Hades... I'm disappointed in you

@Jakobian you aren’t the only one

@SineoftheTime If you'd have $x\in V$, then you can write it as $\sum a_ix_i$ where $x_i\in A$ and $a_i\in\mathbb{C}$. Now we can write $a_i =b_i+ic_i$ for some $c_i, b_i\in \mathbb{R}$. This writes $x$ as linear combination of elements in $A\cup iA$

so its clear that this set spans $V$ as an $\mathbb{R}$-vector space, you just need to prove that its $\mathbb{R}$-linearly independent

that what I did if $\dim V=n$, but I don't know if this works when the dimension is not finite

for that suppose $\sum b_i x_i + \sum c_i ix_i = 0$ for some $b_i, c_i\in\mathbb{R}$ and $x_i, y_i\in A$, then this implies $\sum (b_i+ic_i)x_i = 0$ so that from $\mathbb{C}$-independence of $x_i$ , we have $b_i+ic_i = 0$ for all $i$. Thus $b_i = c_i = 0$ for all $i$.

@SineoftheTime the sums here are all finite

you don't need to worry about them making sense etc. because in the first sum for example, $a_i$ will be equal to zero for all but finite amount of $i$

Sorry, I'm not familiar with not finite vector spaces. Is the base finite?

@SineoftheTime if you have any questions then ask

no, the base isn't finite

a basis is just a set thats linearly independent and spans the whole space (one of equivalent definitions)

so how do you know that a_i is zero for all but finite amount of i?

@Jakobian yes, that's the def I use

@SineoftheTime every $x$ is in the span of $A$

yes

the span of $A$ consists of all finite linear combinations of elements of $A$

it doesn't make sense to speak of infinite ones

that's what I was missing, ty. We still did not study vector spaces with infinite dimension, but this is an exercise about norms so I thought I'd the tools to solve it

But while you might have something like $x = a_1x_{i_1}+...+a_nx_{i_n}$, it's better to just write $x = \sum a_ix_i$, and have $a_i$ defined for all vectors $x_i\in A$

@SineoftheTime this has nothing to do with norms

the last point ofthe exercise is about norms

yeah, that might be true, but this has nothing to do with them

sure, this part has nothing to do with norms

it's just about treating C-vector space as R-vector space (and conversely, if you know complexification of a vector space)

thank you

To confirm: proofwiki.org/wiki/…

np

This theorem does not require choice right?

Whereas countable union of countable sets is countable does?

the latter does require some version of choice

axiom of countable choice is what I saw. which is necessary but not sufficient for AC?

But which is still too strong for ZF without C?

@EE18 yeah. AC implies axiom of countable choice which implies axiom of dependent choice

But there are models of ZF in which those implications cannot be reversed iirc

Gotcha gotcha, thank you!

Just weird cause my book states countable union of countable sets is countable and then only later brings up choice...

to your point, i won't get hung up on this though. will eventually read something that treats all this more rigorously hopefully

without proof?

well, with an "informal" proof using the usual zig zag argument

i posted the proof above somewhere here

well my approach is that, treat everything like its happening in ZFC

and if its not, will think about it then

:) touche

@shintuku At this institution, every faculty is on year-long contracts (or, in reality, either a 9-month or 10-month contract). During your first three years, there is no guarantee that your contract will be renewed at the beginning of the next academic year. These faculty are "on probation", in that they must demonstrate every year that they are doing their job.

Part of this probationary process is that new faculty are observed by several other faculty members every semester. I am on two new faculty committees, hence I have two observations to conduct every semester. I then send a report to the faculty member and the chair of the probation committee.

@EE18 I think the problem here is to write $X_n$ like this. Those follow from existence of bijection from $X_n$ to $\mathbb{N}$, but you need to fix all the bijections and that requires axiom of countable choice

Isn't it just way more intuitive with axiom of choice?

