Mathematics - 2018-04-28 (page 1 of 3)

0celo7

00:49

@ShaVuklia it does some crazy stuff away from the real line

Xander Henderson

01:03

@ShaVuklia It is bounded along the real line, but grows exponentially fast on lines of the form $ib$, where $b\in\mathbb{R}$

remember that $$\sin(z) = \frac{\mathrm{e}^{iz}-\mathrm{e}^{-iz}}{2i}$$

so if, for example, $z$ is pure imaginary (i.e. $z = 0 + iy$ for some $y\in \mathbb{R}$), then $$\sin(z) = \sin(iy) = \frac{\mathrm{e}^{i(iy)}-\mathrm{e}^{-i(iy)}}{2i} = \frac{\mathrm{e}^{-y} - \mathrm{e}^{y}}{2i}$$

if $|y|$ is very large, then $\sin(iy) \approx e^{|y|}$, which is unbounded

Ted Shifrin

@Sha: You need to play more with complex numbers! What's $\sin(ix)$?

@Xander, well $1/2$.

Xander Henderson

01:25

hey, I said approximately :P

and 1/2 is just a silly constant

who cares about those?

maybe I should have used some big "Oh!"s

Obliv

can't there be a bijective function for two sets of unequal size? like $f: X \to Y$ by $f(a) = a^{\frac{1}{2}}$ for $a \in X$ and $X = \{4,9,16,25\}$ and $Y = \{-5,-4,-3,-2,2,3,4,5\}$ every element in $X$ gets mapped to two elements in $Y$ so it's surjective and no two domain elements get mapped to the same codomain element so it's injective.

Xander Henderson

How do you define "size" @Obliv?

though, for most typical definitions, the answer is "No". Two sets are the same "size" if and only if there is a bijective function between them.

Obliv

i know that's why i said size instead of cardinality

i just meant no. of elements

Xander Henderson

well, for finite sets, the number of elements and the cardinality are the same thing

Obliv

well then i'll rephrase the question and ask if the function i defined above counts as being bijective

Xander Henderson

01:38

That function isn't bijective

a function cannot, by definition, map one element to two different things

in that case, it is not a function at all, but a relation

(or, rather, you could make sense of a "multivalued" function by treating it as a relation)

Obliv

ohh

Sha Vuklia

@Xander @Ted yea thanks, I guess I expected $\sin z$ to look like a wave spreading in all directions, but I really had no grounds for that, other than the $\mathbb R$ analogy

Obliv

thanks @XanderHenderson that's very helpful :)

Sha Vuklia

I did not expect this at all (found it online)

but I guess, it is what it is

Xander Henderson

So, like $$A\mathrm{e}^{i\lambda x} + B\mathrm{e}^{-i\lambda x} = A\sin(\lambda x) + B \cos(\lambda x). $$

Sha Vuklia

01:45

actually this is better

Xander Henderson

gah... where is the error???

found it

'cause $A$ and $B$ are just constants

0celo7

@XanderHenderson how did your PDE class go

Xander Henderson

and, for the record, $A \ne A$

it went fine

we solved a wave equation problem

on some kind of rectangular domain

with a silly initial condition

then I "forgot" to erase the whiteboard and gave them the typo-laden quiz that I was directed to give them

(I am going to get AMAZING evals this quarter)

0celo7

@XanderHenderson great!

Xander Henderson

For their homework, they are being asked to solve a heat equation on a circle; their answer is, I think, meant to be the Fourier series of the initial condition (more or less)

which seems reasonable

but (1) I don't know if these guys know much Fourier analysis and (2) even if they do, they are going to write everything in terms of the real Fourier series, which makes me cry :'(

$$f(x) = \sum_{k\in\mathbb{Z}} \hat{f}(k) \mathrm{e}^{ikx}$$

none of this sines and cosines BS :\

0celo7

01:58

what fourier do you even need to know for that

it's just separation of variables

they'll memorize a formula

Xander Henderson

GAH!

That makes me even more sadder-er

They should be more gooder

There is an ice cream truck driving through my neighborhood

the were just playing Swan Lake

and now they are playing Christmas muzak

WTF?

Secret

04:21

Axiom of choice actually puts a very powerful constraint on what kinds of infinite sets are allowed

Alex D

Hi all

Could someone explain this proof here? math.stackexchange.com/questions/1540739/…

In particular, is the convergent subsequence a consequence of the bolzano weierstrass theorem?

