I mean, you've got $x_0\leq x_1+c_1$ and $x_1\leq x_0+c_0$

Which together gives $-c_0\leq x_0-x_1\leq c_1$, I guess.

But yeah, this seems pretty ugh.

I don't think exponential is that much worse than polynomial in this particular case

Obviously there is something you'd have to exploit : the first equations remain the same no matter the case you're lookin at

Do the n=1 case first (that is, x_0 =cste)

An iterative approach, though I have no idea if it's guaranteed to work:

Iterative could work but seems very hard to do in the general case

Hrm. Now I'm not sure that the iterative idea makes sense either.

Yeah it does, bound x_0 from below by the min of the b_i ans c_i

And take the lowest bound each time

Yeah, I guess I was just having trouble formulating it.

Then you can get a lower bound for x_0 (it has to converge)

Apply the same line of thinking on x_1, ... x_i

So start with $x_k:=\text{min}(c_{k-1},b_k,c_{k+1})$

and then update that to $x_k:=\text{min}(c_{k-1}+x_{k-1},b_k,c_{k+1}+x_{k+1})$.

Like the o ly specifkc cases are when x_i + c_i = b_i

Otherwise there is a unique solution for x_(i+1)

(I am not certain at all about my variable name)

Yeah. I can't be arsed to test if that works, though.

Looks np-difficult to me

"Looks <s>np-</s>difficult to me" is enough for me

aww.

<s> Test </s>

Hehe

the example I found online was this:

<s>this is strike through text</s>

which doesn't work :(

hmm. test

oh there we go.

gotta use --- instead

Huh

Found that by googling

chat.stackexchange.com/faq#formatting

nice

hadn't realized it was native to Chat

:-)

test

Oh my god

This opens up a whole new world for me

[super random: Continue] Currently trying to figure out a way to define a basis set, such that none of its vectors can be explicitly written down using said basis set. Guess I need to use axiom of choice somehow...

Yup, it's very handy.

Inspiration of the super random: Months of experience with art and talking to art people seemed to suggest to me that experience and knowledge learnt from art are well defined, but they spontaneously mix in some unpredictable way. So we have some rough language here in terms of my preliminary understanding about the dynamics of art, and thus now the challenge is to construct a structure that model this theory

Adding colour into LaTeX is fun too.

The bizarre thing is this: This structure is expected to be some kind of inverted version of whatever that comes out after applying the axiom of choice, where it is the process of the construction is unconstrucible but both the input and the output is constructible

Further crossfire within the brain of the idea lead to another [random] which is currently explored in h bar, Pictured as two particles that look like colorful balls colliding together, the mental movie is replayed many times. Every collision is identical, but the outcome are always different and unpredictable

This crossfire is currently attempted to made more rigorous by considering probabilistic unitary evolution operators

twitch

Things like that happen painfully often if you use \footnote carelessly in LaTeX

${}^{1}{}_{\text{This is a footnote.}}$

@Secret How about, instead of being a basis for the vector space, have it be a basis for a "large enough" unspecified countable subspace

At the start of the proof, you could probably assume that the subspace is large enough to include everything that will be used in the proof (if not, just add it to the space)

The unconstructability comes from the unspecifiedness

That might work. I have been thinking about letting some unspecified function f, but I never though I can have an entire( in the laymen sense) subspace unspecified

right, and in that case, the superspace should be easy, probably $\Bbb{R}^{\Bbb{Z}}$ would be large enough to contain any unspecified countable subspace

now with something to contain the subspace, the rest of the proof, as you have outlined, can be done inside the subspace and no need to worry about the superspace

This vaguely feels like IST

(Well I guess we don't really need the superspace to be specified, I just tend to do so because of my habit that objects seemed more tractable if it is contained in something concrete)

btw what is IST?

Internal set theory, a type of nonstandard analysis

Hmm... if the basis is constructible, won't a linear combination will allow us to write down all elements of the given subspace. O wait, maybe the countablity and the unspecifiedness might allow us to have vectors expressible as a linear combination but canno be explicitly wrote down (hence unconstructible) let me think...

Let $X_n$ be a sequence of random variables, and let P be some property. Then I want to say that there exists an $m$ such that the probability that all $X_n$ except finitely many satisfy property P is smaller than the probability that $X_m$ satisfies P. Is that even correct?

