Where are you using the first part you told me?

I guess I'm going to use it at the comparison test

How? Where does the $n$ get used?

As far as I seemingly understood today, the convergence of \sum n|a_{n+1} - a_{n}| means the convergence of \sum_{n=1}^{\infty} \sum_{k=n}^{\infty} |a_{k+1} - a_{k}|

If so the ineq I mentioned yields the convergence by the comparison

Aha. So it’s the same idea in reverse. Yes, I think you have it.

thanks for the guidance again

@shintuku I just report what I see. The goal is to improve instruction. One of the observations, I had to work to offer suggestions for improvement. The other, I struggled to identify strengths.

For the latter, they're teachable, I think.

In my career, I did uncountably many teaching observations — grad students, instructors, young faculty. I honestly don’t know how many teaching award nominations I wrote, and plenty of teaching portions of promotion dossiers and teaching letters for jobs. A few grad students needed serious work.

I'm looking to write the equation of a sinusoidal plane wave moving along the z-axis. I got $\psi = A\sin(\vec{k}\cdot \vec{r} - \omega t)$ I don't know how I'm supposed to go any further

I need to find "appropriate expressions for $\vec{k}\cdot\vec{r}$"

the solution is apparently $A\sin(2\pi / \lambda)(z-vt)$

i guess it makes sense

@TedShifrin I was working on the problem we discussed a couple of hours back, but I have no idea, how they wrote, "Then, $f^n(a+\theta h)=f^n(a)+\frac{h}{n+1}f^{n+1}(a+\theta 'h)$ ". Can you please check that out?

If you want, I can repost the image.

Wait, I think I should post it again, coz others may wanna take a look at it, tho.

Here it is (above)

@SometimesMath Oh nice I had no clue

Also no I don't use Python

I hate Python

Cringe embodied, but if it works for you, by all means.

In any case, me like 100 year old mathematician: chalk and intellect

Me no need number cruncher

Ooga booga

fr tho why does $\pi(n) = n$ have to have multiple solutions besides the primes smh

I really wanted to find a prime(n) sooner not later with this thing

Y'know if we're going to solve things more easily without all these complicated hacknslash proofs that require a PhD in math to understand, we need to improve our fundamental set of operations. One way is by creating an isomoprhism within mixed boolean arithmetic (which I'm working on--for my own reasons of course).

Or something simple... like solving $x = 2^n m$ that i mentioned before... that'd be quite a leap... (and very useful for me)

Anyways, I think I've monologued enough here.

@ThomasFinley They applied the formula in case $n$ and also in case $n+1$, set the two results equal and eliminated all like terms.

For the inductive case here I argued that $g$ bijective and $f$ injective imply $h$ injective, which in turn implies the restriction of $h$ to $\{1,...,n\}$ is injective (had to show that it was actually a function from $\{1,...,n\}$ to $\{1,...,n\}$), then used the inductive hypothesis to get that the restriction of $h$ to $\{1,...,n\}$ bijective, which in turn implies that $h$ is onto.

I'm pretty solid on those steps, but from there I guess I can use that $h = g \circ f$ with $g, h$ bijective implies $f$ bijective?

There were a lot of moving pieces so just wanted to make sure I didn't miss something big. I also never would have come up with said function without the hint :/

Computers only have one use to me in Mathematics

People trying to make sure the trillionth value on the critical strip doesn't disprove Riemann Hypothesis be like

that's my favorite sorting algorithm. generate random permutation, apply permutation to list, check if list is sorted, if not, repeat

random.nextInt(): the universal problem solver

you can optimize for speed by generating a random transposition each time instead of a more general permutation. [the more you know logo]

That would improve the worst case. That's illegal according to good programmers who never "prematruely ahptihmajz".

Whenever someone throws me that Knuth quote these days, I swivel around in my non-existent swiveling chair, crack my knuckles, and proceed to tell him how he a) doesn't know what premature optimization means 2) doesn't write good code because I'm a gigachad. Kinda like a bearded Linux user sitting on the lamppost outside the nearest subway.

There's some important details that I intentionally left out of that of course, but I'm sure the real programmers in the room know what I mean

i like spending time writing an algorithm just so i can use it once and never use it again

to check some sequence or something in math when i could have just wrote and record by hand

Life of a programmer. (Don't let the psychologist into a programmer's room.)

where is semiclassical when you need him :((

Do we need him?

