Mathematics - 2024-01-21 (page 1 of 2)

Jakobian

00:00

Vectors of the form $x\otimes y$ generate the tensor product, but not every one of them is of that form

oscarmetal break

I see. I need to read more carefully again..

Jakobian

its okay, tensor products is something that took some time to understand properly for me

my opinion on smoothness is that, just like how I care about continuity I think its important to care about smoothness of maps

Obliv

00:32

is there a way to latex a restriction on a function

or do we use the vertical bar like $f\mid S : S \to B$

Xander Henderson

I believe that the generally accepted notation is $f|_{S}$

Or, if you want something taller, e.g. $$\left.\frac{\mathrm{d}f}{\mathrm{d}x}\right|_{x=a}.$$

Obliv

Ok thank you.

psie

00:49

I'm working the following problem;

> Prove that an inner product on a linear space $X$ is continuous, and that it is uniformly continuous on bounded sets, with respect to the corresponding product norm on $X\times X$.

I think I'm able to show continuity through the sequential definition, but the uniform continuity part I struggle with. I don't really know where to start.

Xander Henderson

01:01

Dang... my brother has a client right now who might actually be innocent.

Ted Shifrin

Is anyone truly innocent?

Xander Henderson

No. But his clients tend to be less innocent than usual.

Ted Shifrin

@psie Start by writing down what you need to prove.

Xander Henderson

Such is the joy of being a public defender.

Ted Shifrin

Yup.

psie

01:09

@TedShifrin I have the definition of uniform continuity (for metric spaces) in front of me. I want to prove that for every $\epsilon>0$, there exists a $\delta>0$ such that $$d_X(x,y)\implies d_Y(f(x),f(y))<\epsilon.$$ Here $X=A\times A$, where $A$ is a linear space and $f$ is the map given by $(x,y)\mapsto \langle x,y\rangle$.

$Y$ is the field which $A$ is a linear space over.

Jakobian

@XanderHenderson he's a lawyer?

@psie should have pinged me, I like this topic

psie

cool :)

Jakobian

alright, so $f$ is an inner product?

leslie townes

some mild potential for confusion here based on the notation alone. in your definition, x and y are distinct points of an abstract metric space X, while in your application, "X" is AxA, and you are [understandably, but in potential conflict with the above] using (x,y) to denote a single point of the metric space

psie

indeed

Jakobian

01:13

for bilinear maps what you usually do is find the "norm", i.e. a number such that $|f(x, y)| \leq C\|x\|\cdot \|y\|$

this is equivalent to continuity

did you ever see the proof of uniform continuity of $x\mapsto \|x\|$?

This is the same

psie

ok

Jakobian

least such number $C$ is called the norm of a bilinear map, just like for linear operators

Xander Henderson

@Jakobian Yes, he is a public defender.

Jakobian

not sure what's the difference between a lawyer and a public defender. I guess its a specific type of lawyer?

Xander Henderson

Yes. Google is helpful: en.wikipedia.org/wiki/Public_defender

Jakobian

01:15

@psie I mean thats to give you general picture, its not a proof

psie

alright

leslie townes

yes it is a lawyer who works for the government whose clients are people accused of crimes

Jakobian

oh, for government

@psie Here Cauchy-Schwartz inequality gives you something like $|f(x, y)|\leq C\|x\|\cdot \|y\|$, it gives you it for $C = 1$

funny thing, Cauchy-Schwartz inequality can be stated as $\|f\|\leq 1$ then

Xander Henderson

In Soviet Russia, Cauchy Schwartzes YOU!

Jakobian

I actually never looked at it this way, but its funny how CS is just a statement about norm of some bilinear map

is this a cursed point of view? Perhaps

Xander Henderson

01:21

Hrm... According to my mother, one of her friend's nephews is up for an academy award for animating an octopus...

That is... strange.

Ted Shifrin

Note that the problem stated on a bounded subset. That surely is relevant.

Jakobian

because of stuff like $x\mapsto x^2$, yeah. Makes sense

@psie What are your thoughts?

