Mathematics - 2018-05-16 (page 1 of 2)

loch

00:00

most likely yes --- technically i have to pass my quals first... but assuming things go well then yes

Leaky Nun

in imperial?

loch

no in the US

Leaky Nun

mind disclosing where?

just curious

would you recommend any book / material to me?

loch

on?

Leaky Nun

algebraic geometry

although I do have a number of interests other than algebraic geometry

algebraic topology and logic and commutative algebra, I suppose

ok, the latter is just a tool in algebraic geometry

loch

00:04

oh. I don't really know anything other than the standard ones e.g. vakil, hartshorne
if you want some more classical things first for more intuition then eg shafarevich

Leaky Nun

I see

loch

i should mention that imperial has a lot of really good algebraic geometers so you should talk to them if possible e.g. by doing a UROP

Leaky Nun

indeed

loch

well some people like algebra for their own sake .. so i wouldn't say they're just a 'tool' in algebraic geometry

Leaky Nun

which one is Hilbert's basis theorem? that R[X] is noetherian or that R[X1...Xn] is noetherian?

I mean, can I say that the latter is also the theorem?

loch

00:07

sure? they're equivalent anyway

Leaky Nun

alright

do you have any advice for me in regards to finding a place to live for year 2?

loch

you might know this already - but kind of the whole significance of nullstellenstatz (i guess for what i'm about to say - technically just weak nullstellenstatz) is that instead of thinking about points (a_1,..,a_n) in k^n, you can instead think of all the functions vanishing at a_1,...,a_n, which is given by the ideal (x_1-a_1,...,x_n-a_n), which is a maximal ideal in k[x_1,...,x_n]. so you can kind of forget this whole thing about k^n and just work with the maximal ideals of your ring.

then when you learn affine schemes you just happen to also include the prime ideals. (then schemes are …

@LeakyNun uhh find friends

Leaky Nun

and then?

I'm planning to live with 3 other people

loch

oh
then it's just a matter of finding a place to live usually through some agency (e.g. RR / Foxton? I don't rmb what they're called etc.)

Leaky Nun

@loch so you're basically saying that the classical affine space $\Bbb A^n_k = k^n$ is the closed points of the modern affine space $\Bbb A^n_k = \operatorname{Spec}(k[X_1,\cdots,X_n])$?

also, do you find issue with imperial only teaching the 1950s algebraic geometry?

loch

00:12

yes when $k4$ is algebraically closed

Leaky Nun

oh right, the modern treatment doesn't need algebraic closure

loch

oh you can also see if you can attend these "TCC" lectures

Leaky Nun

what is TCC?

loch

they usually have grad level stuff - i personally never went to them but i know people who did

maths.ox.ac.uk/groups/tcc

Leaky Nun

too advanced for me...

loch

00:14

otoh you can also see if you can sneak into lectures given under LSGNT.

oh well you asked about non 1950s algebraic geometry - so im guessing you meant schemes and i think at some point there was a scheme theory class under tcc

Leaky Nun

@loch but not in the standard course

loch

yeah

Leaky Nun

do you think that undergrads should not learn schemes, but instead just live in the 1950s?

loch

imperial has an obession of making everything (in y3) accessible to anyone coming from y2

which is ridiculous

imo

Leaky Nun

eh... is that ridiculous?

loch

00:18

i personally think it is :p but maybe im biased

and now that i think of it also dependent on lecturer

Leaky Nun

I mean, the way you stated it seems a normal thing to do

loch

hmm you're right
so maybe let me rephrase that as the depth covered in y2/y1 isn't quite enough to do fun things

Leaky Nun

indeed, I agree with that

I mean, I read a book on group theory before I came, and then it's basically the third year course

the group theory courses are way too slow

ALannister

2

Q: Prove that the origin is Liapunov Stable for the given system

Consider the system $$\dot{x} = y \\ \dot{y} = -4x $$ ($\dot{x}$ means $\displaystyle \frac{dx}{dt}$ and $\dot{y}$ means $\displaystyle \frac{dy}{dt})$ I need to prove that the fixed point $\mathbf{x^{*} = 0}$ is Liapunov stable. For reference, according to my textbook (Strogatz), a fixed po...

