Mathematics - 2024-04-28

12:36 AM

psie this might be one of those times where you need to add a regularity assumption to get a representation like that in general. something coming from a function F with those properties looks like it will be both inner and outer regular, and yet i'm not sure that is automatic on R if you only assume that the measure is finite on compact sets (i think it's automatic if you assume the measure is finite)

this is a general area where almost every book uses different definitions and different routes to the results, and some very well known books sometimes (mildly) fuck up by not including regularity conditions that are actually needed even in simple cases.

if your book hasn't included key words, given the nice enough F, the measure that it determines is sometimes called the "lebesgue-stieltjes measure" associated with F, and another key word that comes up a lot (and is differently and sometimes non-equivalently defined by different authors, and unlike a lot of things is actually worth paying attention to) is "radon measure"

my advice would be to find a book that simply treats the general case and see what they do. it is risky to mix treatments from different authors in this area without checking every definition ten times

Magnus Alexander

1:12 AM

@Bumblebee It will as far as I see, but my smol brain would've gone for sinx sub

Magnus Alexander

1:53 AM

Given the series $\sum\limits_{k=1}^{\infty} \dfrac{\cos{(\sqrt{k}x)}}{2^k}$, how would you proceed in finding the sum?

Xander Henderson

2:13 AM

@MagnusAlexander I'd be surprised if it had a nice closed form.

user 70432

2:43 AM

8

Q: Inner voice when reading mathematics

I am writing some notes for my students, and am curious about how others read mathematics: do you have an inner voice when reading mathematics? What about when you come across a symbol which you don't know the name of? Do you make something up in your head, or just skip over the inner enunciation...

Allie

man i didnt know stackexchange community was pro israel

yikes

user 70432

psycholinguistics

Soumik Mukherjee

4:18 AM

@Allie What happened?

CowperKettle

5:50 AM

Hamilton Walk

The Hamilton Walk from Dunsink Observatory to Broom Bridge on the Royal Canal in Dublin takes place on 16 October each year. This is the anniversary of the day in 1843 when William Rowan Hamilton discovered the non-commutative algebraic system known as quaternions, while walking with his wife along the banks of the Royal Canal. == History == The walk was launched in 1990 by Prof Tony O'Farrell of the Department of Mathematics at St Patrick's College, Maynooth. It starts at DIAS Dunsink Observatory, where Hamilton lived and was the Director from 1827 to 1865, and ends at the spot where he recorded...

Akiva Weinberger

6:16 AM

In my ideal linear algebra lecture, none of the students should be able to agree on what "perpendicular" means

Image description: students observe a linear algebra lecture. They are stretched or squished and rotated at weird angles, and therefore presumably observe the world as stretched or squished and rotated

@AkivaWeinberger At least, not until we fix an inner product

leslie townes

each one can compute inner products in a screen on their head, and they know that at least one of the others is stretched or squished or rotated at weird angles, but they don't know who is who. also the vectors break after being used once, and two of them always lie about what inner product they are using. devise an algorithm that springs them out of jail (they're all in jail, did i mention that) before the end of the month

Akiva Weinberger

6:33 AM

If it's a topology lecture, the students should be melting

Koro

6:45 AM

Why is D\phi an isomorphism here?

Soumik Mukherjee

Proof: Ex hehehe

Koro

no that was for a previous theorem

Soumik Mukherjee

Nice handwriting koro

Koro

it's not mine.

I don't recall such property of covering maps...

is the remark in the image obvious by definition of the covering map?

perhaps some non obvious property of covering maps.

also, I don't see why covering space of Lie group G is also a Lie group.

Is Ted well? I see that he's not participated here in weeks now?

hmm, he's active on mse so perhaps just taking a break from chat.

Ryder Rude

7:11 AM

if $x_n$ is a Cauchy sequence, then it is convergent

we take a convergent subsequence $x_{n_k}$ and call its limit $L$

we find a $N$ such that $n,m\geq N$ implies $|x_n-x_m|<\frac 12 \epsilon$. then we find $n_K$ such that $n_k\geq n_K$ implies $ |x_{n_k}-L|<\frac12 \epsilon$

Soumik Mukherjee

@RyderRude no

Ryder Rude

my book says it is..

Soumik Mukherjee

@RyderRude real analysis book?

Ryder Rude

yes

it gives this proof i was just writing

i think the proof works tho

Soumik Mukherjee

Ok then it is dealing with R which is a complete metric space, so every Cauchy sequence converges there

But this is not true in general

Ryder Rude

7:21 AM

oh.. so only when we r working in real numbers. thanks

Soumik Mukherjee

Not only for real numbers though, for any complete metric space that is true

Ryder Rude

ok so continuing the proof, triangle inequality shud imply that $|x_n-L|<\epsilon$ for $n\geq N$, but the book says it holds for $n\geq n_K$

@SoumikMukherjee oh

or maybe im wrong. lemme think

ok so one of the requirements is choosing $n_K>N$

Soumik Mukherjee

Also it depends on the metric, real numbers with usual metric is complete, but with the metric d(x,y)=|e^x-e^y| is incomplete.