Yup, you are definitely right

After three years, you are put on a "continuing" contract, which automatically renews every academic year (unless you really f*** up). This is the closest thing we have to "tenure" in the Arizona community college system.

thank you Jakobian

:)

@Novice Look at it this way, if $Y < X$ then either $Y\leq \alpha$, or $Y > \alpha$, so that $X\geq \alpha$

thus $\{Y < X\}\subseteq \{Y\leq \alpha\}\cup \{X\geq \alpha\}$

taking probabilities, $P(Y < X) \leq P(Y\leq \alpha\text{ or }X\geq \alpha) \leq P(Y\leq \alpha) + P(X\geq \alpha)$

OK maybe a silly question here but...

when we write down an indexed set $\{X_\alpha\}$, are we always tacitly taking the index function as a bijection with some index set?

EE18: alpha -> X_alpha is just a function, not necessarily a bijection and in general not a bijection.

the whole {X_alpha}_{alpha in A} notation encodes the same thing as a function from A to whatever set the indexed elements lie in

maybe helpful to think of the case of sequences indexed by the positive integers, which perhaps less abstractly are just functions on the positive integers

I guess what I'm not so sure about is that, when we write $\{X_\alpha\}$ are we not tacitly restricting the codomain in such a way as to have this index function be bijective?

So in your case of a sequence of sequences, the codomain are those specific sequences which are indexed by the positive integers?

whoah slow down

constant sequences are sequences. it's fine to define a sequence (x_n)_{n=1}^infty of real numbers by x_n = 1 for all n

that's not going to be an injective function, so it's not going to be bijective, and you can't fix that by playing with what the codomain is

you can make any function f from A to X surjective (if it isn't already) by shrinking its codomain to the subset {f(a): a in X} of X

this might be what you're thinking about? certainly every element of the set of real numbers {x_n: n in N} is going to be in the image of the function x: N to R

I think there's a misunderstanding. EE18 is talking about the index function from $A$ to the collection of sets.

Or maybe I'm missing the point.

sorry, I am being confusing, I'm talking about the "next level up" (maybe I'm misunderstanding you though}. So if we have $\{X_\alpha\}_{\alpha \in \mathbb{N}}$ then I guess what you're saying is that some of those $X_\alpha$ need be the same?

(Having just arrived and butted in.)

Yes, I think that's it Prof. Shifrin

EE18: can be the same, not need to be the same. they aren't required to be different

EE18, you really don't need to call me Prof. ... Ted is fine.

A force of habit when talking to...profs lol. Ted it shall be though

Prof. Ted Shifrin

indexing a collection of things does not require those things to be different

Hmm, OK I will think about that for a little Leslie.

sometimes you do index for that purpose, but it's not inherent in the act of indexing

I even encouraged my students to call me by first name. Some compromised and called me Dr. Ted. :P

you know it’d be so funny if people called me Prof. Hades

Hades is required to be formal, however.

Yeah, never. I’m stayin’ this way forever

ee18: in discrete math settings you often say "okay, let's induct on the number of elements in the set" and write things like S = {x_1, x_2, ..., x_n} and in this kind of context, i think it would be understood that the map {1,2,...,n} to S is a bijection

but indexing is a much more generally used practice and does not generally carry that connotation

Thank you Leslie, I see now. I won't bore with the context but I was using that the index function was a bijection and, to your point, it was in a context wherein the family being indexed consisted of distinct sets (hence the index function is indeed bijective). But I see your point that it's not true in general, so thank you very much for the clarification

@TedShifrin A habit which is hard to break

Yeah, EE18, I agree with leslie (unfortunately). It certainly may happen that $X_\alpha = X_\beta$ for different indices $\alpha$ and $\beta$.

@leslie Have Munchkin's ducks returned?

often you see people defining indexed families of sets without any knowledge or expectation of whether the indexing is bijective or not. e.g. in analysis of a real valued function, sometimes you want to define the "sub-level sets" S_a = {x in X: f(x) <= a}, which will always give you a family of subsets of X indexed by real numbers, although it generally does not give you a family of distinct subsets of X indexed by real numbers

ted: we heard loud goose honking outside around dinner time two days ago. have not seen the ducks at the pool since... maybe june?

