:-)

Copper's humility makes up for my hubris. :)

Wanted more h's ...

Lol......well I mean you've earned the right to said hubris.

Copper's humility restores harmony halting my hubris

Doesn't quite roll off the tongue that well

i'm blushing. more used to abuse :-)

Or tongue :)

We're not all your daughter, copper.

:-). she is sweet to me when she knows i need it :-).

it was snowing in oxford a few days ago

I still want to meet her !

It snowed in Malibu!

me too :-).

global something or other :-)

You haven't met her?!!

i mean soon :-) she was going to return for spring break, but with all the covid fuss it might not happen.

i have yet to visit her in the uk despite being nearby twice last year :-(

Yup. Bad times.

thankfully communication is so much easier & cheaper now.

alliterative people arriving

@BalarkaSen Nice description

Hello

The proof of Smale I have always been most impressed by is that you can essentially canonically isotope a diffeomorphism of the disc to a linear one

AKA, Diff(S^2) ~ O(3)

I started reading Kupers, "Lectures on Diff"

Never expected to see Poincare-Bendixon there but I have never seen a proof of that fact I wasn't impressed by

@TedShifrin Wow, very logically careful.

Really, Poincaré-Bendixson in that?

I guess that's the point.

Yes, to identify structure of flowlines, so that he can move flowlines from left-side to right-side of square to standard horizontal flowlines.

I am trying to move flowlines from one side to the other side currently!

Different context but surprising to see you say that

I should reread that. Is it easily available?

The two other proofs I know use either isothermal coordinates or uniformization of S^2. Both cleaner, but less visually comprehensible.

Smale's proof is harder to read than any modern proof but you will learn more.

Let me find.

@Balarka you need the right current!

@Ted: people.math.harvard.edu/~kupers/teaching/272x/book.pdf

Hmmm. I can imagine proofs in a couple of other spirits but none i have seen written down. There ought to be one with some sort of mean curvature flow.

See II.7

I need to read it, I just know this book has it because Mike told me (I am reading smoothing theory from the book, Chapter V)

It is always hard to find this reference for me, for some reason: ams.org/journals/proc/1959-010-04/S0002-9939-1959-0112149-8

Should be freely available.

reminds me

Unfortunately, I don't have access to that, @MikeM.

@BalarkaSen i had that one coin-flip sigma-field problem

Oh, I do. Apologies.

turns out the chapter notes at the end of the book specifically includes references for those

If PAMS from the 50s wasn't freely available I'd have to write a sternly worded and completely impotent letter to someone.

including a Martin Gardiner article, lol

Always fun to write sternly worded, impotent letters.

@Semiclassical Got it.

"The seething rage in this letter is inversely proportional to its relevance."

run quickly when someone is carrying a big stick

hey

horses

Hi Shmo

not hay, copper

horsas

ted, I forget -- is it always OK too drop the basepoint of the fundamental group?

we will always end up with the same group, IIRC

If you're connected.

Isomorphic groups.

I mean certainly if you're connected

don't you need abelian?

Path-connected, I should have said.

No, copper.

Isomorphic.

this is why i am not a mathematician...

I'm a dead one.

suddenly, a physicist appears

according to Erdos

i only made it to #5

what are you?

#5 of what ?

erdos #

Oh.

half the world is #5

by a different measure my trump number is 3

and my obama number is 2

I'm 2 or 3 because of number theory at UGA.

I don't know wth I am

I want a Tromp number of $\infty$

excellent! i rely on my friends doing well...

I am guessing Cappell knew Erdos?

or how does that work

It's authoring papers, not knowing, Shmo.

well, the person through whom my trump number is 3 is the singularly most unlikely route!

oh, well

@TedShifrin "That's a lot of Trump number" - A Trump supporter

$\infty$ it is

glares at Balarka

you gotta say that in a redneck accent

yeah lol

really balarka? seems just slightly unlikely...

holding an American flag, staring into the sunset

then go storm the American Capitol

in an act of true patriotism

sry, i think i dragged this convo down

off to buy some gamestop...

Are political discussions allowed in this channel?

only if youre tenured

or inured

there's a due process of hazing that has to take place

@politeproofs Trump for Biden, otherwise Trump forbidden

hazing mostly due to climate change

@JoeShmo How does that go?

