Mathematics - 2018-02-05 (page 4 of 6)

Akiva Weinberger

16:00

I'mma just see if there's anything in Chapter 0 that I don't know

Balarka Sen

but its cool i guess

0celo7

@BalarkaSen stop trolling

Balarka Sen

pot calling kettle

Akiva Weinberger

Is the Riemannian category badly behaved?

Or, other side, is it too nice?

Balarka Sen

It's rigid

0celo7

16:01

It's not even a category

He's trolling

Akiva Weinberger

You don't have morphisms?

Balarka Sen

Depends on what you want to be morphisms

0celo7

It's not clear what "the" morphisms should be

And it's not useful in any case

Akiva Weinberger

Weird, for most things it's obvious

0celo7

@AkivaWeinberger Riemannian geometry simply does not benefit from categorification.

Balarka Sen

16:02

a Riemannian manifold is not a categorical object. It has too many informations imposed on it

0celo7

^ this

Akiva Weinberger

Hrm

Balarka Sen

The most basic categorification is to realize that a Riemannian manifold is a smooth manifold $M$ with a section of $T^* M \otimes T^* M$. Then you make it a category by defining morphisms to be the smooth maps $M \to N$ which preserve that section. (This is just fancy language for saying the Riemannian metric is preserved under the map)

But that's at the cost of loosing a lot of local information.

I'm literally assigning isometries to be morphisms

Akiva Weinberger

> Although the necessity of an abstract idea of a surface (that is, without involving the ambient space) is clear since Gauss, it was nearly a century before such an idea attained the definitive form that we present here.

Weird

> One of the reasons for this delay is that the fundamental role of the change of parameters [meaning transition maps?] was not well understood, even for surfaces in $\Bbb R^3$.

Balarka Sen

Transition functions, yes

Akiva Weinberger

16:07

@BalarkaSen Is there a category of metric *spaces?

Balarka Sen

Sure, @Akiva.

Isometries are the morphisms

Akiva Weinberger

But then all morphisms are isomorphisms

Balarka Sen

Well, you mean, metric spaces?

0celo7

So the morphisms are isomorphisms = shit category

Akiva Weinberger

@BalarkaSen Yes whoops

Balarka Sen

16:07

Well a better one is to say the morphisms are isometric embeddings

Mike Miller

Isometric immersions

Akiva Weinberger

Hi

Hm

So like there's a morphism from $[0,\infty)$ to $(0,\infty)$ and vice versa?

Mike Miller

But still, the geometry here is about individual manifolds (sometimes the extrinsic geometry of submanifolds). The way they relate is not so interesting.

Akiva Weinberger

(Cont'd) I guess that's kinda interesting but it does sound like a shitty category. And Riemannian manifolds are kinda just a variation on metric spaces, I think, so it makes sense that they'd make a shitty category as well.

Mike Miller

Some kinds of special maps (isometric immersions, Riemannian submersions) are intetesting

Balarka Sen

16:10

@Akiva The length metric coming from the Riemannian metric is just one facet of the Riemannian metric

Akiva Weinberger

Riemannian submarines?

Danu

o/

Akiva Weinberger

ol

Mike Miller

The most common way I would think about changing Riemannian manifolds is either changing the metric on a fixed manifold or through Gromov-Hausdorff limits

Danu

Riemannian submersions are great!

Akiva Weinberger

16:10

Balarka Sen

Hm, what's a quick example of the same (isometric?) length metrics coming from two different (non-isometric?) Riemannian metrics?

(Bonus question: What's the right question here?)

(Thanks for both asking and answering my question in advance)

Danu

I think maps that are metric-space isometries are Riemannian isometries too, and vice versa

(maybe you want some basic condition on the map)

Mike Miller

I think C^0 is good enough

Balarka Sen

Strange.

0celo7

@MikeMiller It is.

Danu

16:13

Right

Mike Miller

It's why isometries are closed in Homeo

Akiva Weinberger

> Differentiable always signifies of class $C^\infty$.

0celo7

Isometries of metric spaces are continuous anyway

Akiva Weinberger

What happens if we change that to "analytic"?

We lose bump functions.

Does it get too rigid?