Ryder Rude

so for $n\geq N$, we have $|x_n-x_{n_K}|<\frac12 \epsilon$, and we hav $|x_{n_K}-L|<\frac12 \epsilon$. so for $n\geq N$, $|x_n-L|<\epsilon$

so i think it holds for $n\geq N$ rather than $n\geq n_K$

but it holds for the latter too ofc, but the former is more accurate

is this correct

@SoumikMukherjee oh

Bml

8:03 AM

Hi everyone. I have to get a conformal map from the region outside a semicircle, $\mathbb{C} \setminus S_R = \left\{(x_1, x_2):|x_1|^2 + |x_2|^2 = R^2, x_2 \leq 0 \right\}$ to the region outside a disk $D$ of radius $\frac{R}{\sqrt 2}$ centred at the origin, $\mathbb{C} \setminus D$? I should end up with $h(u) = \frac{iR + u + i \sqrt{R^2 − u^2}}{2}$, with $u = x_1 +ix_2$, but although I am familiar with the method, I cannot get the desired result. Could you help me?

ZaWarudo

Consider $\sum_{n=1}^\infty 2^n \sin (1/(3^n x))$ for $x>0$. I proved that the series converges pointwise absolutely because of the inequality $|\sin t| \le |t|$ for each $t\in\mathbb{R}$. For the uniform convergence, I argued that since $2^n>0$ we have $\sup_{x>0} |2^n \sin (1/(3^n x))|=2^n \sup_{x>0} \sin|1/(3^n x)|=2^n$ (for instance, at $x=1/(3^n \pi /2)$) and since $\sum_{n=1}^\infty 2^n$ diverges the conclusion is that the series is not uniformly convergent on $(0,+\infty)$. Is this right?

user 70432

9:28 AM

\o @robjohn

Koro

9:59 AM

How to find the exponential map of G= R^n?

Why is it identity?

Koro

10:11 AM

I wonder if anything could be more boring and less satisfactory than doing Lie algebra.

the most boring subject I've even seen.

Merosity

10:22 AM

maybe learning some of the physics it's used in could help spice it up for you

user 70432

Did somebody mention physics :P

Dirac Conversation: Edward Witten

►

Thorgott

11:31 AM

@Koro it's a local diffeomorphism, so yes

Koro

12:03 PM

yup. I'd understood that.

Koro

12:20 PM

$\exp (X)= \gamma (1)$, where $\gamma$ is the maximal integral curve of X at $0\in \mathbb R^n$.

exp: Lie (R^n) ---> R^n

I know that Lie (R^n) = the set of all linear combinations of partials so it is isomorphic to R^n.

also, $T_0 (R^n)\simeq $ Lie $(R^n)$.

via $X\mapsto X_0$.

to say that exp is an identity in this case is to say that exp is to be considered as $R^n\to \mathbb R^n$.

Let $\phi: R^n\to T_0(R^n)$ be the canonical vector space isomorphism.

We have a commutative diagram: top row is exp: Lie (G)---> G and the bottom row is Exp: T_0 G---> G.

Exp( X_0)= Exp (\phi(y)) for a unique y in R^n.
I define $\gamma (t)= ty$.

I want to show that $\gamma$ is an integral curve of X.

not sure how to take it from here.

not asking it on mse as I asked it to some in my class. None of them knew how to do it.

it's best not asking in class and asking directly on mse.

Jakobian

12:38 PM

@Allie there isn't just one stackexchange community

@user70432 I don't really use inner voice for mathematics, its more of a linguistic thing when I read English

or Polish for that matter

Thorgott

@Koro you know how to differentiate $\gamma$

Jakobian

@RyderRude give an example of a Cauchy sequence in $\mathbb{Q}$ which is not convergent (in $\mathbb{Q}$)

Koro

@Thorgott yes, \gamma : R---> R^n, so \gamma' (t)= y.

Jakobian

@RyderRude what Soumik means is that completeness of metric spaces is not a homeomorphism invariant i.e. it depends on the metric rather than the topology. Another example is $\mathbb{R}\setminus\mathbb{Q}$ which is not complete with the standard metric but can be given a new complete metric inducing the same topology

Koro

(I just had deja vu. I feel like I've asked this before.)

Jakobian

12:54 PM

@ZaWarudo its not correct

say you have $\sum \frac{(-1)^n}{n}$, its uniformly convergent and this argument fails to show it

user 70432

10 hours ago, by user 70432

yep

There is an overlap.

I thought you meant something like that

Koro

1:16 PM

@Koro I think this is wrong.

psie

The above is a proposition in Folland's book, in the beginning on the section of outer measure. Note $\mu$ is a typo for $\rho$, however, what is $\rho$ exactly? Seems kind of abstract to only say it's a function from $\mathcal{E}\to [0,\infty]$ that satisfies $\rho(\varnothing)=0$. Does that define a function?

Koro

The exponential map should be $\exp (a)= (e-1)a$

i.e, multiplication by e-1.

ZaWarudo

@Jakobian Thanks for the answer, so I can only deduce the uniform convergence on $A$ if $\sum_{n=1}^\infty \sup_{x \in A} |f_n(x)|$ converges, and I can't deduce anything if it diverges. Or is this still false?

Koro

@psie If I recall correctly, they call it pre-measure.