Also, what does the D() mean on this line?

"There exists $e>0$ such that $\forall u,v\in D (|u-v|<e\implies |f(u)-f(v)|<1).$"

Ohhhhh D is the set and the implication is just written in parenthesis, right?

Sorry that makes sense

Secret

10:09

[Random worldbuilding]

The most powerful superpower is not omnipotence

It is preorder

Why? Because the moment when you say A is more powerful than B, you have already imposed a preordering into the situation

Those who wield the power of a preorder will overpower everything because they can freely rearrange the preordering

(or not overpower everything because it can be reversed

Which then begs the million dollar question: What happens when preorder encounters omnipotence and logical paradox. Who is more powerful

The answer, requires the entire universe to compute

The Hitchhiker's Guide to the Galaxy (film)

The Hitchhiker's Guide to the Galaxy is a 2005 British-American science fiction comedy film directed by Garth Jennings, based upon previous works in the media franchise of the same name, created by Douglas Adams. It stars Martin Freeman, Sam Rockwell, Mos Def, Zooey Deschanel and the voices of Stephen Fry and Alan Rickman. Adams co-wrote the screenplay with Karey Kirkpatrick but died in 2001, before production began; the film is dedicated to him. The film grossed over $100 million worldwide. == Plot == The film begins with a Broadway-style number, "So Long, and Thanks for All the Fish", sung by...

space, time, reality, mind, soul, power

These are all just that, words

soul > time > reality > mind > space > power

It is all a preordering and is thus subjected to set theory

But what about... $\pi$

$\pi$ is transcendental, thus by definition any circular objects are law breakers

They will not be confined by any categorisation or classification we humans imposed on it and they will break any rules at will

We are part of Cthulhu, and thus the cosmic edritch aabdominon are not subscribed to what we called "rules"

blogs.ams.org/blogonmathblogs/2018/04/22/…

BeginningMath

13:38

can anyone please help me with a simple question?

if i toss a coin 50 times, how can i know what are the odds that i will get 4 times in a row heads?

Secret

(1/2)^4 times the number of permutations in 50 slots where HHHH appears

BeginningMath

14:00

thank you very much for your answer @Secret

so i'm trying to understand - for instance,the chance of getting 4 heads in the first 4 tossings out of 50 is equal to the chance of getting 4 hads in the last 4 coin tossings?

heads*

is equal to the chance of getting 4 heads in the last 4 coin tossings out of 50?

Secret

you only want the probabbility to get 4 heads in a row, not where you get 4 heads in a row, thus the probabbility should be the same regardless of where that string of HHHH are, thus you add those all up as mutually exclusive events

BeginningMath

14:18

thank you very much, i understand

Secret

14:47

[Ordinals]

The 3 Tuple (a,b,c) captures the growth by indicating the easiness of transversing from a starting element x to the destination y. a is the operation that is used to get from x to y, b denotes what is being used each step, and c denotes how many repetitions

For example $1 \to 3$ with the notation $(+,1,2)$ means $(((1)+1)+1) = 3$

As we shall see later, b, c also have dependence on a which is why the nature of a is important

$0\to 1\to 2\to \cdots < \omega : (+,1,1)$

$\omega\to \omega 2\to \omega 3\to \cdots < \omega^2 : (+,\omega,1)$

$\omega^2\to \omega^3\to \omega^4 \to \cdots < \omega^{\omega} : (+,n,\omega)$

Note that in the 3rd case, $c = \omega >$ finite, thus the ordinals in the sequence are additively indecomposable

$\omega^2\to \omega^3\to \omega^4\to \cdots < \omega^{\omega} : (\times,\omega,1)$

ÍgjøgnumMeg

15:28

if $K = \Bbb Q(\sqrt{-17})$ and $K_1 = K(\sqrt{17})$, is there some obvious reason that for some prime $\mathfrak{p}$ of $K$, we have $2 \notin \mathfrak{p}$ and $17 \notin \mathfrak{p}$?

Certainly if $2 \in \mathfrak{p}$ then $17 \notin \mathfrak{p}$ else $\mathfrak{p}$ is not prime, but if $2 \notin \mathfrak{p}$ I can't see how to show that $17 \notin \mathfrak{p}$ (which would be the suggested outcome by the exercise I'm doing)

loch

I'm not sure if I understood your question.. if you're asking for primes in $O_K$ that don't contain 2 and 17 - they clearly exist (by taking any prime lying over a prime not 2 or 17)!