Danu

Then everything is bad :D

Akiva Weinberger

16:14

(Bonus: What happens if we change that to "polynomial"? :P)

Balarka Sen

You get back the theory of algebraic varieties

More or less

Danu

What is the intersection of $SO(4)$ and $SO(2,2)$?

(besides the multiples of the identity?)

Akiva Weinberger

Weirdly, do Carmo uses $\phi$ for the empty set

(Or at least the translation does)

(Rather than $\emptyset$ or $\varnothing$)

0celo7

I think that's an artifact from the early days of TeX

Danu

Right so you just need a bijection (@BalarkaSen @MikeMiller @0celo7) according to Petersen's book.

Akiva Weinberger

16:16

What does $SO(2,2)$ mean?

Balarka Sen

@Mike @Danu @0celo7 Ah OK the point is not that the length metric is not determined by the Riemannian metric (which is false, as you three answered). The point is the subspace theory of the length metric space and the subspace theory of the Riemannian manifold are different

0celo7

@Danu Yes

Danu

Like Ocel07 said

@Akiva preserving the standard symmetric bilinear form of 2,2 signature

Slereah

@AkivaWeinberger orthogonal group for metric of signature $(2,2)$

Akiva Weinberger

Right, so I don't know what that means

Never mind then

Danu

16:17

$dx_1^2+dx_2^2-dx_3^2-dx_4^2$

0celo7

@AkivaWeinberger $(x,y)=-x_1y_1-x_2y_2+x_3y_3+x_4y_4$

Akiva Weinberger

So, not rotations.

0celo7

No, something more sinister

Akiva Weinberger

Weird Lorentz-looking things.

Balarka Sen

So the right question here should be about the space of Riemannian submanifolds of the Riemannian manifold, whatever that means

Danu

16:18

you can still think in analogy with rotations

Just have some hyperbolic stuff to deal with

Mike Miller

These live inside GL(4)?

0celo7

@BalarkaSen It's true that the metric can be recovered from the distance function.

Akiva Weinberger

Like rotating around a hyperbola instead of a circle?

Danu

@MikeMiller Yea

Mike Miller

That's where I'm thinking about the intersection, I mean?

Danu

16:19

yea

Slereah

@MikeMiller As do most lie groups

Danu

I think I found one element that lies in both

accidentally

0celo7

@Danu the identity?

Balarka Sen

@0celo7 Yeah, I did not know that. Should I do the exercise?

Danu

but I somehow thought they shouldn't really intersect except in the obvious part

no

it's the image of $$ \begin{pmatrix} a & b \\ b^* & a \end{pmatrix}$$ ($a$ is real, $b$ is complex, $a^2-|b|^2=1$), viewed as an element of $GL(4,\Bbb R)$ under the natural inclusion of $GL(2,\Bbb C)$.

0celo7

16:21

@BalarkaSen I should probably give you a hint. Given a curve $c(t)$ with $c'(0)=v$ (any vector), show that $|v|_g=\lim_{t\to 0}[d_g(c(t),c(0))/t]$. Then obtain the metric via polarization

Akiva Weinberger

So when the book says "The family $\{(U_\alpha,\mathbf x_\alpha)\}$ [parametrizations or systems of coordinates, where $U$ is a subset of $\Bbb R^n$ and $\mathbf x$ is a function] is maximal", it means that if any potential new $(U,\mathbf x)$ pair that has differentiable transition functions with everything else is included?

That was poorly worded

Balarka Sen

@0celo7 Oh well you gave away the idea

Akiva Weinberger

But like, I can't add anything new, either it's already in there or it's not compatible with existing stuff

is what it means, right?

Mike Miller

@Danu The intersection is 0-dimensional at least

Danu

@MikeMiller because?

Akiva Weinberger

16:23

Something I want to learn at some point is Lie theory. But, Jesus fucking Christ, the list of things I "want to learn at some point" is as long as the list of movies I "should watch at some point"

Friggin' immense

Mike Miller

Write uh $I'$ for the diagonal matrix (1,1,-1,-1)

Danu

I think I messed up some part of the computation, so I guess it's not in $SO(4)$ afte rall haha

Akiva Weinberger

(I recently saw The Social Network. It was good)

Danu

it's definitely in $SO(2,2)$

@MikeMiller Yes..

