all corvids can behave in ways that we think of as mean. they are aggressive.

all corvids can behave in ways that we think of as mean. they are aggressive.

i miss living around blue jays. we have scrub jays here but it's not the same.

@TedShifrin And QUIET :) But Q is far more rare than any of the letters in HOUSE, NOISE, POISE, etc.

The chorp of the scrub jay is pretty cool, though.

it's a happy sound.

@leslie, you can go to Toronto, to watch a bunch of blue jays play baseball!

i like toronto. i visited it once for a math conference, and spent so much time math conferencing that i didn't get to see the city.

i got to know the half block next to the fields institute and the half block next to my hotel pretty well, that was it. seemed nice, though.

i recognized a lot of the downtown streets from american movies, where it is usually a stand-in for a large american city.

@leslie did you get the connection to the Toronto Blue Jays, or are you making the most of my reference to Toronto? They're lower in latitude than parts of other states!

both. i got the connection to the blue jays and made the most of your reference to toronto.

@leslietownes Love it!! :D

the unlicensed cab that took me from the airport to downtown was significantly cheaper than the licensed cab. the driver was from a country that has been in the news lately, ukraine.

one time in chicago i got an uber from my hotel to the airport and my driver was very pro-putin, to the point whether i wondered if i was being recorded or assessed for some kind of intelligence operation. i just nodded, mostly.

@leslietownes smart!

i find it's best to get cab drivers and equivalents to talk about restaurants. politics is its own kettle of fish, especially if they bring it up in the first instance.

If you miss seeing cardinals, St. Louis, in spring and summer, is the place to visit.

i do miss seeing cardinals. a friend of mine owns a bar in st louis. i do plan on going there.

I just went through CHUTE, LUTTE, JUSTE to get one of my worst results (5) in French.

i'm going to play french wordle. cross your fingers.

OK, that didn't work out so well because i played one of your words.

i'll try again tomorrow.

100% "Victoires", i feel so sophisticated!

i feel like emily in paris right now.

Your French is probably as appalling as hers!

@leslietownes We still live in a nation in which men can take risks that a woman can't afford to take... But there are women elsewhere, who have little recourse.

ted i think it might be better. i did study it briefly for the french language exam (which i passed). and i have at least one friend who is actually french. i don't think emily has either of those. (spoiler alert)

amwhy that is a good point. although one time for my work i had to do some research into the local licensing regimes for taxis (these depend on extremely fine geographical distinctions, including city limits) and came to the unfortunate conclusion that some 'licensed' cab companies that could legally pick up at the airport were no better than unlicensed alternatives in terms of driver vetting.

it was a funny case. company A was suing company B, saying "you said you're safer than we are, but you don't require FBI background checks." it turns out that nobody was required to do FBI background checks, and one of the owners of company A had DUIs. i suppose his defense would have been that he didn't drive for his company.

there could be a really good book written about local taxi regulation. some cities/regions have it fairly well regulated, and in other places it's the wild west.

If a is in $Q(\sqrt 2, 2^{1/3}, 2^{1/4},…)$ then can it be said that $a\in Q(\sqrt 2,2^{1/3},…,2^{1/n})$ for some positive integer n?

With that, I want to show that Q(…) is algebraic.

@Koro

@love_sodam thanks :).

So $a$ can't be directly assumed to be in $Q(\sqrt 2,2^{1/3},…,2^{1/n})$.

I'm not sure about that.

I never considered such infinite extension

But in infinite vector space, each vector is a linear combination of finitely many basis vectors so maybe $a\in Q(\sqrt{2},...,2^{1/n})$.

I'm not sure either. I think that in $Q(\sqrt 2,2^{1/3},…,2^{1/n},...)$, there may be a number (not sure) which is something like $1+1/{\sqrt 2}+2^{-1/3}+...$, which may converge to a number x transcendental over $Q(\sqrt 2,2^{1/3},…,2^{1/n})$ for every n.

But this is not required here as the post you shared shows. We simply need the fact that $Q(2^{1/n})$ is a Q-vector space of dimension n.