ÍgjøgnumMeg

15:44

@loch I've asked it as a question on the main site, I may have missed the point of the exercise completely though!

0

Q: Hilbert class field of $\Bbb Q(\sqrt{-17})$

Let $K = \Bbb Q(\sqrt{-17})$. I wish to show that the Hilbert class field $H$ of $K$ is the extension $H = K(\alpha)$ where $\alpha = \sqrt{(1 + \sqrt{17})/2}$ (following an exercise in Cox's Primes of the Form $x^2 + ny^2$). I have already shown that $\operatorname{Cl}(\mathcal{O}_K) \cong \Bbb...

Leaky Nun

Let (P, <=) be a partially ordered set.

Say an element x in P is in the "centre" of P if for every y in P we have x <= y or y <= x.

Is the supremum of a collection of elements in the centre of P also necessarily in the centre of P, assuming that the supremum exists?

Mary Star

Hello!!

Let $\mathbb{R}$ provided with the metric $d(x,y)=|x-y|$. I have shown that the collections of sets $$S_1=\left \{\left (\frac{x}{2}, \frac{3x}{2}\right ): 0<x<1\right \}, \ \ \ \ \ S_2=\left \{\left (x-\frac{1}{2}, x+\frac{1}{2}\right ): 0<x<1\right \}$$ are open covers of $A=\left \{\frac{1}{n} : n\in \mathbb{N}\right \}\subset \mathbb{R}$.

To check if they have a finite subcover, we have to check if $S_1$ and $S_2$ respectively, have a finite subset which is also a cover of $A$, right?

loch

@ÍgjøgnumMeg I think you can use K_1 = K(\sqrt{-17}) = K(\sqrt{-1}), i.e. let u = -1

ÍgjøgnumMeg

@loch well $K = \Bbb Q(\sqrt{-17})$ already but then $K_1 = K(\sqrt{17})$ means $\sqrt{17}/\sqrt{-17} \in K_1$ so $K_1 = K(\sqrt{-1})$? Is this what you mean?

loch

Yes

ÍgjøgnumMeg

15:52

Nice, that's probably (definitely) the way to do it, since it's a similar trick to a worked example given in a previous section lol

Balarka Sen

The Spacelords - Water Planet (2017) (New Full Album)

►

dee oh pee ee

0celo7

@BalarkaSen why should I care about the Poincare-Lefschetz duality

what good is $H(X,\partial X)$

Balarka Sen

It's the right Poincare duality for manifold with boundaries. I don't know what else you want to know.

$H_k(M, \partial M)$ is non-stupid in $k = \dim M$ unlike $H_k(M)$, for example (if $M$ has boundary)

0celo7

So I can prove that $H^k(M)\cong H^{n-k}(M,\partial M)$, both de Rham

but I'm not sure what the RHS is telling me

Balarka Sen

$H^k_{dR}(M, \partial M)$ is made up of $k$-forms which vanish on $\partial M \subset M$. Think of it like compact support; if you had a noncompact manifold you'd want it to vanish at infinity.

0celo7

16:05

I know the definition, thanks

Balarka Sen

I have no interest in helping you with imprecise questions, then. Bye.

Meh, I'm drowned in workload and cracking. Sorry for that, @0celo7, I realize I'm probably being unhelpful. My point was more, having compactly supported cohomology in noncompact manifolds is what gets you the right Poincare duality, eg.

Not a very intelligent point probably.

I don't know what would be a more intuitive description of that group.

0celo7

@BalarkaSen It's ok.

I know why that group shows up in PD, like I said I have a proof in the de Rham case using Hodge-Morrey.

But what's an application of this duality, say.

Balarka Sen

Okay I see. Hm.

0celo7

Hodge-Morrey generalizes the Hodge decomposition to allow for boundary conditions.

Balarka Sen

Ah

I have never seen that.

0celo7

16:18

That's because the proof is quite a bit harder.

Silent

Is there any way that i get points here, as i move x in $(x \cos x, x \sin x)$? @LeakyNun

0celo7

I don't think it's possible to generalize the one in GH or Warner.

Balarka Sen

Strange.