Ted Shifrin

@psie You gave a definition, but you did not write down explicitly what you need to prove.

Jakobian

you've been silent

psie

yeah, thinking everything through you've written

Obliv

01:25

Wow.. PSA: make sure that you're reading the correct textbook before spending 8 hours working through problems that were assigned to a different textbook.

Jakobian

You can prove this as an exercise if you want its not as relevant here

Ted Shifrin

@Obliv Or by a different professor in a different course.

Jakobian

An $m$-linear map $k:A_1\times \cdots \times A_m\to B$ is continuous iff there exists $0 < C < \infty$ such that $\|k(x_1, ..., x_m)\|\leq C\|x_1\|\cdot ...\cdot \|x_m\|$

Obliv

I can't believe I've been trying to work through a graduate level abstract algebra text for the last few hours. The anxiety that's been building and my relief now is palpable..

To be fair, I just clicked the pdf of the first result and it was the same author. It also said algebra on the front shrug

Ted Shifrin

Very different texts.

No, titles are different.

Jakobian

01:28

Either way what I think is that you want to think of the inner product as a bilinear map

Obliv

Yep.. I didn't catch it at first glance. I have been immensely humbled but also I feel a bit disappointed in myself that it was that hard. I feel like I let myself down.

Xander Henderson

@TedShifrin I accidentally bought the wrong Hungerford as an undergraduate. :(

Ted Shifrin

Yes. But please write down what I asked for and forget all this stuff for a moment.

Obliv

@XanderHenderson Omg same, except it was a pdf.. so that's not quite as bad.

Xander Henderson

@Obliv pdfs were a lot harder to get in 2004. It was doable, but a lot riskier than now, and a bit more arcane.

Jakobian

01:30

Pick some $x_0, y_0$ and consider $$|\langle x, y\rangle - \langle x_0, y_0\rangle | = |\langle x, y\rangle - \langle x_0, y\rangle + \langle x_0, y\rangle - \langle x_0, y_0\rangle|$$

Ted Shifrin

The undergrad book is rings first, like mine, unless he changed it.

Jakobian

@psie do you see it yet?

Ted Shifrin

Jakobian. Do not do it for him,

Obliv

@TedShifrin It is indeed.

psie

triangle inequality?

Xander Henderson

01:31

@Jakobian Ah, the analyst's third favorite trick! (Which is really just a special case of the analyst's favorite trick, but with a bit of added flair for the intermediate term).

Jakobian

yep

triangle's inequality is the reason why open balls are open

I might have talked about it here before, but if you weaken metric definition to not satisfy triangle inequality, open balls might fail to be open

Xander Henderson

There's a dirty joke in there somewhere...

Jakobian

@psie This is basically the proof that $x\mapsto x^2$ is uniformly continuous on $[-n, n]$ without using big theorems like "continuous function on compact set is uniformly continuous"

but only algebraic manipulations and inequalities

psie

in my problem, they say only "bounded sets". Why would one expect uniform continuity on these sets?

Ted Shifrin

Do the proof, dammit.

Obliv

01:44

Not sure if I'm just salty, but I'm not sure what we're even doing with abstract algebra as a subject. Why are we developing so many definitions and ideas?

Xander Henderson

Abstract algebra builds healthy teeth and bones.

Drink a tall glass every day.

John Zimmerman

ideas are the source of all things

Obliv

Is hungerford's algebra a reference book? @XanderHenderson

It seemed incredibly dense.. even denser than dummit & foote

Xander Henderson

@Obliv Hungerford has written (at least) two books on algebra. One is a fairly widely used text for undergraduate courses in abstract algebra, the other is a fairly widely used text for graduate abstract algebra.

Obliv

I'm guessing they cover topics in different depth or less topics for the undergrad text?

Xander Henderson

01:52

@Obliv The graduate text is more terse. And generally covers topics in greater depth.

But it has been almost a decade since I've thought about either. Algebra is... not my favorite.

The biggest difference, though, is that one has a black cover, and the other is yellow.

