Mathematics - 2023-04-02

12:00 AM

cotton: if your fornalism seems designed to allow there to be multiple distinguishable edges between the same pair of vertices, then i think your conclusion is correct, but it is a formalism in which an edge is not (in general) determined by its vertices

Cotton Headed Ninnymuggins

I should ask then if we're dealing with simple graphs because as you say, "determined by its vertices" forces graphs to be simple. My instructor also forgets to say when we're dealing with simple graphs or not sometimes

leslie townes

oh. if you've expressly discussed the notion of 'simple graphs,' i'd think that the drawing of G' would place you in a context where graphs are not necessarily simple

Cotton Headed Ninnymuggins

Yes, $G'$ isn't simple

D.C. the III

@TedShifrin I see what you mean when I worked it out just now algebraically. But I was looking down on the projection of the pillow onto the $xy$ plane and it didn't "look" like $0 \leq \theta \leq \pi$...

Cotton Headed Ninnymuggins

I will ask my instructor if we're dealing with simple graphs because it affects things a lot for these proofs

Ted Shifrin

12:04 AM

Then your computer graphics and/or axes aren’t right.

EM4

Hello.

leslie townes

it definitely gets to be a weird family of questions in the non simple context because in generality you'd need to have some extra structure, e.g. a labeling of edges, to determine whether one thing was literally a subgraph of another in the sense you described above (vs. just isomorphic to one)

Cotton Headed Ninnymuggins

alright

Xander Henderson

@leslietownes Oh, indeed. I hadn't even thought about things being "literal" subgraphs. I was thinking something along the lines of "there is an injective graph homomorphism" or something---it has been 20 years since I took a course in graph theory, so I might be messing up the vocabulary.

D.C. the III

@TedShifrin $t = \theta$ and $u = \phi$

Does Wolfram offer a free 3-d graphing calculator?

leslie townes

12:09 AM

xander: i'm guessing the instructor in this case isn't even thinking at this level of formality (from the remark characterizing subgraphness as "obtainable by removing vertices/edges from [another graph]")

Ted Shifrin

Your equations seem messed up, plus bad 3D perspective.

leslie townes

a formulation that seems to require literal subgraphness, although it probably isn't intended

Xander Henderson

@leslietownes Yeah. Probably.

Ted Shifrin

I have to leave in a few.

D.C. the III

Ok....I guess this is why having the ability to verify things algebraically works.

Well I'm done multivariable for the day.....it's now the Stats block

Ted Shifrin

12:12 AM

Oh, the equations are fine.

D.C. the III

it's the graphing that is messed up?

Ted Shifrin

But you need to be able to tilt that picture … look where I put my axes in the book and/or board pictures.

It’s the lack of 3D perspective and your not seeing the $xy$-plane clearly, I think.

D.C. the III

agreed, but I was tilting it in this direction to glean an idea on how to find the linits of $\theta$. I'll play around with it some more. Thanks for the help though

Ted Shifrin

Ugh, did poorly on Wordle (4) but did Waffle optimally today.

D.C. the III

This is how it looks in terms of the $zy$ plane

leslie townes

12:17 AM

makes me nostalgic for hand drawn pictures.

Ted Shifrin

So you see $y\ge 0$, which gives the $\theta$ bounds.

This one would be nontrivial for me to draw without some practice.

you get to see a pillow billowing outward.

D.C. the III

but the $z$ - axis here is the blue one and so I immediately saw the bound for $\phi$ being between $0, \pi$.

Ted Shifrin

Yes, sure.

D.C. the III

$zx$ - axes apologies if this is spamming

I see the "pillowing" effect when I look down on the $xy$ plane or if we angle it

Ted Shifrin

i have the correct Mathematica picture at the end of the errata.

D.C. the III

12:24 AM

Yea I saw the picture.

I was just curious on how to see the limits on $\theta$ through the pictures.

may have to see how much a student version of mathematica is. Will probably be useful beyond this too

Ted Shifrin

ok, I’m out! Have a good night, all.

D.C. the III

Enjoy your evening. And eat a lot

Silly Goose

4:31 AM

Why does Rudin keep saying integer? He means natural number, right?