0celo7

@BalarkaSen I don’t think it’s trivial to show what I claimed

Mike Miller

16:24

Then the Lie algebra of SO(2,2) is X^T = I'X

Danu

Right

differentiate X^T I' X=I'

Balarka Sen

@0celo7 You want to show that $|-|_g$ is a norm, you mean?

on $T_p M$

0celo7

No, I mean that limit gives the norm given by the metric

Mike Miller

Yeah sorry. Anyway this intersecfs trivially with the Lie algebra of so(4)

0celo7

Computing that limit is not so easy, IIRC

Balarka Sen

16:25

Ah ok. I'm guessing you need geodesic coordinates for this computation

Danu

@MikeMiller Oh, cool

Akiva Weinberger

Also, Ocelot, you're becoming a permanent resident of the chat, you've been here so often lately

0celo7

Yes it will involve some sort of expansion

Danu

thanks

Mike Miller

You get an equation just in terms of X and I' which is unsolvable except 0

0celo7

16:26

@AkivaWeinberger Thanks? What do you mean by also

Danu

yeah, makes sense

Mike Miller

I apologise for lack of detail, typing on my phone is a pain right now

Danu

it's clear enough

Akiva Weinberger

@0celo7 "In addition to the completely unrelated thing I said before that"

0celo7

@BalarkaSen there’s a good post by John Ma on expansions of the distance function for small distances

@AkivaWeinberger you mean the removed thing?

Akiva Weinberger

16:28

Nah, the thing on how I want to learn Lie theory at some point

Irrelevant either way

Balarka Sen

I forget geodesic coordinates. It's a chart $(U, (x^1, \cdots, x^n))$ such that the flowlines of $\partial/\partial x^i$ at the origin gives geodesics inside of $U$, right?

(So $\nabla_{\partial/\partial x^i} \partial/\partial x^i = 0$)

0celo7

I’m on my phone

Balarka Sen

Can I also choose "orthogonal"? $\nabla_{\partial/\partial x^i} \partial/\partial x^j = 0$?

I think so

0celo7

@BalarkaSen at the origin, yes

@BalarkaSen yes

But just at a point

Balarka Sen

Right, of course

Otherwise I'll get a flat chart which will break curvature

Speaking of I should prove that

Get an expansion of the metric in terms of curvature etc

0celo7

16:38

My thesis has the computation to fifth order. Ugh

It will have it.

Maybe

@BalarkaSen the easiest way to do the computation is Jacobi fields.

You again use a polarization trick after using Jacobi fields to compute the norm

Balarka Sen

I shall spend some time thinking about the story on the large

BAYMAX

Hi chato!

any help on this1

0

Q: Conversion of Surface integrals to a suitable Volume integral.

While deriving the Euler's equations of motion in case of Fluid dynamics, I came across this part - Here $p$ denotes the hydrostatic pressure(scalar function) I am unable to understand how it transformed this - $\int \int _{\Delta S} - p\hat{n}$ ds $= -\int \int \int_{\Delta V} \nabla p dv$ ...

Silent

Hi

Akiva Weinberger

16:53

$\hat\chi$

BAYMAX

$\hat{\chi}$ \ $\{c\}$

Akiva Weinberger

Translating a math textbook must be easy and hard at the same time

berrygreen

Does anyone have experience with wiki software for math use? If so which software are you using? Preferable one with full latex integration.

Akiva Weinberger

Easy 'cause you probably don't have to deal with foreign idioms, wordplay, or words that have different emotional connotations in the source language than the target language

Hard 'cause you have to know what every bit of jargon translates to, which you won't find in a regular bilingual dictionary

Balarka Sen

I plan to write a mathematical hard science fiction at some point which will be written in $\infty$-categorical language

Akiva Weinberger

17:00

@berrygreen Whatever ProofWiki uses apparently works

berrygreen

@Akiva they use MediaWiki, the same software as proper wikipedia. That is too bulky for my needs.