Hello

If $w\in W^{1,p}(\mathbb{R}^N)$ and $w_{R}=h_{R}w$ where $h_{R}\in C_{0}^{\infty}(\mathbb{R}^N,[0,1])$ $h_{R}(x)=1$ on $B_{R}(0)$ and $h_{R}(x)=0$ on $B^c_{2R}(0)$

How to prove that $w_{R_n}\rightarrow w$ on $W^{1,p}$

where $R_n\rightarrow +\infty$ when $n\rightarrow +\infty$

turns out if you let the interaction to be probalistic, you will end up accidentally left quantum mechanics

Can someone recommend me a good book on Diff Eqs? Ordinary and Partial.

@TedShifrin

(I don't know if pinging you for this is a good idea or not)

If a sequence in $\Bbb{R}$ diverges, will every subsequence diverge?

$0, 1, 0, 1, \dots$

Sorry. I mean the sequence is unbounded, it diverges to infinity.

Are there are only finitely many pairs of 5-smooth numbers that differ by exactly 2?

5,1,4,2,3,3,2,4,1,5,0,6,0,7,0,8,...

And if so, does the same hold for all values of 2?

Then $\lim_{n\to \infty}a_{2n}=0$ but $\lim_{n\to \infty} a_{2n+1}$ diverges to infinity

@user193319 Do you mean the sequence $a_n$ diverges to $+\infty$, to $-\infty$, or $|a_n|$ diverging to $+\infty$ ?

hence $\lim_{n\to \infty} a_n $ also diverges

@SteamyRoot Well, in the problem I am working on I am assuming $x_n$ is unbounded, so I suppose that means I am assuming $|x_n| \to \infty$.

Errr, no, that's very different.

The sequence $0, 2, 0, 4, 0, 6, 0, 8, \dots$ is unbounded.

@SteamyRoot and @Secret Thanks for the example sequences!

Unbounded means that $\forall M > 0: \exists N \in \mathbb{N}: |x_N| > M$

Meaning, for any $M$, at "some point" the sequence must be larger than $M$.

Whereas $|x_n| \to \infty$ means $\forall M > 0: \exists N \in \mathbb{N}: n \geq N \implies |x_n| > M$

"from some point onwards, the sequence is larger than $M$"

Here's a very weird sequence:

1,0,2,0,0,3,0,0,0,4,0,0,0,0,5,0,0,0,0,0 ....

How is that very weird? o.O

@Secret isn't weird if it's in OEIS

You underestimate the OEIS severely

It clearly has a diverging subsequence, but naively the converging subsequence get "more dominant", so will it eventually converge since at some point you will have so many zeros that the next nonzero is unreahcable?

ehhh, yeah. there's some pretty esoteric stuff in OEIS

That's… very naïve

Weird != unknown

Also, it won't "eventually converge"

Lacunary sequences do not work that way.

Converging has a proper definition and it's not about something being more dominant or not.

Fun example: Let $f(z)=z+z^2+z^4+z^8+\cdots$

In fact, I'm not even sure if the cesaro limit of this thing is actually $0$.

@AkivaWeinberger how did you type the i?

@Semiclassical oh, I've heard of this before

very strange

Yeah, it's a weird one

ï î í ì

It's easy enough to show that it's got a radius of convergence of 1.

@LeakyNun On an iPhone you can long-press keys to get dıacrıtıcs

@Secret of course you know the answer to that

@AkivaWeinberger i see

All standard on a european keyboard :^)

But it doesn't converge anywhere on the unit circle, if memory serves, and (here's the kicker) it cannot be analytically continued outside of the unit circle.

There exists a sequence that hits every positive integer exactly once, for which it's an undecidable problem whether one number appears before another

The unit circle serves as a natural boundary of analytic continuation.

(Proof: countably many algorithms, uncountably many bijective sequences)

another weird aspect, if I'm remembering a post on MO right: If $z$ is a nonzero algebraic number in the unit circle, then $f(z)$ is transcendental.

(I have no idea how that's proven.)

@AkivaWeinberger can you help me on functional analysis ? please

I don't really know any functional analysis

I know enough to do quantum mechanics.

Don't you have teachers or classmates who can help you?

(which basically means "Hilbert spaces yay")

does a divergent series allow a string of zeros to be associative? Given how zero is the additive identity, for the above 1,0,2,0,0,3,0,0,0,4,0,0,0,0,5,0,0,0,0,0 .... series, its cesaro limit should be nonzero for any n since any string of zero will be summed away (and the fact that it begins with 1)

The Cesaro limit is the limit of averages, isn't it? That would be zero here

Oh, wait

Wait, you're right… that's not enough zeros for it to equal 0

It should go to 1, I think

Certainly the running average keeps hitting 1.

Hm, right before it hits $1$, it hits $\frac{T_n}{T_{n+1}-1}$

which goes to $1$ as well, I think

($T_n$ being the $n$th triangular number)

So, yeah, it should go to $1$.