Leaky Nun

Cayley-Hamilton and det(AB)=det(A)det(B) isn't proved until year 2

loch

00:19

@LeakyNun it's easy to fall into the trap of being obsessed over learning this abstract new toy called 'schemes', but i don't think its necessary to be an expert on classical ag in order to start learning schemes

ALannister

What are schemes?

loch

former = you have no idea what's happening geometrically
latter = takes a lot of time.. (and a lot of things show up more naturally once you learn schemes)

ALannister

Don't live with people. People suck.

loch

@ALannister if you want to know the definitino you can probably look that up on wikipedia - but vaguely it's a generalisation of varieties in classical algebraic geometry where you look at zero sets of polynomials

Leaky Nun

@loch a functor from the category of open sets in a topological space with morphisms being inclusion to an abelian category :)

loch

00:23

i think that's a sheaf

Leaky Nun

@Daminark hi

loch

*presheaf

Leaky Nun

right, and then you need 100 conditions to make it a scheme

0celo7

5 upvotes

Nice to see there are like-minded souls here

loch

@LeakyNun well once you have a sheaf (presheaf + gluing conditions) you get a locally ringed space, and you just want this locally ringed space to look locally like affine schemes (similar to how manifolds look locally like euclidean space)

Leaky Nun

00:27

@loch I see

loch

so on this note its also a good idea to know some things from differential geometry if you want to learn algebraic geometry

Leaky Nun

:o

0celo7

@LeakyNun ALRIGHT

I'll discord you some good stuff

Leaky Nun

ok

Leaky Nun

00:54

arxiv.org/pdf/1506.08376.pdf

what mess did I just read

a click-bait title (one-line undergraduate proof) which is hardly one-line

"Clearly" "it is manifest" "evident"

but it does have its merits

when I was reading the proof in my text I didn't realize that it is proving that Q is not a finitely generated Z algebra

Daminark

01:14

@LeakyNun hello

Semiclassical

02:19

@ALannister this is probably is a sufficiently different line of thought so as to not be too helpful, but: the trajectories in the (x,y) plane are ellipses, since the quantity $4x^2+y^2$ is conserved by that system

ALannister

They definitely are @Semiclassical I already know that.

I need help doing the $\epsilon-\delta$ proof

Semiclassical

well, here's my line of thinking there

suppose you've got an ellipse $4x^2+y^2=\epsilon^2$; the max distance this achieves from the origin occurs when $x=0$, $y=\pm \epsilon$

ALannister

Does this help me bound $\sqrt{\left( x(t)\right)^{2}+\left( y(t) \right)^{2}}$?

Semiclassical

I think so? if I want to pick an orbit that stays within $\epsilon$ of the origin, then its initial distance had better be at most $\delta=\epsilon/2$

ALannister

I see what you mean. That

That's in line with what the solution manual says, but I wanted to show it with algebra.

Semiclassical

02:27

that's geometric in thinking, and presumably you want to get it down to algebra

yeah

not sure how to do that, which is why I'm not sure how helpful that is

ALannister

But, usually geometry can help us with that

Semiclassical

It at least makes it obvious why $\delta =\epsilon/2$ is a smart choice

ALannister

Right.

Problem is I'm dealing with sines and cosines, neither of which "cancel out" using any kind of nice trig identity.

Semiclassical

yeah, that representation you're using isn't a convenient one

an easier one might be $x(t)=\cos(2t-\phi)$, $y(t)=2\sin(2t-\phi)$

or better yet $\phi=2\tau$, with the idea being that then $(x(\tau),y(\tau)) = (1,0)$

ALannister

Using the triangle inequality, I have that $\sqrt{(x(t))^{2} + (y(t))^{2}} \leq \sqrt{x_{0}^{2}\cos^{2}(2t)-4x_{0}^{2}\sin^{2}(2t)} + \sqrt{\frac{y_{0}^{2}}{4}\sin^{2}(2t)+y_{0}^{2}\cos^{2}(2t)}$

@Semiclassical murp. I'm not loving the idea of parametrization. I think that might make it harder.

Semiclassical

02:31

Well, it'll probably be simpler to bound $x(t)^2+y(t)^2$ directly

otherwise that square root is an arse, as you've seen

ALannister

Then, I can pull factor out the $x_{0}$ and $y_{0}$ from each.