John Zimmerman

02:06

smoothing function

$$ \psi_1(x)=\int_0^x \psi(t)~dt $$ for $\psi(t)$ being Chebyshevs second function. Does anyone know why it is called a "smoothing function"?

fwiw let me define $\psi(t)$

I define $\psi(t)$ as the summatory von mangoldt function

user180328

02:31

Does anyone understand the reasoning for solution 5? artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/…

how does the author jump to "p/q - 1/2" has to be 1/16

Xander Henderson

02:42

@user180328 Well, if $\frac{1}{18} < \frac{p}{q} - \frac{1}{2} < \frac{1}{14}$, and you want to keep $q$ small, what are the possible choices?

The subtraction of $\frac{1}{2}$ is, to me, the more obtuse step. That feels a bit like a rabbit out of a hat.

Jakobian

@psie what do you mean?

compact sets and bounded sets are a bit similar

some things don't hold for arbitrary bounded sets in a normed space, but this one does

user180328

03:01

@XanderHenderson I don't understand how one know p/q-1/2=1/16 minimizes q

Xander Henderson

@user180328 You didn't answer my question.

What are the possible choices for $p/q$, given the inequality?

user180328

any rational number between 10/18 and 8/14

or 5/9 and 4/7

Xander Henderson

If I tell you that $\frac{1}{18} < \frac{a}{b} < \frac{1}{14}$ and that I want to minimize $b$, what are the possible values for $a/b$?

user180328

15,16,17

Xander Henderson

Okay. While there are certainly more clever ways of eliminating some of these, there is always brute force, no?

There are three possibilities. Work through them. What do you get?

user180328

03:15

It simply isn't clicking for me. Maybe I'm really stupid. I've put time into it and I can't see the reasoning. How do you know p/q-1/2=100001/1500000 or p/q-1/2=6125827623/100000000000 isn't what actually minimizes q? How do we just rule out all but one of these infinitely many choices?

Xander Henderson

$\frac{p}{q} - \frac{1}{2} = \frac{2p-q}{2q}$, yes? So minimizing the denominator $2q$ minimizes $q$, no?

user180328

yes

Xander Henderson

So, brute forcing it, start small. The goal is to minimize a natural number, and you already have a lower bound. Start with the smallest possible candidate value, and work your way up until you find something that works.

Or, you know, read one of the other solutions (like I said, this particular solution seems to require a rabbit out of a hat right at the start---it probably isn' the way I would have gone about things, anyway).

Ted Shifrin

@XanderHenderson Nah, cuz both fractions are just slightly more than 1/2.

Xander Henderson

@TedShifrin I mean, sure, but doesn't feel like a very generalizable trick---I don't like it.

Ted Shifrin

03:21

And the numerators are inconvenient.

think about the proof that any positive rational is a sum of distinct reciprocals.

Xander Henderson

@TedShifrin I've honestly never thought about that result in my life.

Ted Shifrin

Oh, it was one of my favorite problems to assign in Spivak.

Xander Henderson

But, ultimately, I just don't find that particular solution to the problem aesthetically pleasing---it is entirely a matter of my own taste.

It is kind of like how I don't like most of algebra---most of Hungerford's proofs, for example, feel a lot like rabbits out of hats to me.

It isn't to my taste.

Ted Shifrin

I haven’t looked at it beyond looking to see what you were talking about.

Xander Henderson

Which is more about me than anything else.

Ted Shifrin

03:27

And Rudin’s proofs? They’re kangaroos out of hats.

Xander Henderson

@TedShifrin Ugh... one of my least favorite proofs of all the terrible proofs in all the books I've ever read is Rudin's proof of the mean value theorem.

But, honestly, rabbits-out-of-hats is one of my biggest objections to Rudin.

Ted Shifrin

Not to mention zero pictures.

Xander Henderson

That, too.

Jakobian

03:48

@TedShifrin agree

leslie townes

you should hear what Rudin's proofs say about you

Obliv

nvm well ordering axiom guarantees non zero smallest number

Ted Shifrin

@leslietownes Do you have a bug to listen?