Thorgott

@Koro no, it was correct

@psie no, he is not defining a function

he is saying this holds for any such function

psie

1:21 PM

ah ok, makes more sense

Jakobian

@ZaWarudo yes this is called Weierstrass $M$-test, and in fact it shows something stronger than uniform convergence, it shows uniform absolute convergence

or absolute uniform convergence, not sure which one should be first and it doesn't matter right now

ZaWarudo

@Jakobian: Thanks, so if I understand correctly I can do two things to show uniform convergence for a series of functions: the M-test, or I can study $\lim_{n \to +\infty} \sup_{x \in A} |\sum_{k=n+1}^\infty f_n(x)|$ and show that this latter limit is $0$. Based on the case, one can be simpler than the other

Jakobian

yes, the latter is just the definition

Koro

@Thorgott Take a left invariant vector field X. Define $\gamma(t)= (e^t -1) X_0$. Then, $\gamma'(t)= X(\gamma (t))$ and $\gamma(0)=0$ so \gamma is the maximal integral curve of X passing through 0.

Jakobian

Here one inequality you have available is $\sin(x) > \frac{2}{\pi}x$ for $0 < x < \pi/2$

Koro

1:27 PM

So exp (X) = Exp (X_0)= \gamma (1)= (e-1) X_0$.

Jakobian

@Koro A pre-measure gets extended to a measure, here we don't necessarily recover the function $\rho$

so its just some function

Thorgott

@Koro not sure how you're getting $\gamma^{\prime}(t)=X(\gamma(t))$, but it's inaccurate

Bml

2:38 PM

No one answers my request?

Jakobian

2:50 PM

@Bml your question makes little sense

oh you're asking about an isomorphism of sorts

1

A: Conformal map from the region outside a semicircle to the region outside a disk of radius $\frac{R}{\sqrt 2}$ centred at the origin

The problem asks to show explicitly a conformal isomorphism between a semicircle exterior and a disc exterior. The main issue here is that while both are "one hole domains" bounded by circle arcs in the extended Mobius sense (so they are conformally isomorphic by Riemann mapping theorem extended ...

> I am familiar with the method, I cannot get the desired result

what does it mean for you to not be able to get a desired result

Bml

3:09 PM

@Jakobian It means that, despite having applied the correct procedure (the one shown by @Conrad), I cannot even come close to the result, which is $h(u) = \frac{iR + u + i \sqrt{R^2 - u^2}}{2}$, with $u = x_1 +ix_2$.

Koro

4:03 PM

@Thorgott So the problem at hand is to find a \gamma which satisfies \gamma'(t)= X(\gamma t). Note that X is left invariant so X is $L_{\gamma (t)}: R^n\mapsto R^n$ (translation by \gamma (t)} related, which means that $X(\gamma (t))= DL_{\gamma (t) (0) (X_0)= L_{\gamma (t)} (X_0) =\gamma (t)+ X_0$

So I have an initial value problem $\gamma'(t)=\gamma (t)+ X_0, \gamma (0)= 0$

Thorgott

rethink what the derivative of a translation is

Koro

I thought it's linear so it is itself :)

yeah I think it is correct @Thorgott

Thorgott

translation is not linear

Koro

not linear in a linear algebra way

but in degree way...

or I should say in affine way

Thorgott

sure, but that's besides the point

what is the Jacobian of the translation $x\mapsto x+v$ on $\mathbb{R}^n$?

Koro

4:15 PM

identity matrix

Thorgott

correct

Koro

So X(\gamma (t))= X_0.

and I desire X_0 = \gamma '(t)

$\gamma (t)= t X_0$

$Exp(X_0)= exp (X)= \gamma (1)= X_0$.

$\ddot\smile $

I like Lie algebra now.

6 hours ago, by Koro

I wonder if anything could be more boring and less satisfactory than doing Lie algebra.

@Thorgott thanks a lot Thorgott.

Thorgott

you got it, np

EE18

5:19 PM

9

Q: Why is complete strong induction a valid proof method and not need to explicitly proof the base cases?

I recently learned about complete strong induction. I am familiar with both strong induction and ordinary induction and make sense. But what particularly confuses me is why we do not to explicit the base cases for complete induction. They seem crucial for modus ponens to work and thus, actually s...

proof-verification logic induction proof-explanation

Jakobian, we were discussing yesterday so am just mentioning for closure: I think Bram's answer here is what gets me out of the loop. Strong induction technically does not need proving of a base case but in practice one usually needs to prove an $n = 0$ base case since the proof that "if for all m<n, P(m) is true implies P(n)" generally requires assuming there are some $m < n$

Hopefully that's a decent understanding of the way out of my confusion^

Jakobian

5:38 PM

@EE18 why are you still confused I thought we went over this yesterday

EE18

I am no longer confused, I was confused at the time

Jakobian

and no, you don't need there to exist m < n to prove that for all m < n, P(m) implies P(n)

EE18

I agree, but see Bram's answer. They are saying that it often so happens that one does in practice use that fact when completing the proof, so it turns out in such cases we need to prove P(0) separately

That's at least what I got from Bram's accepted answer

Jakobian

why does that even matter

EE18

My original confusion stemmed from having seen strong induction only in informal contexts (say, physics or CS books) and, in each of said contexts, always seeing a base case. I was therefore confused about why I see no base case in the proof of strong induction from Enderton. I wanted to reconcile those two facts.

Jakobian

5:45 PM

the explanation that for $n = 0$ this is equivalent to $P(0)$ wasn't enough?