I need him.

I have stupid questions about physics.

How are you Ted?

Cooked anything rad recently?

Nothing worth mentioning.

Last shameless question of the night for me:

zelda.fandom.com/wiki/Dubious_Food

The question and my solution. I am suspicious after reading this (math.stackexchange.com/questions/2189871/…) that I cannot do what i did for the converse

Yeah this is one of those things which is "obvious", but when you get to trying to prove it, you either make a weird jump, or you have to go into the gory details.

Reason is linear moment

Some would say $O(n) < O(n!)$

@anak So my statement that $\{1,...,m\}$ has $s \geq n $ for some $s$ elements is too much of a jump?

The power of mature optimization. Old, I know. N-E-ways, nite!

(It's a good watch tho)

I feel like there might be some underlying circularity. The proposition is effectively saying that the sets have equal cardinality if and only if m=n. Then your argument hinges on this idea that because {1,...,m} surjects onto {1,...,n}, "there are at least n elements in {1,...,m}". But this last statement doesn't actually directly tell you anything about the relationship of m and n. It's kind of a headache because of the notion of cardinality (using the |A| = |B| or |A| \leq |B| notation).

This is why the use of induction is much more natural for addressing the problem.

I see, thank you anak. I will have to think a bit harder about where exactly things are circular, but now I know to look!

I suggest proving that injection from m to n implies $m\le n$. Then injection both ways does it.

You are going for the induction on $m$ argument here Ted?

I don’t know why I need induction. 🤷‍♂️🤷‍♂️

I thought those were flower emojis at first.

Do we know that if $A\subset B$, then $|A|\le |B|$?

I don't think so Ted :( I mean, we know it, but we don't know it.

🤷‍♂️🤷‍♂️

I quit.

Pas de problem. I will attack it tomorrow! Thank you again to both of you for the help n suggestions

@EE18 do you have a definition of $|A| \leq |B|$ yet?

Or is this what you are working on completely silent on the notation of "| |"?

A definition? Yes. Card(A) is well-defined as the $n$ for which $A \sim \{1,...,n\}$. But this is only well-defined if I can give a proof for the claim we are talking about here (the proof in the text used this little lemma)

So I suppose then we must forgo that

I think I will try to carefully follow that link I put above. I think the argument goes through well, and without the dubious circularity you mentioned

$\sim$ being the existence of a bijection?

Yes, sorry for the lack of clarity.

Often times they define $|A| \leq |B|$ as "there exists an injection $A\to B$". Do you have anything like this already?

I do not, no :( this is Chapter 6 of Amann and Escher Analysis I in case you have access thereto/want to consult what I can see, but to that specific question no

Okay, so with your available tools, I feel like the inductive proof is probably the most "solid".

I agree :) will give it a crack tomorrow morning! It's interesting how easy it is to slip in false proofs like I did when it comes to these basic but very delicate questions

I guess I probably need to read Halmos or even some logic book to really be sure of not making these missteps

I think it's more just a mathematical maturity thing.

I don't think it's something you really need to get hung up on if it doesn't make sense right now. Revisit it in a few months or so.

@TedShifrin Haha, prof what are you learning these days?

@LuckyChouhan Nothing. I’m very retired!

Oh, Btw good morning! Do you have any advice for me who is in his undergrad?

I don’t know you, so no :)

I found this recently absolute bomb: math.stackexchange.com/questions/8337/…

@TedShifrin Ugh,

Yes, that is a famous example with numerous proofs.

you are retired, so what do you do these days?

Scalar fields having all partial derivatives defined but this need not imply continuity, what condition on differentiation will definitely make sure it's continuous?

Hi @MathCrackExchange, seeing you after a long time

High

:D

😁😁 I think I proved twin primes (for the Nth time), but this one is bullet proof

It would need some more work around the formula to show that it's valid

:)

I like your username

Thanks, this is my real name

You're blessed :)

:)

what is $\omega(d)$?

According to that link above, I have proven that it is non-decreasing, which means it is non-decreasing on these integer intervals as well. However, I'm not 100% sure about what happens when $x - r \lt 0$. I think it still counts them, but I'm not exactly sure why!