Semiclassical

for instance, one line of thinking: You can express the ellipse as $4x(t)^2+y(t)^2=4x_0^2+y_0^2$

ALannister

So, that becomes $= x_{0}\sqrt{cos^{2}(2t) - 4\sin^{2}(2t)} + y_{0}\sqrt{\frac{\sin^{2}(2t)}{4} + \cos^{2}(2t)}$

Semiclassical

and then $x(t)^2+y(t)^2=\frac14[4x(t)^2+4y(t)^2] = \frac14[x(t)^2+4x_0^2+y_0^2]\geq \frac14(4x_0^2+y_0^2)$

ALannister

You can't express the ellipse here in that case.

I mean you can't express the ellipse like that here. $x_{0}$ and $y_{0}$ are constants

Semiclassical

02:35

Sure? The point is that if $4x(t)^2+y(t)^2$ is constant, then it should be the same as its initial value

so $4x(t)^2+y(t)^2=4x_0^2+y_0^2$ is a perfectly valid representation of the ellipse

ALannister

It's snot constant though. It's bounded, but not constant.

Semiclassical

uh, $4x(t)^2+y(t)^2$ definitely is constant in the above motion

that's why it's an ellipse

$x(t)^2+y(t)^2$ isn't, of course.

ALannister

So, how does this help me with my bound to get $\epsilon/2?$

Semiclassical

yeah, there's the question :/

i mean, you can further say that $4x_0^2\geq x_0^2$

and therefore that last expression becomes $x(t)^2+y(t)^2\geq \frac14(x_0^2+y_0^2)$

ALannister

Great, but it loois like your inequality is going in the opposite direction of what I want.

Semiclassical

02:39

yeah...

well, returning to an earlier POV

If I use the parametrization i suggested earlier $x(t)^2+y(t)^2=\cos(2t-2\tau)^2+4\sin(2t-2\tau)^2$

Meow Mix

hi

Semiclassical

If nothing else, that's easier to differentiate

one can go a bit further and use $\cos^2\theta = \frac{1+\cos\theta}{2}$, $\sin^2\theta= \frac{1-\cos\theta}{2}$

and therefore the above becomes $(1+\cos(t-\tau))+2(1-\cos(t-\tau))=3-\cos(t-\tau)$

bleh, i was being lazy earlier. I need an overall factor of $a$ on each of them, since I don't know which ellipse I'm on

so $x(t)^2+y(t)^2=a^2(3-\cos(t-\tau))$

ALannister

Sorry, I had to leave fora while.

Semiclassical

which is bounded above by $4a^2$, so $x(t)^2+y(t)^2\leq (2a)^2$ when $4x(t)^2+y(t)^2=a^2$

ALannister

Can I say that $\sqrt{\cos^{2}(2t) - 4\sin^{2}(2t)} \leq \sqrt{\cos^{2}(2t) - 4}$?

Interesting what you say there. I'm going to mull.

Semiclassical

02:47

well, $\sin^2(2t)\leq 1$. so $-\sin^2(2t)\geq$, and thus i'd say the inequalty is backwards

ALannister

Right.

Semiclassical

but i'm not sure even that's true, since when $t=0$ the right-hand quantity isn't real

ALannister

Silly of me

What is $a$? The radius of the given ellipse?

Semiclassical

otoh $\cos^2(2t)\leq 4\cos^2 (2t)$

ALannister

It is!

Semiclassical

02:49

no, it'd be the semiminor axis

ALannister

How bou dat?

Semiclassical

since when $x=0$ then $y=a$, while when $y=0$ then $x=2a$

so it'd have semimajor axis $2a$ and semiminor axis $a$

i was being lazy, though. what i had in mind was that I should really have written $(x(t),y(t))=(a\cos(2t-2\tau),2a\sin(2t-2\tau))$

but, ugh, I'm seeing a silly error in an earlier statement of mine. so I'd just ignore my rambling for the moment

anyways. $\cos^2(2t)-4\sin^2(2t)\leq 4\cos^2(2t)-4\sin^2(2t)=4\cos(4t)\leq 4$

ALannister

Yas, children!