All throughout the sequences sections, he says integers when referring to the indices of a sequence, but surely this can't be right (as in he should be saying natural numbers), or am I crazy

leslie townes

what do you mean by "right"? if there is an integer that just happens to be a natural number such that __, "there is an integer such that __" is true.

as to why he isn't tuning these statements to be minimal, who knows. but at least in this example it wouldn't be logically necessary to do that, or incorrect not to do it

Silly Goose

i've just never seen anyone index a sequence by negative integers :P

i feel like this is analogous: I say "it is a type of weather outside" when i really mean "it is raining"

leslie townes

right, and i don't think his definition of "sequence" even contemplates that, but so what?

it could be he is indulging in his favorite pastime, which is saving ink

Silly Goose

xD

leslie townes

math and particularly analysis gets extremely difficult if you require every statement to be as precise as it would be conceivable to be. he's just starting the reader off with that, very early.

e.g. in the limit definition, for every epsilon there is some delta such that blah. the definitions don't care or ask for the largest possible delta that will do for a given epsilon, although in many simple examples it's very possible to do that, and in at least some applications you might even care about that.

so a whole lot of those symbols, particularly deltas and Ns, are generally not going to be tuned to any scale of the 'right size' of what you 'need' to show that a statement is true.

and there's always, if it holds for some delta or N, it's probably also going to hold for any smaller delta' or bigger N'. i see it more in this spirit. in the back of your mind, you know that these kind of conditions are mostly about what happens when something is "sufficiently large" or "sufficiently small" in a way you don't care too much about.

in textbook writing there's a principle, not just used by rudin, which is that within reason, definitions and hypotheses of theorems should be as broad as possible, so that it is as easy as possible to verify them when they are satisfied.

rudin's thing sorta fits into that vibe, too. but my guess is he was just saving ink.

Silly Goose

4:43 AM

hm I was wondering about that. because in class and so on it is "sufficiently small" that matters. so $\epsilon > 0$ is a little bit overkill, though perhaps in this case it is the most general to write it like this

i get to present Taylor's theorem for my analysis course :D quite excited to learn it

leslie townes

cool. that's something that rudin doesn't spend a lot of time on, i think? or have you use at all in his exercises?

user2236

5:02 AM

@leslietownes is that principle chosen for efficiency reasons?

leslie townes

yeah, i think the idea is, make it as easy to verify hypotheses as possible so you can more easily use stuff. minimize the amount of logical rearrangement or case reduction you need to do just to check that you are in a position to apply something.

there is an opposite principle that some people use, which is, state a definition or even a theorem in terms of key features of a key example so they present themselves perhaps most vividly to the reader, and then afterward, include commentary and/or a series of corollaries and more general results that draw out why similar reasoning as used with the motivating example can be used to push the result further.

that might be easier to read on a first approach to some material, but it is a little inconvenient to use a book structured that way as a reference.

i find it actually pretty annoying when a book does a special case, and then has a glib paragraph like "if you look at the above argument, all we really used was [blah], so we deduce the following corollary:[more general statement] [no proof]"

but rudin style writing for posterity can be annoying too.

D Left Adjoint to U

5:17 AM

@Koro, I didn't know you study hatcher / homology. Want to combine efforts into a Weibel book study group? Weibel covers all the questions about the algebra and has plenty of topological examples.

The questions are extremely difficult! We could each work together just one one at a time maybe.

user2236

5:29 AM

@leslietownes indeed, future internet savvy generations will not appreciate "ink" saving.

Silly Goose

Here, $s^* = \text{sup}E$, $x \in \mathbb{R}$. $\{s_n\}$ is some sequence of real numbers. Does $s_n \geq x$ for $n \geq N \in \mathbb{N}$ mean that some subsequence converges to $x$?

leslie townes

some subsequence of s_n? not necessarily, no. what is E?

is E the set of subsequential limits of {s_n}?

Silly Goose

yes with potentially $+ \infty$ and $- \infty$

leslie townes

ok. you will need something like, a bounded sequence has a convergent subsequence. of s_n \geq x for infinitely many n, either every subsequence goes off to infty (which probably means s* is infty and there's no real x larger than s*) or there is a bounded subsequence that is termwise \geq x.

and by the theorem, that subsequence will have a limit, which by a short argument from epsilons and N's, or maybe by rudin's theorem (i forget how/where this occurs), will be \geq x.

you could think of rudin's "y in E with y \geq x" as a limit of some subsequence of a subsequence that is termwise \geq x.

if he doesn't prove that bounded sequences of reals have convergent subsequences by that point in the book, that's maybe a pretty bad flaw in the exposition. so he must have done it sooner.

you need something like that to know that E will behave the way you think it should. some books define lim sup and lim inf differently, in ways that make the notation a little clearer (i.e., as limits of suprema and infima taken over a sequence of subsets of {s_n})

one potato two potato

6:14 AM

hmm... why?