I don't know if it's safe to talk about convergence.

Anyway I believe a is contained in $Q(\sqrt 2,2^{1/3},…,2^{1/n})$ as it's a vector in Q(...)

I think infinite sum is banned.

it's safe to talk about because $Q(\sqrt 2,2^{1/3},…,2^{1/n},...)$ is a field too.

and the infinite sum makes sense.

if the convergence holds.

but none of this is required as the post shows :).

@love_sodam: which text do you follow for fields?

Dummit foote basically

and some other books/lecture notes ...

For convergence, you need to give some topology and I think your convergence is based on subspace topology. I don't know if this is allowed.

Usually for something infinite, it's formal in some sense. they don't care about convergence as far as I know

(somehow the text between ** ... ** didn't turn bold in my last message)

There is no notion of convergence.

But then what does $Q(\sqrt 2,2^{1/3},…,2^{1/n},...)$ mean?

there is no notion of convergence then there is no notion of infinite sums also.

never mind, I got confused.

@love_sodam I think you're right. I was considering subspace topology of euclidean R, which is not given here.

I think I have to spend my entire day calculating inverse matrices.

If $f:S\to\Bbb C$ is analytic where $S = \{z\in\Bbb C\mid -\pi/4<Arg(z)<\pi/4\}$ such that $|f(z)|\leq 1$ on $\partial S$ then can I directly conclude $|f(z)|\leq 1$ on $S$ by MMP?

Done 4-5 hours of calculating inverse matrices. I feel depressed.

Is there not a calculator for those kind of things

gosh, i sure hope so.

if there isn't one, you could code one in 4-5 hours. you might have to lean on pre-written libraries if the matrix entries come from somewhere weird.

@robjohn Not just that, but it's also cheaper to compute than formula 11 from the paper. Now my only issue is whether or not there is an explicit formula of the recursive relation you gave that involves only factors of $z$ such as multiplying by $z^n$ in order to extend from $|z| < 2$ to $\Bbb{C}$.

For obvious reasons, relying on recursive processes or algorithms defined using finite sums is undesirable since the number of terms is proportional to the domain. Even if sums have a minimum upper bound of $O(\log n)$, that still isn't great.

Hence why it would be better if, for example, a partial sum formula is available.

I'm still curious to know what the partial sum formula is for the sum in Gauss's Digamma Theorem, but I don't know enough about sums and products to work with such things. I can only do simple, obvious things such as factoring. I have no techniques besides elementary algebraic manipulation. Very limiting :(

How would I compute the partial sum of $$2\cdot\sum_{k=1}^{\infty} \cos(\frac{2\pi p k}{q})\ln[\sin(\frac{\pi k}{q})]$$ to get something more computationally suitable?

daughter began the day by picking up the cat again.

I don't know why, but I always find it amusing to see a cat in this position.

that's the one.

livvy sometimes bends her legs, so the legs stick out more horizontally, which is also a funny look.

hehe

also sometimes her front paws are like parallel to the ground, like an old style zombie.

we have a bed that resembles a black cat's head, where our black cat can go inside and sit in its mouth. i don't know what that corresponds to.

$f(x) = (f\circ f)(x)$

How I imagine convergence and divergence:

I have this question that I don't have an idea how to start aside from casework: If an electric circuit has five resistors whose resistances are 12 ohms, 10 ohms, 9 ohms, 9 ohms, and 11 ohms, what is the average of all possible total resistances?

Any ideas how to start?

hum. average over all conceivable configurations of those resistors? does a configuration need to include all five?

Not a homework, just a question I made as part of our requirements. I could make another question, but this one is really interesting I need to submit it.

i guess my second question is answered by context.

saying 'has' five resistors rather than uses or employs or something suggests that indeed, all five need to be in there.

@leslietownes Yes, although it seems as if the problem is quite incomplete when it comes to details

i'm trying to think of a good model for the problem, maybe labeled trees with resistors as nodes.

or even symbolic strings of some kind, evaluated at those five values.

hrm.