Also their latex integration really poor

Silent

Suppose $\limsup_\limits{n\to \infty}\sqrt[n] {|a_n|}$ is $+\infty$, does $\sum a_n$ converge? Similarly, does ratio test work for $+\infty$, $-\infty$?

Slereah

If I have a submanifold $S$ of codimension 1 of a connected manifold $M$ such that $M \setminus S$ is disconnected in two pieces $M^+$ and $M^-$, is there a level-set function that exists such that $f(S) = 0$, $f(M^+) > 0$ and $f(M^-) < 0$?

Balarka Sen

Yes.

Slereah

Good

Where might I find a proof for this?

($S$ isn't necessarily connected, by the way)

0celo7

17:08

@AkivaWeinberger I’ve translated some technical French, it isn’t so hard. If you already kind of know what’s going on, it almost writes itself

Something very wordy like metric geometry might be quite hard though

Eric Silva

@BalarkaSen id read it tbh

Balarka Sen

@Slereah Say $N$ is a tubular neighborhood of $S$ inside of $M$. That has a projection map $p : N \to S$ which makes $N$ into a line bundle (in fact it's the normal bundle). Your condition implies $N \setminus N_0$ where $N_0$ is the bundle's zero section is disconnected; this gives a trivialization $\varphi : N \to S \times \Bbb R$ of $N$.
The second component of this trivialization is a map $f : N \to \Bbb R$ which is a homeomorphism of $\Bbb R$ fixing zero on each fiber - that automatically means $f$ is positive on "$N^+$" and negative on "$N^-$", the two components of $N \setminus S$. $…

Slereah

Thank you my dude

Balarka Sen

There's some details to be made clear here

One is that the trivialization of $N$ guarantees the positiveness/negativeness of $f$ on each fiber of $N$ is consistent. Another is to write down an extension of $f : N^+ \to (0, \infty)$ to $M^+ \supset N^+$.

Slereah

i'm trying to show that the Israel junction condition makes sense

Not easy because it's all gobbledigook

Balarka Sen

17:22

The extension can be done by taking the constant function $1$ on a neighborhood of $M^+ \setminus N$ and then using partition of unity on $(M^+ \setminus S, N^+)$ to glue $f$ and $1$ to an everywhere positive function on $M^+$.

Similarly do it with $-1$ on a neighborhood of $M^- \setminus N$ to get an everywhere negative function on $M^-$

These agree on $N$ and is zero on $S$.

Slereah

Thanks!

Balarka Sen

No problem.

Slereah

I was worried there might be weird cases where the function is positive on both connected parts

Balarka Sen

Yeah the point is you can do it locally on a neighborhood of $S$ by the tubular neighborhood theorem

0celo7

@BalarkaSen check discord

Balarka Sen

17:25

In this scenario that local disconnection is in fact a global one

@0celo7 Beautiful

0celo7

@BalarkaSen look at all the norms

I didn't write that shit down

I took a picture for later reference but then thought you'd enjoy it

Balarka Sen

I appreciate the meme

Akiva Weinberger

17:50

@0celo7 Cool!

Dattier

Let $(a_n)_n$ a real sequence such : $\forall n \in \mathbb N, \forall m \in \mathbb N^*,\exists n_0 \in \mathbb N,\forall k\geq n_0, |a_n-a_k| \leq \frac{1}{m} $.

Is it true that $\forall n\in \mathbb N, a_n=a_0$ ?

Is the BGR (Big General Result) lemma true ?

Akiva Weinberger

I also imagine the ratio of words to equations influences things

@0celo7 What did you do if there was a technical word you didn't know how to translate?

Perhaps buy textbooks on the subject in each language and compare, or something

Silent

What does activating notification do?

ok :)

Akiva Weinberger

@Silent Probably something on how, if someone pings you, you might get a notification on the main site

@Dattier $\Bbb N^*$?

@Dattier I think $a_n=(-1)^n$ is a counterexample

mathsta

Is $\mathbb CP^m$ a submanifold of $\mathbb CP^n$?