Ted has to smack you at least a handful of times

Fascinating

the great thing about the change of pres is that the onion is now satire again

A little bit odd too

hahahaha

that's good @copper.hat

:-) i used to like the quayle quarterly

Make the Onion Satire Again

So whoever wants to disclose, who did you vote for?

nothing to do with politics, i voted for any non trump option.

i might have picked pete the unpronounceable if it was an option

Only a handful of US citizens here.

My primary vote was for Liz Warren.

i am a citizen^*

she scared me

proud gringo here

I'm a radical lefty for the most part.

So does that mean Biden overall?

You are?

I'm decidedly center

i am center

Dr. Shifrin is full of surprises

Yes, my proudest vote was for George McGovern (who won only my then-state) in 1972.

left from kensington?

being able to vote against michelle bachmann was my proudest vote previously

The Democratic party is now where the Republican party was pre-Reagan.

:-)

huzzah for the overton window /so much sarcasm

@TedShifrin on economic policies, at any rate

Michelle Chitzpah Bachmann

Like Britain, everything has shifted radically right.

you gotta hand it to michelle bachmann

what in particular would you like to see done?

in terms of social issues (gay rights etc) it's been better

though with the post-trump judiciary...we'll see :S

Well, this idiot from North Georgia makes Bachmann seem sane.

point

well

hahaha

are you talking about senator doucheface?

i dunno. is it that bachmann is saner, or that bachmann didn't have the opportunity to be as batshit crazy

he was just voted out

Kill Pelosi, carry a rifle into Congress, and campaign on anti-Muslim.

i think michelle has some strange notions of what is unconstitutional.

my pelosi number is 2

No, Shmo, Marjorie whatever, just elected.

the republicans have really disintegrated

yeah, i dunno what the next midterm will be like

will the furious trump-nuts seize the primaries, and if they do will they be successful?

yup^^

I am very scared.

no way. that fiasco at the capitol was the nail on his coffin

Nope.

he effectively lobbied for almost 3 months after his loss to undermine democracy

Does anyone know what happened to Biden's promise to make illegal immigration less of a crime (or something along those lines?)

I mean, to your point, 74 million people voted for him

@JoeShmo the nail in Trump's coffin, maybe.

I don't think he can "make it less of a crime" I think he can choose to not enforce the law at the border

And not imprison and lose children.

@JoeShmo I don't remember what was the exact thing he said, but that's not all of immigration. For example, overstaying visas.

hot take: rudy giuliani needs to be institutionalized pronto

Will that still be a lifetime ban from the US?

the incoming biden administration is trying to put a halt to certain deportations

but they're getting hamstrung to some extent by certain things that the trump admin did on its way out

specifically agreements with various states like texas

It's been. One. Week.

i mean, whether said agreements will hold up under pressure is another question

but that's what the recent court case about immigration is about

What Congress can do is dubious.

yeah

i'm only talking about executive stuff to be clear

I really want to know what happens to overstaying visas

as an immigrant i must confess to being stumped by the lack of checking here. it was incredibly difficult for me to come to the us, let alone get my f1 renewed.

I do know he had a reconciling conversation with the Mexican president, where he did say that he wants things to "go back to normal" on immigration

this stuff, to be clear: cnbc.com/2021/01/26/…

but I haven't seen the details

amlo

what do you mean lack of checking?

biden wants a 100-day freeze on deportations, texas says that's not lawful, and judge agrees

@JoeShmo for example, in many countries to take part in society you have to prove you are a citizen (school, etc.)

yes, yes

right

which is reasonable, in my opinion.

where do you come from?

ireland

used to be a lot of illegal irish coming here.

recently?

i wish :-)

early 80's

sorry, i misunderstood your recently

For $y\neq0, y_0 \neq 0,$ is this a fake proof? $\left| \frac{1}{y} - \frac{1}{y_0} \right| = \left| \frac{y_0 - y}{yy_0} \right| = |y - y_0| \left| \frac{1}{yy_0} \right| < \delta \left| \frac{1}{yy_0} \right| < \delta \left| \frac{1}{y_0^2} \right| = \delta \frac{1}{y_0^2} = \varepsilon y_0^2 \frac{1}{y_0^2} = \varepsilon$

I am trying to do *22., but it's hard not to involve delta, which aren't even given.

