@TedShifrin I prefer Go Incremental Progress and Draw a Salary then Retire Peacefully, or go home

It sounds like you're asking: Given a differential equation with boundary conditions and a solution to that, can I conclude that this is the only such solution?

@Semiclassical yes

I see.

Picard-Lipschitz theorem is relevant.

Ugh, Picard-Lindelof

that can get complicated fast, depending on the differential equation and the dimensionality of the space. but in 1D second-order DEs it's not bad

Lol, I mean the function does have to be Lipschitz so we can see where that typo comes from

Freudian slippery

this is where my memory fails me tbh

I'm quite confident that the solution in this scenario should be unique, but I don't recall the relevant theorem

What's the DE here?

u''(R)+R u'(R)=0, subject to u(1)=100 and u(2)=0

so second-order boundary value problem

You should be able to convert that to a system of 1st order dudes

By letting $v = u'$

lol

that's how I had him get the solution in the first place

Ah

true

Is there a way to use P-L on a matrix equation? Or nah

Yep

It works for matrix ODEs

problem is that Picard-Lindelof as stated is for initial-value problems

Rip

I doubt that's a serious problem, but it is a thing

Differential equations is new topic for me. I just learned about this first time today so

Wait what do you mean by boundary value problem? We're in polar?

R is the radial component?

yes, but that's not what I meant

I just mean that the boundary conditions are u(1)=100, u(2)=0

not u(1) = something, u'(1)= something

Anyway here's where my knowledge of ODEs and because I don't know shit, I'll leave it to others who aren't rambling nonsense

@Semiclassical Ahh

Sorry for overlooking that.

np

I suppose one could do something like

That's interesting. That probably changes things

I doubt it's unique anymore

take the boundary conditions to be u(1)=100, u'(1)=c

then deduce how u(2) varies with c

Probably

Yeah you have to do some nontrivial computations

or you can say "f*** it, we're really solving Laplace's equation on an annulus with dirichlet data"

and that's definitely got a unique solution

...probably

I said you have to do some nontrivial computations, not me $:\equiv$

lol

the general solution is $u(R) = A \ln R+u(1)$

someone could explain how this is laplace equation ?

@BalarkaSen I'm gonna hijack that phrase

which has $u'(1)=A$

@Tuki basically, you take the two-dimensional Laplacian and write it in cylindrical coordinates

@Daminark "Save image| Save as... | Hijack this meme"

@Semiclassical oh ok

then you suppose that there's no angular dependence.

if that laplacian is required to be zero on your region of interest, then by definition it satisfies laplace's equation

and when you write that all out what you get is your ODE

so it's a special case

that's not something you'd be expected to know for now, though

So I know in 2D you have that holomorphic and harmonic functions are related, but is there any particularly interesting thing about it in higher dimensions? Like what information does it carry about a function/are there any nifty theorems about it?

Question

(ctd from above) so $u(2)=u'(1)\ln 2+u(1)\implies u'(1)=\frac{u(2)-u(1)}{\ln 2}$

so given $u(1),u'(1)$ you can always find $u(2)$ uniquely.

@Daminark I asked this question in a workshop a few months ago and the answer I got was "subharmonicity"

which I think is enough based on P-L to conclude uniqueness

Or something of the sort

Ted would know

sup chat

inf eric

@BalarkaSen "Only barely"? I thought you were younger than me, or am I misremembering?

gah

Are we both seniors?

\sqrt[n]{n!}: $\sqrt[n]{n!}$?

Yes!

I mean, you are younger than me, but I guess not enough to be in a different grade @BalarkaSen

@AkivaWeinberger Yeah we are. I am a month younger than you I think

Mm When's you're birthday?

$\limsup_{n \to \inf} \frac{\sqrt[n]{n!}}{n}$.

@Semiclassical thanks!

neat. my guess is that you'd want to do squeeze theorem shenanigans

I mean, n! has a pretty obvious upper bound

@orbit-stabilizer My first instinct is Stirling's approximation. My second instinct is, "no, ew"

@AkivaWeinberger Tomorrow, with probability 1/365

Oh wait actually I found something dank. So a diffeomorphism between Riemannian manifolds is an isometry iff it commutes with the Laplacian

@BalarkaSen Happy statistical unbirthday

$n\cdot(n-1)\dotsb1$

@BalarkaSen you didn't take into account the possibility that you may have been born on feb 29

${}\le n\cdot n\dotsb n$

@AkivaWeinberger right.

@AkivaWeinberger Actually it's on February 29th, with probability 1/(360 noscope)

sniped

fuckin hell

But, for you, 1/P(360 noscope)=infty?