Loud goose! Does he have a permit for such honking?

@TedShifrin Everyone in Texas does

apparently

it sounded like a gang of geese

terrorizing the neighborhood

the local population of red-tailed hawks is also through the roof. we see up to a dozen of them on the drive in to day care.

@leslietownes so the singular for cheese is choose then, right?

as the plural for asparagus is asparageese.

The plural for loose is leese

A question because my analysis is, as yet and as for the forseeable future, weak: a continuous function $f$ has $\int_0^x f dx' = 0$ for all $x$ iff $f = 0$, right?

Long time no see :)

@EE18 Yes. Do you know why?

Actually, wait. What you've written is nonsense.

You've used $x$ twice.

My analysis is weak but I guess continuity and the existence of a nonzero point (from $f \neq 0$) gurantees some range over which the function is nonzero, thereby leading to the nonzero integral

oh yes

oh

fixed :)

@冥王Hades Prof. Profesor

Oh, that's gross. Why not $\int_{0}^{x} f(t)\,\mathrm{d}t$?

Prof. $\int_0^x f(t)\mathrm{d}t$

@Jakobian I don’t know why but this is hilarious

a physics horrid habit, i admit

@Jakobian Thanks, that looks legit, but I don't see any use of $\alpha > 0$. Doesn't seem like a necessary assumption, does it?

@Novice its not necessary, but its there. I'm guessing the reason is that $X$ or $Y$ are positive, or something like that

@EE18 in argument we usually index them so called faithfully so that $X_\alpha\neq X_\beta$ for $\alpha\neq \beta$

Hmm, I think they are sample means from sub-Gaussian distributions, so I am not sure. Maybe I am missing something

if you're given such indexing, then it doesn't have to be injective, but in some arguments it can probably be done, and sometimes should be done

Thanks Jakobian, that makes sense

Just thinking about the range of this function here. Can I do any better than "the set of continuous functions which are the antiderivative of some continuous function"?

Leslie and Ted were insisting on it not being necessarily injective and that's true of course. But in proofs you will still do this bijectively a lot of times, even if not mentioned explicitly

Totally agree, that was what I was using in my particular proof but I didn't fully appreciate it was a special (if common) case

I'm revisiting the problem of Find First Occurrence [FFO] for base 2, i.e., given two positive integers $x$ and $z$, if the sequence given by the digits of $x$ exists in the digits of $z$, then $z = 2^n x + y$. Could anyone give me some recommendations as to where I can expand my polynomial and sum capacities in general so I can solve this?

@EE18 well it has continuous first derivative

@MathCrackExchange My highest bounty yet:

@EE18 no. i mean, you might replace' antiderivative' with more explicit language. the set of differentiable functions on [interval] whose derivative is continuous on [interval]

@leslietownes wdym no

oh, is this one of those fucking FTC things

pretty sure those are precisely functions in $C^1$ with $f(0) = 0$

oh, it's just the boundary condition, right. f(0) = 0.

not arbitrary antiderivatives, particular antiderivatives

Is that a characterization of said set @Jakobian My analysis is very weak, but are you saying $f \in \operatorname{range} T \iff f \in \mathbb{C}^1$?

fucking was unnecessary but yeah, it implies that $Tf$ is in $C^1$

ee18: differentiable, 0 at 0, and continuous derivative

OOF. That is a tough question to see in a linear algebra text but maybe they expect more from me than I know :/

They already made you observe that $Tf$ has continuous first derivative, so they done most work for you

That follows from FTC, right?