$\omega(2) = 1$

$\omega(6) = 2$

$\omega(12) = 2$ It's the number of distinct prime divisors of input

okay

Here, because $p_n\#$ is square-free you can replace $\omega(d)$ with $\Omega(d)$ or even $(-1)^{\Omega(d)}$ by $\mu(d)$. That's a lemma somewhere on wikipedia

Under Liouville's function

ooh

$p_n\# = 2\cdot 3 \cdots p_n$.

$p_n = $ the $n$th prime and as you can see it is indeed square-free

This formula actually generalizes the $\pi(x) - \pi(\sqrt{x}) + 1$ formula for counting $0$-separated prime averages (the primes themselves)

That inclusion-exclusion formula on the prime counting page

why doesn't n!= O(n^n) violate Stirling's formula?

nvm

@冥王Hades Thank you for your advice =)

@leslietownes Wouldn't there be a chance that the algorithm might run indefinitely?

@MathCrackExchange nice question

@XanderHenderson nice, i'd be afraid of having to do office politics in a system like this but if everyone is like-minded it sounds efficient

let $X$ be a non empty set and $F$ be a field. construct a vector space V which has a basis with cardinality |X|

I have no idea how to solve this problem.

@ThomasFinley what's the relationship between V and F? and does X have a relationship to V?

@shintuku nothing's given

Nothing

hi chat

consider $C([0,1])$ and define the norm: $ \|f\|=\left(\int_0^1 |f|^2dt \right)^{1/2}$. To prove the triangle inequality, is it sufficient to use Cauchy-Schwarz right?

@SineoftheTime \| or \lVert and \rVert are your friends. :D

$\lVert f \rVert = \left( \int_{0}^{1} |f|^2\,\mathrm{d}t\right)^{1/2}$.

I did not remember the comands and I'm lazy, so I wrote || :)

>:(

@SineoftheTime yes

@ThomasFinley Given a fixed field $F$, it's impossible to construct a vector space with arbitrary non-zero cardinality. For example, if $F$ is a field with $3$ elements, you can't construct a vector space with two elements

If $F$ is a finite field, then finite vector spaces over $F$ will be of sizes $|F|^n$ for $n = 0, 1, ...$

@Semiclassical I am used to dealing with probability dists where you have some probability density $\rho\,d\mu$ and then to find the prob. of an event $A$ you use $\int_A\rho\,d\mu$. When it comes to QM and density ops, what's the analogue for finding probabilities? I have not found a clear answer yet, so maybe there is no analogue? I have seen $\text{tr}(\rho A)$ to signify a probability where $\rho$ is a density op, but I am unsure of what $A$ is?

The direct sum of $\kappa$ copies of $F$ will be a vector space of size $\kappa$ when $\kappa$ is an infinite cardinal

For infinite fields however, they can have size either $1$, or at least $|F|$. If $\kappa$ is an infinite cardinal $\geq |F|$, then the direct sum of $\kappa$ copies of $F$ will be a vector space of cardinality $\kappa$

Summarizing, if $F$ is an infinite field, then vector spaces over $V$ can have sizes $1$ or any size $\geq |F|$, but if $F$ is a finite field, they can have any infinite cardinality, or $|F|^n$ for $n = 0, 1, ...$

Thus if you fix a field $F$, then there will be no vector space of size $2$, unless $F = \mathbb{Z}/2\mathbb{Z}$

But for this $F$ we don't have a vector space of size $3$ on the other hand

to get vector space of every possible positive size, you need to consider vector spaces over different fields $F$

sorry my bad this won't get you all possible positive sizes

for finite vector spaces you only get prime powers

there's no vector space of size $6$

But yeah, if you consider different fields you can get $(\mathbb{Z}/p\mathbb{Z})^k$ as a space of size $p^k$ with $k = 1, 2, ...$

and $\bigoplus_{i\in I} \mathbb{Q}$ for $I$ infinite or empty, as a space of size $|I|$ (over $\mathbb{Q}$)

@Jakobian Your answer is great. But, unfortunately, being just a beginner in Linear Algebra, some portions of your argument, goes above my head :?). But, the main takeaway is that the claim is not true. One example, is: If $F\neq\Bbb Z/2\Bbb Z$ is a field then there is no vector space of size $2.$

I have just one question:

The thing is, how to prove, "If $F\neq\Bbb Z/2\Bbb Z$ is a field then there is no vector space of size $2.$". Did your arguments in the sections say, chat.stackexchange.com/transcript/message/64449324#64449324 actually proved this claim intrinsically?