A cheeky way to try it would be to find a function with that is the power series coefficients and use complex analysis :P

Gtg

Well, this is the coefficient of some power series. Gotta find the radius of convergence

@Daminark most things true of holomorphic functions stay true for harmonic functions in arbitrary dimension so that's a nice thing

they're also good for approximation

@orbit-stabilizer Akiva had the right of it, though it looks like that bound isn't an effective one

Gotta converge FASTER FASTER FASTER

SONIC X

wow, is that show even on still

Oh I guess that makes sense why you'd be doing this problem then

@AkivaWeinberger finelly i don't understand what is the relation between $\frac{a-b}{2}\in K$ and $||\frac{a-b}{2}||^2\leq ||x-a||^2-||x-\frac{a+b}{2}||^2\leq0$

numerically (i.e. according to mathematica) the number is around 0.37

please

@EricSilva nice

@Semiclassical thanks, I'll see what i can do

there are like billions of reasons that geometers would care, there are more reasons analysts would care

etc etc

@EricSilva Yeah but isn't there's a nice PDE condition that completely determines the real parts of holomorphic functions in higher dimensions?

@orbit-stabilizer try to see if doing ratio test is at all easier

Calculus question:

Is this going to be a rick roll

For x > 0

@Rick_Roll I don't trust you

x = 1+1+1+1 (x times)

no

it doesnt work

@Balarka what do you mean by "holomorphic functions in higher dimensions?"

diffeerentiating it

x^2 = x+x+x+x+x (x times)

Wait

this is a "find the error problem"

WHAT'S THE PUNCH LINE. I swear it's all gonna be a farce.

mfw I think it's gonna be a Rick Roll youtube.com/watch?v=dQw4w9WgXcQ

is the error that you can't differentate it?

it continues like this

you can't use the linearity of the derivative

@EricSilva I mean a holomorphic function of several variables

d/dx(x^2) = d/dx(x+x+x (x times))

is this the error?

Can you take the derivative of this?

you can't write it as d/dx(x)+...+d/dx(x) [x times]

What if x were equal to 1.13425435245?

they're harmonic + extra relations

what does x times mean in that case?

Why not @anon?

@Daminark Never gonna Rick you up, never gonna roll you down ... youtube.com/watch?v=Gc2u6AFImn8

@Rick_Roll does that make sense?

as orbit says, "x times" really only makes sense for integer x, and when you take the derivative x takes values on a continuum, including noninteger points

Oh I remember that stumped me so hard in calculus

Yeah I also thought about this at some point

It's a subtle point.

Ok thanks! :D

I thought that was the error

but everyone else in my class thought it wasn't

Turns out it was a subtle rick roll. @Rick_Roll is actually Terry Tao in disguise.

Where's the rick roll tho?

sniped

come on dude

Sorry that's just my unfortunate name

I love it

Hi, what software do you use for symbolic math?

I used Maple a lot, but non open source

I'd like to use an open source one

Chalk is my favorite

microsoft word

it is the best

@orbit MS paint tho

I use notepad

doesn't support non-integrable functions

@Daminark paint is for plebs

Really I think Wolfram Alpha is gonna be way less effective than Maple but I think that's common for simple stuff

I'd like to move for an opensource one

ugh, hate hate hate random connection outages

(that I can recommend to students)

Wolfram is going to noscope us with A New Kind of Science anyway

so even if Maple is cool (and Mathematica probably too), I can't

I've heard of one called GAP that's supposed to be good for discrete math

sigh... if you really want to read an interesting derivative article tho here's one I read recently: latlmes.com/world/the-turbulent-history-of-derivatives-1

i dont know if a good open source option exists that's as flexible as maple or mathematica

@Daminark how does ratio test help me?

I want to find the value of the limit

go old school and give your students each a quill and an inkwell and tell them to go crazy

@Rick-Roll I knew it

I fucking knew it

@EricSilva: that's also the conclusion I found...

probably cheaper than one copy of mathematica tbh

@orbit you can use the ratio test to determine the radius of convergence of the power series instead of evaluating this limit

Sometimes

I looked at Sage (big 3 GB package, but no easy IDE out of the box!) and many others, but none of them would be easy to use for undergrad in economics (my students) who have zero knowledge about programming

Does your school not have some sort of licensing deal going on?

latlmes.com/breaking/trumpgetsimpeached-1

@orbit-stabilizer arghh

Here we get mathematica for free

"free"

@Daminark ah

@Eric I mean sure but like, once you're already a student it is a sunk cost. Plus if you're on enough aid it's actually free

yeah maybe i should look at this

some costs arent monetary

I was thinking in strictly monetary terms, like that's all I care about here, but what costs are you talking about?

im being very not serious

Oh I thought you were dead serious and I was like, Eric u ok?

it's hard to tell when eric is serious and when he is facetious

i got him

but otoh i do the same thing

even when Im serious im being faceitious

More generally, which language / software do you use the most for your math?

