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8:07 PM
Yo @Daminark
What would you guys call the section of a torus ?
The circle ?
 
it can be a pair of circles too
 
Yeah that's my question
In a physics exercise I was asked to compute the magnetic flow through the section of a torus of wire (dunno how those are called)
 
I dunno. They probably mean a not so effed up section
 
Mmm probably
Boring exercise anyway
 
i got my finger a tiny bit dislocated today :p
not my best day
 
8:13 PM
Ugh I had that some weeks ago, not too keen
 
Weird
 
I'll have to go see a doctor tomorrow probably
 
Also sad reacts @Balarka
 
it's ok but nontrivially swollen now
 
Balarka'! Your a mess!
How was boot camp, Demonark?
 
8:18 PM
that's a feature, not a bug
 
Bob
@Typhon My question has been answered and I thank you.
 
the lecture i attended was quite good
much better than last year @Daminark
 
Cool ...
 
Hello again chat.
 
Hi, Fargle.
 
8:21 PM
rehi Ted
Hi Fargle
 
Rehi.
 
The infinite products lecture was also pretty fantastic
That one had more interruptions because the original lesson plan basically had one computation at the beginning, so he was always to give examples of stuff
 
Hi @Fargle
 
infinite products are fun
 
@Astyx Related to the multiplicative thingy
which apparently was proven by Erdös,
I guess you could use that fact to prove the theorem contained in this answer of mine.
(Using the bottom half to show multiplicativity.)
(And I'm pretty sure that the increasingness isn't hard to show from the definition, though I haven't actually tried to show that yet.)
 
8:31 PM
Are these continued fractions ?
 
@Astyx Did you see why it's trivial for integer codomain?
 
No, just infinite sums
 
@BalarkaSen yeah when $f(2)=2$
 
Like $\frac11+\frac12-\frac23+\frac14+\frac15-\frac26+\dotsb=\ln3$
 
Erdös is more general than that
 
8:32 PM
Semi knows about this - this was essentially a large part of the first question I asked on this site, which he answered.
 
Ok I don't know off the top of my head how to do it if I do not have a fixed point
 
howdy, DogAteMy
 
It's not necessarilly the identity. But Erdös proved that a multiplicative (not necessarilly totally) strictly increasing function $\Bbb N \to \Bbb R$ is a power law, and that if you force $\Bbb N \to \Bbb N$ you even have an integer power law
 
Got it.
 
8:35 PM
Right, so just $e$ raised to the power of the function I want to show is the natural log
 
@AkivaWeinberger Right, that was written at the beginning of the answer ... I guess I'm tired
 
Oh, yeah, I guess it does use the same notation
I just made it up for this
 
Where did you learn about this? I wish I had a textbook on this flavor of number theory/combinatorics/prettymuchliterallywhatever
 
@Akiva Typo ? $[\overline{1,1\dots, 1, 1-n}]$ ?
 
because it seemed like the best way to write it
@Astyx Oh, yeah, thanks
I didn't notice that
 
8:36 PM
So you're saying that result is a corollary from Erdös ?
 
Yeah.
Well, you need the fact that $\sum(-1)^{n+1}/n=\ln2$ to jumpstart it.
Otherwise you just know it's a log function, not necessarily the natural log.
 
Taking $n\mapsto e^{[\overline{1,1,\dots, 1, 1-n}]}$ ?
 
Yeah
And now I realize that I don't actually have a proof of increasingness :P
Um, you know what: Exercise. :P
 
@AkivaWeinberger rekt
 
it being multiplicative is lcm stuff ?
@BalarkaSen This again ..
 
8:39 PM
@Astyx It being multiplicative is the thing on the bottom half of the answer (shaded in yellow)
 
Increasingness seems straightforward (emphasis on "seems")
 
I gave two special cases but it generalizes
OK, I really need to run now. Bye!
 
Yup, with lcm stuff
Seeya, I'll give it some thought
 
And DogAteMy vaporizes as fast as he materialized ...
 
@Astyx golden stuff
 
8:41 PM
sighs So many books to read, so little time, so little focus.
 