@anak In the notation you've given there, $A$ is usually an (self-adjoint) operator representing an observable, so $tr(\rho A)$ would actually give an expectation value.

I should add that $tr(\rho A)$ giving the expectation value associated with the observable represented by $A$ is usually given as a postulate. If you want probabilities you would need the relevant projectors $E_A$, and then $tr(\rho E_A)$ is the probability of the even corresponding to the projector $E_A$.

This latter statement is not quite a theorem, nor a postulate. Given our earlier postulate, if one makes assumptions about the correspondence rule between observables and the operators representing them (in particular, that the rule preserves functional relations) then you can derive what we said above as a theorem. But of course there is that assumption, strictly speaking.

Chapters 1 and 2 of Ballentine go over these details somewhere between the level of mathematical rigor associated with a mathematics text and a physics text (but closer to the latter)

I am confused @Jakobian so perhaps I misunderstand @ThomasFinley's question. For $|X|$ finite, we have $|X| = n$ for some $n \in \mathbb{N}$. Can I not then take $F^n$ as the requisite space (if $n = 0$ then I can take the trivial vector space over $F$)? If $|X| = \infty$ then I can take the space of all polymoials over $F$? I think an assumption of $F$ having characteristic 0 may be slipping in there though...

@shintuku do you mean the Parabolic group or the alternating sum that's always non-decreasing?

@Jakobian I think the question asks to construct a vector space whose basis size is of the given cardinality

@PrithuBiswas No problem. Make sure to find 2 of the same 4GB RAM sticks at the exact same speeds and latency. If you’re unsure of what these are, head into your BIOS and write it down somewhere

and place them in alternating slots on your motherboard if it has 4 slots. If it only has two slots then just place them normally

@ThomasFinley do you know about the notation $F[X]$ where $X$ is an indeterminate?

@shintuku

Another answerer in a related post has come up with also an alternative proof already using reverse inclusion-exclusion principle

I noted that in the comments section

It is related to twin primes, however, you would need to show that the lower-bounding trend line of the function has value $\gt 1$ in the integer interval $[0, ..., p_{n+1}^2 - 2]$. Once I prove that the formula validly does count twin prime averages in that interval, and that has the $\gt 1$ value in that restricted domain, then the twin primes will be proven.

Let $v,w\in\mathbb R^2$ be linearly independent vectors, and let $a(t),b(t):\mathbb R\to\mathbb R$ be non-vanishing functions such that the norm of $a(t)v+b(t)w$ is always equal to $1$. By choosing an appropriate section $\sigma$ of $t\mapsto e^{it}$, is it true that if $a(t)/b(t)$ is strictly monotone (and if necessary of the same sign), that $\sigma(a(t)v+b(t)w)$ is strictly monotone?

The fact that it needs to be $\gt 1$ is because I think $0$ is always counted a twin prime average though it isn't one.

@SoumikMukherjee sorry, but no

The reason behind my question is that I want to argue why the plot of $\gamma(t)=a e^{\lambda t} v+ b e^{\mu t}w$ ($\lambda,\mu\neq 0$) looks the way it does. I've argued what must happen with the direction of $\gamma(t)$ in the limits of $t\to\pm\infty$, but I also want to argue that $z(t)$ monotonically goes from the direction of one vector to the other (depending on $\lambda,\mu$)

maybe I could look at the second derivative

since the $x$ and $y$ component of the second derivative changes sign at most once, but I have to think if this takes away my doubts

@ThomasFinley ok, do you know about the set of polynomials with coefficients from $\Bbb{R}$?

This set forms a vector space under polynomial addition and scalar multiplication

ah, I think it's even true in my case that $b(t)-a(t)$ is strictly monotone

in which case the answer to my question is yes

@SoumikMukherjee yeah

@SoumikMukherjee yes...

