Mathematics - 2016-05-21 (page 1 of 2)

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12:42 AM

evening chat

1:16 AM

hi @SemiC

Carry on Smiling

1:46 AM

what is wrong with this argument?

0

A: Is there such a thing as an equation with noncomplex quaternion solutions?

Complex numbers have a very special property, they are an algebraically closed field. This means that any polynomial has complex solutions. If $P=a_nx^n+\dots +a_1x+a_0$ is a polynomial with complex coefficients then there exist $n$ complex numbers $r_1,r_2\dots r_n$ (not necessarily distinct) su...

3 hours later…

4:30 AM

Hello, guys could anyone link me to what $\int_{C} f \cdot dc$ means?

@Jake1234 Is $C$ the region of integration? What's the context?

curve

Okay, so it means you're integrating $f$ along curve $C$?

user147690

$dc$ or $ds$? Or I mean those $c,C$'s are unrelated surely

Is this a curve in $\mathbb R ^2$?

4:37 AM

I presume it's the same thing as ... taking the integral of f dot product tangent vector of length 1 with hausdorff meausre

user147690

It's a line integral for a scalar field

Could be a vector field too.

i'm sorry, both c are small

user147690

Sure

but i can't find a rigorous definition of it

yes, in R^2

user147690

4:39 AM

It is truly $\int_c f\cdot dc$?

yes, it's that.

Here's a definition for you. mathworld.wolfram.com/LineIntegral.html

it's equal to $\int_{c} rot f d lebegmeasure$

Is that rigorous enough? Parametrize your curve $C$ as a function $\sigma(t)$.

that's great thanks

i was about to ask something about that after this question - what's there on the link on the second line/

4:41 AM

Where?

$\int_{C} F_1 dx + F_2 dy$ = $\int_{C} F_1 dx$ + $\int_{C} F_2 dy$ /

?

I'm not sure what the thing means basically

no, I'm not just not sure, I have no idea.

Well, $F = (F_1, F_2)$ is a vector-valued function. Consider each of the "outputs" as a function in its own right, ie. $F_1(t)$ is the $x$-coordinate, $F_2(t)$ is the $y$-coordinate.

right I understand what $F$ is, but I don't understand what the expression with the integral is.

Let $\sigma(t) = (x(t), y(t))$. Write the dot product $F(\sigma(t)) \cdot \sigma'(t)$ as $F_1(x(t), y(t))x'(t) + F_2(x(t), y(t))y'(t)$.

And notice that $x'(t)dt = dx$.

So you can write the integrand as $F_1 dx + F_2 dy$. I know it looks funny.

Of course this is all just symbol manipulation. What does this conversion actually mean?

so what I wrote above - $\int_{C} F_1 dx + F_2 dy$ = $\int_{C} F_1 dx$ + $\int_{C} F_2 dy$ is non-sense?

4:56 AM

No, it's not nonsense. But you have to clarify what the meaning of the integral bounds on the right side is. Also, it's only true under certain conditions on the path/field.

Because on the left side you're still integrating along the curve, but on the right you're integrating first in $x$ and then in $y$. As if you changed the curve to be just a horizontal and then a vertical leg.

Is the value of the integral still the same? Yes, if the field is irrotational.

As for the rigorous definition of the expression $\int_C F_1 dx + F_2 dy$ -- I'm not sure. Could somebody else pitch in?

@AlexClark?

Thanks, I'll have to digest all of that, and it will probably take me some time. What $\int_{C} F_1 dx$ could mean, would be nice. It could just be some set, not necessarily $C$ that it's integrated across...

$\int_C F_1 dx$ is just your plain old Riemann integral. Actually you should write something like $\int_{x_0}^{x_1} F_1(x, y_1) dx$ instead.

$y$ is fixed for that one.

but if $F_1$ is a function of 2 variables..

It is, but $y$ is fixed for that integral.

Just as $x$ is fixed for the second integral.

That is, you're integrating along $x$ holding $y$ constant, then integrating along $y$ holding $x$ constant. That's why I said it's like you're integrating over a new rectangular path.