leslie townes

i don't know any of the relevant definitions, but is 'the divisor of omega_1" a finite set of points? is "the set of differentials vanishing at [a given point]" a codimension one subspace of H^q(M)? if so, that would explain it, and if not, what's the closest thing to that that stands a chance of being true?

one potato two potato

6:30 AM

Yes I think one can view the divisor of omega_1 as a finite set of points. The part I don't understand is codimension one.

leslie townes

is "evaluation at [a given point]" a linear map from H^q(M) to C or whatever your scalars are? if so, its kernel is at least codimension one in the sense of linear algebra

i dunno about in any geometrical sense

one potato two potato

Ah kernel... I think that's the right one. Thanks

btw I answered my own question because of mysterious upvotes

robjohn

7:25 AM

About 30 minutes after we passed through the parking lot in front of our local Trader Joe's, there was a fatal shooting.

leslie townes

yikes! good to hear you are OK.

but the next time your counterparty is short on cash, see if you can work out something more peaceful, like a payment plan. helps you avoid having to file a claim in probate court.

robjohn

@leslietownes But I didn't like the way they looked at me.

leslie townes

OK. sometimes setting an example is more important than collecting on a debt.

robjohn

It is a bit jarring to realize this happened just 30 minutes after we drove right through that parking lot.

Node.JS

7:49 AM

Can two total orders be compatible with each other without them being equal?

leslie townes

yeah, that's spooky. i live by a main street where pedestrians get hit by cars fairly frequently, maybe one or two deaths a year. it's spooky to think about when just driving by normally. not the same as gun violence, exactly, but still.

node: what does it mean for two [total] orders to be compatible with one another?

Node.JS

For all $x,y \in A$ , if $x<y \in (A, <)$, then exists $x<y \in (A', <)$

I think this is a common definition of compatibility

Summerday

Can someone help me here?

0

Q: How can I justify this step in bounding conditional expectation of Brownian motion?

Let $B$ be a standard Brownian motion. I have showed that $$\sup_{t\in [0,1]}|B_t|\leq \sum_{n=1}^\infty \sup_{0\leq k\leq 2^n-1}\big|B_{\frac{k+1}{2^n}}-B_{\frac{k}{2^n}}\big|$$ now I want to deduce from this that for $p\geq 1$ $$\Bbb{E}\left[\sup_{t\in [0,1]}|B_t|^p\right]^{1/p}\leq \sum_{n=1}...

Silly Goose

@leslietownes ah thank you he does prove the theorem that bounded sequences in the reals have a convergent subsequence it earlier. but this explanation helps !

leslie townes

8:09 AM

silly: rudin definitely does not do this, but it might help to consider what would happen with rudin's definitions if applied to rational numbers, i.e. the sequence s_n is rational, and E now defined to be the set of rational subsequential limits. this set could be empty, even for a bounded sequence, or if nonempty, could still break the logic of that part of the argument.

e.g. if s_n is 0 for n odd and an increasing sequence of rationals that converges to sqrt(2) for n even, the set E will be {0}, s* as defined there will be 0, and there will be a number x > 0 and infinitely many n for which s_n >= x, but there will be no corresponding y in E that is >= x.

Silly Goose

oh by the theorem i meant 3.6.b here!

oh wait

leslie townes

yeah, hooray for rudin. all i'm noticing is that he puts it before 3.17, which makes sense - as you need it, or something like it, to handle 3.17 as glibly as he does in the text you quoted above.

without it, i'm not sure how you justify the "there is a y in E such that" in his argument for part (b) of 3.17

Silly Goose

8:36 AM

i got it i think :D thanks so much

user2236

9:01 AM

@robjohn that's a bizarre twist of fate, considering there was a high speed car chase involved which could've went in any direction and lasted for who knows how long...

Glad you're safe pal.

one potato two potato

Lots of facts (especially advanced) or literature about Riemann surface are written in terms of Algebraic geometry...

I guess it's useful to describe using AG, especially if one has to study several variable holomorphic functions. So viewing Riemann surface as a special case of higher complex manifolds

schn

11:20 AM

Suppose we have the following series $$\sum_{k=1}^\infty \frac{k}{\sqrt{k^4+17}}.$$ Now, I'd like to determine its convergence/divergence by comparing it with another series. My first step would be to bound the terms, i.e. $$ \frac{k}{\sqrt{k^4+17}}\geq \frac{k}{\sqrt{k^4+k^4}},$$ but is this a valid inequality? After all, $k\geq 1$ and so this inequality would only hold if I were to compare the series starting from $k=3$.

robjohn

11:41 AM

@schn It is true for $k\ge3$, but it is the tail of the series that determines convergence/divergence.