Akiva Weinberger

17:55

It sounds like you want every point in the sequence to be a limit point of the sequence

Balarka Sen

If $m < n$, it sits as a submanifold, yes @mathsta

mathsta

ok, thx

Dattier

@AkivaWeinberger choose m=2

0celo7

@AkivaWeinberger that's rather rare

Ted Shifrin

In lots of ways, often, @mathsta. I presume you mean just as a "linear subspace."

Hi DogAteMy, Balarka.

mathsta

17:58

@TedShifrin What do you mean with often? On what does it depend?

Ted Shifrin

@Mathsta: For example, you can embed $\Bbb CP^1$ into $\Bbb CP^2$ as a line or as a conic (and there are lots of ways to do each).

s.harp

Hi, is the Verdier dual of a sheaf usually defined as the right derived functor of $Hom_{Sheaves(X)}(-, S_X)$ where $S_X$ is the complex of singular chains?

This is how my text introduces it but it seems a bit too on the nose

Balarka Sen

or P^1 in P^3 as twisted cubic, say

just to give a fancy example

mathsta

@TedShifrin So you say, that we can find CP^m in many different ways in CP^n, but not all of them are actually submanifolds

Balarka Sen

No these are submanifolds

Ted Shifrin

18:00

No, I'm giving embeddings, so the images are submanifolds.

You could take a map of $\Bbb CP^1\to\Bbb CP^2$ whose image is a nodal cubic (so it would not be a submanifold).

s.harp

We are doing topology and the text is trying to motivate how a dual needs to look and then says well its just $RHom$ with the singular chains of closed support, which somehow as a sheaf is too topologically relevant to be believable as a candidate for an algebraic idea

Ted Shifrin

There are interesting embeddings, called the Veronese embeddings, of $\Bbb CP^m$ into $\Bbb CP^n$ if you pick a particular value of $n$ in terms of $m$.

mathsta

Sorry, I am having troubles to follow

Ted Shifrin

@mathsta: Where did your question come from?

Akiva Weinberger

@Dattier Oh, I see. Sorry

Ted Shifrin

18:03

DogAteMy: I'm not paying attention, but I sure would guess that it needs to be a constant sequence with those crazy nested quantifiers.

Akiva Weinberger

@Dattier Right, I think you're right, then, it would be a constant sequence

Sorry

mathsta

@TedShifrin Actually I was wondering if $CP^1#CP^1$ is a complex submanifold of $CP^2#CP^2$

Ted Shifrin

If you said $\exists k\ge n_0$, then it would be about subsequences converging to any given element of the sequence.

Balarka Sen

Is there a complex structure on CP^2 # CP^2?

Ted Shifrin

@mathsta: First of all, connect sum does not stay in the holomorphic category.

<--- yields to Balarka :)

Akiva Weinberger

18:05

$\forall n_0\in\Bbb N,\exists k\ge n_0$, yeah

Balarka Sen

CP^2 # \bar{CP^2} does

Akiva Weinberger

The reverse of what it actually is

Balarka Sen

But those are different manifolds

user193319

Dumb question: if I take the arbitrary union of a collection of path-connected sets, all of which have at least one point in common, will the union be path-connected?

Ted Shifrin

Sure.

Akiva Weinberger

18:06

Yes.

Balarka Sen

@mathsta This question, for example, says $\Bbb{CP}^2 \# \Bbb{CP}^2$ does not even admit an almost complex structure.

Akiva Weinberger

To connect two points, draw a path from the first point to the common point and from the common point to the second point

mathsta

Oh, ok.

Ted Shifrin

@Balarka: Make sure you send Mike a royalty check.

mathsta

But without requesting a complex structure, is it a submanifold?

Ted Shifrin

18:08

So you're just staying in the topological/smooth category ... What's the big deal? Is there a manifold structure on the connected sum? Sure.

mathsta

And we can embed it in the connected sum the manifolds these are submanifolds of?

Balarka Sen

Sure, CP^1 # CP^1 is just S^2. You can embed S^2 in CP^2 # CP^2 in many ways

Dumb way: Take a chart in the manifold and embed S^2 inside that ball.

