to boil it down: I know that my PDE has the right structure for the IST to matter. But that's usually presented in terms of the evolution of an initial value problem (i.e. complete information at one time) and I've got a boundary value problem (partial info at two different times)

so it's not clear that it's actually useful in an analytic sense :/

Not that I understand any of that, but fair enough.

(could still be useful for numerics, but i just got done doing that and i don't want to do any more :P)

Eh, in physics terms it's the difference between finding the trajectory of an object based on 1) knowing that an object starts at a certain position with a certain velocity, versus 2) knowing its position at two different times

The second is inherently harder than the first.

ok, i see

In that case, both are pretty easy. But in mine the difference is more substantial.

how can you tell what the trajectory is just based on the information of two points on it?

that doesn't seem right

Well, you need one more thing (for both of them, really): the acceleration as a function of time.

but if you assume we're just talking about projectile motion under a uniform gravitational field, then the trajectory is parabolic

...hmm. but is that actually enough info.

i think so, yes

a parabola is quadratic, no? two points should determine it fine

wat

Eh, there are a lot of parabolas through the points $(1,0)$ and $(-1,0)$

I think @Balarka means three

Or I'm missing something, like that it should go through 0 or something

more prosaically: If i know the initial/final positions and times, then I can work out the (constant) x-velocity

for the y-coordinate, i've got $y_f=y_i+v_{yi}t-\frac12 gt^2$ with $y_f,y_i,t,g$ known

and that's enough to find $v_{yi}$ as well. so the trajectory is determined, yeah.

but, of course, the intended lesson still holds: If I knew position and velocity at $t=0$ then I'd definitely know them at later times. But for the other problem I had to stop and think about it :)

@Semiclassical Yeah, that was a fine nonsense.

It's the classical mistake a lot of my students of linear algebra make: the dimension of the space of polynomials of degree $n$ is not $n$.

No, of course it's not.

:(

Of course not, yet so many students make that mistake

It's hilarious

I was thinking of something else. Anyway, how many points do you need to determine a parabola?

well, what's the dimension of the space of quadratics?

5.

In projective.

I always had trouble understanding why $S^1$ is the circle and $S^2$ the sphere and not $S^2$ the circle and $S^3$ the sphere

But parabolas are given by quadratic equations $b^2 - 4ac = 0$.

simpler than that: a quadratic is a polynomial of degree 2, so there are 3 coefficients

Huh?

(The other aspect of my problem is that the IST involves a kind of nonlinear version of the Fourier transform, and it's not obvious how my boundary conditions carry over once I do said transform.)

$ax^2 + by^2 + cxy + dx + ey + f =0$

$6$ coefficients.

In projective, you get one less, 5.

That's the dimension.

ah, you meant a general parabola

Ehhh. Is a circle a parabola?

No, @Krijn, that's the problem.

What does the moduli space of parabolas look like?

I meant one of the form $y=ax^2+bx+c$

Same, @Semi

which is what it'd be in the projectile motion case

It's cut out by $x_1^2 - 4x_0 x_2$ for one, so it's a quadratic hypersurface inside $\Bbb P^5$.

a parabola whose axis of symmetry is vertical, i guess

Anyways, I'm gonna time myself reading so I'll be gone for like, ten minutes.

Bottom line for me is: Initial value problems are hard enough. Boundary value problems are even worse :)

I am forgetting the argument for showing 5 points determine a conic.

Oh, now I remember. The space of conics passing through one point is a linear hypersurface and 5 of them intersect at a point in $\Bbb P^5$. OK.

5 of what intersect?

5 linear subspaces of P^5.

I want to translate that argument to prove something analogous for the parabola.

Ah. You mean in the sense that those are enough to span it?

I don't know what you mean. I just mean the space of conics passing through some fixed pt $P$ is a linear hypersurface. So if I have a bunch of points $P_1, \cdots P_5$ and look at space of conics passing through each, they intersect at a unique point.

That's the unique conic passing through $P_1, \cdots, P_5$.

mmkay

So space of parabolas is a quadratic hypersurface $H$ in $\Bbb P^5$. Space of parabolas passing through a single point is a codimension $1$ subspace of $H$.

Would that mean four points for a parabola instead of a generic conic?

Asking how many pts determine a conic is the same as asking how many of those codimension 1 subspaces of $H$ intersect at a unique point.

@Semiclassical I am not sure.

I don't have a good idea of what $H$ looks like.

Well, here's how I'm counting it.

If I have two points, that determines a line and a family of vertical parabolas through them. picking a third point fixes a unique vertical parabola.

that would seem to indicate that three points don't fix a generic parabola. but five points do determine a generic conic, so it should be four points to fix a parabola.