EE18

I think the extra point Bram made is that sometime the proof that $(m<n \implies P(m)) \implies P(n)$ often makes use (tacitly) of $n \neq 0$

So yes, strictly speaking you're right

Anyway, I won't waste your time today. Just wanted to mention I got closure on this

Bml

@Jakobian So did you understand what I wanted to say?

Ryder Rude

6:04 PM

@Jakobian yes.. i got it

any rational sequence which converges to an irrational is a non convergent Cauchy sequence within rationals

Pizza

7:45 PM

Hi. I was reviewing the proof where I had to show that $\lim_{n \to +\infty}(1+\frac{1}{n})^n= e$ has a finite limit. So I had to prove that $a_n$ is monotonically increasing, thus $a_n \geq a_{n-1}$. By performing some calculations, I arrive at the point where $(1-\frac{1}{n^2})^n \geq 1- \frac{1}{n}$. So I need to prove this statement using the Bernoulli inequality. What I don't understand is why we need to stop at this point. Thanks.

leslie townes

pizza, the way you've framed the problem is a little misleading. you don't "need" to prove that the sequence is increasing in any particular way. if you've found, however, that a_n >= a_{n-1} is equivalent to (1 - 1/n^2)^n >= 1 - 1/n, and you've found that you can prove the latter statement using the bernoulli inequality, then you've proved that the sequence is increasing.

so it's not clear whether your problem is about the logic of this particular proof (i.e., is one of those implications not clear to you), which can be answered without investigating where this particular proof might have "come from," or whether your question is more generally about where this argument "came from."

if it is the former, can you get more specific about which of these [maybe more than one] is unclear: (1) what "bernoulli's inequality" is, (2) how "bernoulli's inequality" implies that (1 - 1/n^2)^n >= 1 - 1/n for all positive integers n, (3) how (1 - 1/n^2)^n >= 1 - 1/n holding for a positive integer n implies that a_n > a_{n-1} for that integer n.

Pizza

@leslietownes the book tells me to prove that $a_n$ is monotone increasing

Xander Henderson

@Pizza As @leslietownes, it is not clear what you are saying, and the "need" statements are confusing.

Assuming that you want to show that $$\left( 1-\frac{1}{n^2}\right)^n \ge 1 - \frac{1}{n}, $$ this seems to follow from Bernouli almost immediately.

leslie townes

if it is the latter, my first pass at a response is, i don't actually know how someone would have come up with this argument if they weren't already aware of bernoulli's inequality. if they were aware of bernoulli's inequality, they might have come up with this argument about a_n by algebraically playing with the inequality a_n >= a_{n-1} and finding that it was equivalent to something where maybe they could use that inequality

stepping back further, you might ask, how would one even dare to guess that the sequence a_n is monotonically increasing. you probably wouldn't just "know" that in any elementary way. you would dare to guess it, maybe from numerical experimentation

Pizza

I mean why I have to perform certain calculations to get to this (now I'll indicate what)

@XanderHenderson this.

if I treated the calculation differently and therefore arrive at a point that is different from this. is a mistake?

Xander Henderson

7:57 PM

I am not familiar with the name "Berouli's inequality". However, Wikipedia indicates that it is the proposition that $$ (1+x)^r \ge 1 + rx$$ for any integer $r \ge 1$ and $x \ge -1$.

Pizza

yes

Xander Henderson

On the LHS of your inequality, you have $$ \left( 1 - \frac{1}{n^2}\right)^n, $$ which looks like $(1+x)^r$ for $x = -1/n^2$ and $r = n$.

leslie townes

if all you were asked to do is prove that the sequence is monotonically increasing, you are presumably free to do that using any tools available to you, and not just via one particular sequence of algebraic inferences.

Xander Henderson

Bernouli would then say...?

leslie townes

this is a very high level thing. i'm not sure it's what you're confused about, but it's very important. if two similar proofs depart from one another in various algebraic particulars, it doesn't mean that one of them is "wrong"

Pizza

7:59 PM

read above pls

@XanderHenderson If I arrive at an inequality other than this, am I making a mistake?

Xander Henderson

2 mins ago, by leslie townes

this is a very high level thing. i'm not sure it's what you're confused about, but it's very important. if two similar proofs depart from one another in various algebraic particulars, it doesn't mean that one of them is "wrong"

Pizza

no not this

its not 2 proof

wait

Xander Henderson

The point is that I have no idea if you are making a mistake, as I don't know how you got to that inequality in the first place. The fact that you have arrived at that inequality doesn't tell me anything about whether or not you have made a mistake.

Pizza

yes i will write it

now

sorry

Soumik Mukherjee

Maybe it is not about proving something, maybe it is all about the inequalities we made along the way.

Pizza

8:15 PM

$(1+\frac{1}{n})^n >= (1+\frac{1}{n-1})^{n-1}$
$(1+\frac{1}{n})^n >= (1+\frac{1}{n-1})^{n} \frac{1}{1+\frac{1}{n}}$
$\frac{1}{(1+\frac{1}{n-1})^n}(1+\frac{1}{n})^n >= (1+\frac{1}{n-1})^{n} \frac{1}{1+\frac{1}{n}}\frac{1}{(1+\frac{1}{n-1})^n}$
$\frac{(1+\frac{1}{n})^n}{(1+\frac{1}{n-1})^n} >= \frac{n-1}{n}$
$(\frac{n^2-1}{n^2})^n >= \frac{n-1}{n}$
$(1+\frac{n^2-1}{n^2}-1)^n >= 1+\frac{n-1}{n}-1$
$(1-\frac{1}{n^2})^n >= 1-\frac{1}{n}$

Xander Henderson

Are all of those "if and only if" statements? or what?