Soumik Mukherjee

04:58

Find the number of permutations $a_1,a_2,...,a_8$ of the sequence $1,2,...,8$ satisfying the condition $a_1\leq 2a_2\leq ...\leq 8a_8$.

I can see that any product of disjoint consecutive transpositions would work. By consecutive transposition, I mean transposition of the form $(n,n+1)$. But are they the only type of possible permutations?

Ted Shifrin

05:30

Is this interesting for some reason?

I guess you need to start by showing you can only have $a_1=2$?

leslie townes

06:27

2

Q: Number of permutations $a_1,a_2, \ldots,a_n$ of the sequence $1, 2, \ldots, n$ satisfying $a_1\leq 2a_2 \leq 3a_3 \leq \cdots \leq na_n.$

How would you start solving, Let $n$ be a positive integer. Find the number of permutations $a_1,a_2, \ldots,a_n$ of the sequence $1, 2, \ldots, n$ satisfying: $$a_1\leq 2a_2 \leq 3a_3 \leq \cdots \leq na_n.$$

although that question was closed for lack of context, someone posted a sketch of a proof that there aren't any others

for n = 8 one can also verify by brute force that there aren't any others

Soumik Mukherjee

@TedShifrin It was asked in an exam

@TedShifrin 2 or 1

@leslietownes Thanks

But this counting approach still seems like brute force

Oh wait, not brute force

I can't get rid of the silly mistakes:/ I wrote 33 as I forgot to count the identity

leslie townes

07:10

heh, that is the kind of mistake where if i had given the exam, i might wait to see how everybody did before deciding whether to penalize it

Soumik Mukherjee

07:28

No possibility for partial markings in this exam, questions are either fill in the blanks type or true/false type.

leslie townes

that sounds perfectly horrible :)

Soumik Mukherjee

Though last time I did a worse silly mistake, there was a question about writing all the generators of a certain group. I wrote $\pi/2$. The correct answer was $\pm \pi/2$

@leslietownes Indeed

Of course if $x$ is a generator of a certain additive group then so is $-x$. But as they asked for ALL generators, my answer was wrong:(

leslie townes

08:07

there's nothing quite like making education as much like navigating an automated telephone system as possible

did you try pressing 0 over and over, or saying "operator" or "representative"

user 85795

08:26

even worse, there's a trend of subtracting wrong answers from right answers in these types of "exams"

leslie townes

ah yes, one of the all-time great practices

user 85795

and leaving it blank is an automatic -1

Soumik Mukherjee

08:40

There was no negative marking for leaving blank or incorrect answer in the fill in the blanks section. The true/false section had negative marks for incorrect answers.

leslie townes

i sort of understand that, to discourage random guessing

... that is, the random guessing that you encourage by asking only fill in the blanks or true/false type questions in the first place

Soumik Mukherjee

Yes, equal negative marks in the true/false section is fine.

Though people still guess a random number(mostly 0 or 1) in the fill in the blanks section.

user 85795

were there any multiple choice questions if so how many choices were given

Soumik Mukherjee

@user85795 No

Sine of the Time

"you can't fail an exam if you don't do it" - Sun Tzu

Soumik Mukherjee

08:50

@leslietownes Though there will be an interview in the 2nd stage. So the lucky random guessers can't get away.

@SineoftheTime True. Magnus has never defeated me.

Sine of the Time

nice

Soumik Mukherjee

@user85795 Actually there was an mcq in disguise. There was a fill in the blanks asking to write down all the correct options.

user 85795

I see.

Soumik Mukherjee

This was the imposter

Ryder Rude

10:52

Rethinking the real line #SoME3

►

what is the relation between the two bubble diagrams in this video?

one is the continued fraction approximation diagram. other is the Dirchilet bubble diagram

it's at 4:34 and 9:12

Sourav Ghosh

13:52

Q. Let $f\in C[0, 1]$ and $f\ge 0$ and suppose that $\int_{0}^{\infty} f $ is finite. Does this imply $\int_{0}^{\infty} f^2$ also finite?

Jakobian

14:21

$f$ is on $[0, 1]$ or $[0, \infty)$?