Quite heuristic, of course, but it seems sensible.

I don't see how you're getting that 3 points do not fix a generic parabola.

yeah, that's the weak point

...

oh, wait.

would you agree that, once I pick three points, that there's one unique vertical parabola?

$H$ is a dimension $4$ thing, so I'd expect at most $4$ points to determine a parabola, myself. Because intersection theory: it's a manfold of dimension $4$, so cup square cannot be more than $4$. So an upper bound is $4$

o/

What's a "vertical parabola"?

axis of symmetry is vertical i.e. $y=ax^2+bx+c$

(i'm assuming no x-coordinates of the chosen points are the same)

Sure.

Assholes :P

There are three coefficients. Three points determine three coefficients.

@Danu no u

okay. suppose i now rotate the plane through some angle such that the x-coefficients are still distinct

I've still got three points, so there's once again a unique parabola which is vertical relative to the rotated axes.

(Note that I have proved an upper bound is 4. If you can prove 3 points do not determine a parabola, you're through that 4 is an answer)

but the axis of symmetry of the second parabola is definitely different from the first. so they can't be the same parabola, so I'll have two different parabolas through the same three points.

What do you mean by 'rotate the plane'?

What is the new parabola, explicitly, which you claim to share the same three points with $y = ax^2 + bx + c$?

explicitly, i mean to introduce coordinates $(u,v)=(x\cos \theta+y\sin \theta, -x\sin \theta+y\cos \theta)$

Eh

whose axes will be at an angle $\theta$ with respect to the original axes

Just tell me the parabola. sin and cos aren't allowed if you're doing algebraic geometry.

pff.

actually, here's a simpler statement (though it requires the y-coordinates all be different as well.)

I pick three points to define a vertical parabola.

@Semiclassical I mean, are you trying to introduce them as coefficients, or what?

They are not algebraic functions.

i'm saying that you pick some particular $\theta$ and use those coordinates

I don't understand what you mean by 'coordinates' in this context. All you need to do is to just figure out another parabola, with the very same coordinates, which share three points as $y = ax^2 + bx + c$.

I am asking, what is your 2nd parabola?

Let me try once more. My argument is geometric.

OK

I pick three points. Then there is some unique parabola whose axis of symmetry is vertical, yes?

Sure. Aka, of the form $y = ax^2 + bx + c$.

Mmkay. Now, suppose I've drawn those three points on a piece of paper. I rotate the entire piece of paper through some small angle $\theta$ and draw new axes.

So you literally move those three points by the angle $\theta$ along a circle?

Right.

OK.

So far so good.

In these new coordinates, I once again have three points. So there'll be a unique parabola whose axis of symmetry is the new vertical axis. Yes?

Agreed.

Okay. I now rotate back to my original frame.

The parabolas will then coincide.

Why? They have different axes of symmetry.

The two vertical axes will differ by the angle $\theta$.

Eh?

I'm really not seeing what's confusing about this. The second parabola was drawn to be symmetric across the new vertical axis, not the original one.

Look, you had three points $P, Q, R$ you rotated them to $P', Q', R'$ and drew a parabola. The axis of symmetry of that is axis of symmetry of the first rotate by the angle $\theta$

So when you rotate back, they'll be the same. I think you're confused by your coordinates.

@BalarkaSen No, it's not.

I've got a new vertical axis. There's no reason for it to coincide.

How can you choose an arbitrary vertical axis and also guarantee that parabola will pass through your given three points?

I've got three points and an axis. That's all I need.

I am afraid you'd have to tell me an explicit equation of two such parabolas to convince me now, I don't see your picture.

Sorry.

Ah, but it's easy to see examples of two parabolas in $\Bbb R^2$ which intersect in 3 points.

I guess my point is really that there's nothing special about that vertical axis of symmetry. Three points and a line determine a parabola whose axis of symmetry is parallel to that line.

That follows simply by choosing coordinates so that said line is the vertical axis.

@Semiclassical Sure.

But then if I pick three points and two skew lines, I'll have two parabolas through those same 3 points but with different axes of symmetry.

The same parabola can't have two axes of symmetry, after all.

Ah, yes.

The rotation by angle was confuzzling me.

That's what I was getting at.

E.g. take $y^2 = 2x$ and $x^2 - 3x = y$.

I just constructed one on desmos.

Ahh. I was trying to work through Geogebra, but it doesn't seem to have the right tools for that.

Eh, those pass through 4 points though.

i mean, the rotation thing basically just amounts to saying: "If I draw this so that one of the two parabolas is vertical, I could rotate the system through some angle so that the other one is vertical."

I am confused now: clearly even 4 points do not determine a parabola on the plane.