Semiclassical

Which is true but I think sorta useless, since the first inequality is only saturated when $\cos(2t)=0$ and the second when $\cos(4t)=1$

ALannister

Then, perhapses we can do something similar with the other doo-dad

Semiclassical

02:55

and since $\cos(4t)=2\sin(2t)\cos(2t)$, you can't have both conditions at once

so I'm just confused

actually, why $\cos^2(2t)-4\sin^2(2t)$ not plus?

that'll be negative when $\cos(2t)=0$

ALannister

Because of what the solution is

Semiclassical

hmm

geometrically, i see how this all works

if $4x(t)^2+y(t)^2=4a^2$, then $a^2\leq x(t)^2+y(t)^2\leq (2a)^2$

that's the correct version of what I was saying incorrectly earlier

ALannister

Isn't $\frac{1}{4}\sin^{2}(2t)\leq \sin^{2}(2t)$?

Semiclassical

sure, with equality at $t=0$

a quarter of a nonnegative quantity can't be bigger than that quantity

ALannister

Okay, so the second square root then $\leq y_{0} \sqrt{\sin^{2}(2t)+\cos^{2}(2t)} = y_{0}\cdot 1 = y_{0}$

Semiclassical

03:01

one easy way to do the inequality I just said

ALannister

So, overall, we have $\leq 4x_{0} + y_{0}$?

Semiclassical

$x(t)^2+y(t)^2\leq 4x(t)^2+y(t)^2$, and $x(t)^2+y(t)^2\geq x(t)^2+y(t)^2/4=\frac 14(4x(t)^2+y(t)^2)$

so $\frac14[4x(t)^2+y(t)^2]\leq x(t)^2+y(t)^2\leq 4x(t)^2+y(t)^2$

ALannister

If you let $x(t)^2 = x_{0}$?

soni

hi

Semiclassical

What?

ALannister

03:04

Nevermind. I'm confused.

Semiclassical

ok

ALannister

How does this all relate to epsilon, though?

Semiclassical

I think the delta-epsilon proof should amount to showing: If $x_0^2+y_0^2<\delta^2$ with $\delta=\epsilon/2$, then $x(t)^2+y(t)^2< \epsilon^2$

ALannister

Yeah, I'm working on the algebra. Hold on.

Yeah, this works out great.

Semiclassical

neat.

ahah, yeah: $$x(t)^2+y(t)^2 \leq 4x(t)^2+y(t)^2=4x(0)^2+y(0)^2\leq 4x(0)^2+4y(0)^2<4\delta^2$$

So if $\delta=\epsilon/2$ then $x(t)^2+y(t)^2<\epsilon^2$

ALannister

03:13

You end up with $\sqrt{4x_{0}^{2}(\cos^{2}2t + \sin^{2}2t) + y_{0}^{2}(\sin^{2}(2t) + \cos^{2}(2t))}$

$= \sqrt{4x_{0}^{2} + y^{2}_{0}}$

$\leq \sqrt{4x_{0}^{2} + 4y_{0}^{2}}$, yup.

Hey, we make a pretty good team @Semiclassical

Semiclassical

lol

ALannister

I"m going to print this convo out because my glasses keep falling off of my face, and unless I hold my computer screen right up to my face, I can't see anything.

Hence all the typos.

all the scotch probably didn't help either...

but that was hours ago, so.

You know what they say though, dontcha?

Ye olde mathe joke ---> DON'T DRINK AND DERIVE ;P

ba dum tiss.

Semiclassical

for me right now it's more like my brain is tapped out

i'm smart enough to do geometry still, but not algebra :P

ALannister

I think it's just a different part of your brain

Semiclassical

yeah

geometry is more directly intuitive for me, i suppose

0celo7

03:29

"Another useful tool for computing mass are Bartnik's harmonic coordinates. "

How do I make this sentence grammatically correct

skull

~~are~~ is

0celo7

harmonic coordinates are plural

skull

Ok, ~~Another~~ Other

tools

Secret

Last night dream shows a very strange maths thing (which in the dream it is a solution of a problem whose nature has drifted from my memory)

Basically, there's a function consists of off diagonal terms, and they are self similar in some way

And as the solution goes, we reach to higher and higher terms so that the previous terms become relatively insignificant in an exponential or hyperbolic manner

In addition, for every jump the function take, the oscillation became more damped, eventually after some number of jumps, the amplitude of the oscillation get damped so much that it stablised into a sinusodial thing

The solution is thus given by the total area occupied by each of those off diagonal terms (which forms a pair of black rectangles in a 2x2 rectangle formation)

As for the possible inspiration. I am suspecting:

10 hours ago, by Mary Star

is the base material

and Ted's highly unexpected reply

10 hours ago, by Ted Shifrin

@MaryStar: I don't understand enough to say anything.