Leaky Nun

@Silent desmos.com/calculator/0lkm2r7pyq

@BalarkaSen hi, are you familiar with filters?

Balarka Sen

Nope, @Leaky. I think @Akiva is though.

Leaky Nun

16:19

also, I have a question in chain homotopy

Silent

thank u so much!

0celo7

@BalarkaSen If you knew their proof you wouldn't find it strange!

Leaky Nun

@BalarkaSen why do we require that $g-f = dP+Pd$ even though the $Pd$ is not used?

I mean, can't we set $g-f=dP+2Pd$?

Alessandro Codenotti

@LeakyNun Ah, by the way, I looked for that pdf but I can't find it anymore... What's your question about filters?

Leaky Nun

@AlessandroCodenotti well I was not getting the generalization of limit (as in analysis) via filters

Mike Miller

16:22

Ocelot: H(M, dM) is the homology of M with a cone attached to it's boundary. You most often use it if you want your intersection-theoretic pairing to be non singular.

Or whatever you want to call the pairing that gives you Poincare duality

Balarka Sen

I think you can use the Lefschetz duality to prove a version of the Poincare-Hopf theorem for manifold with boundaries, say. (With vector fields vanishing on the boundary)

0celo7

@BalarkaSen Oh, that would be very nice.

Balarka Sen

Or at least that's one way to do it.

Mike Miller

Lefschetz duality eg can be used to prove the half-lives-half-dies principle for 3-manifolds: if M has a surface boundary components S, then the map HS -> HM has half of the homology in the kernel

Balarka Sen

@MikeMiller Wow that looks like Lefschetz hyperplane theorem

Well, "looks like"

Semiclassical

16:25

fun fact: my thesis secretly has an example of the Picard-Lefschetz formula in it

Balarka Sen

Maybe only superfically though

Waiting

@MikeMiller Where is the mistake here? If there is one math.stackexchange.com/questions/2756119/…

Silent

@LeakyNun Why this is not workinf for negative t's?

Waiting

@Semiclassical maybe you can help too

Leaky Nun

@Silent it is

Semiclassical

16:26

0celo7

@MikeMiller I see.

Semiclassical

is an image in my thesis

Mike Miller

I do not draw images

0celo7

@MikeMiller @BalarkaSen I'm looking for a quick application of $H^k_\text{dR}(M)\cong H^{n-k}_\text{dR}(M,\partial M)$.

Mike Miller

Because I am an unreadable author

Yeah I saw

Balarka Sen

16:27

I can confirm ^

Mike Miller

I'm just throwing shit

Waiting

@LeylaAlkan maybe you know where I'm wrong, math.stackexchange.com/questions/2756119/…

@LeakyNun or you.

Mike Miller

I have used that duality n+1 times but it's hard to immediately name examples

Oh, Alexander duality for submanifolds of S^n.

Alessandro Codenotti

@LeakyNun Well a filter $F$ converges to a point $x$ if every open nbhd of $x$ is in $F$

Waiting

Wise @BalarkaSen, where is my mistake?

Leaky Nun

16:29

@AlessandroCodenotti why?

Alessandro Codenotti

If you think about sequences as functions $f:\Bbb N\to X$ you can recover the usual notion of convergence, $f(n)\to x$ is the same as asking that the pushforward of the Fréchet filter on $\Bbb N$ through $f$ converges to $x$

Leaky Nun

as in, what is the intuition behind?

what is the Frechet filter?

Balarka Sen

@Waiting Please don't spam pings. I am sure whoever knows can help. I certainly don't!

3

Alessandro Codenotti

the cofinite filter, it is made of all the cofinite subsets

Waiting

@Secret It's you!

0celo7

16:30

@MikeMiller You mean I can get a proof of Jordan curve theorem from it?

Leaky Nun

@Waiting and who told you you're wrong? (not that I'm saying you aren't)

0celo7

For smooth spheres at least (lame but ok)

Waiting

@Secret When I posted this answer here I though it's a brilliant one, but the no one voted it. Maybe it's wrong? math.stackexchange.com/questions/2756119/…

Leaky Nun

@AlessandroCodenotti ok more context. They defined $\displaystyle \lim_{x \to c} f(x) = L$ as "the filter around c tends to the filter around L"

Mike Miller

@0celo7 For smooth embedded curves, since you're going to use the normal bundle. For these I think it's more or less already trivial.