Ted Shifrin

Connect summing with $S^2$ sure is boring :(

mathsta

ok, thx

Balarka Sen

@Mike shows empty pocket; cricket.avi

Slereah

18:14

There should be a big sign for connected sum

$$\underset{i}{\large{\#}} T^2_i$$

Balarka Sen

$\large \#$

$\huge \#$

Ted Shifrin

Hmm, I am not sure I know how to do that in chat.

@Balarka: You're preoccupied with the empty connected sum. A new meme.

Balarka Sen

Use \large and \huge

Hahah

I'll star that just to reverse psychology this room into making that a meme

Ted Shifrin

I guess I hadn't thought those font controls would work in ChatJax.

Balarka Sen

Me neither

PabloZ392

18:17

Hello :) I am looking for example linear function but not continuous ..

Akiva Weinberger

What's $X\#\emptyset$?

:P

I imagine undefined

Ted Shifrin

I wonder if Demonark and @AlessandroC figured out that question about $\ell^\infty\to c_0$.

Slereah

How do you remove a ball from the empty set

Ted Shifrin

It's defined, DogAteMy.

You do it vacuously.

Akiva Weinberger

But you can't puncture the empty set

Slereah

18:18

Wouldn't $M \# \varnothing$ just be $M \setminus B$

Ted Shifrin

Hmm ...

Since it's vacuous, you don't puncture $M$ either. :)

Balarka Sen

Nah it's M

Slereah

Can you remove a ball of radius 0

Balarka Sen

Yeah @Ted

Slereah

I dunno

Akiva Weinberger

18:19

I guess $d(p,\emptyset)=\infty$ as well :P

Ted Shifrin

Yale will be so proud of how we've trained little DogAteMy :P

Fred

why can't you find iqr in ungrouped frequency table using cumulative frequency?

Ted Shifrin

This is not a statistics room, Fred.

Fred

why only in grouped data

oh vell

its still math

Ted Shifrin

We only have one chatter who knows any of this stuff, and he's rarely here.

Fred

18:21

so what do you talk about

Ted Shifrin

More mainstream math from middle school level to graduate level and more.

Some of us know a little probability, but that's very different from technicalities of statistical inference.

Fred

i'm only doing my gcse , i'll post it to the forum but everyone will probably vote it down becuase I need to upload a picture of the table

Ted Shifrin

I have no idea what you just said, but OK.

Fred

and no-one likes pictures ,it has to be in latex

Ted Shifrin

Sorta like turning in scribbled and crossed out homework exercises — frowned upon.

Fred

18:24

no its straight from the textbook so we'll see

Abcd

How do I find the maxima of $\dfrac{x^2}{1+x^2}$?

Ted Shifrin

Do you think you know what the graph looks like?

Abcd

I am trying to solve this question: $\arctan{\dfrac{x^2}{1+x^2}}$

Till now I have reached $\arctan y\ge 0$

I am trying to find the maxima of y to get the upper bound.

@TedShifrin Yeah, it's tending to a maximum but not reaching it.

Ted Shifrin

You're right. It approaches $1$ asymptotically.

Xander Henderson

heh

do you know about derivatives?

Ted Shifrin

18:29

You don't need calculus for this.

Abcd

@XanderHenderson Yeah.

Ted Shifrin

You didn't finish, @Abcd. What is the question you're trying to solve?

Abcd

2 mins ago, by Abcd

I am trying to solve this question: $\arctan{\dfrac{x^2}{1+x^2}}$

Ted Shifrin

That's not a question.

Xander Henderson

@TedShifrin Indeed, you don't, but that is a pretty quick tool for showing that it is monotonic on $[0,\infty)$

Abcd

18:30

Find the value of $\arctan {\dfrac{x^2}{1+x^2}}$

Ted Shifrin

Just write it as $1-1/(1+x^2)$.

Xander Henderson

Though, come to think of it, the direct approach might be easier

Ted Shifrin

What does that question mean, @Abcd? Find the value??

Abcd

@TedShifrin Such tricks don't strike me :/

@TedShifrin It means the range...

Ted Shifrin

It's elementary algebra. Like with your complex number thing. You mess up on easy stuff because you're too busy trying to be fancy.

Xander Henderson

18:31

...

Ted Shifrin

Oh, the set of values ...

So $0\le \frac{x^2}{1+x^2}<1$.