Pizza

this is how I arrived at the thesis to prove

Xander Henderson

@Pizza That doesn't answer my question. How does each line relate to the lines before and after it? Are they "if and only if" statements?

Pizza

wait

Xander Henderson

@Pizza No, that is how the LHS of each line relates to the RHS. I am asking about how one LINE relates to the LINES before and after it.

Here is a statement: $$\left(1+\frac{1}{n}\right)^n \ge \left( 1 + \frac{1}{n-1}\right)^{n-1}. \tag{1}$$

leslie townes

8:18 PM

if the background question is "how does a string of inequalities prove anything" the answer is 'a string of inequalities might or might not prove anything; what it might or might not prove, and how it does that, would depend on logical statements on how the inequalities relate to one another"

Pizza

@XanderHenderson 1 + 1/n etc

Xander Henderson

Here is another statement: $$ \left(1+\frac{1}{n}\right)^{n} \ge \left( 1+\frac{1}{n-1}\right)^{n} \frac{1}{1+1/n}. \tag{2}$$

How are these two statements related to each other?

Pizza

its $\geq$

Xander Henderson

@Pizza No, that doesn't make sense.

Pizza

no wait im confused

whats the first question

>= is $\geq$, but i didnt remember the syntax

Xander Henderson

8:21 PM

There are two statements, which I have numbered (1) and (2). What is the relation between those two statements?

They way that you are presenting your argument doesn't make sense to me...

Pizza

$$\left(1+\frac{1}{n}\right)^n \ge \left( 1 + \frac{1}{n-1}\right)^{n} \left(1+\frac{1}{n-1}\right)^{-1}$$

isnt this?

aa ok yes

i missed n-1

sorry

Xander Henderson

I don't think that you understand the question that I am asking you. How is statement (1) related to statement (2)? What is the verb which joins them together? Is it the case that (1) $\iff$ (2)? or (1) $\ge$ (2) (which is what you keep saying, but which doesn't make sense)? What are you doing with those statements?

Pizza

but I don't understand the first one is wrong

the second one sorry

Xander Henderson

I have said nothing about whether or not any of the statements are "wrong" or "right". I have asked you to explain how they are related to each other. For example, if I want to solve the equation $3x + 4 = 0$, I would write something like \begin{align} 3x + 4 = 0 &\iff 3x = -4 \\ &\iff x = -\frac{4}{3}. \end{align}

At each step of the process, I am invoking some theorem or axiom (in this case, additive and multiplicative cancelation) in order to get from one statement to another.

I am not just writing $$3x +4 = 0 \\ 3x=-4 \\ x = -\frac{4}{3}. $$

In this latter presentation, I have no idea how one statement is related to the next, or if, in fact, those statements are related to each other in any way at all.

Pizza

8:36 PM

ah ok clear

im fixing

Xander Henderson

What I *think* you mean is \begin{align} &\left( 1 + \frac{1}{n} \right)^n
\ge \left(1+\frac{1}{n-1} \right)^{n-1}
= \left( 1 + \frac{1}{n-1} \right)^n \left(1 + \frac{1}{n-1}\right)^n
= \left( 1 + \frac{1}{n-1} \right)^n \left(1 - \frac{1}{n}\right) \\
&\qquad\iff \left( \frac{1+\frac{1}{n}}{1+\frac{1}{n-1}} \right)^n
= \left(\frac{n^2-1}{n^2}\right)^n
= \left( 1-\frac{1}{n^2} \right)^n
\ge 1 - \frac{1}{n} \\
\end{align}

Is that correct?

(In both lines, the equalities are algebraic manipulations; the second line follows from the first by multiplicative cancelation.)

Pizza

yes

Xander Henderson

Okay, so you have now convinced me that $$\left( \left( 1 + \frac{1}{n} \right)^n \right)_{n\in\mathbb{N}}$$ is monotonically increasing IF AND ONLY IF $$ \left( 1 - \frac{1}{n^2} \right)^n \ge 1 - \frac{1}{n}. $$

Pizza

yes

Xander Henderson

You have suggested that the final inequality it true by Bernoulli's inequality. Bernoulli states that $$(1+x)^r \ge 1 + rx$$ for $r \in \mathbb{n}$ and $x \ge -1$. So if you want to invoke Bernoulli, what are $x$ and $r$?

Pizza

8:45 PM

i will say x = -1/n^2

instead of r, I would use n

Xander Henderson

Okay, so does that finish the proof?

Pizza

yes

what I don't understand is why we stop here

$$ \left( 1 - \frac{1}{n^2} \right)^n \ge 1 - \frac{1}{n}. $$

because is it something that I can prove then?

so i stop here?

Xander Henderson

Who said that you stop there?