What makes it difficult is that nested circuits (if nested is a correct term) makes the computation harder. Something like R1 and R2 are parallel, R3 and R4 are parallel, and R2 is in series with R3 and R4 (the parallel circuit made by R3 and R4)

Also, R5 is in series with R1

casework is very time-consuming in this problem.

i think i've seen some version of this in some kind of programming contest.

I think I'll change the resistances of the resistors to be the same, and is equal to 10 ohms. That way, computing will be easier.

Combinations work in this case, unlike before which focuses on permutations

When I think of it, it seems as if the question is asking what's in between a harmonic mean and an arithmetic mean

It's the geometric mean. Maybe the answer is the geometric mean of the resistances?

soupless: did you try making cases like this: 1) cases when anode and cathode each are connected to one resistor each, 2) cases when cathode is connected to 2 resistors, and anode to 1 etc.?

This should easily cover all the cases.

@Koro Anodes and cathodes? I think that's not involved for us

if you're finding equivalent, that's always there I think.

What we just have are basic components of a circuit, like batteries, resistors, switches, and such but not anodes and cathodes

good morning, @Ted

Changing cathodes and anodes may change calculation of equivalent resistance.

Good morning, @robjohn.

@Koro Can you provide a brief explanation, please?

@soupless you have a battery :) consider positive and negative terminals of the battery. and forget that cathode/anode words were used.

I think the issue is to consider all possible series/parallel arrangements. Yuck.

The initial cases are the cases without nesting, and there are five configurations. The nesting will add complications.

@soupless Consider square ABCD (each side to be treated as a resistor). Connect the battery to AB and find the equivalent resistance in this arrangement.

Then, connect the battery to AC and then find the equivalent resistance of this arrangement.

(assumption: the wires connecting battery to A/B/C/D have 0 resistance).

Is the equivalent resistance the same in both the cases?

$\sin (1^o)$ is algebraic over rationals. :)

@Koro Does each side have the same resistance?

@Koro it's an algebraic integer

or perhaps it's half of one

It really seems as if the geometric mean is the answer since the smallest possible total resistance is obtained if all resistors are parallel to each other which is the harmonic mean divided by $n$, and the largest possible resistance is if all resistors are in series and the total resistance is equal to the arithmetic divided by $n$

if all the sides are the same resistance, connecting to two adjacent corners gives a resistance of $\frac34$ of the resistance of a single side. Connecting to opposite corners gives the same resistance as a single side.

@robjohn I had gone for dinner. Yes, that's what I meant. :)

@robjohn indeed :). I didn't know this term before. Thanks :).

Are $\mathbb R^{\omega}$ and $\mathbb R[[x]]$ basically the same thing? I've come across the former in the context of point-set topology and the latter in the context of ring theory, but it's just crossed my mind that they might be isomorphic.

I guess that might explain why the Wikipedia page on formal power series rings has a section discussing topology.

Isomorphic in what sense? A formal power series is given by its (infinite sequence of) coefficients, hence an element of $\Bbb R^\omega$. But you're right that one is a topological space and the other is a ring/algebra.

"Isomorphic in what sense?" Well, I'm probably not knowledgeable enough to try to answer that, but I guess I was just thinking that they're both sequences, so they might be the same in some sense (and AFAIK "isomorphic" is the word math people use when they want to say two things are the same).

To be clear I'm more or less entirely self-taught in math so I might be missing something foundational here.

I have to go in a few minutes but I'll come back if someone replies (or if I think of something else to ask or say)

As I said earlier, @Novice, the two sets are in an obvious bijection. But the word "isomorphic" is inappropriate.

@Koro $e^{i\pi/180}$ and $e^{-i\pi/180}$ are roots of $x^{180}+1=0$, so they are both algebraic integers. Their sum is also an algebraic integer. This requires a bit of proof. Also shown there is that a product of algebraic integers is an algebraic integer. Thus, $2\sin(1^{\large\circ})=-i\left(e^{i\pi/180}-e^{-i\pi/180}\right)$ is an algebraic integer

@soupless Actually, nested circuits are still very simple. In general, you can have circuits that are not series-parallel DAGs, such as:

You might put your 5 resistors at R0,R1,R2,R3,R6.