Either way consider $1/\sqrt{x}$

John Zimmerman

14:48

$\pi(x)$ is integrable actually... wow

that resolves the question about the continuity of $J(x)$

$J(x)$ is continuous

the kicker is to just view $\pi(x)$ as a step function and integrate over the rectangles

then it's clear that said integration of $\pi(x)$ is a continuous, and strictly increasing function

Sourav Ghosh

@Jakobian Sorry $[0, \infty) $

@Jakobian It's not continuous at $0$.

Ted Shifrin

15:43

@Jakobian Not either way.

Thorgott

16:17

yeah

I feel like it should be wrong anyway, though

Jakobian

@SouravGhosh oh right good point

it still should be no

Ted Shifrin

16:34

It’s yes. For large $x$, $f^2(x)<f(x)$.

Jakobian

I see

Continuity makes it so that the situation resembles more of that of $\ell^p$ than $L^p[0, 1]$

Thorgott

well, I don't think that's true

you can have $\int_0^{\infty}f$ finite and $\limsup_{x\rightarrow\infty}f(x)=\infty$

Sine of the Time

Thorgott

ah, I was thinking of that example, but was too lazy to estimate the area of the squared triangles

if $f$ is uniformly continuous, we should be able to conclude that $f(x)\le1$ for large $x$ and then the answer is yes

Ted Shifrin

Yeah, I’m wrong. Never mind.

Jakobian

16:44

and I followed in being wrong

Sha Vuklia

WOT and SOT are just the funniest abbreviations. I keep thinking of the Harry Potter meme: "Yer a wizard herry" "I'm a WOT?"

Jakobian

16:59

@ShaVuklia I guess $2$ is a generalization of Banach-Alaoglu theorem

Sha Vuklia

hm, but Banach-Alaoglu is about duals, so I don't see how this would be a generalization?

Jakobian

I just used a wrong word

Thorgott

yeah, wot and sot are pretty funny

Thorgott

17:33

the identity $A/A'\otimes B/B'=(A\otimes B)/(A\otimes B'+A'\otimes B)$ holds in any closed monoidal abelian category s.t. the monoidal product is bilinear

I think these are the right conditions to generalize the expected behavior of tensor products in module categories

psie

18:00

I'm reading these notes on measure theory. There is a definition I struggle with.

> Definition 4.8. If $f: X\to \overline{\mathbb R}$ is a measurable function, then $$\int f d\mu=\int f^+d\mu-\int f^-d\mu,$$ provided that at least one of the integrals $\int f^+d\mu,\int f^-d\mu$ is finite. The function is integrable if both $\int f^+d\mu,\int f^+d\mu$ are finite, which is the case if and only if $$\int |f|d\mu<\infty.$$

> Note that according to Definition 4.8, the integral may take the values $-\infty$ or $\infty$, but it is not defined if both $\int f^+d\mu,\int f^+d\mu$ are infinite.

So the integral is allowed to be infinite, but a function is not called integrable if this is the case? Seems so...strange.

Xander Henderson

The definition of "Lebesgue integrable" is that the value of the integral is finite. That is what is means to be integrable. Yes, integrals can take on the value "infinity", but such functions are not "integrable".

It is a definition.

psie

ok, I need to get used to that definition then...phew.

Xander Henderson

Note also that the condition that one of $f^+$ and $f^-$ is integrable (i.e. has a finite integral) ensures that you can meaningfully give a value to the integral of $f$. Otherwise, consider something like $\sin$---what is $\int \sin$?

$\int \sin^+ = \int \sin^- = \infty$. Oops...

Jakobian

@psie Did you finish showing that inner product is uniformly continuous on bounded subsets?

psie

yep :)

Jakobian

18:09

nice

psie

I've written it down in latex

Xander Henderson

If either $\int f^+$ or $\int f^-$ is infinite (but not both), then it is meaningful to write $\int f = \infty$ or $\int f = -\infty$, since $\infty - \text{(something finite)}$ is infinite, and $\text{(something finite)} - \infty = -\infty$. These are meaningful values which can be assigned to the integral, even if we don't consider the function itself to be integrable.

psie

makes sense

Jakobian

psie

hmm, that I haven't tried

maybe this is also simply triangle inequality and CS?