Just make them intersect transversely.

Huh.

I should have realized that before :P

5 is the answer, then.

hmm

My cup-square thing is nonsense because $H$ is not a 4-manifold, it's a complex manifold of dimension 4.

aka dimension 8.

I guess the point is that 4 points still determines a family of parabolas.

Here's an interesting question, then.

That's obvious though, because if you have a conic $ax^2 + by^2 + cxy + dx + ey + f = 0$ that passes through 4 points you are left with 6 - 4 = 2 degrees of freedom in the coefficients. That is to say, 2 - 1 = 1 dimensional subspace of P^5.

Suppose I pick 4 points. What is the locus of points P such that the conic drawn through those five points will be a parabola?

what are you guys talking about?

You have a pencil of conics.

i'd think that said locus should be some curve, but i don't know what it looks like

@Semiclassical You can probably write down an equation from the fact that a parabola is a conic with $b^2 - 4ac = 0$. But I am done for today.

though, playing around in geogebra

If I plot the two parabolas you gave earlier, their 4 intersections, and drop down a fifth point P

then the conic determined by including P seems to only be parabolic when it falls on one of the two parabolas

@Semiclassical Isn't just three poitns requried? The focus and the two points of the directrix.

This gives a unique parabola.

@Physicist137 Points on the parabola.

Not outside.

Although I have no idea of what you talking about.

What you guys are trying to do?

Three points determine a parabola. But they don't determine a parabola through those three points.

The question was, how many points determine a parabola through them? The answer is obviously 5, we just took a long time to realize that :P

We're just being silly, don't listen to us.

I see.

Why were you trying to answer this?

Just for kicks.

@SemiC probably had a physics reason which brought it up

you're the one who brought it up :P

I was perfectly happy just doing $y=ax^2+bx+c$

I see.

@Semiclassical I do think there's only 2 parabolas passing through 4 points.

I was about to say

Is there an obvious reason for why that's the case?

you're still discussing the parabola? :D

I noticed it in Geogebra but that's merely a heuristic

If the Greeks were happy to discuss conics, why shouldn't we be?

geogebra

are you an expert in it

Nope.

Yes, @SemiC.

I've got it on my laptop, but I barely touch it.

I thought it might be useful in my geometry course at high school after summer

but I don't know

I hardly know it

$ax^2 + by^2 + cxy + dx + ey + f = 0$, $b^2 - 4ac = 0$. Plug in four points in the first equation to get five of the six coefficients in terms of the last, assume the last is $b$.

Then the quadratic equation $b^2 = 4ac$ gives two values for $b$.

So there's two equations.

hmm.

what's not clear to me is the relation between the two parabolas.

From this it's even more obvious why 5 points determine a parabola :D

Lesson for all of us (at least me): think simply.

obviously there is one since there's two signs for b.

but the geometry isn't obvious.

@BalarkaSen or very profoundly and cleverly

@Semiclassical You're thinking of axis of symmetry, etc, Semi?

though, glancing at this link, the example you gave is even nicer than i thought

@BalarkaSen i guess.

No idea, but I guess I am not going to think about that anymore.

i don't think the parabolas need to have orthogonal axes for that to happen

Me neither.

I can just perturb one a bit.

A non-orthogonal example is visually clear.

right.

might be a bit annouying to construct, but perturbatively it's obvious

Hey, G(n,m) = 3^n . 6^ m is a one to one function right? n are positive integers. Just want to confirm that I got the correct answer.

Mhm.

m and n both are positive integers.

@aste123 Number of factors of 3 is n+m, number of factors of 2 is m.

so if you know both n+m and m, then you also know n.

Yawn.

@Huy Next topic: hyperbolas.

That was a hyperbole.

Taking a good sleep seems like a better option in this very moment.

Hmm! This problem seems to have a nice pedigree: www2.washjeff.edu/users/mwoltermann/Dorrie/45.pdf

I'm out.

I have a question.. a small one.

Let $S = [0, 1). Define $x~y$ iff $x-y\in\mathbb{Q}$.

Let $A\subset S$ such that $A$ contains exactly one element of each class of equivalence of the relation $x~y$.

What does he means by that? What exaclty is $A$?

No two elements of $A$ are equivalent, that is.

Hmm

How simple... but how have you concluded this? Hm.. What is an "equivalence class"?

as an example with S being positive integers instead, suppose the equivalence is having the same final decimal digit

e.g. 5 is equivalent to 25, which is equivalent to 125, etc.

@Physicist137 An equivalence class is precisely collection of all elements in your set which are equivalent to each other.