$\sum\limits_{k=3}^\infty\frac1k$ diverges.

schn

2:18 PM

@robjohn okay, thank you very much.

Mats Granvik

2:37 PM

In the E8 lattice, shouldn't the Euclidean Distance between all basis column vectors be equal to each other?

Or put differently. Is it possible, since the basis matrix is not unique, to form a basis matrix for the E8 lattice in which all column basis vectors have the same Euclidean distance to each other?

Or am I showing my confusion about what basis is in respect to lattices?

ILikeMathematics

Is there any way to avoid having to write \left( and \right) all the time in LaTeX to make the brackets fit what is inside of them?

schn

3:02 PM

In this answer, why $n\geq 3$? Clearly the inequality also holds for $n=2$ and $n=1$.

Franklin

4:19 PM

Can anyone please help me with this : math.stackexchange.com/questions/4671380/… ?

mick

4:54 PM

0

Q: $f(n) = \frac{n^2 + n + 4}{2}, g(f(n)) = f(g(n))$ such that $g(n)$ is an integer.

Let $n$ be a strict positive integer. Lets define an integer sequence $f(n)$ : $$f(n) = \frac{n^2 + n + 4}{2}$$ so $$f(1) = 3$$ $$f := {3,5,8,12,17,23,30,38,47,...}$$ $$f(17) = 155$$ etc Notice $$3+2=5$$ $$5+3=8$$ $$8+4=12$$ $$12+5=17$$ etc and $f(R)$ has no real fixpoint for real $R$. Now we w...

commutative-algebra function-and-relation-composition functional-inequalities nonlinear-dynamics integer-sequences

for fans of composition and integer sequences

How do we know if a question is homework or an exercise from a book ? Because one is ok and the other is not ??

robjohn

5:34 PM

@schn Of course, it's true for $n\ge1$, but it's also true for $n\ge3$. My guess is that they looked at $\frac{n!}{n^n}=\frac1n\frac2n\underbrace{\frac3n\cdots\frac nn}_\text{each term is $\le1$}$

I would have then said that $n\ge2$, but they might have been uncomfortable with the empty product.

schn

@robjohn thank you for confirming!

leslie townes

7:30 PM

@schn you didn't ask (in fact you explicitly wanted to use comparison), but for this type of series, the extra care that is needed to come up with inequalities is sometimes a reason to prefer the limit comparison test (where with a quotient of polynomials, you can just limit compare with the quotient of leading terms, irrespective of the amount of nonzero coefficients of other terms, their sizes/signs, etc.).

schn

8:43 PM

@leslietownes thanks for the reply. I have another series that I am working on. Maybe we can do limit comparison test on it? $$\sum_{k=1}^\infty \frac{(2k)!}{k^k}$$

Xander Henderson

@schn Are you familiar with Stirling's formula?

leslie townes

schn: looks ratio-testy to me. without the full force of stirling it can sometimes be helpful to note/remember that n! is between (n/3)^n and (n/2)^n for large n.

schn

@XanderHenderson a little bit, but please, elaborate

leslie townes

of course it's asymptotically some mess with (n/e)^n in it and a sqrt( )thing and other stuff you can refine in various asymptotic ways.

but you rarely need all of that.

Xander Henderson

@schn My first intuition is to hit that thing over the head with Stirling, then apply the root test. My guess is that the ratio test is going to be inconclusive (but I could be wrong).

Or, rather, I want to apply the root test, and Stirling gives me a way in.

schn

8:50 PM

I just did $\sum_{k=1}^\infty \frac{k!}{k^k}$. They are not related in any way, or?

Xander Henderson

I mean, $2k$ is not the same as $k$.

schn

Indeed :)

leslie townes

mm, maybe for unitarians.

Xander Henderson

@leslietownes Like of the universalist sort?

leslie townes

i dunno the particulars. feels like there oughta be some philosophical or religious tradition in which 2k and k are one

2k-itarians

Xander Henderson

8:59 PM

@leslietownes Maybe they life in a world of characteristic 2?

leslie townes

wow, i think so, yes. unitarians deny the trinity, or assert that the three trinitarian beings are just one being. so they maintain that 3k = k, or that 2k = 0

at least for spiritual beings k

Xander Henderson

Heh.