Secret

16:32

it's just 2 hours, just wait for it

it is natural to have new answers not get lot of votes just yet

Gabriel Romon

@Waiting where's your book ?

Alessandro Codenotti

"The filter around" means the neighbourhood filter?

Waiting

@LeakyNun Just kidding people! ;)

0celo7

@MikeMiller Yeah.

Balarka Sen

@LeakyNun Hm, it's not immediately clear to me if writing $f - g = dF + 2Fd$ doesn't harm anything. What if you're in $\Bbb Z/2$?

Leaky Nun

16:33

@AlessandroCodenotti well they permit other filters. For example, there's the "top" filter representing the filter around infinity

Waiting

@GabrielRomon What book? You mixed the users.

Leaky Nun

and I find this unification quite intriguing

Secret

meanwhile, on that integral question, I don't know how to visualise $\Gamma_{\epsilon}, \Gamma_{\infty}$

Leaky Nun

unification as in a unified theory behind everything

Alessandro Codenotti

@LeakyNun I guess that agrees with the neighbourhood filter of infinity in some compactification though

What's the source?

Gabriel Romon

16:34

@Waiting Aren't you Chris's sis ?

0celo7

I'm looking at Bredon, all of his applications require more machinery than I'd like. No point in giving an application of some analysis that requires a book of topology behind it.

Waiting

By the way, just out of curiosity, does any understand my answer here? math.stackexchange.com/questions/2756119/…

Leaky Nun

@AlessandroCodenotti it's in a definition inside a library of a proof assistant

Alessandro Codenotti

How to they define convergence of a filter to another filter? I've always seen convergence of filters to points

Waiting

@GabrielRomon I'm Waiting.

Leaky Nun

16:35

/-- `tendsto` is the generic "limit of a function" predicate.
  `tendsto f l₁ l₂` asserts that for every `l₂` neighborhood `a`,
  the `f`-preimage of `a` is an `l₁` neighborhood. -/
def tendsto (f : α → β) (l₁ : filter α) (l₂ : filter β) := l₁.map f ≤ l₂

Secret

how does the floor thing in the sum become a harmonic number?

Leaky Nun

the ≤ is reverse inclusion @AlessandroCodenotti

Secret

5

Q: Arc contribution in $\int_{-\infty}^\infty \mathrm{d}z \frac{e^{-z^2}}{z-1}$

Consider an improper integral with a pole on the integration contour at say $z=1$, $$ \tag{1} I = \int_{-\infty}^\infty \mathrm{d}z\ \frac{e^{-z^2}}{z-1+i\epsilon},~~~~~\epsilon>0. $$ Let $$f(z) = \frac{e^{-z^2}}{z-1+i\epsilon}$$ then $$ \sum_{residues~inside~\Gamma} = 0 = \oint_\Gamma f(z) ...

integration complex-analysis improper-integrals contour-integration residue-calculus

Leaky Nun

and map is defined as follows:

/-- The forward map of a filter -/
def map (m : α → β) (f : filter α) : filter β :=
{ sets            := preimage m ⁻¹' f.sets,
  exists_mem_sets := ⟨univ, univ_mem_sets⟩,
  upwards_sets    := assume s t hs st, f.upwards_sets hs (assume x h, st h),
  directed_sets   := assume s hs t ht, ⟨s ∩ t, inter_mem_sets hs ht,
    inter_subset_left _ _,  inter_subset_right _ _⟩ }

Secret

I need help to visualise the contours of this one

Leaky Nun

16:36

@Secret thanks for splitting my message in three parts

four

Secret

lol

Waiting

@Secret I used simple inequalities with $\lfloor x\rfloor$.

Letting the joke apart, this question should be the easiest one for getting a job in a researching institute for example. To be done under a minute.

Secret

Waiting

It takes 5 seconds to guess the limit.

And at most 1 min to finish it easily not with fancy tools.

Secret

what does $\Gamma_{\infty}, \Gamma_{\epsilon}$ even look like?

Alessandro Codenotti

16:42

Uhm ok, so $\lim\limits_{x\to l_1}f(x)=l_2$ if for every filter $F$ with $F\to l_1$ we have $f^*F\to l_2$ (where $f^*F$ denotes the pushforward filter)

Vrouvrou

hello,

Waiting

@Secret No one finished that question with an inspired contour integration. Still waiting. :-)

Vrouvrou

how are you?