@ThomasFinley Now if you consider the space of polynomials of degree at most $n$, then you still get a vector space. Right?

What will be a basis of this vector space?

actually, I'm not sure about my problem anymore

I have a proof for this that uses that there are finitely many subsets of a finite set

I love tea

Obviously this is true, but would that follow from induction?

The inductive step and what I can take for granted have me uncertain after my discussion with anak last night

ee18: putting that aside, how do they define the binomial coefficient in the first place? their sketch of a proof is only one of many possible ways to get there. maybe not the easiest way, depending on the definition.

Binomial coefficient is defined as the quotient $$\frac{n!}{m!(n-m)!}$$, which is to say the unique $k \in \mathbb{N}$ such that $n! = k{m!(n-m)!}$.

We will give a partition of the bijections of $\{1,...,n\}$ onto $N$, $B(\{1,...,n\},N)$, as follows. Define $A_M : = \{f \in B(\{1,...,n\},N) \mid f(\{1,...,m\} = M)\}$. We claim that $B(\{1,...,n\},N) = \cup_{i=1}^k A_{M_i}$, and then using disjointness and the corresponding result about sum of cardinalities to show that $k$ as so defined obeys $n! = km!(n-m)!$, so that $k$ is that binomial coefficient. But writing that as a finite union presupposes what I mentioned

I'm just quoting the start of my proof for the proof strategy^

that was my sense for what they were hinting at doing anyway (given the allusion to Prop 6.3 which, essentially, says there are $n!$ bijections between two sets of size $n$)

How can I argue that the image of this parametrization has this form? I was able to deduce the limit behaviour (both magnitude and direction) as $t$ goes to $\pm\infty$, but how do I know that nothing strange happens in between? I would kind of like to argue that the angle of the solution changes monotonically, but I'm not sure how to do that

of course I can prove that there can only be one turn by looking at the derivatives of the $x$ and $y$ components

but I would like to prove that there are no wiggles or some slight/damped oscillatory behaviour, and I was hoping to achieve that by arguing how the angle behaves

or maybe not so much the angle of the solution, but rather the angle of the derivative

(which shouldn't matter for the problem, since it yields something comparable)

Oy, I'm an idiot. Ill have to check but pretty sure I've proven that the powerset of a finite set has size 2^n

If so, this is immediate, because any subset of a finite set is itself finite

@EE18 if you haven't done this, this is easier than proving the binomial thing (at least, as the binomial thing has been defined).

Nope, I have done it, just checked :) I did it earlier when I didn't even have a rigorous notion of size, but i can see how I would "rigorize" (word?) it in retrospect.

So I think I can go ahead with this current proof sketch/skeleton given that

sha: this you? wolframalpha.com/…

[one way of analyzing the plot is noting that if x = -e^(-t) + 2 e^(-2t) and y = -e^(-t) + 3 e^(-2t), then [treating e^(-2t) and e^(-t) as independent unknowns and solving the linear equation] y - x = e^(-2t) and 2y - 3x = e^(-t), so that y - x = (2y-3x)^2 eliminates t]

@SoumikMukherjee you're right, I misunderstood, or rather misread

the question asks for a basis, not size of vector space

@leslie ah, that is a great insight, as I'm only interested in the shape of the plot. let me see if it helps my case

so if i'm not messing up, dy/dx - 1 = 2 (2y - 3x) (dy/dx - 3), so dy/dx = 1 at (0,0) gives you the slope at the origin.

yes, I've already computed the slope at the origin

I've also computed the slope in the limit $t\to-\infty$

@ThomasFinley I'm sorry, I didn't read "basis" in your question. If $F$ is a given field, then direct sum $\bigoplus_{x\in X} F$ works perfectly well for an $F$-vector space with basis of size $|X|$. You can construct this vector space as the space of functions $f:X\to F$ with $f(x) = 0$ for all $x\in X$ except for finite amount. The functions $f_x$ defined as $f_x(y) = 1$ iff $x = y$ and $0$ otherwise, form a basis of this space

and it wouldn't appear as though there is a limiting slope/angle as t goes to -infinity.