Ted Shifrin

9:25 PM

Here’s to atheistic mathematics.

robjohn

9:45 PM

@schn Try this one: $$\sum_{k=1}^\infty\frac{(2k-1)!!}{k^k}$$

Jakobian

@leslietownes they work in characteristic 2

Xander Henderson

@robjohn By $k!!$, you mean $k(k-2)(k-4)...$, yes?

robjohn

yes

Node.JS

I posted these two questions, I appreciate your feedback / help: math.stackexchange.com/questions/4671524/… and math.stackexchange.com/questions/4671533/…

I am just wondering if I am on the right track

mick

10:32 PM

added my first picture

0

Q: Is there a simple fixed point iteration for the Delian constant (cube root of $2$) in base $10$ or $16$?

Im looking for a simple iteration that computes the cube root of $2$ in base $10$(decimal) or base $16$(hexadecimal). Probably the simplest way is iterating a small polynomial with simple rational coefficients of the form $\frac{n}{10}$ or $\frac{n}{16}$. Newton and Householder methods do not pro...

polynomials numerical-methods dynamical-systems radicals

Amit

10:57 PM

Is there a theorem or a known property of a rotation matrix $R(\theta)$ that $$\left( \frac{d}{d\theta}R(\theta)\right)R(-\theta)$$
Gives something "nice"? I see I'm always getting an off diagonal matrix, for example $$\begin{pmatrix}0 & -1\\\ 1 & 0\end{pmatrix}$$ for two dimensional rotations... but I can't see why it has this property

Ted Shifrin

Yes. The Lie algebra of the orthogonal group is the skew-symmetric matrices.

True in far more generality.

Amit

@TedShifrin But it's not sufficient to know that $R^{-1} = R^{T}$ to show that right?

Ted Shifrin

Yes, it is. Differentiate $RR^\top =I$.

Amit

I went in that direction... so $$R'R^{T} = -RR^{T}{'}$$
Also note I want to be able to show not only skew symmetry but this particular form of the matrix I've mentioned...

Ted Shifrin

For $2\times 2$ that’s all there is, up to multiples. I don’t know what you mean.

Amit

11:12 PM

Ahhh I see @TedShifrin yes, I also realize now that the derivative commutes with the transpose so it's definitely a skew symmetry relation there

Right.

Thank you!

Sure.

11:50 PM

@geocalc33 want to study some math shits together?

geocalc33

take a look at this question @DLeftAdjointtoU

2

Q: function "converges" to prime counting function

Consider $f(x)=\ln(\pi(e^x))$ (blue) and $g(x)=\ln(li(e^x))$ (green). $\pi(x)$ is the prime counting function and $li(x)$ the logarithmic integral. I plotted these on SageCell and what surprised me is that the differences of $f$ and $g$ seem to tend to zero (at least in this interval) (red): Wha...

elementary-number-theory prime-numbers asymptotics graphing-functions approximation

D Left Adjoint to U

Show me, son

geocalc33

@DLeftAdjointtoU yes i'm down to study tomorrow

D Left Adjoint to U

Notice your post score :>

geocalc33

I don't quite understand why the differences are tending to zero so fast

D Left Adjoint to U

11:52 PM

I don't know enough AnNT to comment though

geocalc33

this is in stark contrast to li(x) and pi(x) where the differences increase!

of course there's skewe's number and li and pi change sign ifnintely often

D Left Adjoint to U

Oh, they'll increase at first I think, I think the appriximating function is sporadic at first then for large enough inputs starts to converge (in the case of $\pi$ I thought)

@shintuku

Do you study HoAlg

geocalc33

no that's wrong @DLeftAdjointtoU

shintuku

no

D Left Adjoint to U

@shintuku what do you study these days?

shintuku

11:56 PM

linear algebra and real analysis

D Left Adjoint to U

ha ha

:D

They're forcing you to, right?

shintuku

no

i'm an econ major

D Left Adjoint to U

Oh dang

Good going then

Well, do you know what the precursor to HomAlg is?

Ted Shifrin

Yup, required for graduate school in econ.

D Left Adjoint to U

It's linear algebra

It's also abstract algebra, but you know a little bit of both. If you know how to quotient abelian groups, I dare to say you're fully prepared for Weibel.

Some category theory would be nice, but the concepts needed are simple: product e.g.

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$Mathematics$ Mathematics