In other words, if the proof above works, then it's easier for me to prove it from scratch, rather than use the given (implicit) delta by Spivak...

you need to upper bound $| {1 \over y y_0}|$, the above does not do that. you need to limit $y-y_0$ in some meaningful way.

It's upper bound is $\infty$ as either one of them are near 0

That's what I did, no?

I have that $0 < |y - y_0| < \delta$

Which I used in the next inequality

Why is $|{1 \over y}| < |{1 \over y_0}|$

Oh oops, where did a $y$ go...

Oh wait, no I know.

as tuco said, shoot, don't talk...

Hello everyone!

I have a question please if you can help!

42

no. no

908

😂

I'm struggling on finding a solution to a variant of the knapsack problem

@politeproofs try adding $\delta \le { 1 \over 2} |y_0|$ into the mix.

oh.. you might need to go to cstheory for that

new jersey?

nobody here would know

@copper.hat Yeah, that's one of his minimums, but where does he pull that from?

you need to bound $y$ away from zero.

I am in NJ. How did you know? That's creepy

908 is a nj area code

Actually , I have a list of numbers { 1, 5, 5, ... N } with duplicates ( not necessarily) , and n is not known and can change

@Mohtaa wrong audience and i am going to bed shortly...

Oh ok

ha! I made that up. but probably subconsciously put 908

Thanks anyway!

@politeproofs if, for example, $\delta \le {1\over 2} |y_0|$ then $|{1 \over y}| \le 2 | {1 \over y_0}|$.

@Mohtaa sorry, didn't mean to shoo you away, just didn't want you to waste your time.

But where is that coming from?

Not a problem :)

@politeproofs draw a little diagram.

show $|y| \ge {1 \over 2} |y_0|$.

Diagram

But this is kind of cheating! I am not supposed to know that this is the limit of 1/x

I just happen to know that, because it's obvious

Oh oops

It should be "y" where it says delta, of course

Well

This is an interesting picture copperhat

I can see why it's clear from this

But I also don't know where I would get the picture with a) knowing that I am, in fact, computing a limit b) thinking of drawing it

at the very least, a good picture is a good plausibility argument

it shouldn't necessarily seal the case but it can convince you to pay attention to part of it

Slightly better picture, by the way.

But I can't help but feel like this is pulling a rabbit out of a hat for this problem. Oh well.

amusingly

one of the problems I helped a student with today was, in effect, showing that $(y\pm \delta)^{-1}\approx 1/y\mp \delta/y^2$

(this was in the context of uncertainty calculations, e.g., you know that a period of oscillation is $1.6 \pm 0.2$ seconds and want to figure out the frequency $f=T^{-1}\approx 0.63\pm 0.08$ Hz. no real analysis in sight)

@politeproofs i meant to convince yourself that for the above $\delta$ that $|y| \ge {1 \over 2} |y_0|$. Inverting gives the desired result.

Wow, that is not great to read.

Slightly better. Do I have the right formula?

@copper.hat Thanks, but I'll do it later. I don't like limits very much. Especially when they're not even limits!

I think you need a sum somewhere...

Why is that?

sry, looks good.

Thanks, onto part b) then. This exercises derives the Lagrange interpolation formula, which is nice

(Or so Spivak claims at least)

i think there is a better notation (divided differences) but i am going shortly.

Hmm. Is the Lagrange interpolation formula $$a_i \frac{\prod_{\substack{j = 1 \\ j \neq i}}^{n } (x - x_j)}{\prod_{\substack{k=1 \\ k \neq i}}^{n} (x_i - x_k)}$$ for some given $a_i$?

that looks right

Oh, it's so simple!

well

Well?

hmm

That's not what I like to hear.

well, compare with the wiki article: en.wikipedia.org/wiki/Lagrange_polynomial

though actually you can combine top/bottom there

Oh, why is there a sum?