I think there is

maybe not for the slope

but for the directional derivative divided by its norm

oh, hrm

oh!

but this is a quadratic equation in 2 variables

these solutions are known

so we have either a parabola or a hyperbola

in either case no weird wiggles going on

so thank you!!!!!!!

your insight was really great. I wouldn't have thought in a million years to let go of the parametrization and just look at the equation we are dealing with

it's a parabola

yeah, no weird wiggles in a parabola. citing the ancient greeks or somebody

loool, my students are also taking a course called "Intro to geometry" now which deals with quadratic equations, so I will refer to that course :)

@ShaVuklia Part of one, anyhow.

sure sure

oh man, this is GREAT. last week I struggled giving a proper explanation regarding the shape of the solution (they had to draw a phase portrait for a linear system of ODEs with constant coefficietns), and it really bugged me that I didn't manage to do that

History question: does anyone know when the quotient topology was first defined? @Jakobian

now I'm really glad I revisited this, because my explanation next problem class will convince everybody

famous last words

@EE18 So both $\rho$ and $A$ are self-adjoint operators? Does Ballentine go over these projectors, E_A that you mentioned?

Correct, both are (again, there may be a much higher level of mathematical rigor which says I'm wrong about this). And yes, he does

The $\rho$ have some extra conditions on them too I should add. namely, unit trace and nonnegative

Gleason's theorem is related and potentially of interest

hm, I do wonder if I could still argue the general case

which would yield an equation of the form $(ax+by)^\lambda=cx+dy$

defining E_A can be subtle when A is unbounded (or even when A is bounded i guess) but is not too bad and it is hard to write wrong formulas unless you really try

Thanks @EE18!

i think even in physics rho is always bounded

Pfft, who even checks boundedness.

These are the details which are above my pay grade @anak Ballentine pays only some (but still more than most physics texts) heed to these details (talks briefly about rigged Hilbert spaces for instance, as well as von Neumann's standard rigorization of QM)

But if you want all the gory details you should consult Moretti, Galindo/Pascual, or even Reed/Simon (or so I hear)

I will certainly have a look at Ballentine first, and then check out these other ones if needed.

Is this your first brush with QM?

I only wanted the rudimentary idea of how to make an analogy with probability measures in the usual case. I was working with the Bures metric, and I needed like a tl;dr on density operators. You provided a pretty good one.

hm, I guess $\lambda$ has a nice shape

it's of the form $\lambda_1/\lambda_2$ where $\lambda_1,\lambda_2$ are solutions to a quadratic equation

Oh gotcha. I hope I'm not too misleading because it sounds like you're working on something a bit more, er, serious mathematically than what I've offered up. But if it helped that's great :)

@AlessandroCodenotti by Baer and Levi in 1932

@EE18 I will quickly retreat to the realm of pure geometry after I mine QM for ideas. :D

Wow. Well, it had to happen before Whitehead invented CW complexes.

Who were Baer and Levi?

Oh, Levi I know.

Please don't forget to return though :) @anak

although he was separate from Levi-Civita. :D

baer worked some topology into the middle of his boxing career

I never heard of them

And starring in Beverly Hillbillies?

A classic case of a hyphenated name causing some confusion about credit!

I did not know for a while that Lévy-Leblond referred to...one guy

Lévi-Strauss

"Stetige Funktionen in topologischen Raumen." by Baer, R. and F. Levi

There ya go, wouldn't have guessed that a priori either. And the sad bit is, my last name is hyphenated too!

link.springer.com/article/10.1007/BF01180580

oh, found it on Springer

@leslietownes Just referring back to this question, the last bit of the proof I can't stitch together is showing that the set of bijections on an $n$-element set which, in particular, map the first $m \leq n$ elements to some fixed $m$-element subset $M$ has size $m!(n-m)!$

Any potential for a hint here? I mean, intuitively it must be true, because any bijection in that set can be "built up" out of the set of bijections restricted to $\{1,...,m\}$, plus a bijection on the last $n-m$ elements

Those two sets have size $m!$ and $(n-m)!$ factorial, respectively, which I know. But I'm not sure what the right notion is for "pasting" sets of functions together in this way is

en.wikipedia.org/wiki/Friedrich_Wilhelm_Levi en.wikipedia.org/wiki/Reinhold_Baer

those are links to the authors in wikipedia

For context, it's claim 3 that I have outstanding. $B$ is the set of bijections between the two argument sets

Interestingly, both were primarily algebraists.