Or is the statement of FTC i guess

if you are not already using the notation "C^1", i would discourage introducing it a class where it was not previously in use. a grader might think you are burying the problem in notation that is 'unexplained' for purposes of the textbook

I don't study math in school so this is all just for me :)

okay, well mentally i would discourage thinking "oh it just means C^1" unless you are expressly familiar with the definition of C^1 as you do that

yes, it falls out of the FTC

In mother russia, math studies you!

$C^1$ just means set of functions with continuous first derivative no?

my reference to the f-word FTC things up there is that if you want to consider extensions of that operator to larger domains, it sometimes becomes more difficult to characterize the range and to define the operator

I see how $f \in \operatorname{range} T \implies f(0) = 0$ (and of course $f \in C^1$), but I'm not sure I see the converse. Should that be obvious to me?

ee18: it is not a universally defined notation (hence i would not use it without comment outside of a textbook that provided the context) but generally yes. note that the notation suppresses potential detail about what the domain is, and what "derivative" means at endpoints if the domain has them

@EE18 $Tf' = f$

if $f$ is a $C^1$ function with $f(0) = 0$

Jakobian: To confirm, you are saying that if $f \in C^1$ then $f'$ well-defined and continuous, and so by the "second FTC" (this is corollary to FTC in Spivak's calculus IIRC) we have that, since $f$ is an antiderivative of $f'$, $\int_0^x f' = f(x) - f(0) = f(x)$?

Sorry for writing it out in gory detail, just wanna make sure I follow

Gotcha Leslie, I will file that away. Now that you mention it, I can't even remember how the Hubbards define it, and that's how/where I'd seen it

@EE18 yeah

Thank you all. No way I woulda came up with that characterization but good to know now

Both halves of the FTC, in other words!

What do you mean both halves?

One half says that when $f$ is continuous, the derivative of $\int_a^x f(t)\,dt$ is $f(x)$. The other half says that when $f'$ is integrable, we have $\int_a^b f'(t)\,dt = f(b)-f(a)$.

I don't quite see how these feel like "opposites" but maybe I need to reread Strogatz's infinite powers or watch some 3blue1brown. Anyway, all good for now -- FTC is hundreds of pages away in my analysis text :)

but you know it's out there. what an ominous feeling.

I didn't say opposites, although I suppose one could. I said two halves. One is about differentiating the integral; the other is about integrating the derivative.

You might mumble something about inverse operations rather than "opposites." :D

This is true...I was jumping ahead and doing exactly that mumbling which one so often hears about :)

and now, paul mccartney and stevie wonder will sing "ebony and ivory"

contradistinction $\neq$ contradiction :>

Now why is it that something as simple as my question here is still so difficult to solve? Is it all because it's diophantine?

The idea of dealing with all possibilities of a particular kind always comes to mind, but I don't know enough about such relations in general to use them how I intend.

Simple like Fermat’s last?

Something like that I guess XD

For example, consider here with something like $2^n m$ which I mentioned here I think last time I was here, then something which has all possible $2^n m$ up to some point would be given by $2^{\frac 1 2 k (k + 1)} k!$.

That makes me wonder about this: for those who have been "in the mathematics game" for long enough and to the extent that an answer can be given...was the proof of FLT the most remarkable thing to happen in mathematics in the last, say, 50 years?

That is to say that $2^n$ is a factor of $2^{\frac 1 2 k (k+1)}$ and $m$ is a factor of $k!$

I don't know what professional mathematicians think about it, only what one hears from the pop sci hoopla

The problem, though, is that while I might be able to give an inverse for $2^{\frac 1 2 k (k + 1)}$ or $k!$ in this context, when given either individually or together, they don't account for unique permutations of factors wherein we'd have some sequence $a(n)$ and inverse $a^{-1}(n)$ which maps a lexicographic set as our codomain to the positive integers $n$ and vice versa.

ee18: i guess it depends on what you mean by remarkable. i don't think there's any more 'famous' result from that time period, but that might be just because most math problems aren't capable of being phrased in such an accessible way. it was famous before it was solved, perhaps largely because of that.