Balarka Sen

Hm, the number of subsets of $\{1, 2, \cdots, n\}$ such that any pair of elements of such a subset differ by at least two is exactly the Fibonacci sequence.

(The recurrence is easy to derive)

Alessandro Codenotti

I guess checking the neighbourhood filter of $l_1$ is actually enough

Secret

16:43

@Waiting I don't even understand the question, what do the two other contours look like?

Waiting

The integral on the big arc doesn't vanish. In fact, the integral on the big arc has 2 components that blows up at infinity, but wth opposite sides, which together leads to a finite value. So we also view that integral on the arc as a CPV integral.

(sorry I cannot edit my messages at al)

Secret

try pressing "edit" and post again, the chat sometimes does that lag thing, you modified message shoudl be in there

Waiting

The proposed contour doesn't lead to the extraction of the value of the integral on the real line.

Secret

Anyway, I don't know what $\Gamma_{\infty}, \Gamma_{\epsilon}$ will look like if plotted in the diagram of the function, which is what currently stumped me

Vrouvrou

for any $r>0$ we have $card\{n\in\mathbb{N}, (\frac1n,1)\in B_r\}=+\infty$ where $B_r=\{(x,y)\in\mathbb{R}^2, (x-3)^2+y^2<r^2\}$

Waiting

16:49

@BalarkaSen You meant you certainly don't understand such solutions?

Not sure I got your point.

Secret

@Waiting Put it simply, I have the diagram of the function $e^{-z^2}/(z-1)$ plotted above. What shape does $\Gamma_{\infty}$ look like cause I need to be able to visualise the contour said by the OP to make sense of the question?

Waiting

Anyway, that's less important.

Balarka Sen

I am bad at English, I meant I certainly can't help you.

Vrouvrou

$(\frac1n,1)\in B_r\Rightarrow (3-\frac1n)^2<r^2-1$ for $r>1$ we have $|3-\frac1n|<\sqrt{r^2-1}$ how to continue please

have you an idea @Secret?

Secret

uh what is the question?

Vrouvrou

16:58

for any $r>0$ we have $card\{n\in\mathbb{N}, (\frac1n,1)\in B_r\}=+\infty$ where $B_r=\{(x,y)\in\mathbb{R}^2, (x-3)^2+y^2<r^2\}$

Secret

but what is the question?

Vrouvrou

find the adherent value of $(\frac1n,1)$ in the toplogy $\{\mathbb{R}^2,\emptyset, (B_r)_{r>0}\}$

Waiting

Almost every time I come in here I have some fun. ;)

No more fun, I have to go now.

Secret

It has to be r=$\sqrt{1^2+3^2}$?

Vrouvrou

yes i see it with a picture $\sqrt{10}$

i want to see it with algebra

Secret

17:14

In your first step, you let $n \to \infty$ and then you will be at the limit point of the set $(\frac{1}{n},1)$

which gives $3^2 +1 < r^2$

Vrouvrou

i don't understand

why let $n\to \infty$ ?

we are not in $(\mathbb{R},|.|)$

Secret

any smaller than that and your open set $B_r$ is not a neighbourhood of $(\frac{1}{n},1)$?

The circle $r = \sqrt{10}$ contains the point $(0,1)$ and for any $r \geq \sqrt{10}$, they are neighbourhoods of $(0,1)$ and since $\lim_{n \in \Bbb{N}}\frac{1}{n} \to 0$ it must contain countably many points in $(\frac{1}{n},1)$. Thus $(0,1)$ is an adherent point of the set $(\frac{1}{n},1)$

Vrouvrou

but you speak about limits in $(\mathbb{R},|.|)$

Secret

Isn't the set $(\frac{1}{n},1)$ by definition: $(1,1), (\frac{1}{2},1), (\frac{1}{3},1) ...$?

So $n \to \infty $is only along one dimension and the limit will took place in $\Bbb{R}$

Keep these mind

Is it correct to say that 1/x converges 'logarithmically' for x to infty?