Something like a Cartesian product for pasting these sets of functions together?

@Jakobian the year makes sense, but do you happen to have a reference?

@TedShifrin I think the motivation for the quotient topology was to study quotients of topological groups

Then it was extended to all spaces

@AlessandroCodenotti yes, of course

link.springer.com/book/10.1007/978-94-017-0468-7

volume 1

Of course, Poincaré and others had been talking about gluing to make surfaces, etc. So that was earlier.

page 271

Ah, I looked into that very same book without finding it

Thanks!

np

(I was curious because I was trying to read a 1925 paper by Moore which uses very weird language because he's proving stuff about quotients of the plane before the quotient topology was even invented lol)

Well, that's sorta my point. It appeared casually quite a bit before this.

I guess the question becomes: how to describe the plot of $t\mapsto a e^{\lambda t}+b e^{\mu t}$, and I think I can figure that out

if a a wiggle (inflection punt) occurs then the second derivative must change sign I believe. since the second derivative changes sign at most once, we are good

@anak Sorry to raise this again, but I came back to our discussion this morning and i am no longer convinced of the soundness of the argument suggested in the linked question (it's this math.stackexchange.com/questions/2189871/… one just so you don't have to go looking again)

I mention it in a comment to the answerer, but I'm not sure the argument proves that which it claims to. Ostensibly, the argument proves $Card(\{1,...,m\}) = Card(\{1,...,n\})$ but there are two problems with that: (1) We are using this theorem/lemma to prove that $Card$ is well-defined in the first place! Thus we cannot make mention of the notion which is not, as yet, well-defined.

(2) Even if (1) were not a problem, $Card(\{1,...,m\}) = Card(\{1,...,n\})$ just means, by definition, that both have a bijection to some $\{1,...,k\}$, which is to say there is a bijection between $\{1,...,m\}$ and $\{1,...,n\}$. But that was our hypothesis in the first place, and is certainly not the claim that $m = n$!

I am suspicious that perhaps my issue here is with the dubious ellipsis in the $\{1,...,m\}$ etc. set definitions, and that if I could make that more precise then I would be able to show that which I want to show

@TedShifrin Hello Ted. We discussed about the fact 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. I thought if the inverse sentence is also true, but I provided a counter-example for this. Currently I'm wondering what if we also assume (a_n) is monotonic. That is supposed to work, right?

Solved me little prime counting function adventure (begins line 137) desmos.com/calculator/mysaguadng

Not sure how useful this is tho

@anak seems like you got some other help on this but: the most obvious analogue of an indicator function in QM is a projection operator. So you get probabilities as trace of $\rho A$ for an appropriate projection operator $A$

The trickier question is “probability of what?”

It’s easiest if A is a rank-one projection onto some vector v. If this v is a non-degenerate eigenvalue associated to an observable M with eigenvalue $\lambda$, then the probability of measuring $M$ and finding $\lambda$ as the result

There’s some additional subtleties when you have degenerate eigenvalues or if the rank isn’t 0 but that’s the basic story for probabilities of projective measurements (aka measurements you can repeat and get the same result)

There’s also cases where the measurements aren’t projective. Most generally you could have A as some observable (aka a self-adjoint operator) such that A and id-A are both positive

That ensures $\rho A$ is a real number between 0 and 1, hence is not nonsense as a probability. But then the challenge is making sense of what this is exactly the probability of!

One way it can arise is if you make a projective measurement and then assume that there’s some probability to read the result wrong

In which case you could repeat the measurement and not necessarily get the same result

But there’s a few other ways to make sense of such measurements so it’s a good deal hairier

@TedShifrin Oh sorry, I found a counter example too, which means the monotonicity isn't sufficient

(Formally such observables are elements of a POVM, which generalizes the notion of quantum measurement beyond projective measurements. Interesting stuff but you only really see this stuff in quantum computing contexts)

Well the the example was wrong, eh

graph is edge disjoint union of trees and cycles.

what does this mean?

That you can write every graph as the union of subtrees and subcycles that have pairwise no common edge

(but they might share vertices)

@X4J What precisely are you trying to prove or disprove?