Having such would then permit, in this case, finding the factors $2^n$ and $m$ for some $2^n m = a$.

I guess I mean "most impressive to a professional" rather than "most impressive to the layperson", to your point

or as 'famous' as any result can be. there were lots of amateurs 'working' on FLT in the way that there were not amateurs working on, say, the poincare conjecture.

Maybe, EE, for number theory flavor. I might put Yau’s proof of the Calabi conjecture on the geometric analysis list. Important in complex geometry and math physics.

If you picked someone off the street and asked what was the most remarkable thing to happen in math in the last 50 years, you'd have a chance of hearing FLT and probably nothing else. If you did the same thing at the AMS conventions or whatever, what would folks say?

I will look that one up, thanks "Ted" :p

ee18: i see. that's a good question. i wonder if a number theorist would answer it differently from someone who was in math but not in number theory. people in a field often have expectations/beliefs that mathematicians outside of the area do not.

Yeah, Perlman for Poincaré conjecture for topology and geometric analysis.

...and now I know why I hear "Calabi-Yau manifolds" sometimes

Has the infamous subspace conjecture gotten any traction, leslie?

Wasn't there a purported solution last year?

per enflo announced something a while ago, which means, hahaha. who knows what it means.

I don’t follow any longer …

maybe I am misremembering, to quote our old president...

oh yes that was it!

it was not something that most people seemed to be working on when i was active. you know how sometimes you can tell from someone's research that they're sort of "secretly" working on a famous problem? i didn't even get that sense.

is the general consensus that it's not a valid proof then?

i don't know that there's any consensus about it at all. enflo's work is incredibly difficult to read.

in a bad/mochizuki way?

no, just incomprehensible.

it took people like a decade to figure out his first "big" thing that he's famous for, to the point that at least one person who came up with a simpler example many years later can claim, more or less with a straight face, that he was the first to do it.

huh, i guess I would have thought that tons would be putting in all the effort to verify the proof for such a famous problem (at least given that i've heard of it) from a serious person

So... curious... is $p_n$ the only solution to $\pi(n) = n$? For some reason I tried solving something and ended up with $$\pi(2^{\frac 1{t(1-\frac 1 {x(t)})}}) = t$$

but perhaps i fail to understand how professional mathematics work

ee18: that's sometimes how it works? it depends on social currents. functional analysis is not as popular of a field as it once was, so it's already a small group of people who might be an audience. nobody is getting NSF grants to work on anything related to the invariant subspace problem. nobody is going to get tenure for finding a flaw in that stuff, either.

spending large amounts of time on a famously unsolved problem has a reputation for being career suicide. there's a reason wiles (who had less to worry about than most, career-wise) worked secretly.

i do actually remember one guy with a deserved reputation, not enflo, who was working on the invariant subspace problem as an emeritus professor and was kind of cranky about it at conferences.

that's the time of your life in which you can publicly work on the famously unsolved problem. when your house is paid off and people know enough about you that they don't shoo you away.

Putting aside that there's no other practical possibility (that I can see) for how things would work given frictions like...money...do you think that in a vacuum that's the best way for things to go? I gotta imagine we'd be further ahead if there were some mechanism to support serious people taking moonshots at things like that, at Collatz, at RH (forgive me for mentioning only the things I know of)

This has unironically been easier for me to work on than $z = 2^n x + y$ (sorry for the mess): desmos.com/calculator/ixh5vgvxlt

I guess I'm divulging that my priors for the possibility of solving those are nonnegligible. If they were negligible then maybe the answer would be no, that's not the right allocation of resources

if eccentric billionaires want to spray money at something, they certainly could do far worse than to spray it at unsolved math problems. (a couple of them do this, maybe not directly in a "bounty for a problem" way, but indirectly through creating or supporting math institutes, research fellowships, etc.)

e/math, as it were

fair enough, thanks for the discussion Leslie

@EE18 my impression as a not-quite-scientist is that scientists are sometimes incentivized to focus on little problems rather than big problems

There's a long passage in my textbook which I have problems understanding. I think I have summarized the main points in the following:

(I am not directly involved in pure math)

how "best" to direct resources at stuff like that, in some values-based sense, is a really hard question (even assuming you have the resources to direct, which is its own really hard question).