Well, that's -one- Lagrange polynomial. You need a basis of such when doing interpolation

$$\prod_{\substack{j = 1 \\ j \neq i}}^{n } \frac{x - x_j}{x_i-x_j}$$

is the simpler version

Oh, yes it is. But wait, the sum..

point is that this polynomial equals 1 at $x=x_i$ and 0 at all other $x_j$

Yes, but then we multiply by a_i

So it should satisfy what we want

so if you want to represent an arbitrary set of values, you need to get $f(x_1)$ at $x=x_1$, $f(x_2)$ at $x=x_2$, etc

right

with $a_i=f(x_i)$

Right

Doesn't mine do that? Why the need for the sum?

i mean, where do -all- $a_i$ enter in?

If you don't have the sum, you're only using one $a_i$

Well, if you'd like, we can write it as $f(x_i,a_i) = a_i \frac{\prod_{\substack{j = 1 \\ j \neq i}}^{n } (x - x_j)}{\prod_{\substack{k=1 \\ k \neq i}}^{n} (x_i - x_k)}$

Is kind of my point

do an example. take $x_1=1$, $x_2=0$, $f(x_1)=1$, $f(x_2)=0$

oh, that won't work. need at least three points

(since for two points you're just doing a linear fit)

I think I see your point though, if we have $x_1 = -1, x_2 = 0, x_3 = 1$, then $f(x_i) = a_i (x-1)x(x+1)$

And suppose that "x_i" in this example is $0$, but $a_i = 5$ or something. Then $f(0) = 0,$ not 5

?

not quite. if you're taking a product from $j=1$ to $3$ and omitting one index, then that's only two factors in the product

so you'd have, for instance, $$f_1(x) = \frac{(x-x_2)(x-x_3)}{(x_1-x_2)(x_1-x_3)}=\frac{x(x+1)}{(1)(2)}=\frac12 x(x+1)$$

that's the first interpolating polynomial

Oh yeah, oops

the second is $$f_2(x) = \frac{(x-x_1)(x-x_3)}{(x_2-x_1)(x_2-x_3)}=\frac{(x-1)(x+1)}{(-1)(1)}=-(x-1)(x+1)$$

Wait, but that works though!

$f_2(0) = a_2$

If we multiply your latest one by some constant $a_2$

last is $$f_3(x) = \frac{(x-x_1)(x-x_2)}{(x_3-x_1)(x_3-x_2)}=\frac{(x-1)x}{(-2)(-1)}=\frac12 x(x-1)$$

$f_2(0)=1$, sure

but $f_2(1)=0$

so $f_2(x)$ doesn't do anything at $x=x_1$ (or $x=x_3$)

You can use each of the three Lagrange polynomials to achieve -one- of the three points

Okay, I see now. Thanks.

but combining them as $\sum_{i=1}^n a_i f_i(x)$ works yadda yadda yadda

I regret using $f_i(x)$ for the interpolating polynomials tho

better to do $\ell_i$'s to distinguish it from the function $f$ you'd be fitting

but the full formula would be

$$f(x)=\sum_{i=1}^n f(x_i)\prod_{\substack{j = 1 \\ j \neq i}}^{n } \frac{x - x_j}{x_i-x_j}$$

there's also linear algebra ways of looking at this, which the wiki article discusses

with all the products being related to "hurr, invert the vandermonde matrix"

Why does this work?

Given 3 points when interpolating $x^2$, you get exactly $x^2$.

because the interpolating polynomials each have degree which is one less than the number of points used

I suspect it's something to do with that a degree 2 polynomial has at most 2 complex roots, but I don't know how to reason through it completely

so three points -> quadratic approximants

Yes, but I mean why does it work?

because you're trying to approximate a quadratic polynomial using quadratic polynomials

you already have one quadratic polynomial which goes through your three points in that case; you're not going to suddenly come up with another

ultimately, what Lagrange interpolation gives you in that case is a polynomial $Ax^2+Bx+C$ which passes through the three points

but you already have a quadratic polynomial which does that, namely $x^2$ itself

How are the $a_i$ decided?

$a_i=f(x_i)$

those are the coefficients you give to the basis polynomials, of course

Oh durr

to put it a bit differently, Lagrange interpolation gives you a basis of polynomials which can describe $n$ fit points

but you already start off with the basis $1,x,x^2,\cdots,x^n$

so if the function you pick happens to have a nice representation in the latter basis, then Lagrange interpolation will (after a lot of work) just reproduce that

more generally, an $n$-point Lagrange interpolation of a degree-$(n-1)$ polynomial is always exact. it just reproduces the original function in a different basis

the nontrivial part of this is really that the interpolating polynomial is unique

If you had two such polynomials $f(x), g(x)$, then $f(x)-g(x)$ would be a degree-$(n-1)$ polynomial with $n$ zeros. No go.