Right, what I don't get, is what happens if i'm not going across x, but across some curve.

5:06 AM

Check out this video, it might help clear things up. ocw.mit.edu/courses/mathematics/…

You are going across $x$.

You were going along the curve in the original integral. But now you've rewritten it as two integrals, one across $x$ and the other across $y$. This is only valid under certain conditions, because in effect you've changed the curve.

It's hard to explain without a diagram.

maybe maybe $\int_{C} F_1 dx = \int_{x_0}^{x_1} F_1(x,y(x)) dx$ would make sense?

By the way, it turns out the notation $\int_C F_1 dx + F_2 dy$ doesn't have any "special meaning". It's just another way of writing a line integral.

ooh alright.

No, it wouldn't make sense, because $y$ is not necessarily a function of $x$ along your curve.

For example, it could double back or go in loops and so on.

ok, but let's say it was?

5:14 AM

No, I'm not convinced that would work. I suspect it wouldn't.

I just still don't understand what $\int_{C} F_1 dx $ - you wrote it's riemann integral - but if the y changes across C, how can it be riemann integral, or in what sense would it be a riemann integral?

You're not listening to me! $y$ does NOT change.

You have a NEW curve.

This new curve has a horizontal leg and a vertical leg.

Its endpoints are the same as those of the original curve.

So we are integrating along the first leg of this new curve, and $y$ is constant. Then we add the integral over the second leg, along which $x$ is constant.

Ooh ok.

And the $y$ that is constant... is the one at the start of the curve? (maybe depending on orientation, it could be the end ?)

Yeah, exactly.

You could do it either way.

Essentially you're using the fact that the curl of the field $F$ is everywhere zero, and that implies that there's a potential field of which $F$ is the gradient. So integrating along a curve just gives the difference in potential, and so the exact curve doesn't matter.

It's a very convenient result, because we can just say okay, curve doesn't matter, only endpoints, so let's pick the simplest possible curve with those endpoints.

Alright, I have a basic idea of what the things are meant to be used for, thanks a lot.

5:22 AM

If we're working in cartesian coordinates, the rectangular curve is usually the easiest to integrate along because we can split up the integral into two regular one-dimensional integrals.

Okay, that's good. You're welcome.

Is $x+2 \cong \langle x+2\rangle$? (isomorphic)

What do the triangle brackets mean?

$\langle x + 2 \rangle = \{ \alpha(x+2), \alpha \in \mathbb{R}\}$ ($\mathbb{R}$ is an example) OR simply, all linear combinations of $x+2$ in $\mathbb{R}$

Okay.. what is $x$ in this example?

$x \in \mathbb{R}$

5:27 AM

But then the set is just the set of all reals, isn't it? That is, unless $x = -2$ in which case the set is ${0}$.

I'm confused. What do you mean by isomorphic?

Oh, right.

Thanks, I think I made an error

Okay.

this is going to kill me

5:49 AM

@SAWblade That's a neat formula for the Fibonacci numbers. But there's a cleaner form which is easier to remember: $$\sum_{k=0}^n \binom{n-k}{k} = F_{n+1}.$$ Of course the terms where $k > \lfloor n/2 \rfloor$ vanish.

Ah yes, but my less clean formula popped up in my research and is why I was so surprised by it. xD But I did not know that identity, thank you for sharing. :)

No problem. :)

It's amazing where the weirdest things pop up, tbh ...

Yeah? Where did it pop up?

I'm counting the number of self-avoiding walks connected the bottom left corner of a $2 \times n$ grid to the top right corner of said grid.

5:54 AM

Oh, neat.

Mhm. It's part of my larger project to count the SAWs connecting an $m \times n$ board.

I'm aware it's not the most grand mathematical undertaking, but it's cool to me. P:<

Have you solved the $m=2$ case ?

No, that sounds like an interesting problem.

Yeah, it's $12$ for $n = 2$.

Oh, I misunderstood your question!

I'm currently in the midst of trying to solve $m =2$. My bad. xD

Wait, how is it 12 for $n=2$?