Secret

17:28

looks nothing like logarithmic, since $\frac{d}{dx}\frac{1}{x} = -\frac{1}{x^2}$

jaslibra

Hey, sorry never used chat before, so not too sure about the etiquette. I thought my question was pretty simple, but I got pretty weird responses from member with reputation 63.1k. math.stackexchange.com/questions/2756359/…

Its just about really understanding the terms in the infinite series solution to $y' = y$

Keep these mind

@Secret Yet: en.wikipedia.org/wiki/Rate_of_convergence#Basic_definition

Secret

ah ok, so logarithmically has a specific meaning, not just it behave like log

it is definitely sublinear since the stuff in the limit are both polynomials of deg 1 and hence $\mu =1 $. Meanwhile the other limit gives 3 as the limit thus it is not logarithmic (computed by plugging 1/x into the given limit in the wikipedia page)

Keep these mind

The other limit gives 3??

Secret

ah sorry typo

so it gives 1

hence by definition, logarithmic

Keep these mind

17:39

Thanks for the sanity check.

Vrouvrou

is there a relation with my problem, because i don't understand

Keep these mind

My question was unrelated.

Vrouvrou

ok thank you

Secret

2

Q: Understanding solution to $y' = y$ and exponential distribution

My Understanding: I would derive the exponential random variable as follows: I consider an experiment which consists of a continuum of trials on an interval $[0,t)$. The result of the experiment takes the form of an ordered $n$-tuple $\forall n \in \mathbb{N}$ containing distinct points on the...

probability differential-equations exponential-function

Why will equations that describe exponential growth will have something to do with bernoulli trials?

so if this connection is not just a mathematical coincidence, then it will mean that the dynamics of y'=y (hence exponential growth) can be approximated by a bernoulli process of adding things into the set?

I wonder if anyone have checked whether a monte carlo solution to y'=y actually give a bernoulli distribution...

feynhat

17:56

Is the relation "is a dense subset of" transitive?

Vrouvrou

18:07

Bonsoir @TedShifrin

Alessandro Codenotti

@feynhat what do you think?

Vrouvrou

@TedShifrin j'ai besoins de votre aide pour ca math.stackexchange.com/questions/2757794/…

s'il vous plait

feynhat

@AlessandroCodenotti I think yes.

Alessandro Codenotti

Why?

feynhat

@AlessandroCodenotti Say $A$ is dense in $B$, and $B$ is dense in $C$. Then any open subset of $B$ intersects with $A$, now this open set is intersection of $B$ and an open set of $C$. So, any open set of $C$ which intersects with $B$ also intersects with $A$.

^ Any comments on this proof?

jaslibra

18:16

@Secret, perhaps the question reduced to understanding each term in the series solution to $y' = y$

Nicholas Roberts

@Vrouvrou what is adherent value?

Keep these mind

Moving sofa problem

The moving sofa problem or sofa problem is a two-dimensional idealisation of real-life furniture-moving problems and asks for the rigid two-dimensional shape of largest area A that can be maneuvered through an L-shaped planar region with legs of unit width. The area A thus obtained is referred to as the sofa constant. The exact value of the sofa constant is an open problem. == History == The first formal publication was by the Austrian-Canadian mathematician Leo Moser in 1966, although there had been many informal mentions before that date. == Lower and upper bounds == === Lower bound...

Can somebody please solve this? Thanks

feynhat

I needed this result to show that $\mathcal{C}[a, b]$ (the space of all bounded, continuous, real valued functions defined on $[a, b]$) is separable. By Weierstrass approximation, set of polynomials is dense in $\mathcal{C}[a, b]$, and the set of polynomials with rational coeff. is dense in set of polynomials. So, set of polynomials with rational coeff. is dense in $\mathcal{C}[a, b]$ (this is where we use the result). And the set of polynomials with rational coeff. is countable.

Vrouvrou

@NicholasRoberts @NicholasRoberts we say that $(x,y)\in \mathb{R}^2$ is an adherent value of $u_n$ iff $\forall V\in \mathcal{V}_x, card\{n\in\mathbb{N}, u_n\in V\}=+\infty$

Silent

Please help me with this: 'Put a metric $d$ on $\Bbb R$ such that $|x_ n − x| → 0$ if and only if $d(x_ n , x) → 0$, but that ${x_ n }$ is a Cauchy
sequence in $(\Bbb R,d)$ when $|x_ n | → ∞$.'

Vrouvrou

18:39

@Astyx bonsoir

ca va?

avez vous une idée sur cette question: math.stackexchange.com/questions/2757794/…

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