What does it mean for a positive integer n to be powerful but not a perfect power?

@Rick take 36 for example.

36 = 2*2*3*3

Clearly, it is "powerful", but not equal to x^n for some integers x,n.

@Rick 72 is powerful but not a perfect power. 72 = 2 * 2 * 2 * 3 * 3, so p divides 72 implies p^2 divides 72. But it's not a power of another integer.

@anakhro Wait 6^2 = 36.

Oh whoops

Bad example!

Good point :P

I think 72 is the smallest example.

Hmm, let me re think this

Since I am spouting nonsense

Yeah, 72 is the smallest, you are right.

108 is the next?

I think so

Okay, so then I guess you just can phrase it in terms of the prime decomposition $n = p_1^{e_1}\dotsc p_k^{e_k}$ as $\gcd(e_1,\dotsc,e_k) = 1$?

And then because it is powerful, $e_i \geqq 2$ for all $i$.

Yeah, I think that works, @Rick

If both of these conditions are satisfied, yes.

Do these have a name?

Apparently!

"Achilles numbers"!

HEH

"powerful but imperfect", alike to Achilles.

String diagram for a pie recipe

HOW MANY EGGS

(Strings are objects and boxes are actions, which is kinda the reverse of my instincts)

@anakhro Four

The nice thing is, if you have a string diagram for a subroutine (like "fill crust"), you can fill it in

as long as it has the right number of inputs and outputs

Exercise of the reader: draw one of these for Euclid's gcd algorithm

@AkivaWeinberger how do you do if statements

@anakhro I'm not actually sure

I asked the guy I got the picture from. I didn't get a response yet

I don't think you really can.

While maintaining the spirit of the diagram

You'd have to output a boolean string :(

Aw

So maybe this is best for big-picture structure

Did you try your euclidean algorithm?

I couldn't come up with a way to escape an if statement

I thought about it but I came across the same problem that you did

(Σn)^2=(Σn^3)

Basically 2 strings, $a>b$, which go into the first box and do division to output $b,r$ such that $a = bq + r$ and $r<b$, then you have to check for $r=0$ which returns $b$ if we are done, otherwise inputs $r,q$ into the division box..

Yeah I asked the question without thinking it through

I don't understand how do we get $x|\bar g(x)$ and $x|\bar h(x)$

@AkivaWeinberger Sorry, I wasn't really reading/following what you wrote

@Ultradark Look up Gauss-Bonnet formula for domains with corners.

If you assume each stage of the sieve of Eratosthenes kills roughly $1/p_k$ of the remaining numbers then this would imply the Goldbach conjecture @BalarkaSen

I dunno exactly what bounds we would need on that to make it still work

I think, that, in my above post, since $x$ prime, hence $x|\bar g(x)$ or $x|\bar h(x)$. But how do we get and there?

or how we would prove those bounds

If you assume the sides of the triangle are geodesic, then the total interior angle of the triangle is related to the curvature of the surface.

but if we're sufficiently nonrigorous and fast-and-loose then Goldbach falls out immediately

@Akiva k

@Silent $\deg g, \deg h < n$

i am not really thinking about it but i believe ya

There's someone on Reddit who claims he has a proof of the Goldbach conjecture and is saying "Why won't anyone look at my proof :("

link pls

Ah nah wait, I'm thinking of another guy

hahaha

i wont look at his proof because i dont want to

and the answer is "No one can tell what you're saying 'cause you write in broken English and because you don't know what anything's called"

Hi @ÍgjøgnumMeg, a @Balarka, DogAteMy ....

"or how to communicate math"

Hi @Ted!

Hey @Ted

but also really "You don't have a proof, you have some heuristics and numerical experiments and that's not enough"

@ÍgjøgnumMeg So, this implies \bar g, \bar h can't be constant. I think

There was a guy at my university who was convinced he had proven the Collatz Conjecture even tho several lecturers had told him otherwise, and he sent his paper (written on Microsoft Word) to some journal citing the names of various lecturers at the university

without their permission

Oh boy.