Jakobian

18:12

its not going to be CS

did you ever see proposition like this one? A linear operator $f:A\to B$ is continuous iff $\|f(x)\|\leq C\cdot \|x\|$ for some $C > 0$?

psie

I haven't, but it's reminiscent of Lipschitz continuity, maybe that is totally unrelated

Jakobian

ah. You will see it eventually, for now forget what I said about $m$-linear maps

psie

I have another measure theory question though :)

Jakobian

maybe do that exercise when you see how to prove the statement for $m = 1$ (which is the one that people most often use)

Xander Henderson

@psie I really don't think that is the right way of thinking about it at all. Everything is linear. In finite dimension land, anything linear is "obviously" continuous. Things break down in higher dimensions.

Jakobian

18:15

@psie its not unrelated, its a good observation

Xander Henderson

@Jakobian I agree that it is a reasonable question, I just think that the answer is "no".

Jakobian

question?

Xander Henderson

Or, rather "yes, it is pretty much unrelated (in the sense of pragmatics)".

Jakobian

@XanderHenderson That's vague and confusing

Xander Henderson

@Jakobian The question "Is this related to Lipschitz continuity?" is vague. I gave an answer which is at a similar level.

It was not meant to be a precise answer.

Jakobian

18:19

Someone asking a question often asks it in a vague way

its the responsibility of the one answering it to get rid of the vagueness

Xander Henderson

I disagree, because I don't think that precise answer is helpful here.

On another front, what do you gain by constantly nitpicking the statements that I make in chat?

It is incredibly irritating.

Jakobian

I disagree with you because its totally a valid question and has a totally valid objective answer

Xander Henderson

It is a valid question. I agree.

Jakobian

$\|f(x)\|\leq C\|x\|$ can be stated as $\|f(x)-f(y)\|\leq C\|x-y\|$ which is precisely that $f$ is Lipschitz

Xander Henderson

It also has a good, technical answer for someone who is already familiar with the basic ideas, which @psie is struggling with.

Jakobian

18:22

not seeing psie struggling anywhere

Xander Henderson

Holy crap. You really just can't help but nitpick, can you?

Jakobian

its like jenga

Xander Henderson

I'm really not interested in this conversation. Someone who is just learning about Hilbert spaces is not going to get anything useful out of noting that "Oh, continuous in Hilbert space means the same thing as Lipschitz".

Jakobian

you're the one making the conversation complicated

Xander Henderson

It isn't a terribly useful or practical observation, because Lipschitz continuity is ultimately about the boundaries of what is differentiable, which is completely orthogonal to the goals set out when first learning functional analysis.

It isn't wrong, it just isn't a very helpful way of thinking about things.

Which is exactly what I said in the first place: I don't think that it the right way of thinking about it.

G-d... I am so tired of hearing "NO! YOU'RE WRONG!" from you whenever I make a statement about pedagogy...

Sine of the Time

18:31

hi, I'm struggling to solve the following exercise: let $H$ be an Hilbert space and $\{e_n\}_{n\in \Bbb N}$ be an orthonormal base of $H$. Define $d_n:=e_{n+1}-e_n$, prove that for every $v\in H$, $\lim_{n\to +\infty} \langle v, d_n\rangle =0$

Jakobian

@XanderHenderson I don't think about Lipschitz continuity like that and its bold to assume others think the same as you. For me, for example, Lipschitz continuity is a simple condition which allows us to preserve Cauchy sequences

As for if its helpful, its certainly not a problem to ask questions that allow you to explore new concepts

Xander Henderson

@Jakobian At any point, did I ever say "This is a bad question and you shouldn't have asked it"?

Sine of the Time

now

Jakobian

@SineoftheTime Can you show that $\langle v, e_n\rangle$ has a limit?

Sine of the Time

since $\{e_n\}$ is a base, I can write $v$ as a linear combination of the vectors of the base

Jakobian

18:44

in what sense?