And after you've gone through n terms, the gaps between 1s are about sqrt(n) terms long.

(NB one cannot make a sum of zeros to become nonzero without causing a contradiction. And as far I know, unlike what I did a year ago and for 5 months, nothing can be nuked to get around it.)

@Secret ... what the

Do you know how sum to infinity is defined?

$\lim_{m\to \infty} \sum_{n=0}^{m}a_n$

@Secret exactly

Can someone help on strong convergence in $W^{1,p}$ please

and you can prove by induction that finite sums of zero must be zero

and then your problem becomes a limit of the zero sequence

which I hope you'll be able to figure out without any guidance

The limit of the zero sequence has to be zero, otherwise some cardinal number arithmetic will be violated since all cardinals has to be absorbed by zero

we can also prove that with transfinite induction (I think)

what...

Since any finite sums of zero is zero, and summing to $\omega$ is the supremum of the sequence of all the finite sums of zero, it follows that the sum to $\omega$ is zero

stop conflating everything with cardinals/ordinals

we can then repeat this induction as far up the ordinals as we want and thus proving that the sum of any number of zeros is zero

Otherwise the limit of the zero sequence is zero because it is a constant sequence

@SteamyRoot hello

I haven't dealt with sums of transfinite sequences of real numbers before...

hello @Astyx

someday, I need to read into a book about infinite sets as obviously the henious amount of mistakes I have made, means just reading wikipedia, proofwiki and discussing with mathers is not enough

Pretty sure you can prove that the limit of 0, 0, 0, ... is 0 without invoking cardinal numbers or transfinite induction at all.

@TannerSwett Indeed, $0$ will be a limit of that sequence regardless of the choice of topology even

@TobiasKildetoft it might not be the only one if you mess with the topology though

nevermind, I should really learn how to read before answering, you wrote "a limit" :P

My ignorance of point-set topology means that my brain breaks at the suggestion that there can be more than one limit. Presumably it makes sense but it sounds like a contradiction in terms.

@AlessandroCodenotti Right, exactly

this happens only in spaces that are not Hausdorff

Ah. So 'here be dragons' territory.

The plane with two origins is an easy example

the trivial topology is probably the easiest, every sequence converges to every point of the space

I think, at least.

I like non-Hausdorff spaces.

The Hausdorff condition pretty much just asserts that any two points can be distinguished from each other.

Ugh. Anyone know a non-terrible way to get the radius of convergence of the following series? My suspicion right now is that the radius is zero so that it's valid only in an asymptotic sense.

$$\sum_{k=0}^\infty \frac{(2k)!}{k!^2} \Gamma(1-2k,s) x^k$$

where $s>0$.

I don't really care about that. I'm fine with spaces having "indistinct" points in them.

...I'll take that as a no, for now :P

@Semiclassical Say we have a sequence in $[0,1]$

If $a_0\in\{0\},~a_1\in\{0\},~a_2\in\{0\},~\dots$, clearly the limit equals $0$

But if I say $a_0\in(0,1],~a_1\in(0,1],~a_2\in(0,1],~\dots$, you don't know; it could be in $(0,1]$ or in $\{0\}$.

Now, consider taking the topological space $[0,1]$, and quotienting $(0,1]$ down to a point.

That's the Sierpiński space, a non-Hausdorff space with two elements.

(Usually they elements are just called $0$ and $1$; here, $1$ is the point that $(0,1]$ got shrunk down to.)

The constant sequence $1,1,1,1,\dots$ has two limits: $0$ and $1$.

The constant sequence $0,0,0,0,\dots$ has only one limit: $0$.

Yeah, this is sufficiently weird to me

The reason is that, if you "undo the quotient", you can't tell where the limit would end up being

That's what the first two examples with the $a_i$ were demonstrating

I believe that all non-Hausdorff spaces are quotients of Hausdorff spaces, actually

ctrlv.in/980291

Well, the point seems to be that the definition of limit doesn't behave the way I'm used to in the non-Hausdorff case

Could somebody give me a hint for this area please? Seems no matter what I do, I would either undershoot or overshoot the area to subtract

For example, the countable cofinite topology is $\Bbb Q$ where the equivalence classes are the sets of things with the same denominator.

(Every open set in $\Bbb Q$ hits all but finitely many denominators.)

@JoeStavitsky Use polar integration. Break it up into three integrals

What are the angles of the points where the two curves intersect?

@AkivaWeinberger Right, I got that far. Angles of intesection are +/- pi/6

What portion of the curve would $\int_0^{\pi/6}$ be?

(Also, do you have MathJax on?