Here is one part of the Peter-Weyl theorem: Let $\rho$ be a unitary representation of a compact $G$ on a complex Hilbert space $H$. Then $H$ splits into an orthogonal direct sum of irreducible finite-dimensional unitary representations of $G$.

lol

@Silent I was actually thinking that if you remove all of the factors of $x$ from one of $ \bar g$ or $\bar h$ then there will be some left over

What exactly does it mean for $\rho$ to split into finite dimensional unitary representations? Does it mean that $\rho = \oplus_{i \in I} \rho_i$, where $\rho_i$ is a finite dimensional unitary representation?

@Silent If $x\not\mid h$ then $\deg\bar h=0$ which means $\deg\bar g=n$ which means $\deg g=n$ which means $\deg h=0$

Yes @user193319.

hey i have a question

I guess I didn't remember that you decompose into finite-dimensional guys.

given a function f: A -> B and a function g: C -> D, with A,B,C,D arbitrary sets

Thanks @TedShifrin

what can we say about the product fg

fg: ... -> ...

It makes totally no sense, @S.Crim

or can we not say anything about that

what do u mean

The domains have to be the same and the ranges have to be the same and an object where you can multiply.

hi @Eric

hmm

alright let me change the question

hlo hlo @Ted

given a function f in L2(A) and a function g in L2(B), what can we say about fg

@AkivaWeinberger why $x\not\mid h$ then $\deg \bar h=0$? Why can't $\bar h$ be something like $(x-a)$?

or does it still not make sense

@anakhro so 36 is considered an Achilles number because it's decomposition is [2,2,3,3] and it's gcd = 1

That didn't help, @S.Crim.

fair enough

@Silent 'Cause $\bar h$ is a factor of $\bar{a_n}x^n$

Is this supposed to be a function on $A\times B$ defined by $(fg)(a,b) = f(a)g(b)$?

its not given, im asked to prove that given $K \in L^2(M)$ and $H \in L^2(K \cdot M)$, then $HK \in L^2(M)$

and $x-\bar a$ is not a factor $\bar{a_n}x^n$

What is $K\cdot M$?

one sec, let me type it out

ok. Where do we use the fact that $Z_p[x]$ is a UFD?

$(K \cdot M)_t = \int^t_0 K_s dM_s$

This makes absolutely no sense to me, @S.Crim. $L^2(X)$ should be $L^2$ functions on a space $X$. You wrote down a function.

Well thats what that \cdot thing si defined as here, I can't do anything about it

is defined here*

here being in these notes

I suggest you talk to your professor.

Alright, I will

thanks for the effort anyway!

@Silent That's how we know the only factors of $\bar{a_n}x^n$ are $x^k$ (up to constants)

because you can factorize it as $\bar{a_n}\cdot x\cdot x\dotsb x$

so any factor is made up of those factors

OK!!

Wow

Thank u so much

@S.Crim Is it possible that $M$ and $K$ are measures on the real line, and you mean to write $K*M$ (not $K \cdot M$)?

Hmm no, sadly not, I typed it correctly

but this thing

$(K \cdot M)_t = \int^t_0 K_s dM_s$

is confusing me in general

so I need to ask my professor

Because if $K \in L^2(M)$ then what does $K_s$ even mean

for that matter, what is $M_s$ supposed to mean?

@Rick it's not. The first Achilles number is 72

@anakhro so the gcd of the prime decomposition won't work, because I even tried it when removing decompositions whose values were less 2

Anybody know what surface the quadrilateral above should be on. It doesn't match up with any of the pictures

It looks like it's half spherical and half hyperbolic or something

*by removing decompositions which were less than 2

@PaulPlummer!

Hi @BalarkaSen What is up?