@Abcd: Saying the values would have been clear. I'm sorry such tiny details in English make such a big difference in understanding.

Xander Henderson

If all you want to do is find the range, note that $f(x) = \frac{x^2}{1+x^2}$ is continuous, $f(0) = 0$, $\lim_{x\to \pm\infty} f(x) = 1$, and the function is monotonically increasing on $[0,\infty)$, and that $f$ is even

this implies that $f(\mathbb{R}) = [0,1)$

Ted Shifrin

NO, Xander.

Xander Henderson

what am I missing?

Ted Shifrin

$f(\Bbb R) = [0,1)$.

Xander Henderson

18:34

oh, right.

ALL REAL NUMBERS ARE EXTENDED REAL NUMBERS!

Ted Shifrin

smack

Abcd

what is happening.

Xander Henderson

the only reason that I care about monotonicity is that I want to make sure that $0 \le f(x) \le 1$ for all $x$; you can also do this directly

Ted Shifrin

Yes, you could easily invert the function on $[0,\infty)$.

But IVT is just lovely.

Xander Henderson

and it isn't really that hard: $f(x) \ge 0$ for all $x$ as it is the ratio of positive numbers

and $x^2 < 1+x^2$ for all $x$, so $f(x) < 1$

in either case, as @TedShifrin points out, there is an application of the intermediate value theorem going on, which is why continuity is important

Bennett

18:43

Does anyone know the proof of $\sum a_i + \prod (1-a_i) \ge 1 $ via derivatives?

$a_i \in [0,1]$. It considered the LHS a polynomial, iirc.

Tuki

Hello chat

Bennett

I can't find it online anywhere.

I think it said $ f= \sum a_i + \prod (1-a_i) $ is a constant polynomial or something.

Kasmir Khaan

@TedShifrin Ted :D

@MatheinBoulomenos How did it go on your exam? :D

Tuki

your function is discrete @Bennett ?

Ted Shifrin

That doesn't sound believable, @Bennett.

Hi, Kasmir

Bennett

18:48

@Turi yeah, @TedShifrin I must be misremembering it. It first defined a family of sequences (subsequences?) or something.

I even know where I saw the proof first.

But the website is down.

Ted Shifrin

Bennett: So think inductively. It's clear with one $a_i$.

Abcd

If $\alpha = e^{i\dfrac{8\pi}{11}}$, then find $Re(\alpha+ \alpha^2 + \alpha^3+\alpha^4 + \alpha ^5)$

Bennett

@TedShifrin Yeah, I know how to prove it via induction. It's just that particular proof left an impression on me.

planetmath.org/node/34744/pmgroup

That's where it was.

Abcd

Its so simple, I don't know why I don't get the right answer.

Using euler's formula and cosine summation series:

Bennett

planetmath.org/sites/default/files/texpdf/34744.pdf and here too.

Ted Shifrin

18:50

Now when we go from $n$ of them to $n+1$ of them, we add $a_{n+1}$ and subtract $a_{n+1}$ times a number that's between $0$ and $1$. So it works.

Bennett

Both are unreachable.

Abcd

$Re(\alpha...\alpha^5)= \dfrac{\sin 20\pi/ 11}{\sin 4\pi/11}\times \cos(24\pi/11)$

MrAP

Can anyone help me prove this: arctan 4/3 +arctan 12/15=pi- arctan 56/33.

Abcd

But answer is $-1/2$

Alessandro Codenotti

Hi chat

Ted Shifrin

18:56

Hi @Alessandro

Alessandro Codenotti

Hey @Ted! Did you reach a conclusion regarding that $\ell^\infty$ question?

Ted Shifrin

I was asking if you did? :)

Kasmir Khaan

Ted, in order to have a vector space

we need those axioms and a scalars from a field

but my comfusion is what field is it?

Ted Shifrin

You have to declare what field it is when you define the set of vectors itself.

Alessandro Codenotti

Nope, I didn't think about it, I studied for an exam today

Ted Shifrin

18:59

Well, that's a crummy excuse, @Alessandro :P

Kasmir Khaan

if we take a Z/pZ

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$Mathematics$ Mathematics