Is what trigger it to go into the dream possibly because it triggers Mechanism 2 (Shocking event have a higher chance to become a dream element)

This presumably mixed with step functions, voting patterns in elections where oscillations eventually stablised, functional analysis and matrix algebra to give the weird curve seen in the dream

Zee

05:17

Dead room?

Tobias Kildetoft

06:45

262 pages of LaTeX in just a few minutes :)

Leaky Nun

@TobiasKildetoft hi

I just learnt Nullstellensatz

I excited

Tobias Kildetoft

@LeakyNun Cool

Leaky Nun

so if $\alpha_i$ are >not< indeterminates, can I breakdown $k[\alpha_1, \cdots, \alpha_n]$ into $k \subset k[\alpha_1] \subset k[\alpha1, \alpha_2] \subset \cdots \subset k[\alpha_1, \cdots, \alpha_n]$?

if all of them are fields, do I have tower law?

if the last one if a field, are all of them fields?

because I'm thinking what if $\alpha_2 = 1/\alpha_1$

Poline Sandra

Hello, $|2x+3y|$ is not a distance on $\mathbb{R}$ because the first condition is not satisfied right?

Leaky Nun

first condition?

Poline Sandra

06:56

$d(x,y)=0$ iff $x=y$

Leaky Nun

right, this is not satisfied

Poline Sandra

if i suppose that $|2x+3y|=0$ then $x=-\frac{3}{2} y$

and for $d(x,y)=|arctan(x)-actan(y)|$ it is a distance

$|arctan(x)-arctan(y)|=0$ it is equivalent or just implies that tan(arctan(x))=tan(arctan(y))

@LeakyNun

Leaky Nun

right

Poline Sandra

we have equivalence directly

d(x,y)=0 iff arctan(x)=arctan(y) iff x=y

Leaky Nun

right

tan is surjective

arctan is right inverse

arctan is injective

Poline Sandra

07:12

how arctan is a right inverse ?

Leaky Nun

tan(arctan(x)) = x for all x in R

you write arctan on the right of tan

Poline Sandra

but actan(tan(x)) is not x ?

PrithiviRaj

Consider the points $A=(5,7) , B=(8,2) $ and $C=(0,0) in $\mathbb{R^2}$ . How many parabolas will pass through A,B and C.

Leaky Nun

@PolineSandra so it's not left inverse

PrithiviRaj

anyone please.

skull

07:28

have you plotted the points?

Poline Sandra

@LeakyNun $|x^4+y^4|$ is not a distance right?

Leaky Nun

@PolineSandra I don't know

right, it isn't

Poline Sandra

$|x^4+y^4|=0 $ then $x^4=-y^4$ it is impossible sine $y^4, x^4$ are positive

Leaky Nun

right

skull

@PrithiviRaj it's a good idea to plot the points because if they lie on a straight-line you'll know the answer right away.

Poline Sandra

07:38

if i consider $|ch(x)+ch(y)|$ it is also not a distance because if $x=y$ then $|ch(x)+ch(y)|= 2ch(x)$

@LeakyNun

Leaky Nun

ch?

Poline Sandra

cosh

PrithiviRaj

@skull they are not co-linear , it means infinite ??

parabolas

skull

Nope

Look for an axis of symmetry.

Leaky Nun

@PolineSandra right

PrithiviRaj

07:41

these three points form a triangle

so any median can work as symmetry axis right?

mercio

why would you say that

PrithiviRaj

sorry its not an equilateral triangle , let me think one more time

skull

The symmetrical points will have the same y-coordinates.

Poline Sandra

the last one $|sinh(x)-sinh(y)|$ is a distance and for the first property $|sinh(x)-sinh(y)|=0$ iff $sinh(x)=sinh(y)$ iff $arcsinh(sinh(x))=arcsinh(sinh(y))$ iff $x=y$ @LeakyNun

Leaky Nun

right, sinh is a bijection

skull

07:52

A=(5,7) so A'=(a',7). B=(8,2) so B'=(b',2). C=(0,0) so C'=(c',0) @PrithiviRaj

Poline Sandra

@LeakyNun please if i have that $f$ is non decreasing and $x>0$ then $f(x)\geq f(0)$ or $f(x)>f(0)$

Leaky Nun

$f(x) \ge f(0)$

skull

Can you sketch a parabola that fits those points?