Sky

lame joke, nvm

How's it going

going okay, wrapping up a couple of small projects

Coolio

I guess Mike has left us

Kafkaesque

How are you @BalarkaSen

@PaulPlummer Yeah he deleted his MSE account

I suppose that's what's best for him

I'm good, I attended a workshop on probabilistic methods in negative curvature a week ago that I enjoyed a lot

Oh cool

Haissinsky, Kaimanovich, Gouzel, etc came. I suppose you have heard of them

I have not heard of Gouzel seems like he does cool stuff

I attended only first week of the two week long workshop, and Gouzel's talks were in the second week, so I couldn't attend them. I'll watch the uploaded talks sometime

But I know that these three people laid down the foundations of the theory and whatnot

Hmm, Mike didn't even say goodbye to me :(

Hi everyone

@Mike that's messed up ^

"empirical" question as a math student: how long should one stay in a problem?

obviously, a problem with known solution. specifically a theorem or important result of what you're studying

@Lucas: Hard to answer.

Sometimes my hint to my students used to be: "Hint: You're making this way too hard." Sometimes you overthink. Other times it's a truly challenging result and it takes a while to discover the right approach.

@chandx: Write down the linear equations. You've done this before.

(I think it was you.)

Well, assuming poly is of form $ax^4 + bx^3 + cx^2 + d$

Oops.

equations are $\frac{32}{5} a + 4b + \frac{8}{3}c + 2d + 2e = 4a + 3b + 2c +d = 0$

which doesnt give me much i guess

Sure it does. But, once again, I think you have sloppy mistake(s).

Oh sweet lord, not this again

Are the two linear equations scalar multiples of one another?

@chandx missing $x$ part

Yeah, I think that was just a typo .. he meant $dx+e$.

But there's still a mistake or two.

But it's irrelephant.

Oh. I didn't see the other equations

yes, i meant $dx+e$

@TedShifrin I don't know how he'd get that differential there... :P

smacks Lucas

@anakhro I got a response from the string diagram person

> I'm not aware of a way to use string diagrams as a full graphical programming language, but that's not what I want anyway -- I just use them for sketching module boundaries & abstractions

Oh oh, @Dair is goofing off again.

@TedShifrin How am I goofing off?

You're here, aren't you?

I'm strategically relaxing.

And i'm just bored lol.

Ha ha @"strategically relaxing."

"Strategically relaxing," I like that term.

@TedShifrin Surprisingly enough, I say it only semi-jokingly.

k, going slowly $\int_0^2 (ax^4 + bx^3 + cx^2 + dx + e)dx = \frac{32}{5}a + 4b + \frac{8}{3} c + 2d + 2e$

right

Yeah, that one looked right.

I feel like whenever I see a missing $dx$ i wonder if they forgot it, or if they're just reading from Spivak.

If you ever write $\int f(x)$ rather than $\int_a^b f$, you're just wrong.

painful to stare at the two "dx"-es there.

Once the $x$ is in there, you must put the $dx$ ... or else, nine chances out of ten, you'll mess up integrals by substitution. Indeed, if you read my blue book, you discover that it really only makes sense to integrate forms in the first place :P

What if $f$ is a function that returns a function?

and $P'(x) = 4ax^3 + 3bx^2 + 2cx + d$ so $P'(-1) = -4a + 3b - 2c + d$ oh jesus, here was my mistake

Then $f(x)$ might denote a single variable function... :P

@TedShifrin I was about to ask that.

That's the value of the function at $x$, not the function, @Dair. You make my point very well.

@chandx: The moral of the story is — don't talk to Ted :P

Too weird for me to think that derivatives and integrals can be linear transformations

$f : \mathbb{R} \to (\mathbb{R} \to \mathbb{R})$ and $x : \mathbb{R}$ then $f(x) : \mathbb{R} \to \mathbb{R}$... But I'm only kidding, no one sane would type $f$ and $x$ like that...

Oh my

Anyways @TedShifrin how come one is multiple of another

That's actually very important in more advanced mathematics, @Lucas.

I'm saying that is NOT so, @chandx.