There is a standard way to solve this kind of circuit, at least in the finite case, but it's much more complicated than for series-parallel DAGs. You might want to think about it.

@user21820 With reference to the topic containing your comment here, I am currently working to get something my father wrote into publishable form. It was written as part of his PhD research proposal, back in the late 70s. However, I think that it remains relevant today (I'm working with one of my cousins on it---he actually has a doctorate in a relevant field).

It is never too late to publish. :D

@robjohn algebraic integers terminology was new to me :).

(There is also a paper on the Mecham impeachment which my father wrote in the 90s, but never published out of deference to House leadership at the time---I'm trying to convince my brother to get that submitted to a law review somewhere...)

@XanderHenderson Wow that's a long long ago, ... =)

Indeed.

Thinking about it more, I guess part of the problem is that in a topological space there is no inherent meaning to the addition (or scalar multiplication) of points.

@XanderHenderson I don't know whether it's polite to ask what your father thinks about this.

@user21820 He doesn't have an opinion, being dead and all.

@XanderHenderson I see, that's what I suspected, hence my hesitation. Good to know that his knowledge isn't going to be lost!

Indeed.

Alright, I'm off now, so see you around!

Time to teach.

Just nice! =)

I am stuck on this problem. Bring example of infinite group that all elements of that group have finite order and all subgroups are cyclic

I was thinking about set of n roots of $1$ but certainly it's not infinite group

What do you know about $\mathbb Z/p\mathbb Z\times \mathbb Z/q\mathbb Z$?

@unit1991 What if you don't bound the order by $n$?

How to approach if we don't bound?

what do you mean?

Take just union of all roots of unity?

and generator will be $cos(\frac{2\pi}{n} + isin(\frac{2\pi}{n}))$

Given a (smooth, connected, paracompact) manifold M^n, is it possible to embed it as a proper subset into a (smooth, connected, paracompact) manifold N^n?

The embedding need not be a proper map, but I want the image to be a proper subset of N.

M^n should also be noncompact.

There's a cool generalization of Banach fixed point theorem by Nadler

ngl, the fact that Mathematica has the ability to specify the chart for a Laplacian is handy

makes it easy for me to be lazy and not look up the Laplacian in spherical coordinates :P

Let $BC(X)$ be the space of bounded closed sets of a complete space $X$ with Hausdorff distance as metric. A set valued function $F:X\to CB(X)$ which is $\alpha$-contractive has a fixed point.

The usual Banach fixed point theorem is an easy corollary

(Public Service Announcement:)

As you were!

Let $a,n\in\mathbb N, n>0$. For a proof I need that $\operatorname{lcm}(a+n,n)\cdot a/(a+n)$ is a multiple of $n$. Examples seem to confirm this. Does anyone know why (if) this is true?

Yes, it's $\dfrac{an}{\gcd(a,n)}$.

lcm(a+n,n) = (a+n)(n)/gcd(a+n,n) and gcd(a+n,n) is a divisor of a+n.

or what ted said.

oh! oeps, thx, lemme see

ted's always there when we need him.

oh man xD

Well, we have to point out that $\gcd(a+n,n) = \gcd(a,n)$. That's the principle of the Euclidean algorithm :P

I had completely forgotten about lcm(a,b)gcd(a,b)=ab

And then we both hit you over the head with it.

x''D

no shame in it. sometimes some really great stuff leans on elementary number theory.

I only know this stuff because I made my abstract algebra students do it numerous times.

Working with the different definitions of gcd, for example.

@leslietownes He is xD

@leslie Munchkin can celebrate that I am about to have my fourth tooth extraction and implant in the last few years. Sigh.

ted: oh god, i've had some of those. i hate them.

oof

well, hate is a strong word. implants are better than bridges and they last a lot longer than they used to. but i've needed bone grafts and gum stuff on top of all that, which is no fun. it's also expensive.

i thought about getting in a fight for the last one. as i understood it, my medical insurance would cover reconstructive surgery if it took place in a hospital after a fight, but neither it nor my dental insurance would cover if i elected to do it at a scheduled time.