@PrithiviRaj

PrithiviRaj

I found the equation of a symmetric axis as $13y=9x$ .

Leaky Nun

43 mins ago, by PrithiviRaj

Consider the points $A=(5,7) , B=(8,2) $ and $C=(0,0) in $\mathbb{R^2}$ . How many parabolas will pass through A,B and C.

can we bash this by Bezout's theorem after embedding our affine plane into the projective plane?

PrithiviRaj

08:03

I haven't read that theorem , I'm an undergraduate student.

skull

High school?

PrithiviRaj

no

skull

sorry, my apologies.

I was using a high school approach

the sketch is a good check :-)

PrithiviRaj

I'm working on that

@skull yeah

tell me the next step after finding symm axis

skull

the question only asks for "how many parabolas"

PrithiviRaj

08:15

oh got it ! it means only 1 or 2 .

skull

yes

PrithiviRaj

so if they are two , then I have to find one more symm axis , right ?

skull

no

They face up or down, right?

provided we are talking about functions of the form y=f(x)

Leaky Nun

34

A: Ring of polynomials over a field has infinitely many primes

You can copy Euclid's proof. Let $p_1, \dots, p_n$ be a finite collection of prime polynomials in $F[X]$. Consider $f=p_1 p_2 \cdots p_n +1$. Let $p$ be a prime factor of $f$. Then $p$ cannot be any of $p_1, \dots, p_n$ because otherwise $p$ would divide $1$. Hence, no finite collection of prime ...

still upholding the tradition not to check that the set of primes is non-empty :P

Leaky Nun

08:48

@AlessandroCodenotti hi

Alessandro Codenotti

Hi

Leaky Nun

@AlessandroCodenotti any response to ^^^?

Alessandro Codenotti

It's clear the set of primes is nonempty, I don't think there's any need to state it explicitely

mercio

09:06

lol

Leaky Nun

09:27

@AlessandroCodenotti X?

Alessandro Codenotti

Depending on how much you assume known about the structure of $F[X]$ pick any nonzero element, it must have a prime factor

Or degree one polynomials, those are irreducible, hence also prime

Tobias Kildetoft

@AlessandroCodenotti You need to start with a non-unit of course

Alessandro Codenotti

@TobiasKildetoft oh, right, let's just stick to degree one polynomials to be sure then

mercio

proof that there is no prime in F[X] : I suppose there is no prime, and I consider the empty product plus 1 which is 2, and 2 has no prime, so there is no new prime, qed.

Leaky Nun

09:53

so can we generalize it to any euclidean domain that is not a field?

in praticular they must be infinite?

Alessandro Codenotti

How can they have infinitely many primes if they are finite?

anyway any finite euclidean domain (any finite integral domain really) is a field so it doesn't have many primes

Leaky Nun

oh... right

Gabriel Romon

@LeakyNun What do you think about UCL in terms of reputation, compared to Imperial ?

Leaky Nun

@GabrielRomon no idea

ccorn

10:32

There is a thread where the comments had almost led the asker to believe that the problem had no solutions. Fortunately, there is a correct answer.

blat

11:34

hi

If I have a manifold $M$ with a connection $\nabla$ and a metric $g$, how is $\nabla g$ to be understood?

because $\nabla$ usually takes two vectorfields and maps them to a third

blat

11:48

nvm

Topologicalife

14:31

Let $P$, $Q$ be two points on the sphere of radius 1, centered at the origin. Let $L(t) = P + t(Q-P)$ with $t \leq 0 \leq 1$. If there exists a value of $t$ in $[0,1]$ such that $L(t) = 0$ show that $t= \dfrac{1}{2}$ and that $P=-Q$.

What should I do there? Shall I simply use subtitution? I don't understand how to "prove" it. I would simply subtitute the values they gave me.

user193319

Let $e_n \in \ell^p$ denote the vector with a $1$ in the $n$-th entry and zeros elsewhere. Is $\{e_n\}_{n \in \Bbb{N}}$ a basis for $\ell^p$?

