You refered to landau as a constant, the link is cut at the half to refer to (coincidentally) a district iin germany. that's the funny part.

I wonder

Desmos suck at automating it, but graph is pretty

The radii of each circle are the integers that are in the sequence

are you not missing some?

Desmos cannot do recursive sequences, and the above are all handtyped so... errors

fair

I don't have a mathematica copy with me thus I cannot wrote the recursive function

wanna see something pretty? I wrote a program that generates hairy balls: i.sstatic.net/ZGnr8.png

Someone's @Salt-y

Also hey @Secret and @Mathein!

hi Daminark

yeah, some of the hairs are tangled

it's a result of doing a bunch of specific but randomly chosen operations to a random graph

One thing I noticed about number theory is it often give pretty diagrams, despite I understand absolutely nothing on how it works

@Salt what is the significance of the bolded heads?

color is the out-degree, blue to yellow for few neighbors to lots of neighbors

size of nodes is the in-degree

oh, if you're referencing the fat parts of the edges and not the nodes...that's just how networkx draws arrows

because it's simpler

it's a directed graph

I see

anyway, i got a folder with lots of pictures of hairy balls now

so, no shame

no shame

but mathematicians love hairy balls, so i'm in good company, i think: en.wikipedia.org/wiki/Hairy_ball_theorem

physicists may soon be converted, dunno: livescience.com/…

blogs.ams.org/phdplus/2017/12/27/…

Hey @Ted!

Hey guys!

I'm currently trying to write my personal statements for some REUs and I have no idea where to start...

writing personal statements should be frowned on in the math community

What experiences have prepared you for this REU? Well, honestly none! I'm just a simple lower division community college student who's barely began multivariate calculus! I'd love to do research but none of my professors do, so I'm kind of a hopeless candidate!

@salt seriously!

@mick how did you get $ln(\sqrt z)^{-4}$ ?

oh gone ...

this text game brings me back to my childhood

Hi Demonark.

@Cookie: My general advice (although I think you're at least a year early to be applying) is to write about challenging exercises you've spent hours and hours being absorbed by. Or interesting mathematics that you've read about because you're so interested.

@TedShifrin how's it going?

Not doing too much. Just thought about a question on main I'd never seen before. Sorta cute. Prove that if $f$ is a polynomial of degree $n$ with $n$ real roots, then $(f')^2>ff''$.

that's neat

Yeah, it is. The proper proof was not a method I'd used before in this setting.

link?

I don't want to give it away. Think about it first.

So, I know if you have a polynomial with all real roots, there's something about the derivative having roots that interlace?

Well, that's true, Demonark, but how does that bring in the product $f f''$?

i mean, i'm curious, but my mind is on other problems atm

Yeah double interlacing might not help

Uh

Maybe you could pull a log trick?

That's a good start, Demonark.

We should be able to assume everything is monic

@TedShifrin Ok! There was this one problem from an old putnam exam in my textbook; $\int_{0}^{\frac{\pi}{2}} \frac{dx}{1+tan^{\sqrt{2}}(x)}$. The answer was a simple but obscure $u$-substitution, but I instead spent half of the semester evaluating the interval by showing that it was symmetric with respect to the point in the direct center of the region of integration using infinite sums and generalized binomial theorem :) I could talk about that!

$\log(f)' = \frac{f'}{f}$

But we can write $f(t) = \prod (t-c_i)$

Specific things that you got engaged about are great, @Cookie. BTW, I have used that problem in my own multivariable course. There's another great way to do it.

Demonark: You mean $(\log f)'$, not what you wrote.

@TedShifrin what's the other way? Now I'm really curious :)

Well, I do mean what you said but that's usually how I write it

Put a variable $y$ in there instead of the $\sqrt2$, @Cookie. How does the integral vary as you vary $y$? [There's a powerful theorem that says under reasonable circumstances you can take the derivative of $y$ by differentiating under the integral sign.]

I do NOT approve of how you wrote it, Demonark. To me that's too much like $\log(f')$.

Fair, that could be confusing

I really dislike the sloppy things people write like $(x^2)'$, even.

But that, at least, is unambiguous.

@TedShifrin Differentiation under the integral using this method? math.uconn.edu/~kconrad/blurbs/analysis/diffunderint.pdf

Yes, Cookie.

dunno, i prefer $\log(f)'$ as well, not that it matters, i haven't done any calculus in some 10 years

@Ted Ah crap! Going to go try it now!

Please remember that you write $g'$ for differentiating $g$, and in this case $g=\log f$.

@Cookie: It's really cool, but, yeah, the Putnam folks intend you to use symmetry cleverly.

But yeah so $(\log f) = \sum \log (t-c_i)$, so $(\log f)' = \sum \frac{1}{t-c_i}$

Demonark: With perhaps an additional $\log c$ out in front of the $\log f$ sum. But it disappears.

Yeah I'm assuming everything's monic

Because I think for the inequality it doesn't figure in

IT doesn't, but you should comment that to start.

True. But yeah so $(\frac{f'}{f})' = \frac{f''f - (f')^2}{f^2}$

So we just need to show that $f^2 (\log f)'' \le 0$

Which we can just compute

Yup. It was just commented that the correct thing to prove is $\ge$, not $>$.

No, you have a sign wrong.

Whoops, I thought the inequality was the other way

Well, did you just change it? Now it's right.

Yeah I just did

But yeah $(\log f)'' = (\sum \frac{1}{t-c_i})' = -\sum \frac{1}{(t-c_i)^2}$

That's automatically negative, and you're multiplying it by $f^2$ which is automatically positive

So yeah that does it

That's pretty sick

Along with the justification of what happens where things are undefined.

True, well, where things are undefined, you have a $0$ of $f$, so the inequality just holds

sup chat

At first, it reminded me of the problem I put in Spivak that an old student gave me. $f$ poly of even degree $n$ with $f\ge 0$. Show that $f+f'+f''+\dots+f^{(n)}$ is always $\ge 0$. [I think I'm remembering it correctly.]

Taking the limit of $>$ gives $\ge$, you mean, Demonark?

Hi, Eric.

@Ted I just meant that the only places where the stuff is undefined is where $t-c_i = 0$

But then look at those points directly

But you could have equality at those points.

True, I mean the non-strict inequality

Well, how was I supposed to know?!

hi @TedShifrin

@TedShifrin @TedShifrin

hi Karim

Stop that.

sorry :P

hi everyone

I was going to university to get something from there I found some professor running up and down the stairs at university

haha

fitness minded prof, nothing wrong with that...though stair running has to be bad on the knees, yeah?

@TedShifrin ok, so I'm a little stuck. I differentiate w.r.t. $y$ and get $\int_{0}^{\pi/2} \frac{ - \tan^{y}(x) \ln( \tan(x) ) }{(1 + \tan^{y}(x))^{2}}dx$, but I'm not sure where to go from here, unless its some really shitty by-parts or $u$-sub

@TedShifrin I don't understand something small about constant sheaf. So it is an assignment for every U connected we assign the single group G ?

Um, it's been years since I did this, @Cookie. Give me a minute.

Haha no problem!

For every U open and connected and subset of X we assign only 1 group G ?

Yes, Karim. A section on a connected open set is a constant element of $G$.

Here's a fun question to ponder about:

The physical meaning of the distributive law in many algebraic systems that represents physical systems

Oh okay so we fix G and sections are choices of elements of G ?

man im sooo bored

do math @EricSilva

i was until a couple minutes ago

oh okay that would sense why we would denote such a sheaf by G @TedShifrin