I thought you were referring to the $2 \times n$ board and said the number of SAWs on a $2 \times 2$ board. xD

5:58 AM

Aren't there only $2$ ways for a $2 \times 2$ grid? There's something I'm not getting.

Could you define SAW for me?

Ah, right, I think of it differently. I'm traveling along the lines, not the squares of the grid.

You might think of it as a $3 \times 3$. xD

Haha, okay. I was considering the grid of points.

Everyone else does. xD I stubbornly insist on viewing it my way. xD

So what answer did you get for $n=1$?

And that's why all of my entries on the OEIS are off ...

An $m \times 1$ board has $2^{m}$ SAWs connecting its diagonals.

6:04 AM

Makes sense.

Mhm. Not enough room for there to be 4-directional paths, so the formula for 3-directional and lower paths is enough there. xD

Seems so much harder for $m=2$!

It is! T^T

I was thinking "the number of odious numbers between $0$ (included) and $2^{n+1}$ (excluded). So $2^{n+1}/2 = 2^n$.

Yeah, just one extra direction and so much more can go on!

Tell me about it. xD

What do you work on, Yakov? :)

6:10 AM

Just studying for now. I'm working through Knuth et al's Concrete Mathematics.

Oh, I love that book. :0

Very solid, dabbles in a lot as well. :)

Yeah, I'm really enjoying it. I saw the one question you posted on MSE a year ago. It's hilarious reading through the answers. Why are people introducing all that complicated machinery?

What was wrong with the first answer?

Nothing, but there's nothing wrong with different approaches to a problem. Though the rest do look ... waaaaay too involved. xD

Yeah, always good to have an extra tool. Even if it's one you'd rather avoid using if at all possible.

So the Fibonacci numbers really come up for the $m=2$ problem?

Yup!

All SAWs on a $2 \times n$ board can be expressed as a placement of certain objects on the board that can be connected to form a unique SAW.

6:23 AM

Cool. You have a candidate OEIS number?

The number of allowable placements of these objects is $$2^{n} - \frac{1}{2}[F_{n} - \frac{2}{\sqrt{3}} \sin (\frac{\pi}{3} n)]$$.

Candidate? :0

Ahh, I see, you posted that above already before I joined.

Mhm. :)

I figured out the answer to my question and thought I would post it in case someone was interested. P:

Oh, I thought you hadn't solved it yet.

No, I was asking how to simplify a summation. P:

It simplifies to that. xD

Or really, the half after the minus sign.

6:28 AM

Wait, I don't get it. Is that expression equal to the number of SAWs on a $2 \times n$ board?

Or is there more than one way to connect each placement of objects?

There's more than one way, but each way is unique.

I don't get it. Unique in what sense?

By the way, are the objects "blocks" that take up the entire breadth of the row?

Yes. Each connection forms a unique SAW that cannot be made with another placement of objects.

I see.

So you don't know yet how many SAWs correspond to each placement?

I do not. :x

That's my next step tbh. xD

6:40 AM

Haha, okay.

Seems very difficult. Have you tried to solve it in any other way?

No, I couldn't think of any. xD

Hi all, is it possible to form a quote-unquote "subring" of the Complex field out of a subset of $\mathbb{C}$?

I have to show whether or not $\mathbb{Z}[\sqrt{3}] = \{a + b\sqrt{3} : a,b \in \mathbb{Z}\}$ is a subring of $\mathbb{C}$

I think I found another way.

oh, hit me up

Wait, give me a few minutes to verify it.

6:46 AM

(it's not a subfield, but I can make a commutative ring with unity out of that set)

6:57 AM

@SAWblade Yes, it works (I had to fix something.)

Hit me up!

@BenjaminR I don't know. Maybe if you stick around someone will answer, though it's pretty slow right now.

@SAWblade A recurrence.

Okay ... :0

no worries. I will hang around, eventually I will make it into a question if necessary

I don't know how to solve the recurrence but I have a recurrence which at least gives the correct answer $12$ for the $2 \times 2$ case.

6:59 AM

show it to me P:<

Okay, give me a couple minutes.

alright but after this i must leave

as it is 3:00 a.m. here

ok wait

Did you get the sequence 1, 4, 12, 38, 125, ...?