@MikeMiller i can verify

I usually have a distinct persona when I am serious and when I'm not

I use small caps when I am in meme mode

I rarely find myself using much beyond Wolfram Alpha, and even then I haven't found myself using it often

I know how to use the Weierstass-M test to prove uniform convergence, is there a way to use something like it to show non-uniform convergence? In general how do I prove non-uniform convergence?

@Daminark hey that worked! Got that x has to be less than 1/e

series I have is $\sum_k \dfrac 1{1+k^2x}$, I proved its uniformly convergence on $[1..2]$ by using $M_k = 1/n^2$

now i have to show non-uniform convergence on $(0..1)$

just bound it from above by a convergent thing that depends on x

@GFauxPas it's somewhat situation dependent. You could try direct stuff, for example by showing that its maximum is bounded away from the function in question over all $n$. Sometimes you can do so by showing that it breaks certain rules such as the integrals converging nicely, or that the limit is not continuous

@Semiclassical you said it was around 0.37 and 1/e is 0.3678, so I take it that my answer is right

@orbit sick

@GFauxPas using Knuth notation

but I don't know what it converges to

just that it converges by the ratio test if $x \ne 0$

@orbit-stabilizer yup

so I can't check if the limit preserves integration or differentiation

I'd have said 1/e but didn't want to give the game away

hey Semi

heya

I gave up on that extra credit

figured it was a better use of my time to work on the regular credit

right

so, now I have this creature

I proved by Weierstrass-M it's unif. conv. on $[1..2]$, and now i have to prove its not unif. conv. on $(0..1)$

I shall think about some hard induction problems now

from this Russian book that I have

@BalarkaSen Prove that the sun will come up tomorrow, based on the fact that it's risen for every day up to now.

actually I proved it's uniformly convergent on any interval $x > 1$

true communist mathematics

@Semiclassical problem of induction

and possibly $x < 1$ I didn't think about it because I didn't need to

a hint please?

@Semiclassical I will disprove it, because it contradicts my sleep schedule

lolok

I give this a 5.5/10

I remember a guy like that in college, who would always do yer-mom jokes

and I think I did a similar turnabout-is-fair-play maneuver on him as well

It's a good comeback

yeah

A friend of mine does "ur mom" jokes but related to math every now and then

have you tested the fine print as well?

I have a friend who does it when people say nice things

He only recently started doing it because he made general "ur mom" jokes to his brothers over the summer and it stuck

But now it's stuff like "ur mom is cyclic"

lmao

Your mom is simple

your mother is simply laced

Your mom is nilpotent

(i dunno)

Your mom is real :P

your mum vanishes in the limit

reminds me of a non-mom related joke

Your mom is like a non-abelian group of order p^2, unexistent.

(Is this gonna be like that time we beat the living shit out of the "Geometric Approach" jokes?)

I liked to joke that one of my undergrad prof's office hours were complex

insofar as they contained both real and imaginary parts

hah

Question

Askaway

Let $E = \mathbb{Q} \cap [0,1]$. Let $f: E \rightarrow \mathbb{R}$. If $f$ has two continuous extensions $g,h$, show that $g(x) = h(x)$ for all $x \in [0,1]$.

mhm

What do you know about $g(x)-h(x)$? Just immediately

So: Show that there's exactly one continuous extension of $f$ on $[0,1]$

Sniped in Socrates

just take limits

We say $f$ has a continuous extension $g$, if $g(x) = f(x)$ when$ x \in E$, and $g$ is a real valued continuous function from $[0,1]$ to $\mathbb{R}$.

My eyes are bleeding

sorru

it's still profusely bleeding

Question, anybody: If I take the derivative of a solution of a PDE w.r.t. to a parameter, to be precise $\nabla_m u$ where $(\omega m + \partial^2/{\partial x^2})u = q$, where $u$ is the solution, and $m$ the parameter, can I just "take the derivative" of the PDE?

ITS BETTER NOW OK

g(x) - h(x) is cts

You can say a lot more than that...

Now note that both $g$ and $h$ are extensions of $f$, so what else can you say?

my book using $\{a_n\}$ to denote a sequence and it's a bad notation

It's 0 for at least countably many points

Not just countably

$\{\cdot\}$ always implies the order of elements are immaterial

Q has a special thing Z doesn't

That's not even the statement's final form

I want to show that it's 0 for all points, but I know just from the defintion that it's 0 for countably many