You can be like our president and just not read.
 
I don't want to be president. I want to be a nerd.
I certainly don't want to be our current president.
 
Build a ten feet wall around the publishing houses of your books and make them pay for it.
 
@Akiva Totally legit proof : $[\overline{1,1,\dots, 1, 1-n}] = [\overline 1] - {1\over n}[\overline 1] - {n-1\over n}[\overline 1] = 0$
On to serious maths
@TedShifrin He still wrote a book
 
No, he did not. He got interviewed, and someone else wrote the book.
 
8:46 PM
the guy who wrote the book doesn't like him either iirc
 
Really ? Doesn't really surprises me
 
Not for a second.
I could swear there was a question on multilinear algebra the last couple of weeks that has now reappeared in disguise, but I can't find the original. I actually thought about it, but someone else beat me to a good answer.
Ah. I just remembered. It was posed as a diff geo question about maps preserving $d$-dimensional volume for $d<\dim M$.
 
maybe i will do differential geometry for a while before sleeping
 
@Balarka i say the same thing to myself every day
 
well you're a geometer
 
8:52 PM
im also a lazy person
 
clubs everyone over the head
 
Ow! I didn't even do anything that time.
 
what did i do!
 
@AkivaWeinberger Taking the partial sums up to $Nn(n+1)$ of each series, you get $H_{Nn(n+1)} - H_{N(n+1)} \lt H_{Nn(n+1)} - H_{Nn}$
 
eh. riemannian geometer is not my favorite branch of mathematics ... mostly because it's too out of my comfort zone
 
8:53 PM
Taking the limit as $N\to\infty$ you get increasingness of your functino
 
You used to say that about multivariable calculus.
 
and my comfort zone is a narrow gutter down the topology lane
 
But not strict
 
@Ted I love multivariable calculus, though
 
That's a significant change, @Balarka' ... I'm just making a point.
You have to work to get things into your comfort zone.
 
8:54 PM
2
Q: Eigenvalues of truncated binomial matrix

beckoThis question is closely related to Eigenvalues of binomial matrix, but is a bit more general. I am trying to determine the eigenvalues and eigenvectors of the following matrix: $$M_{ij} = 4^{-j}\binom{2j}{i}$$ where it is understood that the binomial coefficient $\binom{m}{k}$ is zero if $k<0...

any ideas??
 
or i can just hole up further down my comfort zone :D
 
I remember when algebra was your only comfort zone, @Balarka'.
 
or so i thought it was lol
 
learning broadly can let you make connections to stuff in your comfort zone and you're better off for it tbh
 
maybe i'll learn higher category theory with my man @Daminark there
 
8:56 PM
@TobiasKildetoft thank you
 
barfs
 
and just ditch riemannian geometry i dunno
 
Now $H_{N(n+1)}-H_{Nn} = \sum_{k=Nn+1}^{N(n+1)} {1\over k} \ge {N\over N(n+1)} = {1\over n+1}$ so you can never have equality, thus you get strict inequality @AkivaWeinberger (sorry for intensive ping)
 
starts to craft an ignore note
 
lol
ook, onto math
 
8:56 PM
(I deliberatly want to avoid using analysis on the harmonic series cause that's overdone, I'm a hipster)
Algebra will always be my one and only confort zone
 
Interestingly, @EricSilva, the OP hasn't mumbled one word at me regarding my sketch of an argument on that $\int \tau/\kappa\,ds$ thing. Plus, never responded to my point about needing closed curves.
You haven't even studied much algebra, have you, @Astyx?
 
No :)
 
i said i was gonna think about that more and never did
 
Not as much as I'd want to anyway
 
You're good at promising me things and then welching, Eric :D
 
8:58 PM
rip me
 
You're still young, @Astyx.
 
Hopefully
 
I do a lot of work @Ted, just not very focused work usually
i need to finish Bryant oh geez
 
LOL ... You have too much on your plate, for real, Eric.
Maybe we should just do a seminar on cooking/eating for the rest of the summer :D
 

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