Just to verify.

Is this for $2 \times n$?

Yes.

7:06 AM

That seems familiar, yes.

Okay. Here's the recurrence. I have to translate from my "points" to your "squares", subtracting one everywhere, I'll try to not make a mistake. For $n=0$ we get $1$, denote the result for $2 \times n$ by $P_n$.

I also use points, it's fine. xD

Let $P_0 = Q_0 = R_0 = 1$ and
$$P_n = P_{n-1} + Q_{n-1} + R_{n-1} + R_{n-2} + \cdots + R_0 + 1,$$
$$Q_n = P_{n-1} + Q_{n-1} + R_{n-1},$$
$$R_n = P_{n-1} + Q_{n-1} + R_{n-1} + P_{n-2} + \cdots + P_0,$$
for $n > 1$.

2

Sorry, I don't know if I can use the align* environment here.

Wow, that's crazy. :0

Also the $\cdots$ in the last equation for $R_n$ seem ambiguous, I mean that everything "hidden" is decending $P$ terms.

7:13 AM

@YakovShklarov It's fine, I have my answer:

user image

Yeah, I had to break it down into three interdependent sequences. $Q$ is the number of ways to end up in the "middle point", $R$ is the number of ways to end up in the "lower point" (I mean on the same side of the row instead of the opposite side.)

I will look into it more tomorrow, but I really appreciate the help. :)

Hey, my pleasure! It's just a matter of breaking it down into cases: to get to the corner we still have to get to either here or here, how many ways are there for each of those?

I mean "get to either here or here first".

As for a closed form for the recurrence -- well, I don't know! But it at least seems approachable.

@BenjaminR, good!

@SAWblade I'd imagine there are standard methods for attacking these sorts of recurrences (though I don't know what those methods are.)

Also, it seems you would be able to generalize this solution to arbitrary $m$.

It's kinda cool that you can have a subring that doesn't require all the properties of the supering, but I guess that makes sense, you can do the same thing with groups from from rings, sets from groups

Wait, which property is "missing" in your subring?

Oh, I think I see what you mean -- you don't have to check all the ring axioms?

The thing is, the subring test is actually a theorem: it states that if these conditions hold, then the subset is a ring in its own right. It's not that the ring axioms don't hold for a subring, it's just that it's unnecessary to verify them.

Anyway I'm going to bed. Good night!

1 hour later…

8:36 AM

Hello!!

Could you explain to me the meaning of a short exact sequence? Which is the difference of a short exact sequence and an exact sequence?

user147690

@MaryStar Have you read the definition on wiki?

user147690

(Specifically that of an exact sequence)

I have understood the definition of an exact sequence... At the short exact sequence is the difference that it is always of the form $0\rightarrow A\rightarrow B\rightarrow C\rightarrow 0$, but at an exact sequence it can be of the form $0\rightarrow A_1\rightarrow A_2\rightarrow \dots \rightarrow A_n\rightarrow 0$ ? @AlexClark

user147690

@MaryStar Yes to the former. A short exact sequence of objects is given by two maps (The inner two), where the sequence is exact

user147690

With exact sequences, starting and ending in $0$ is a restriction. They don't need to start and end with $0$

8:52 AM

Ah ok...
I have shown that if the sequences $0\rightarrow A\rightarrow B\overset{f}{\rightarrow}C\rightarrow 0$ and $0\rightarrow C\overset{g}{\rightarrow}D\rightarrow E\rightarrow 0$ are exact, then the sequence $0\rightarrow B\overset{gf}{\rightarrow} D\rightarrow E\rightarrow 0$ is exact.

So, each exact sequence can be implied by short exact sequences as above, right?
But how could we prove this? Could you give me a hint? @AlexClark

2 hours later…

user147690

11:09 AM

@TobiasKildetoft What do you think about having two or three supervisors for Msc programs and such?