You have made the claim that the sequence is monotonically increasing if and only if the inequality you just stated is a true statement. You still have to argue that the statement is true, so you don't just stop there. But you claim that the statement is true by Bernoulli's inequality, so after you invoke Bernoulli, you are done.

Pizza

my book gets to that point and then says that "this thing" is equivalent to the thesis to be proven and then uses bernoulli

@XanderHenderson yes, clearly, but why in the inequality we arrive precisely at this $$ \left( 1 - \frac{1}{n^2} \right)^n \ge 1 - \frac{1}{n}. $$

Xander Henderson

I don't understand what you are asking.

Pizza

8:56 PM

$$ \left(\frac{n^2-1}{n^2} \right)^n \ge \frac{n-1}{n}. $$

could I stop here and invoke bernoulli?

for example

by stop I mean in the calculations

So i was thinking that perhaps we get to that point with the calculations because it then becomes easier for us to use Bernoulli's inequality

Xander Henderson

9:13 PM

Personally, I would probably present the proof more like the the following:

Bernoulli's inequality states that
$$ (1+x)^r \ge 1+rx $$
for all $r \in\mathbb{N}$ and $x \ge -1$. Therefore, by Bernoulli's inequality,
$$ \left(1 - \frac{1}{n^2}\right)^n \ge 1 - \frac{1}{n} $$
for any $n\in\mathbb{N}$ (by taking $x = -1/n^2$ and $r=n$). Via a little algebraic manipulation, the left-hand side of this inequality can be written as
\begin{align} \left(1 - \frac{1}{n^2}\right)^n
&= \left( \frac{n^2-1}{n^2}\right)^n \\
&= \left( \frac{n^2 - n + n - 1}{n^2 - n + n} \right) \\
&= \left( \frac{n(n-1) + n-1}{n(n-1) + n} \cdot \frac{\frac{1}{n(n-1)}}{\frac{1}{n(n-1)}} \right)^n \\

I would probably leave out a lot of the computation in most contexts, but with a group of students, I might show all of the annoying details.

The point is that I want to start with a true statement (e.g. the statement obtained from Bernoulli, and work from that towards the statement I want to prove (i.e. that the sequence is monotonically increasing).

Suppressing a bunch of tedious computation, the argument that I would write up would likely look more like

Bernoulli's inequality states that
$$ (1+x)^r \ge 1+rx $$
for all $r \in\mathbb{N}$ and $x \ge -1$. Therefore, by Bernoulli's inequality,
$$ \left(1 - \frac{1}{n^2}\right)^n \ge 1 - \frac{1}{n} $$
for any $n\in\mathbb{N}$ (by taking $x = -1/n^2$ and $r=n$). Via a little algebraic manipulation, the left-hand side of this inequality can be written as
\begin{align} \left(1 - \frac{1}{n^2}\right)^n
&= \left( \frac{1+\frac{1}{n}}{1+\frac{1}{n-1}} \right)^n,
\end{align}
hence
$$
\left( \frac{1+\frac{1}{n}}{1+\frac{1}{n-1}} \right)^n \ge

(The reader can fill in the gaps.)

EE18

One has the following theorem (which seems to sneakily come up a lot):

Let $A$ and $B$ be sets with equivalence relations $R$ and $S$, respectively, thereon, and let $F: A \to B$. Then $F$ is compatible with $R$ and $S$ iff there exists a unique function $\hat{F}: A/R \to B/S$ such that for all $x$, $\hat{F}([x]) = [F(x)]$.

It comes up a bunch in wanting to define functions between quotient sets

Regarding compatibility: Let $A$ and $B$ be sets with equivalence relations $R$ and $S$, respectively, thereon, and let $F: A \to B$. Then $F$ is said to be compatible with $R$ and $S$ iff for all $x,y \in A$, $xRy \implies F(x) S F(y)$.

My question is what goes wrong if $F$ not compatible?

Can we still define an induced function between quotient sets, perhpaps not uniquely? That is, what stops us from "taking a representative of a given equivalence class" and mapping that equivalence class to the equivalence class of wherever that element maps?

leslie townes

9:34 PM

EE18: nothing would stop you from picking out a representative from each class and defining your hat(F) [x] to be F(the representative that we chose from [x])

just formally speaking, though, this isn't as nice of a formula, is it (there's this "the choice we made from [x]" that appears on the "right hand side")

and when F doesn't respect the relation, it's probably potentially confusing (to someone who isn't too careful about what they're doing) to think of hat(F) as "coming from F" when there might be tons of F(s) for s in [x] (i.e., for the representatives of [x] that we did not choose) that have nothing to do with what hat(F)[x] is defined to be

EE18

I see. In the case of compatibility there is, in a very real sense, no choice to be made

Is AC at all at play here?

To choose an element of each equivalence class?

leslie townes

that's a good point, if F is compatible you would maybe not need AC to define hat(F) in terms of representatives, because [F(s)] is actually the same for any s in [x]

where if your hat(F) actually depends on which representatives, you would definitely need AC to pick them (in general, for arbitrary sets)

you sometimes see books doing something kind of like this (i.e., expressly defining in terms of choices of representatives, even when the choice wouldn't matter if you framed it in terms of an equivalence relation) because it can allow you to avoid having to introduce the abstract notion of equivalence relations at all

like in very elementary number theory books, or books that do number theory type stuff for a CS audience, where "mod n" might call to mind not an equivalence relation on the integers, but a unary operation that takes an integer to an element of {0,1,2,..,n-1} [i.e. a natural but specific set of choices of representatives of the equivalence classes mod n]

EE18

9:50 PM

This is awesome, thank you Leslie!