Sine of the Time

is it not correct?

Jakobian

its correct in some sense

I'm just trying to see if you're not thinking about this wrong

Xander Henderson

In the mathematical literature, Lipschitz continuity is most often invoked when working with PDEs and SDEs. It is an "almost smooth" condition, which can be used to push around some estimates in helpful ways. It is not a very relevant condition in the context of functional analysis, where continuity is a guarantee of Lipschitz continuity, and where "smoothness" and "differentiability" aren't really the main thrust of introductory material.

Pedagogically, the observation that a linear map is Lipschitz isn't terribly helpful. It is "obvious" in finite dimensional cases, and doesn't really get you much of anything new in infinite dimensional cases.

Sine of the Time

$v=\sum_{n\in \Bbb N}\langle v,e_n\rangle e_n$ (?)

Xander Henderson

So, again, "Sure; it is related to Lipschitz continuity. But I don't think that this is a very helpful or useful way to think about it."

Jakobian

18:48

yeah sure. Infinite linear combination

Sine of the Time

ok, one minute let me see if I can make progress

Jakobian

Someone beginning functional analysis, thinks of continuous linear maps as Lipschitz continuous, they just don't call them that

You're arguing that the name should be restricted to certain subjects, but I find that irrelevant

Xander Henderson

No, that is not at all what I am arguing.

What I am arguing is that the word "Lipschitz continuity" simply does't appear very often (or at all) in functional analysis, because Lipschitz continuity isn't really relevant in those contexts. So adding the extra cognitive burden of thinking about Lipschitz continuity in a context where it buys you nothing that you don't already have is not a very useful or helpful way of thinking.

Sine of the Time

@Jakobian I've no idea on how to procede :(

Xander Henderson

You are arguing about the mathematics and I am making a statement about pedagogy.

Jakobian

18:54

@SineoftheTime inner product is continuous

what does it mean for $v = \sum_n a_n e_n$?

Sine of the Time

Is it related to the tail of the series?

Jakobian

convergence

its a limit

$v = \lim_{N\to\infty} \sum_n^N a_n e_n$

Sine of the Time

yes

Jakobian

and inner product is continuous

Sine of the Time

so I can bring $\lim_n$ inside the inner product

Jakobian

18:59

$\langle v, e_n\rangle = a_n$

Bessel's inequality

In mathematics, especially functional analysis, Bessel's inequality is a statement about the coefficients of an element x {\displaystyle x} in a Hilbert space with respect to an orthonormal sequence. The inequality was derived by F.W. Bessel in 1828.Let H {\displaystyle H} be a Hilbert space, and suppose that e 1 , e 2 , . . ....

there's such thing

it implies that $\sum |a_n|^2$ is convergent

did you see that before?

Sine of the Time

ok, so $a_n \to 0$

yes, I saw it

Jakobian

maybe my talk with limits was not that relevant

Obliv

19:16

to solve $x^3 + x^2 = [2]$ in $\mathbb{Z}_{10}$ I have to put it in the form $x^3 + x^2 - 2 \equiv 0 \pmod{10}$ right?

Jakobian

why write $[2]$

If you want to be pedantic and write $[2]$ then you should also be pedantic and write $x = [y]$ for some $y\in\mathbb{Z}$

and only then consider $y^3+y^2-2 \equiv 0\pmod{10}$

copper.hat

our kitchen has had a sudden influx of pedants because of recent rains

Obliv

it's just the way it was given lol. i copied it down verbatim

actually wondering if there was a typo in the textbook. shouldn't it be in the modular arithmetic notation

Jakobian

19:32

No, you misunderstood

If you write $x^3+x^2 = [2]$ then you can't write $x^3+x^2-2\equiv 0 \pmod{10}$ because $x$ is an equivalence class

Sine of the Time

@Jakobian thank for your help :)

Jakobian

@Obliv The book is trying to be less negligent about distinguishing between $2$ and its equivalence class in $\mathbb{Z}_{10}$ than usual. Normally one would just write "Solve $x^3+x^2 = 2$ in $\mathbb{Z}_{10}$" without caring much about distinguish between them.