@sunny Let $f:I\to\mathbb R$ be uniformly continuous on some compact interval $I\subseteq \mathbb R$. Let $x_{\delta}: I\to\mathbb R$ be a sequence of continuous functions on $I$. Let $\delta>0$ and suppose $\varepsilon(\delta)$ is defined by $$\varepsilon(\delta)=\sup\limits_{|t'-t''|\le\delta}\left|f(t',x_{\delta}(t'))-f(t'',x_{\delta}(t''))\right|,\quad t',t''\in I.\tag1$$

Then consider the following inequalities: \begin{align}&\left|x_{\delta}(t'')-x_{\delta}(t')-\int_{t'}^{t''} f(s,x_{\delta}(s))ds\right|\leq \varepsilon(\delta)|t''-t'|, \tag2 \\ &|x_{\delta}(t'')-x_{\delta}(t')|\leq B|t'-t''|,\tag3\end{align}where $B$ is a constant. From these identities, is there anything we can conclude about $\lim\limits_{\delta\to0}\varepsilon(\delta)$?

Why is $\delta$ enumerating a sequence?

It's an approximation to a solution to an ODE. I guess you could change it to $n$.

And your domain of $f$ is incorrect.

It’s not for me to change things. You have to pose a well-defined question.

You’re using $\delta$ in two completely different senses.

has the dirac delta made an appearance yet?

I have a presentation today, and this time I also have to do the talking as well, in front of a large class

I’m going to screw up I’m sure of it

Practice in front of a few discerning friends.

And smile/laugh internally, if that makes sense. It will put you at ease. Good luck Hades!

for a presentation $5 \leq n \leq 30$ minutes long, I practice at least $n$ times, which has served me well

@Shaun my Parabola group post went through the roof. It's weird that every now and again some weird post of mine gets highly upvoted, though not much work went into it :|

And my other hard-to-write posts get -1 voted

@Shaun the parabolic group is an algebraic group according to Qiachu Yuan

@Novice i don’t believe you practice more than 5 times!

You don't believe I do that, or you don't believe one should do that?

I can’t see the gain in repeating a 30-minute talk more than 5 times. Not to mention not having 15 spare hours.

@MathCrackExchange What’s the link?

@TedShifrin of what?

@AMDG Have you tried to use SymPy and NumPy in Python. It can preform some crazy computations. If you know some basics of python it can do more advance stuff than Desmos. Also, if you want all the feature of LaTex use Jupiter Notebook in VS code!

Your parabola post.

Yep, I'm a pro :)

at MSE

I figured it was analogous to the circle, but the missing point changes multiplication to addition. The elliptic curve case is deeper, because that’s working over an algebraically closed field, in projective space.

Yes, Qiachu Yuan linked a paper titled: "Conics - the poor man's elliptic curves". Idk how he knew that I am poor -_-

But my parents are wealthy and I live with them, so...

Money is hard to come by these days, math is free

Qiaochu is back? Since when?

Since a year ago, welcome to outside the rock you were under

How is math free? It is one of the most expensive subject to buy used books for.

Qiaochu Yuan is a hero of math IMHO

407k rep, he doesn't mess around

@SometimesMath Lots of free books available online.

And by free, I mean actually given by the author for free.

@SometimesMath I'm speaking roughly. E.g. you don't have to go to college to learn math, college being more expensive than a Porsche

@anak Where?

@SometimesMath What topic are you interested in? I can send you links to a few.

Any ideas on the problem? I am not even sure where to start with this one

We share an alma mater (with leslie and copper, too). I think Qiaochu gave up on academia, but certainly not on math.