If not, look for where it says "LaTeX in chat" in the room description on the top right)

@AkivaWeinberger yea I put it in my toolbar :)

ty

Test: $\displaystyle\sum_{n=1}^\infty\frac1{n^2}$

Is that rendering?

yea works fine, ty

Problem is I dont know how to type it yet :/

Yeah whatever

But, that integral is 1/2 the whole petal

If we wanted the entire petal, not just the piece of it, what would the bounds of integration be?

+/- pi/6

And what are the angles of the points where it intersects the circle?

Where $\cos(3\theta)=\frac12$?

3(theta)=pi/3

theta = pi/9

Right. Plus or minus pi/9

(sorry, I had tht wrong before, ain't I)

What part of the bulb would $\int_{-\pi/6}^{-\pi/9}$ be?

And $\int_{\pi/9}^{\pi/6}$

If you put those two areas together, you get the area you want minus a sector of the circle, I'm thinking.

@AkivaWeinberger, OK ty so much. It just didnt occur to me to chop up the inner area

You should say $V$ and $W$ are subspaces of $\Bbb R^4$, and $T:\Bbb R^4\to \Bbb R^4$.

Presumably you also have a condition on the dimensions of $V$ and $W$?

Hint: pick bases for $V$ and $W$ and extend them both to $\Bbb R^4$.

@anon Did that, but I dont know how to continue

map the basis elements of V to 0, map the other basis elements to those of W. done.

I'm thinking on creating a system of equations

Is the domain of $T$ $~\Bbb R^4$? Do we know anything about the codomain?

Also $~\Bbb R^4$

It was $T: ~\Bbb R^4 \to ~\Bbb R^4$. My bad

Hello

The condition you want is dim(V)+dim(W)=4 (rank-nullity theorem)

If e.g. V=span{v1,v2}, extend {v1,v2} to a basis {v1,v2,v3,v4} of all of R^4, and if W=span{w1,w2}, extend {w1,w2} to a basis {w1,w2,w3,w4} of all of R^4, then for ker(T)=V and img(T)=W you can set T(v1)=0, T(v2)=0, T(v3)=w1, T(v4)=w2

reasoning is similar if dim(V)=1,dim(W)=3 or dim(V)=3,dim(W)=1

if dim(V)=4,dim(W)=0 use zero map, if dim(V)=0,dim(W)=4 use identity map

What happens if $dim(V) + dim(W) > 4$

Then its not possible to generate a $T$ with said conditions?

Yeah

indeed, if T is a linear operator on R^4 then dim(ker(T))+dim(img(T))=dim(R^4), that's the rank-nullity theorem

With that kernel, the image wouldn't be able to become big enough

Thanks :) !

Always pick an appropriate basis

If the sum of dimensions is greater than 4, the intersection is a subspace of dimension $\geq 1$

no one want to help me ?

@Axoren Not sure what you're looking for specifically, Axoren. For qualitative ODEs, Hirsch/Smale (new edition with Devaney) is a superb book on dynamical systems. For an intro to PDE, look at Fritz John's book (Springer).

hi @anon, @Maks, DogAteMy

@TedShifrin Thanks, Ted. I'll take a look at Hirsh/Smale's book next.

@Kirill: To give you a further hint, suppose I'd asked you for the initial condition $y(-1)=0$ instead of $y(0)=0$? What about the two solutions we had already?

@TedShifrin Does this seem like an interesting question for MSE to you? ^

i think you should put it on MO

Hi @Te

@TedShifrin

done

@MikeM: I just met an old friend of yours.

hi @Ted

Hi @Alessandro

hi

people

@Ted Dustin told me you were in his office when it happened.

Not exactly, but ... I'm confused about names ...

Another problem

Give me some insigth :D

Given $T(x,y) = (y,x)$

I'm trying to find a linear combination

That diesn't make sense, Maks.

$T(x,y) = (y,x) = aT_1 + bT_2 + cT_3 + dT_4$

Oh .... I see.

What is wrong ?

Coordinates with respect to ... is wrong. Express as a linear combination is ok.

So what have you got?

A system of equations

Hi chat

Ok, so?

HI, PVAL..

No.

Oh, that's the problem

My variables are $a,b,c,d$

This holds for all x,y.

why not ?

What did I do to deserve the trailling periods..

Hush, PVAL..

@TedShifrin heyo

Hi, Danu!

Yippee! Ausgezeichnet!

You happy?

Yeah, I'm very relieved mostly to have found something :P

Well, upward and onward!

The research group is really nice and a lot bigger than the one I'm currently in, which is great.

Hopefully a less dysfunctional set-up :)

And they seemed excited to welcome me into the group

Ooohh, grats :O

I should look and see if I know anyone.