Oh, that really makes sense

I'm aware of that, @Ted... but I'm simply still a baby for stuff like that. :P

Oh. I have a question about stuff that my prof stated but I couldn't prove:

Even though I can't find a way how to find dimension of it, I know, maybe it's just late hour or my ADHD interrupting again, but daym still don't know

Using the recursive definition of the determinant (cofactors), and letting $\operatorname{det}(A) = \sum_{j=1}^n \operatorname{cof}_{1j} A$, how do I prove that the determinant is independent of the choice of the line?

Of course you can, @chandx. Each independent homogeneous linear equation lowers the dimension by $1$. You really need to learn some basic theory.

is independent of the choice ... @Lucas

Did you look in my book? :P

thanks, @Ted.

I didn't. I didn't end the basis stuff..

Let $M$ and $N$ be $\mathbb{Z}$-module and $H$ be a subset of $N$. Is it possible that $M \otimes_\mathbb{Z} H$ to be a submodule of $M\otimes_\mathbb{Z} N$ even if $H$ is not a subgroup of $N$ but $M\otimes_\mathbb{Z} H$ is additive subgroup of $M\otimes_\mathbb{Z} N$ and $rt \in M\otimes_\mathbb{Z} H$ for all $r\in\mathbb{Z}$ and $t \in M\otimes_\mathbb{Z} H$?

@Lucas: Have you proved there's a unique function satisfying the multilinear criteria?

I did, but I think I can't use it, I'm allowed to use only theorem that for linear $F: V \to W$, where $V,W$ are linear spaces, $\dim( \ker (F)) + \dim ( \mathrm{Im} (F)) = \dim (V)$

Well, you have an $F$ and you know the dimension of its image.

Gee, what's my $F$ here?

You tell me.

Well, I guess $W$ should be some kind of simple stuff

Let's be a bit more specific. What are $V$ and $W$ and what is $F$?

oh wait

I got an idea for $F$

sweet lord

How can you have an idea for $F$ without an idea for $V$ and $W$ lmao.

@TedShifrin nope. The definition was (unfortunately) a uglier one.

Oh, the permutation definition is what your prof used?

Also: are determinants generalized to all zero char fields?

I disapprove, but it's OK.

@TedShifrin not that ugly, LOL

They're defined for any characteristic.

here: chat.stackexchange.com/transcript/message/49595816#49595816

So what is the definition you have?

$\{ P \in \mathbb{R}_4[x] : \int_0^2 P(x) dx = P'(-1) = 0 \}$ this space look like god daym k e r n e l of some suspicious $F$

Oh, he's defining it by expansion by cofactors. Then it's very non-obvious, without proving the multilinearity properties and uniqueness, that you can switch rows.

@chandx: You're not answering my question. Tell me $F$, $V$, and $W$.

$V = \mathbb{R}_4[x]$

@chandx It's easy now... :p

So, a question I've been thinking about: How difficult would it be to get a paper published on a topic which isn't exactly motivated by anything real? Like integration in the sedenions, or something.

you have 2 equations, 4 variables. use the augmented matrix associated with this system so you get the simplest solution and then the rank is your dimension

You haven't specified whether the paper had any correct content or whether anyone would be interested. Both of those are more important than "motivation" by "something real."

No augmentation, @Lucas. It's a linear map and he wants the kernel.

oh, alright

nor have you specified the writers ability to be persuasive...

Well, there are lots of horribly written papers that get published, @Dair.

@TedShifrin Horribly written papers that have no particular motivation?

actually there probably are...

I think probably a significant percentage of papers that are published aren't really worth publishing.

Good thing I don't read.

sounds reasonable to me. I remember not being impressed by many CS papers...

@TedShifrin what I've proved so far: row is sum of vectors $\alpha X + \beta Y$ is the sum of $\alpha$ time that matrix where the associated row is only X, same to $\beta$ and Y

Well, assuming that the paper is all correct (or at least to a reasonable point). I guess what I'm asking would really be "how much does 'motivated by real world application' affect whether people would be interested in the contents of the paper?"

but at the same time, I had like literally no idea wtf any of the math papers were trying to say lol.