Makes sense

A perpendicular vector of that plane is a multiple of $(1,2,3)$.

So $t(1,2,3)$. The length has to be 5, so $|t|\sqrt{1+4+9}=5 \Rightarrow |t|\sqrt{14}=5\Rightarrow |t|=\frac{5}{\sqrt{14}}$

Ahhh so there only two vectors that satisfy these conditions, one with $t=\frac{5}{\sqrt{14}}$ and one with $t=-\frac{5}{\sqrt{14}}$

right?

If I think of that scenario geometrically, I don't see how you could have a third vector

Take the case of length 2 and the plane z=1, say

the vectors of length 2 form a sphere

but to be perpendicular to z=1 a vector has to point in the +z or -z direction

so that limits you to either (0,0,2) or (0,0,-2)

that's what you have, tho, so i agree

Great!!

A vector is perpendicular to a plane if and only if it is multiple of the vector of coefficients of the equation of plane, right?

Or isn't it "iff" and just "if" ?

if and only if, so long as you require it to be a nonzero multiple

Aaa ok!! Thank you!! :-)

@Semiclassical Hi

hi

It is been a long time when you were in h bar?

yeah

I am struggling with Qunatum field energy can u help?

no

:(

If p is a polynomial of degree 4 with complex coefficients and $z_0\in \mathbb{C}$ is a root then it doesn't mean necessarily that the conjugate $\overline{z}_0$ is also a root, right? That would hold if we have real coefficients, right?

yep

e.g. $(z-z_0)^4$

welche einige Lösung ist $z_0$

Great! Thank you!!

I have also an other question : If $z\in \mathbb{C}$ such that $1+2z+(2z)^2+(2z)^3+(2z)^4=0$ then it follows that $|z|=\frac{1}{2}$ ? Do we have to use a formula for a sum? Or how do we have to do that? @LeakyNun

hint: geometric series

I thought about geometric series but then I thought that since we don't know that $|2z|<1$ we couln't use that. @LeakyNun

you're thinking about infinite geometric series

don't forget the basics (finite geometric series) after you learn the advanced (infinite geometric series)

Ok.... So we have that $$1+2z+(2z)^2+(2z)^3+(2z)^4=0 \Rightarrow \frac{1-(2z)^5}{1-(2z)}=0 \Rightarrow 1-(2z)^5=0 \Rightarrow (2z)^5=1$$ @LeakyNun

yep

That means that we don't get $|z|=\frac{1}{2}$ but $|z|=\frac{1}{2^5}$, right? @LeakyNun

why?

Ah no... It is: $(2z)^5=1 \Rightarrow |(2z)^5|=|1| \Rightarrow |2z|^5=1 \Rightarrow |2z|=1^{1/5}\Rightarrow |z|=\frac{1}{2}$ @LeakyNun

yep

In space if line $\ell_1$ is perpendicular to line $\ell_2$ and line $\ell_2$ is perpendicular to line $\ell_3$, then $\ell_1$ is parallel to $\ell_3$, right?

@MaryStar $(0,0,1),(0,1,0),(1,0,1)$, so not in $\mathbb{R}^3$.

But, in $\mathbb{R}^2$ or $\mathbb{C}$, yes.

@robjohn Ok! Thank you!! :-)

I am trying to prove that if $f(y) - f(x) \le (y - x)^2$ for all real $x,y$ then $f$ is constant.

I seem to be stuck, however. I could do it if I knew that $f$ was differentiable, but without that, I am not sure how to proceed

Well, is f differentiable?

We don't know?

What does differentiable mean?

That $\lim_{h \to 0} (f(x+h)-f(x))/h$ exists

Ok. Can you not find that limit from the above equation?

I would be assuming that it exists, would I not?

Well, before taking the limit, you can make the correct terms appear

Also, it's not quite an equation, it's an inequality

True

Several proofs and disproofs of Riemann Hypothesis vixra.org/author/shekhar_suman