Was just reviewing the proof in the case of $F$ compatible for why choice doesn't creep in

Jakobian

@XanderHenderson no wonder you don't understand... you don't TALK to a pizza, you EAT a pizza!

Something something... call the doctor, grandpa didn't take his meds again?

@leslietownes what even is periodicity of a K-theory? How is it defined I mean

Thorgott

@EE18 note that if $F$ is not compatible with the relation, then there is no function $\overline{F}$ s.t. $\overline{F}([x])=[F(x)]$ for all $x\in A$

EE18

Oh yes, absolutely @Thorgott

We have to throw that out the window

I'm just trying to confirm that we can still have some notion of induced function here

Thorgott

10:06 PM

the construction you discussed with leslie works insofar as it does produce some function, but I would not call this an "induced function" in any meaningful sense

EE18

That's fair too, insofar as there is no unique procedure for obtaining said "induced function". I guess just one of those instances where I'm curious what happens, but doubtless it's a totally useless scenario

Thorgott

you're essentially using that any surjection splits, it's instructive to observe that this logic does not work in other categories: for example, consider the identity homomorphism $\mathbb{Z}\rightarrow\mathbb{Z}$ and consider the equivalence relation "congruent mod $2$" on the domain and the identity relation on the codomain. the identity is not compatible with these relations and there is no homomorphism $\mathbb{Z}/2\mathbb{Z}\rightarrow\mathbb{Z}$ whatsoever.

Lukas Heger

@Jakobian for any compact Hausdorff space the (reduced, to be precise) K-theory of degree $n$ is isomorphic to the one of degree $n+8$ (or $n+2$ for complex K-theory)

Thorgott

I think I've mentioned this before, but a slightly less loaded (albeit equivalent) way of thinking about these induced maps is that if you have an equivalence relation $R$ on $A$, a map $f\colon A\rightarrow B$ induces a (necessarily unique) map $\overline{f}\colon A/R\rightarrow B$ s.t. $\overline{f}([x])=f(x)$ for all $x\in A$ iff $f$ is constant on the equivalence classes of $R$ in the sense that $xRy$ implies $f(x)=f(y)$

(this corresponds to the case that $S$ is the identity relation in the theorem you stated, but I leave it as an exercise that it also implies your theorem)

EE18

Hmm, interesting recapitulation of the theorem

I guess it's more general insofar as it doesn't demand any equivalence relation on $B$?

Lukas Heger

10:14 PM

@Thorgott this is a very useful idea to wrap your mind around. You will see variants of this in topology and abstract algebra @EE18

@EE18 a priori, it's actually less general (but as Thorgott said, it implies the general case by a little argument)

EE18

Hmm, I will have to think on that then. I'm not sure I totally see how to go from thorgott's to mine but will think

Thorgott

the reason to "prefer" (if there is such a thing) this version is that it characterizes $A/R$ purely in terms of other sets (in terms of gibberish: it exhibits it as a representing object of a certain functor)

the reason the statements are equivalent is cause we can always interpret any set as a set with the trivial equivalence relation (in terms of gibberish: these representing objects constitute parts of an adjunction), but in other situations in life, you may not have such luxury

Lukas Heger

for the other direction you apply Thorgott's version to the composition $A \xrightarrow{f} B \to B/S$ and work from there. (Where $B \to B/S$ is $b \mapsto [b]$)

(of course that's all included in the talk about adjunctions, but maybe it's worth being explicit here)

Thorgott

10:31 PM

yes, it's definitely worth being explicit

the general nonsense doesn't mean anything to EE18 at this point

actually, I never thought about it, but this scenario is quite rich

EE18

Thank you all for the comments above^

Thorgott

for the category $\mathbf{Rel}$ of sets with relations and functions compatible with those relations and the category $\mathbf{Sets}$, there is an adjoint quadruple $-/-\dashv triv\dashv U\dashv cotriv$

EE18

I will think on it a little more (but not too much more ;))

leslie townes

@Thorgott ... gesundheit

Thorgott

where $-/-$ takes the quotient of a set with relation, $triv$ equips a set with the identity (minimal) relation, $U$ is the forgetful functor and $cotriv$ equips a set with the maximal relation

the second and fourth are also fully faithful

this ought to be some topos-y thing

EE18

10:42 PM

57?

Xander Henderson

11:09 PM

@Thorgott I'm calling the doctors, now. You clearly need help.

leslie townes

is this "adjoint quadruple" in the room with us right now?

John Zimmerman

what would a mellin transform of the riemann zeta function imply?

if it converged

I am well aware that it doesn't converge

psie

I'm reading about Caratheodory's extension theorem in Folland's. We have a sequence of sets $A_i$ in our algebra $\mathcal A$, but it's not stated anywhere that this is an increasing sequence of sets. Continuity from below of a measure only holds for increasing sequences of sets. What's going in the yellow highlighted bit?