So you're trying to prove the 4 main properties, @Lucas? If you argue uniqueness with that, then you're done.

@Rithaniel: Is this an applied math paper?

oh, uniqueness is guaranted by induction

I thought it was an obvious property with this definition...?

Uniqueness of an alternating multilinear function ?

No, you're not addressing my question.

Sorry, I don't think I get what you're talking about exactly

The cofactor expansion does, however, prove existence ...

@Rithaniel $2 + 2 = 4$ is a true statement. Would you publish that in a paper? Maybe... On the surface it seems dumb, but if you can convince me the proof is actually hard... then maybe I would reconsider.

Although not the only route, can you tell me something contrary to what I expect?

I don't see what uniqueness the cofactor definition has. It's just a formula.

would that be just $V = W = \mathbb{R}_4[x]$?

Nah, nothing specific. If anything it would be a paper on non-associative magmas, so definitely not applied math.

I mean that a function $\Delta$ such that $\Delta([a]) = a$ and (the cofactor thing) is well defined.

By well defined, I mean that it's unique.

It's a formula. There's no question of well-definedness.

I'm making the claim that there's a unique function with the 4 multilinear properties. If you prove that your formula satisfies those (with any row), then it follows that they all give the same answer.

OH. Yeah, I didn't prove that.

@Eric: I just made a frangelico hazelnut-chocolate torte :P

@TedShifrin Damn I miss chocolate.

@TedShifrin that's hard. :(

I thought I could proceed with induction like I did with the multilinear property of linear combination

I miss tasty food.

By induction, you'll prove the permutation definition, most likely, @Lucas.

Why are you denied tasty food, @Dair?

@TedShifrin Acne. :(

Oh, you shouldn't have bad, greasy food. That's not what I call tasty.

I need to make it low glycemic. No dairy. And a lot of meat causes issues... and I break out after even dark chocolate.

it's like more restrictive than vegan.

Wow.

Well, a lot of meat isn't good for us, anyhow.

What about fish?

:/

Nah. Even then... it's a lot about how it is prepared...

as in, why would you want to each just cooked fish?

I understand. But herbs, spices, and citrus are not an issue ... and they make for flavorful food.

I need to figure out how to better incorporate seasoning into things...

Anyhow, back to math(s).

Seasoning is all the art of cooking.

But where do you learn the art of seasoning? Also, it sucks 'cause butter is out of the picture...

I use almost no butter (except in baking the cake).

Use olive oil.

Do you like cooking, @Ted?

Garlic. Onions. Herbs, middle-eastern spices, lemon and lemon zest, orange and orange zest.

LOL, @Lucas, what do you think?

You don't look too surprised I discovered your secret. :P

It's because he isn't surprised? I'm confused.

@Dair wut

</sarcasm>

Nvm. I misread it. Sorry.

Before going to uni I never cooked at all.

I discovered I love cooking. Is it a mathematician thing? :P

@LucasHenrique I think I'm a counterexample to that conjecture. :P

Maybe it's a Brazilian thing, @Lucas. Talk to @Erico.

@Dair dammit. Maybe the measure of the set of counterexamples is zero.

No, I know plenty of mathematicians who can't cook. There are even a lot of foodies who can't cook at all.

:(

Just because a conjecture was proven false doesn't mean that the conjecture was bad :D

LOL

People should try it. It's so relaxing, and so realizing.

@LucasHenrique This implies that I want to be alone with my thoughts which is definitely not the case.

(I don't think that's the right word, nor it's a noun)

@TedShifrin I have a relatively more interesting question (than the linearisaion one)

I'll be the judge of that, @anakhro :P

someone somewhere told me that somehow residues (in complex analysis) are residues of forms.

Yes, that's correct.