There can be a lot of equivalence classes, however.

then the class of elements equivalent to 5 would be {5,15,25,...}

^

(i should probably have just done all integers, but i didn't want to muck around with the minus sign)

Oh... I see. =).

but if i want to define the class, all i need is one of its elements

because everything else is, by definition, everything that's equivalent to it

Ohh... Makes sense.

Thanks!! =).

so I could just say "the set of positive integers whose last digit is 5 is an equivalence class"

@Semiclassical sorry, I didn't understand it. How does it help in finding if it is one to one or not?

well, what's an example of an output of $G(n,m)$?

it's 3^n+m . 2^m. But what about it?

i meant a concrete example. suppose G(n,m)=144.

how many factors of 2 does that have?

yes, we can find that.

2^ 4

and 3^ 2

right. so we'd need m=4 and n+m=2. so you'd need n=-2, which means that was a bad example

i needed one which had more factors of 3 than 2.

Then we can't take 144 because it is 9 * 16

Maybe 9 * 4

36?

right.

sure. then m=2 and n+m=2, so n=0. so G(0,2)=36.

yes

more generally, suppose I give you some number which only contains factors of 3 and 2 and which has more factors of 3 than 2.

then i can divide by 6 once for each factor of 2, and the resulting integer will be some power of 3.

so each possible output has a unique input. therefore, it's a one-to-one function.

however, G(n,m) doesn't map to every positive integer---you'll never get G(n,m)=17, for instance.

so it's not onto.

Hmm yes. What I did was a proof by contradiction. I assumed different values of n and m exist for which same result is given by G. Then equated both and finally got 3^something = 2^ something. Since the bases are prime numbers. Taking their positive exponent wouldn't ever make them equal. And they are postive integers so I equated both to 0. That gave me: n1 = n2 and m1= m2.

Thank you for your method.

:)

@Semiclassical Why does dividing by 6 and getting integer of power 3 prove it to be one-to-one function? What would have been the result if it was not one-to-one? Maybe your method is obvious but it is not clicking for me..

@Balarka, have you read Don Quixote?

@Krijn Hah, no, but I have heard of it a lot.

Isn't that like a giant book?

Yeah, it's two books really, but they are almost always sold together nowadays.

Book I is like 450-ish pages and Book II is 700?

:S

Way too long.

No wait, 500/600 is better for my edition

Monte Cristo is more and isn't too long

There are great works that are long and there are also failures that are way too long, really

Agreed.

I mean, I have no problem with such books, but it's hard to get the time to read it.

I don't really like to bookmark stuff and read from where I ended last.

It's about priorities I think

Well, I guess.

I'm just gonna read Book I for now, and save Book II for later, though

Seperated the length is okay, I guess

Hi again @TedShifrin

@Ted's here, let's pretend we were talking about mathematics

So, conics in $\mathbb{P}^5$, crazy, amirite

Hi @Balarka, hi fake @Krijn

:P

@Krijn even better are lines in $\Bbb P^5$.

You're starting to play with Grassmannians, @Balarka?

They were in my lecture yesterday, of course.

Along with quadrics.

Ah, right, it was your lecture yesterday. How did it go?

All three of the faculty talked a lot, and about 3 of the 8 students participated. It seems it was well received.

Nice.

So @Pedro would be proud of me for pulling off more pretense.

I also talked a bit with two of the students about their research projects. I actually need to sit down and think about that a little bit. I promised.

Nice

Well, you should get sleep and I'll go off elsewhere ...

Have a good day. I gave you an example of your topology question above, btw.

Oh, right. I need to track that down. It wasn't my question. It was someone's.

To check closed you need to check all closed sets.

But I need to read what you wrote.

Hmm, yes, true.

Not sure how to thicken a Cantor set, actually.

Take an epsilon-neighborhood.

For some fixed epsilon.

Not clear how disconnected that is!!

It may be the whole interval, in fact?

For small enough epsilon it'd be pretty disconnected.

I doubt that.

I know how to make Cantor sets of positive measure, but that's far from what you're doing.

OK, right, that can't work.

There's got to be something far, far simpler.

Oh, I have it.

Duh.

It's the first example that I thought of, but I didn't see why it wasn't open.

Sign function? Map to $\Bbb R$.

Take a real number, spit out it's sign ($0$ has positive sign)

The usual convention is to send 0 to 0.

OK, I'm sending it to +1.

Why?

I guess it won't matter.

But, yeah.

Right.

I was trying to keep the image simple :P

{+1, -1} is simpler than {+1, 0, -1}

Good thing I wasn't here to hear your answer.

@BalarkaSen Bah

@TedShifrin Hmm?

Your first answer.

Oh yeah. In my defense I was tired, had to do a lot of horrible stuff in school today.

Sorry about that, hope my second answer compensates.