Tobias Kildetoft

@AlexClark no idea really. It sounds either great or terrible

user147690

One of the academics at my uni said approximately: '1 supervisor is fine, 2 supervisors is better, 3 supervisors is excellent, 4 supervisors is too many'

Tobias Kildetoft

I suppose it depends a lot on how it works precisely

user147690

As in, I have just one project in mind, they all work in areas that intersect over my project choice, and I meet with all of them individually each week

user147690

Seems good if I can handle the work load right?

Tobias Kildetoft

11:14 AM

Sure

user147690

(just theoretical atm)

Tobias Kildetoft

as long as the scope is fairly fixed from the start, or it can easily spiral out of control with so much different input

user147690

How did you choose your advisor?

Tobias Kildetoft

@AlexClark Only guy who did finite group theory at the institute, and that was my main interest at that point

user147690

I think Peter Mcnamara is a really good advisor, but Masoud Kamgarpour is giving out a really interesting project too

Tobias Kildetoft

11:16 AM

(as it turns out, this was not a great choice in terms of future prospects)

user147690

@TobiasKildetoft Oh really?

Tobias Kildetoft

@AlexClark mainly due to various bureaucratic stuff that was impossible to predict

user147690

@TobiasKildetoft Oh damn :S

Tobias Kildetoft

As in the institute not offering any PhD positions for a short while unless you could get external funding of some sort

user147690

Btw why does your paper here: arxiv.org/abs/1605.01373 have written "rep\-re\-sen\-ta\-ti\-ons"?

Tobias Kildetoft

11:24 AM

Hmm, probably LaTeX kept splitting the word incorrectly across lines (at some earlier point), and it got copied to the abstract there even though MathJax does not seem to handle it properly

It was Mazorchuk who did the submission to arXiv

user96977

if i have P(a|b) and i marginalize out P(a), what am i left with?

@TobiasKildetoft Hey
Are you familiar with Hölder's inequality? I have a question.

Tobias Kildetoft

@Evinda no

Ok

Tobias Kildetoft

11:45 AM

@AlexClark I submitted my updated paper now, with you in the acknowledgements and the exception of order 768 removed.

(plus the other improvements suggested by the referee)

user147690

@TobiasKildetoft Very cool :D. This is the first time I've been mentioned in an academic paper haha

Tobias Kildetoft

@AlexClark Get writing so you can get properly cited instead of just acknowledged

user147690

@TobiasKildetoft For sure :D

Tobias Kildetoft

You know what they say, citation is the sincerest form of flattery (I am pretty sure that is the correct saying at least :) ).

user96977

what is P(a|b)P(c|d) ?

user96977

11:50 AM

is it P(a,b|c,d) ?

user147690

@TruthSerum Why?

0

Q: Existence of operator

I want to show that for $ s> \frac{1}{2} $ there is a bounded linear operator $ T: H^s(\mathbb{R}^n) \to H^{s-\frac{1}{2}}(\mathbb{R}^{n-1})$ following the below steps: Consider that $ u \in C_0^{\infty}(\mathbb{R}^n) $ and show that $ \widehat{Tu}(\xi')=\int_{\mathbb{R}} \widehat{u}(\xi', \...

inequality fourier-analysis sobolev-spaces

user96977

P(a|b)P(c|d) = P(a,b)/P(b) * P(c,d)/P(d) = P(a,b,c,d) / P(b,d) = P(a,c|b,d), is that correct?

Does anyone have an idea?

user96977

12:19 PM

hello?

12:37 PM

hi

In each part of this question a proposition p is defined. Which of the statements that follow the definition correspond to the proposition ¬p? (There may be more than one correct answer.)

p is "All people in my class are tall and thin".
(i) "Someone in my class is short and fat"
(ii) "No-one in my class is tall and thin"
(iii) "Someone in my class is short or fat"
(You may assume in this question that everyone may be categorised as either tall or short, either thin or fat.)

What would your answer? I think it should be (ii)

user147690

Anyone have any experience plotting 3d polytopes?