Topologicalife

(chat.stackexchange.com/transcript/message/44640399#44640399 Solved)

user193319

I don't think it's going to be a basis, since, in general, an element is a infinite linear combination of the vectors in $\{e_n\}_{n \in \Bbb{N}}$. However, is it true that $\overline{\mbox{span} \{e_n\}} = \ell^p$?

user193319

14:51

Yep. Here's the proof: Given $x = (x_k) \in \ell^p$ and $\epsilon > 0$, there is an $N \in \Bbb{N}$ for which $$\epsilon^p > \sum_{k=N}^\infty|x_k|^p = \sum_{k=1}^\infty |x_k - b_k|^p = ||x-b||_p^p,$$ where $b = x_1 e_1 +...+x_N e_N \in \mbox{span}\{e_n\}$.

Secret

bar is conjugate?

Leaky Nun

@Secret completion

closure

span{e_n} is dense in the whole space

Secret

I see

Topologicalife

15:36

Hi. I am trying to prove the root criterion for sequences. What I have so far is:

By the limit definition, if $\displaystyle \lim_{n \to \infty} \dfrac{a_{n+1}}{a_n} = L $ then $\left | \dfrac{a_{n+1}}{a_n}- L \right | < \epsilon$ for all $n \geq m$. Therefore, it is also true that

$L - \epsilon < \dfrac{a_{m+1}}{a_m} < L + \epsilon$

$L - \epsilon < \dfrac{a_{m+2}}{a_{m+1}} < L + \epsilon$

$L - \epsilon < \dfrac{a_{m+4}}{a_{m+3}} < L + \epsilon$

$\quad \vdots$

$L - \epsilon < \dfrac{a_{k}}{a_{k-1}} < L + \epsilon$

Multiplying all inequalities and simplyfing:

$(L - \epsilon)^{k-m} < \dfrac{a_{k}}{a_{m}} < (L + \epsilon)^{k-m}$

It is correct? What confuses me is the last step

Leaky Nun

@Topologicalife entonces que intentas probar?

que $\displaystyle \lim_{n \to \infty} \dfrac{a_{n+1}}{a_n} = L$ implica $\displaystyle \lim_{k \to \infty} {a_{k}}^{1/k} = L$?

Topologicalife

Sí.

mercio

the last step needs something like $0 < L- \epsilon$

Topologicalife

Lo tengo probado si logro justificar el último paso, pero me confunde.

I've proved it if I can justify the last step.

mercio

puedo leer español

Topologicalife

15:51

When I apply the limit $k\to \infty$ I am not sure what does it happen with $a_k$

Oh, ok.

mercio

also you are not saying how you come up with $\epsilon$ and $m$

Topologicalife

Si aplico el límite $k\to \infty$ de $(L - \epsilon)^{1-m/k} < \dfrac{{a_{k}}^{1/k}}{{a_{m}}^{1/k}} < (L + \epsilon)^{1-m/k}$ no sé cómo se comporta ${a_{k}}^{1/k}$

$m$ is fixed, and $\epsilon > 0$.

mercio

maybe you should multiply by $a_m^{1/k}$ ?

Leaky Nun

maybe don't use the same epsilon?

use the epsilon/2 trick, I think

Topologicalife

That is not the problem.

If I apply the limit I would get: $L - \epsilon <\displaystyle \lim_{k\to \infty}{a_{k}}^{1/k} < (L + \epsilon)$ instead of $L - \epsilon <{a_{k}}^{1/k} < (L + \epsilon)$

Leaky Nun

15:56

right

Topologicalife

That is confusing me.

mercio

do you have that $L > 0$ in your assumptions somewhere ?

Leaky Nun

the intuition is that the limit is within $L-\varepsilon$ and $L+\varepsilon$ for each $\varepsilon > 0$

so the limit must be $L$

but then you need something more to justify it

Topologicalife

Uhm I see.

I don't think we need something more. Since $\epsilon$ is random it would be justified, yeah.

mercio

ooh I have never seen anyone say "let there be a random $\epsilon$" in a proof before

Topologicalife

16:02

What I mean is "for all $\epsilon > 0$"

Tug' Tegin

16:21

Hi!

I would like to know if the problem below is solvable or it is incorrect:
Several timbers of the length 2 m were cut into small logs of 50 sm in 4.5 hours; how long wolud it has taken if they were cut in 40 sm?

the answer is 6 hours according to the book but

I can only get 5.5 hours

mercio

16:49

how long would it take to cut them in small logs of 200 sm ?

Tug' Tegin

@mercio they are 200 sm themselves.

i meant 'timbers each of the length 2m', sorry

mercio

yeah but how long would that take ?

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$Mathematics$ Mathematics