Not surprising, because they arise as an integral.

In several complex variables, there are generalizations to meromorphic $n$-forms.

You realize that's the case when you try to decide what the residue at infinity should be.

That's exactly when I heard it

$Res(f;\infty) = -\sum_{k=1}^m Res(f;a_k)$, where $a_k$ are the isolated singularities.

Well, that's a theorem, not a definition.

In fact, if you have a meromorphic $1$-form on a compact Riemann surface, then the sum of all its residues is $0$. That's the theorem.

Conway defines it that way.

Conway sucks ...

Amen.

What's your fave book for complex. I need to take my comprehensives in September. :(

It's old-fashioned, but I've used Ahlfors. I tried Stein/Stakarchi and disliked it a lot. I was going to use Gamelin's book, but I ended up with cancer and didn't teach the grad complex course that time.

Lang's book actually has some good things. I like things in Narasimhan's book, but it's pretty sophisticated.

With Nievergelt?

I do strongly recommend Forster's Riemann surfaces book.

Oh, maybe that's a second edition.

Probably. I will take a look at Forster's.

So how exactly do the residues manifest themselves geometrically with the forms? Or is it just the naive integration of forms?

I know next to nil about complex geometry, so I don't know if there is any nuances here.

You define the residue to be $1/(2\pi i)$ times the integral around any suitably small smooth curve around the singularity. Of course, then you can calculate $\text{res}_0\big(\sum a_nz^n\,dz\big) = a_{-1}$ and check this is independent of coordinate system.

@TedShifrin it's holiday! :P

Heya @Leaky. It's Islam New Year.

well I'm not a Muslim

LOL, well, my boyfriend is and it's also his birthday today :P

oh ok

If anybody knows variational calculus, I could really use some help

That's why I knew that.

I have a problem posted here math.stackexchange.com/questions/3155772/….

@TedShifrin well you seldom mention him here :P

@TedShifrin amazing

@Leaky: It's really not the place for such discussions ... but I threw it in there because you said holiday.

ok :P

@TedShifrin more paperwork for my year abroad...

@A.Hendry: It looks pretty sophisticated, so I don't know the answer(s) off-hand. The things on $u$ at endpoints look like the dual boundary conditions. I vaguely remember this from teaching the material 30+ years ago.

@Eric: If you go eastward, we'll never cook! :(

I'm also making a spinach soufflé tomorrow — I don't think I've done that in 30+ years. Crazy ridiculous.

oh man

another aspect to toughen this choice lol

LOL, hardly a serious one, but ...

all the little considerations add up u know

@TedShifrin Thanks for the help! Dual boundary conditions, eh? I'll look that up. I'm mostly concerned about $u(a)=0$ in the term $u'/u$ appearing in $h'-\frac{u'}{u}h$ (and also for $w=-\frac{u'}{u}P$)

Or maybe adjoint conditions. I forget @A.Hendry.

@TedShifrin It seems to me like $u$ can't be zero, or else $w$ would be infinite.

@TedShifrin I know the Jacobi accessory equation is a type of Sturm-Liouville problem, from which Fox demonstrates in his book that $u$ and $u'$ cannot simultaneously be zero, but that doesn't stop $w$ from blowing up when $u(a)=0$ in the denominator

I haven't thought about that auxiliary Riccati equation thing. But, to play devil's advocate, if $u'$ also vanishes, then you don't have a problem.

@Ted i think i’m gonna wait until i’m back in chi and able to talk to my letter writers before i think any harder about my decision, i’m getting nowhere on it

@TedShifrin Unfortunately, no, since Fox demonstrates both cannot simultaneously vanish ($u$ cannot have a double root)

LOL, poor @Eric.

@A.Hendry: I have no further thoughts and am too lazy to work on it for hours.

im incredibly lucky and in a good position i’m just bad at choosing things

@TedShifrin Hahaha! I don't blame you! Thank you so much for your help.

@TedShifrin any good vegetarian recipes?