1:04 PM

@AlexClark With a pen and paper, yes.

user147690

@BalarkaSen :P

user147690

@BalarkaSen You'd have trouble plotting with pen and paper what I have in mind

user147690

Or rather, what I don't have in mind :P

@AlexClark Did you learn blowups?

user147690

1:32 PM

@BalarkaSen I have, but I am not 100% happy with doing them yet, other stuff has been getting in the way(mainly that we outright skipped the chapter in class :\ )

hello, one question

$$|\sin x|\leq x$$ and since increasing function retains inequality we get$$f(|\sin x|)\leq f(x)\,\,\,\,\,\,\,(1)$$

similarly $$f(|\sin x|+3)\leq f(x+3)\,\,\,\,\,(2)$$

then $(2)-(1)\implies f(|\sin x|+3)-f(|\sin x|)\leq f(x+3)-f(x)$

is this correct?

1) first inequality is valid for $x\geq 0$. 2) Is $f(x)$ assumed to be an increasing function?

yes $x\ge 0$ and function is increasing

but is my subtraction of equations correct?

mmkay. unfortunately, i don't think you can subtract inequalities in that way

user147690

@BalarkaSen I have blown up $\Bbb A^3$ along $V(x,y)$, $\Bbb A^2$ at the origin, and an elliptic curve at the origin, but nothing else. No idea if the last one was correct though, and calculated the exceptional sets. I have also been thinking about affine schemes

1:39 PM

adding them is fine, but your number two is equivalent to $-f(|\sin x|+3)\geq -f(x+3)$

and adding that to your initial one isn't going to work

one can probably find an explicit counterexample, i.e. a case where $A>B$ and $C>D$ but $A-C < B-D$

actually, here: 2>1 and 3>1 but 2-3 < 1-1 =0

hey but $|A-B|\ge |B-D|$, correct?

similarly $f(|\sin x|+3)-f(|\sin x|)=|A-B|$

so inequality should hold, shouldn't it?

How do you call the angle where angle $ACB$ is half of angle $AOB$ because $O$ is the center of the circle? How do you call the $ACB$ angle or how is the theorem called? (There's a name in my first language).

user image

Sounds like the inscribed angle theorem

Thanks!

2 hours later…

4:15 PM

morning

l' - ' - ' - ' - 'l

Good morning!

Well, not morning here anymore.

o well

l' - ' - ' - ' - 'l

so it goes

hi chat

how it be?

4:29 PM

pretty good. i left a calculation running overnight, and i arrived about a few minutes ago to find it has about 10 minutes left

l' - ' - ' - ' - 'l

What were you calculating?

(not one calculation, really. more like 35*16 individual calculations)

an algorithm to solve a certain pair of coupled nonlinear PDEs

for 16 different parameter choices, iterated 35 times each in order to get convergence

l' - ' - ' - ' - 'l

cool, let us know how it went, although I guess things shouldn't go awry in that situation?

the choice of 35 was an estimate so that i'd get as many computations done in the time between my leaving for the evening and arriving back this morning

god i hope not :/

l' - ' - ' - ' - 'l

Seems like you've estimated pretty well

4:32 PM

well, i thought i'd be in here a half-hour ago

so not perfect, but close enough

l' - ' - ' - ' - 'l

a true analyst

estimating his time of arrival to create more estimates at the office

hah

it's probably the most 'experimental' thing i've done in research in my time as a physics grad student, which tells you how much of a theorist i am :p

l' - ' - ' - ' - 'l

I've seen you around here quite a few times and never realized you studied physics

yeah

though to a great extent it's applied math as much as anything else

but if he didn't study physics, I don't know how I would poke fun :)

4:37 PM

has that ever stopped you with Balarka? :P

where there's a will there's a way.

l' - ' - ' - ' - 'l

@MikeMiller I remember you recommending Lee's Topological Manifolds to me. Do you know if there are any selected answers in the back? They're usually helpful when studying alone.

@SemiC you got me there

@Impossibletotype: more or less no math books approaching or at the graduate level will have exercise solutions. You're approaching the point where there's signidicantly more value in being able to look at your proof yourself and determine if it's a good proof.

l' - ' - ' - ' - 'l

@MikeMiller good point, I guess I am slowly reaching that point. I'm free from school so I have so more time, I'll see how it goes (if at all).