@AlessandroCodenotti how to prove this? I don't know 'blocks'.

I'm thinking of writing the graph as U of its connected components.

If you have no further requirements on the decomposition it is trivial by dividing the graph into its edges and isolated points. You don't even need cycles

Then I write every connected component as union of minimum spanning tree + its complement.

I conjecture now that minimum spanning tree U its complement should do the desired decomposition but I'm not sure.

@Ted, @Leslie, I think I finally have a solution

maybe you have no idea what I'm talking about lol

but I wanted to prove that there would be no "wiggle" in a certain parametrization

which for the sake of rigour I defined as the vector $\dot x/\vert x\vert x$ having a monotonic angle

(I believe I can apply a linear change of variable, and then argue in the case of perpendicular vectors, but it's ok if you have no idea what I'm talking about)

@TedShifrin I've already proved the following statement: if (a_n) is a sequence of reals such that \lim_{n \to \infty} a_n = 0 then the convergence of \sum k|a_{k}_a{k+1}| implies the absolute convergence of \sum a_k. My next question was if the inverse is also true, namely, is it true that for a sequence (a_n) s.t \sum a_n converges absolutely the sum \sum k|a_{k}_a{k+1}| is finite? I found a counter example for this

i think some of those _s are supposed to be -s

Yes, I meant to write \sum k|a_{k} - a{k+1}|

iPad floating keyboard is annoying

Thanks tho

$a_n=1/n\log^2n$ surely is a counterexample or close.

I wouldn’t expect a converse/inverse.

I wondered this since i know the absolute convergence of \sum a_n implies the convergence of \sum |a_{k+1} - a_{k}| but now I understand that the convergence of \sum k|a_{k+1} - a_{k}| is a significantly "stronger" property

Making even terms go to $0$ and odd terms all $0$ will give lots of countetexamples.

@X4J Well, you learned that where we started this whole discussion!

@TedShifrin Indeed, but I slowly understand that this is way much more stronger that I thought

I guess those counterexamples do not apply where we assume the monotonicity

It means terms need to decay $1/n$ faster.

I’m typing on my iPad too and now I see why Ted doesn’t like it

@ShaVuklia there are sometimes cool results in ODE/PDE that get at stuff like this from aspects of the form of equation alone (i.e. without any parametrization of any solution). although sometimes they are difficult to motivate in a first course without background theory.

Ted is generally irrational (probably not transcendental).

you have to be just right in how you formalize what 'wiggle' ought to mean.

What does wiggle mean to a functional analyst?

@TedShifrin My intuition is still vague in the sense that this is not true if we use log(n) instead of n. For example, taking (a_n) to be (\frac{1}{n}) the sum \sum log(k)|a_{k+1}-a_{k}| is finite but \sum a_k does not converge.

@leslietownes if you have a reference of thing being treated, I'd be curious!

I just finished writing out my proof

I'm not sure I am going to share it with my students

I do wonder if it's correct, so if anyone cares to proofread it, I'll post it here

@X4J Remember that log grows slower than any (fractional) power of $n$. What if you replace $n$ with $\sqrt n$?

@Sha Funny how much you learn when you teach (at least the first few times through).

o lol, yes especially with this course, because I hardly work with differential equations

That’s a course I TAd my first quarter in grad school and never taught during my career.

@TedShifrin It still provides a counter-example, so what is so "special" with n

Roughly, it’s because $1/n$ is (at least among $n^{-p}$ series) the precipice between convergence and non-convergence.

also, whenever I see/hear/think of the word "wiggle", I think of: youtube.com/watch?v=novfcrFVHPM

I use “wiggle” a lot talking differential topology and stability.

Transversality, etc.

epic

I always used $\epsilon$ fences teaching limits and uniform convergence. We all have our pet phrases.

lol, nice pun, if intended

Cat wiggles ….

precisely xDD

@TedShifrin which graphically can be described as the terms of 1\n becomes more and more "closer" in a greater rate than the rate it approaches to zero?

@TedShifrin I fixed it for those who use ChatJax

Huh? @robjohn

@X4J I have no idea what this means.

@TedShifrin There was a & instead of a $ and it was messing it up

Oh. Sorry. Damn iPad,

Yeah I guessed so, at least I tried