Yup, I've needed bone grafts as well. And it is expensive.

My dental insurance does cover it, but I only have \$1500 or \$2000 of coverage for the year, so it disappears in a hurry. Plus, of course, the crown after the implant.

that's maybe bone stuff for one site? if that?

i'll need more soon, the bone has gone away from under my implants. what a scam. i thought lawyers were bad.

my last doctor let me watch the operation, which was pretty cool.

Oh, I haven't watched. Oy, the bone goes away and you have to redo everything for a mere $6000.

@ShaVuklia There's some nice gcd stuff with the Fibonacci sequence. See en.wikipedia.org/wiki/Fibonacci_number#Primes_and_divisibility Eg, "every kth number of the sequence is a multiple of Fk".

ted the first time i did these i was making $40,000 a year and it was almost my entire after-tax income, start to finish.

Oh, you're definitely exaggerating.

or i was ripped off. could be both.

I'm paying 2020 San Diego prices.

mine was, bone grafts, gum grafts, followed by implant implants, and then the crown.

at two sites.

Oh, two sites. And I haven't had gum grafts. OK, you win.

when i was 11 i had twelve teeth removed from my mouth on the same day. that's where it all started.

Oy vey.

AND they confirmed justice thomas during my operation.

to make things topical.

Which was the more barbaric?

in retrospect, probably the latter.

i asked my mom "did it happen?" and she said "don't ask."

justice thomas is the only supreme court justice i have seen in person.

I can no longer say that he's the most immoral one.

with gum grafts they can do some kind of futuristic polymer that your skin grows into, but the bone grafts are still "this comes from cadavers," or at least they were the last time i got them.

they only get worse.

they called it the 'bone bank,' which was euphemistic. i asked my periodontist, who makes deposits at the bone bank?

I was smart enough not to ask these questions.

i like it, it's pretty cool. it's like being part zombie.

i'm an organ donor on my driver's license, but they should put a star next to it or something so they know that some of the materials are not first-use.

Did munchkin take Olivia up to the roof and drop her for fun today?

@TedShifrin who's surpassed him

At least two of Tromp's appointees.

i judge him and Alito as pretty much equivalent

she did pick her up as her first act as an awake person. i was in there getting her up, livvy came in and meowed, i said hi to the cat, and munchkin said "i can get you for her, dad" and picked her up and dumped her in my lap

@leslie I figure it's only a matter of weeks until there's a cat-apult in play.

kavanaugh and barrett are pretty embarrassing but you make a good point, i think alito is potentially worse.

he'd be at home during the spanish inquisition. except worse.

as bad as Kavanaugh and Barrett are as representations of politics

you can actually find decisions where they've come down on the sensible side. Alito/Thomas...

thomas is actually fairly sharp on IP law. we cite his decisions all the time. the odd one or two is incoherent, but mostly they are great.

better than RBG, i'm sorry to say.

hm, fair enough

What is IP?

intellectual property.

intellectual property

Ah.

mostly interpretation of the patent act

also copyright

I can see that that was far from RBG's main interests.

RBG was absolutely horrible for copyright. she wasn't convinced that the public domain exists.

fair use etc

alito does something that many judges do, which is, in any appeal from a criminal case, they include as lurid as possible a description of what the criminal supposedly did, before denying them what they were appealing. i think catholicism has something to do with it. it's super gross and unseemly.

viewing himself as an arbiter of decency rather than of the law

it's just perverted. the implied message is that the constitution stops working if you hate someone enough.

right

in the same way that an object floating in water doesn't mean lacks weight; it just means the bouyancy from the water is stronger than the weight

he would love to work in a system where he could throw people in a bottomless pit at the end. he has sadistic tendencies.

ted: she did try to throw the cat down the stairs once. this may be an early test run of the cat-apult.

it wasn't much of a throw. livvy can do a flight of stairs in one second.

@leslietownes Indeed.

@leslietownes Maybe you're just projecting.

@Semiclassical Can you list the decisions of the latter you deem "insensible"?