Obliv

ohh ok

Jakobian

Then I'd accept writing something like $x^3+x^2-2\equiv 0\pmod{10}$. But since your book is insisting on being formal, I think you are ought to follow it and write $x = [y]$ for some $y\in\mathbb{Z}$

Thus the equation becomes $y^3+y^2-2 \equiv 0 \pmod{10}$

actually they made a typo because it should be $x^3\oplus x^2 = [2]$ since they're using a different notation for addition and multiplication

(again, one would just write $+$ instead of $\oplus$ without care in the world)

Thorgott

@Jakobian you really don't

$x$ is a variable

Jakobian

19:40

a variable in $\mathbb{Z}_{10}$

Thorgott

indeed

Obliv

which just means its an equivalence class of the form $x=[y]$ for some $y \in \mathbb{Z}$

Thorgott

it does, but you don't need to say that out loud

Jakobian

I think I've adressed it already

Obliv

so it's okay to write $x^3 + x^2 - 2 \equiv 0 \pmod{10}$?

Jakobian

19:41

I think they do need to say that if they're writing $[2]$ instead of $2$

Thorgott

@Obliv not if $x$ is an element of $\mathbb{Z}_{10}$

Jakobian

Thorgott, what do you mean

Thorgott

@Jakobian there is no reason for them too, they've (presumably, at least) already explained how $\mathbb{Z}_{10}$ is defined

Jakobian

do you disagree with anything I wrote

Thorgott

yes

Jakobian

19:43

which part

Obliv

Isn't $Z_{10}$ the set of all congruence classes mod 10

Thorgott

I disagree that they have to repeat what elements of $\mathbb{Z}_{10}$ look like

you can write $x\in\mathbb{Z}_{10}$ and call it a day. of course, there will be a $y\in\mathbb{Z}$ such that $x=[y]$, but that follows by definition and you don't have to repeat it.

Jakobian

You need to write it to say $y^3+y^2-2 \equiv 0\pmod{10}$

how else are you supposed to write something like this?

Thorgott

that I don't disagree with

but there's a difference in saying "you (the reader solving it) have to write this" or "the book has to write this in the exercise statement"

Jakobian

then what are you disagreeing with

Obliv

19:46

aren't there an infinite number of solutions for this equation? Since there are infinite integers congruence classes like $[1] = [11]$ in $\mathbb{Z}_{10}$

Jakobian

I said you and not book

you as in Obliv

Thorgott

ah, sorry, then I don't disagree

Jakobian

its alright

Obliv

nvm I guess it doesn't matter which representative you choose.

Jakobian

@Obliv solutions are elements of $\mathbb{Z}_{10}$ which is only $10$ of them

or did you mean $y^3+y^2-2\equiv 0\pmod{10}$?

Obliv

19:49

Don't know the difference unfortunately. I'm just wondering if I have to factor to solve $x^3 + x^2 - 2 \equiv 0 \pmod{10}$ or not lol

Jakobian

are you disagreeing that you need to write $x = [y]$ for some $y\in\mathbb{Z}$?

We could just work with $x^3 \oplus x^2 = [2]$ directly instead

oh wait

are you considering $\mathbb{Z}_{10}$ as a ring or a group?

Obliv

i'm following hungerford intro to abstract algebra chapter 2 we haven't done groups or rings yet. rings are soon though

Jakobian

a ring okay. You can just factor $x^3\oplus x^2\ominus [2] = [0]$

Obliv

@Jakobian if it was labelled like this i'd be a lot less confused. idk why the textbook wrote it this way.

Jakobian

and then factor this polynomial, but you need to be careful after factoring it

because $x\odot y = [0]$ doesn't imply $x = [0]$ or $y = [0]$

okay maybe we really should use notation with modulo

$y^3+y^2-2\equiv 0\pmod{10}$

first thing first is that this is $y^3+y^2-2 = (y-1)(y^2+2y+2)\equiv 0\pmod{10}$

Obliv

19:59

indeed.

so we have $x = [1]$ as a solution?

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