It's also worth mentioning that there are many cheap textbooks now available through Dover publishing. They republish many gold textbooks.

@MathCrackExchange Lol, College is free if you work hard!

I abhor hard work

I'm a mathematician, I'm lazy

@euvageloskazazis Start with equality?

@euvageloskazazis I would try $(1 + x)(1 + y) = ?$

That's just a guess though, probably not the right approach

@TedShifrin so set x = y = a + t, t > 0 ?

@anak Any topic that will get me proficient enough to understand this stuff.

@SometimesMath what stuff?

Universal mathematician is impossible unless you have NZT-48

And it doesn't exists

@MathCrackExchange Indeed, it has. Congratulations.

@SometimesMath you have to pick a problem that interests you or a goal to work toward, and then study related material

@MathCrackExchange The same thing happens with my posts. It stands to reason really. Easy stuff is far simpler for people to verify.

And for after you do any of these, if you wanted to continue on, I just realized Ernst has a page dedicated to open source textbooks: danaernst.com/resources/free-and-open-source-textbooks

Of any number of topics.

Your parabola question is, conceptually, quite beautiful too.

Well, I knew something could be done, I just had to figure out how by constructing lines

@euvageloskazazis No. Replace the > with a =.

I bet there are other groups via geometric construction as well, they're just way too deep to construct blindly via drawing, so algebra is needed to discover them

That is very cool! Thanks for sharing about the open source world of math books.

@Shaun I still haven't figured out my integer homology question $(pq|\cdot) \mapsto (q|\cdot) -(p|\cdot)$ etc (it's a boundary map to a chain complex); according to the answerer it is an exact sequence meaning homology is zero everywhere, but this always happens in the standard complex. So I don't know enough weibel to go further

I should also convert it somehow to cohomology side - haven't figured it out yet though

So the primes become vertices analogous to simplicial complex stuff in topology

I also showed that reverse homology can be defined when $\ker \partial \subsetneq \text{im } \partial$

When they're equal it's the same thing as exact sequence. The reverse homology is functorial and everything just as in usual homology. Still haven't found an app for it though

@SometimesMath If you ever want suggestions for resources, I pride myself in collecting and sharing resources. Especially on the pedagogical side of things.

A question. Suppose I have a subspace of $R^3$ which is characterized by the condition that $(a,b,c) \in W \iff c-a+b = 0$. Now obviously that subspace has dimension 2 since I have "1 condition and 3 parameters". But what would be the way someone would verify that rigorously? Would one setup the corresponding homogeneous system $$[-1 \, 1\,1 ]$ in matrix form, "row reduce", and then...

... extract the basis (as seen in a theorem I know) which comes from setting each free variable to 1 and the others to 0?

I ask because this feels like a long-winded way of justifying it but, as I thought about it, I realized I didn't know a faster way

Yes. That’s the approach. Nullity-rank is the fancy theorem.

How would you work nullity-rank in here?

I thought that needed a particular linear $T$ transformation to deploy?

@TedShifrin Would a reasonable consequence of the property we discussed earlier be that for each sequence (a_n) s.t the sum \sum n|a_n - a_{n+1}| is finite, if (a_n) approaches zero at infinity then \sum a_n converges absolutely?

How is this related to the earlier discussion?

@EE18 You have one, just as you found a matrix.

If I am not mistaken, we can deduce for each n: |a_n| \leq \sum_{k=n}^{\infty} |a_{k+1} - a_{k}|

@XanderHenderson can you bribe faculty with pastry and increase your chances this way?

if m > n then |a_n - a_m| \leq \sum_{k=n+1}^{m} |a_{k-1} - a_k| and finally take m to infinity

I see it now Ted, thanks very much!

@X4J OK, good. Now what?

Can we just use the assumption about \sum n|a_{n+1} - a_{n}| with comparison test? If not, then I apologize, I might need to review my understanding again