John Zimmerman

And I am defining this $\zeta(x)$ for real $x>1$

leslie townes

the sequence E_n = union from 1 to n of the A_j is increasing, irrespective of what the sets A_i are

so he's using that, and just the definition of A as the full union of the E's

psie

11:18 PM

@leslietownes ah, ok, I see I think, so we can construct a disjoint sequence of sets

leslie townes

psie: one sometimes does things like that (i.e., from an arbitrary sequence, constructing a disjoint sequence with the same union), but AFAIK there is no need to do that specifically in this case

it's just, the union of the first n elements of any sequence sets will, as a function of n, be an increasing sequence of sets

if you like, we have here that E_{n+1} is E_n union something (namely A_{n+1}), but it isn't necessary for this union to be disjoint for the sequence to be increasing

psie

ok

Thorgott

@XanderHenderson if only you knew what math I was actually doing right now

psie

yeah, disjoint-ness is not needed at all, my bad

leslie townes

yeah, you just need that E_{n+1} is E_n union something :)

folland is a pretty good book for this material :) some of the older books are fine as logical presentations of the subject, but there was a lot of variation in how people presented measure theory in, say, the 50s-60s, which is not very helpful to how most people use it now

psie

11:25 PM

yeah, what I was asking about the other day (yesterday about Lebesgue-Stieltjes measures), I haven't been able to find it in some other book, hence I'm reading Folland's :)

leslie townes

i have never worked through or taught from more recent works than folland, but i think terence tao has a measure theory book and i would bet that it is good

bartle had a good book on integration but it's out in so many editions (i think he might have more than one book by this name) i don't remember which one is the good one

psie

I'd like to see Folland's book in a new edition (1999 is some time ago now)

leslie townes

nah, as i'm sure some members of the channel will remind you, 1999 is only, i dunno, five years ago, tops

ZaWarudo

Let $\{a_n\}$ be a convergent sequence of real numbers and let $\{f_n\}$ be a sequence of functions satisfying $\sup_{x \in A} |f_n(x)-f_m(x)| \le |a_n-a_m|$, $n,m\in\mathbb{N}$. Show that $\{f_n\}$ converges uniformly on $A$. I tried this: let $\epsilon>0$ and let $x \in A$. Since $\{a_n\}$ converges in $\mathbb{R}$ and $\mathbb{R}$ is complete, $\{a_n\}$ is Cauchy and so there exists $N_\epsilon \in \mathbb{N}$ such that for each $n,m>N_\epsilon$, $|a_n-a_m|<\epsilon$.

So if $n,m \ge N_\epsilon$ then $|f_n(x)-f_m(x)| \le sup_{x \in A}|f_n(x)-f_m(x)| \le |a_n-a_m|<\epsilon$. Now I have a …

leslie townes

apparently a new edition of royden came out (some new person authoring new material, royden passed away in the 1990s). i wonder if they fixed the errors in it

John Zimmerman

11:33 PM

$$\prod_{p~ \mathrm{prime}} \int_{0}^1 \bigg(\sum_{n=1}^\infty e^{\frac{n\log p}{\log x}}\bigg)~x^{-s}~dx= \prod_{p~\mathrm{prime}} \sum_{n=1}^\infty 2\sqrt{\frac{n\log p}{-(s+1)}}K_1\bigg(2\sqrt{-(s+1)n\log p}\bigg)$$

@XanderHenderson do I need help?

leslie townes

zawarudo that is correct. to say f_n converges uniformly on A is to say that given epsilon, there is N with n, m larger than N implying |f_n(x) - f_m(x)| < epsilon [for all x in A]. the one expository thing i would add is a quantifier, after your "So if n, m >= N_epsilon" i would add "and x is in A"

Xander Henderson

@Thorgott as far as I can tell, it is some kind of sick perversion, and I want nothing to do with it.

ZaWarudo

@leslietownes Thank you for checking it and for the suggestion about the quantifier, I will correct that part. My lecturer, after proving that $f_n$ converges pointwise, uses the continuity of absolute value and takes the limit as $m\to+\infty$ in $|f_n(x)-f_m(x)|<\epsilon$, so I wasn't sure if my proof was correct as well :)

Thorgott

@XanderHenderson I think you accidentally pinged me instead of John Zimmerman

Xander Henderson

@Thorgott nope. I meant to ping you.

I even linked to a specific comment.

John Zimmerman

11:46 PM

I posted a question about that^^^^^^^

leslie townes

thorgott may be jokingly implying that this equation john just posted is more indicative of perversion than his humble category theory

John Zimmerman

but I don't wanna dump on anyone

It is perverted what I am doing

But it's fun

so I don't care

leslie how is category theory humble and a simple product over the primes perverted

leslie townes

i don't know, i did not introduce the term "perversion" to this discourse, like many things it often just a matter of taste

i will say, i have less of a doubt that the equation is in the room with us right now than i did that the adjoint quadruple was, and i don't like it

thorgott's alien symbols left a lot of the cosmic horror behind them to the imagination, your thing is right here in the room

John Zimmerman

leslie my confidence interval (statistical concept) for the equation being correct is 90%. I am so happy with that.

One day I will try and learn category thoery

Xander Henderson

@leslietownes Ph'nglui mglw'nafh Cthulhu R'lyeh wgah'nagl fhtagn.

John Zimmerman

11:59 PM

phocantie morum absudarme tengo cantreatis mudomus

leslie townes

🐙

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