = AM - MA$. Prove that the rank of this operator is at mo...

linear-algebra matrices eigenvalues-eigenvectors

Tobias Kildetoft

@BalarkaSen Honestly, I don't remember much about the topic, only that there are essentially two options (I wrote a review of a paper that did some diagram stuff for studying automorphisms of cross product algebras some years ago)

Balarka Sen

Ah I see

ms._VerkhovtsevaKatya

So far, we've learned that for two square matrices AB of the same size in general, rank(AB) is less or equal {min rank(A), min rank(B)}, but I'm not sure if I understand the general case proven in the above thread.

Tobias Kildetoft

@BalarkaSen Not even sure if I have the review anywhere myself now, and I don't have access to MathSciNet any more.

ms._VerkhovtsevaKatya

15:15

My task is slightly different. I have to prove that an operator AM-MA is singular for any A in M_n. It is obviously singular from the first sentence in the post I found, but I don't know how to deal with a general case when A isn't diagonalizable.

Anjani Gupta

Hi! I wanted to find this improper integral - $\int_{0}^{\infty} \frac{x^{-a}}{x+1}$ where $0<a<1$. The book that am reading is converting this to complex contour integration by choosing the branch $0<\theta<2*pi$. But even then its using the residue theorem despite f not being analytic (not even defined) on the branch cut. Why?

By $f$ I meant the numerator and hence the whole integrand not analytic there.

EnjoysMath

@Shaun that's not good

that page 6 is missing

Masterphile

i am trying to bound some product and to show that it is absolutely bounded and to find the best possible bound, if someone is interested let me know

EnjoysMath

15:33

@Shaun I've read the prospectus

of the book

@Shaun you might want to read this supplement on Kripke Semantics (referred to in prospectus of Goldblatt): en.wikipedia.org/wiki/Kripke_semantics

Sebastiano

@Knight yes....an happy day for you and your parents :-). Yes of course for your answer but is missing $|\cdot|$ in the answer of the user. For my humble opinion i have put a comment. Hiiiiiiiii :-)

EnjoysMath

@Shaun I think we should read about everything mentioned in the prospectus :D

Sha Vuklia

anyone here who knows how to work with ordinals?

Tobias Kildetoft

@ShaVuklia Depends on what you need to do with them

Sha Vuklia

I want to prove that $\alpha+\omega$ is a limit ordinal. but I'm not that good into the material, so I could appreciate a hint

Alessandro Codenotti

15:47

What's your definition of $\alpha+\omega$?

Sha Vuklia

the supremum of $\alpha+n$ for $n<\omega$

Alessandro Codenotti

Ok, so suppose that $\alpha+\omega$ is not limit, let $\gamma$ be its maximum, can you derive a contradiction?

Sha Vuklia

So say we have $\gamma+1=\alpha+\omega$. Then $\gamma+1\geq\alpha+n$ for each $n$. Also, $\gamma+1\geq \alpha+n +1$ for each $n$. If I can argue that it follows from here that $\gamma\geq\alpha+n$ for each $n$, then I've found a contradiction.

I have to check the properties of addition, one sec

Alessandro Codenotti

Well but if $\gamma+1=\alpha+\omega$, then $\gamma\in\alpha+\omega$. What do the elements of the latter look like?

Sha Vuklia

ah

Then $\gamma=\alpha+n$

and hence $\gamma+1=\alpha+n+1$, which is smaller than $\alpha+\omega$

thx :)

Alessandro Codenotti

15:57

Right

The same argument really shows that $\alpha+\beta$ is a limit ordinal whenever $\beta$ is

Sha Vuklia

ah, true that

Alessandro Codenotti

Also you can think about $\alpha+\beta$ as "a copy of the ordered set $\alpha$, followed by a copy of the ordered set $\beta$" (formally let $\alpha+\beta$ be the order type of the order $(\alpha\times\{0\})\cup(\beta\times\{1\})$ with $(a,n)<(b,m)$ if either $n=0$ and $m=1$ or $n=m$ and $a<b$), then it is obvious that $\alpha$ followed by a thing without a maximum is an order without a maximum

Leaky Nun

@BalarkaSen til $H^\ast(\Omega S^n) = \Bbb Z[x,x^2/2,x^3/6,\cdots]$ if $n$ is odd although I'm not sure when I will ever use this fact

Alessandro Codenotti

this definition agrees with the inductive one but that requires a (not hard) proof

Balarka Sen

@LeakyNun Yes, it's a divided polynomial algebra.

Leaky Nun

16:01

so why do we care?

Sha Vuklia

ah right, ye my teacher told us we could look at it like that

I'll try to show it at some point

the intuition helps for now, in any case

Balarka Sen

@LeakyNun Who's "we"

Sha Vuklia

lel ^

Leaky Nun

@BalarkaSen let's say it's you

Alessandro Codenotti

Are you doing an intro to set theory course or something similar? @sha

Balarka Sen

16:06

$\Omega S^n$ is a fun object, I can run Morse theory on it using the energy functional $E : \Omega S^n \to \Bbb R$, $E(\gamma) = \int \|\gamma'|^2$, passing to the subspace of smooth closed loops on $S^n$ that the full loopspace deformation retracts to.

It's a toy model for infinite dimensional Morse theory

Sha Vuklia

@AlessandroCodenotti yes, exactly that

Balarka Sen

Indeed, you can compute the cohomology groups of $\Omega S^n$ using Morse theory like in usual Morse theory. The ring structure seems like a natural follow-up.

I only know how to compute the ring structure using SSS though

Knight

@Sebastiano That’s great. How do you doing at Physics.SE ?

Alessandro Codenotti

@ShaVuklia cool

Shaun

I'm self-learning topos theory, @AlessandroCodenotti. It has nothing to do with my formal studies in group theory, unfortunately.

Alessandro Codenotti

16:08

I know nothing about topos theory, I'm mostly interested in set theory

Shaun

I'll make a start on the Wikipedia article now, @EnjoysMath.

topologicalorientablesurface

You probably do not want to ask what you asked, because if $a,b$ are units, then $ab$ is a unit and no irreducible element can divide it. — GreginGre 15 mins ago

EnjoysMath

@Shaun it veers off deep into logic which is out of the scope of Topos theory, but the first part of article is

*I mean readable for me

Anyway, I mostly read it because I thought it would save the idea of having logic in BananaCats, but it just got way too complicated (the article)

I'm on the first few pages of Goldblatt. It's just talking about set builder notation and set extensionality principle: sets are equal precisely when they have the same elements

Mike Miller

Balarka is a topos theorist, but he mostly knows about (omega, omega-1)-topoi

4

Shaun

Let's use our chatroom, @EnjoysMath; it'll help us keep track of what we've done (without searching through other people's posts). I've read up on a fair bit of logic over the last few years.

Leaky Nun

16:12

what on earth is omega-1

Mike Miller

One less than omega

EnjoysMath

@Shaun started a room

Sebastiano

16:41

@Knight Very bad :-) See the image....:-( It is not a nice place.

@Knight physics.stackexchange.com/users/140404/sebastiano

Transcript for

$Mathematics$ Mathematics