My username starts with an L if you're wondering, but I should probably change it lol.

I'm tempted myself to change mine to just Semiclassic

given that that's what ted shortens it to anyways :P

"I'm not quite a classic yet"

l' - ' - ' - ' - 'l

4:50 PM

In 50 years stack chat will still be active and semiclassical will finally be able to call himself a classic.

not enough rigor to be modern, not enough imagination to be classical

what makes this data frustrating is that i'm looking for a specific point in the parameter set

problem is, it's at that specific point that things converge the slowest.

user image

what i would ideally see is the blue curve, showing the relation between input parameter value and the measured output. but depending on when i stop iterating each input parameter, i'm likely to instead see the orange, green, or red curves

that captures stuff in the positive just fine, but not so well at or below the transition point at zero

add to that that i don't actually know the point at which that occurs :p

l' - ' - ' - ' - 'l

Iterate for yet another day?

Never stop iterating.

something like that

good way to not get published

therein lies the challenge. have to stop somewhere

i can hopefully improve things somewhat by taking each sequence of iterations and extrapolating

i.e. arguing that it only went to Y but it's 'really' converging to Y'

l' - ' - ' - ' - 'l

4:59 PM

how do you justify that?

as a way to somewhat evade the critical slowdown near the transition

@l'-'-'-'-'l glaring at people who ask questions really hard

by arguing that it accounts for what is seen. it's an experimental/phenomenological argument, to be sure.

but hey, physics

l' - ' - ' - ' - 'l

"So, how exactly do you argue for convergence to Y', Semiclassical?"

"ಠ_ಠ"

no, i do sympathize with that, for all the fun i'm poking

5:01 PM

what's amusing is that these are exactly the kinds of issues experimentalists run into all the time

l' - ' - ' - ' - 'l

What is the context of this PDE?

you can only measure for a certain amount of time, and the closer you are to a phase transition the longer everything takes

l' - ' - ' - ' - 'l

As in what does this PDE describe.

it's a short-time form of the KPZ equation

oh, cool, the bank is open today and i can get laundry money

l' - ' - ' - ' - 'l

5:03 PM

I'm not exactly sure why I asked, considering I've never solved a PDE myself. But thanks!

we all solve PDEs in here

it's used to describe fluctuations, in space and time, of the height of a 1D interface

under a certain choice of boundary conditions

l' - ' - ' - ' - 'l

@MikeMiller you're telling me I've been unknowingly solving PDEs all along?
I'll take that.

so you ask things like: hmm, if I was to measure that the surface has a height of X at one spot at a certain time, what height do i expect to see at another time at that same spot

this is the first PDE thing I've done in a long while, i should note

l' - ' - ' - ' - 'l

@Semiclassical Ahhh I see, thanks.

5:06 PM

plus, i didn't derive the form of the equations i'm solving. that's something I was given as part of the suggested project

at least SemiC and I do

I guess everyone else is a mystery to me

i should also point out that the graphs i showed above are guesses for how it should look. they're not predictions

nor do I know that it behaves like $x^{1/2}$ as it was in the blue graph

so there's quite a lot of mystery

user image

this is what my last numerical run actually gave me

the red points are the last iterates, and the blue shows the previous iterates (all 34 of them) at each parameter value

and if you're wondering how I know that the red point is a good estimate for the final converged value..."ಠ_ಠ"

what i plan to do for the next few hours is start a run that scans from 6.02 to 6.08 and does more iterations.

while i try to analyze the data set above

@SAWblade Looks like it's not so easy to generalize to $m > 2$ after all.

5:24 PM

I did not expect it to be. xD

Have you written any code to compute answers?

what I'd really like is a good way to distinguish time series of the form $a+b e^{-n c}$ for $a$ being zero versus nonzero

wouldn't be a problem if $c$ was larger, but it's small enough that things iterate painfully slow

I have not. I have been attacking it the theoretical way. P:

5:41 PM

It be useful to be able to check solutions.

5:57 PM

Fair enough!

But that's what the OEIS is for. xD

oeis.org/A006192

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