@user76284: Are you requiring the vectors to be linearly independent?

Hey there! Can someone explain me what cross-boundary directional derivative means in point-normal interpolation?

They don't have to be.

Then it's clearly false.

One example where the $\ell_2$ norm fails to satisfy this property is where $b$ is exactly between two points $a$ and $c$. Then clearly $b = 0 \cdot a + 1 \cdot b + 0 \cdot c$, but also $b = \frac{1}{2} \cdot a + 0 \cdot b + \frac{1}{2} \cdot c$, and the latter coefficients have smaller $\ell_2$ norm.

@TedShifrin Is there a similar example for $\ell_1$?

I was just going to take $b=2a$ and note that $a=1\cdot a + 0\cdot b$ but also $a=\frac12\cdot b$.

The coefficients must add up to 1 though?

Oh, oops.

OK, let $c=0$ and add $\frac 12 \cdot c$. You can make $c$ nonzero with something similar.

@user76284 I guess you want nonzero vectors then

Hmm, if they're going to add up to $1$, we're going to need a negative to get something interesting.

Right, I'm allowing the coefficients to be negative. It doesn't have to be a convex combination, just an affine one.

But if the numbers add up to $1$, the absolute values will add up to $1$ or, likely, more, so we can only get minimal $1$-norm if you have all positives, adding up to $1$. So I guess you're right if vectors are nonzero.

Yes I agree that the coefficients adding up to $1$ is the real constraint here, otherwise it should be easy to make the sum as small as you want

Also, the trivial combination doesn't have to be the unique solution. The example I gave before (but with $\ell_1$ norm rather than $\ell_2$) shows that there are at least 2 combinations with minimal $\ell_1$ norm (they have the same norm $1 = \frac{1}{2} + \frac{1}{2}$).

Yes, sure.

So I should say $\mathrm{e}_j \in \operatorname*{argmin} \cdots$.

a rather trivial observation: If $v=av+\sum_k c_k u_k$ where $a+\sum_k c_k=1$, then $$(1-a)v=\sum c_k u_k\implies v=\frac{\sum_k c_k u_k}{1-a}=\frac{\sum_k c_k u_k}{\sum_k c_k}$$ so long as at least one $c_k$ is nonzero

So if $v$ is a nontrivial affine combination of itself and the uk's, then $v$ is an affine combination of the $u_k$'s.

Not sure if this helps, though, since the quantity to be minimized is still in terms of the original coefficients.

Unless $a=1$?

nontrivial

Can there a nontrivial combination where $a = 1$ but the $c$s cancel out?

I saw that. You could still have $\sum c_k = 0$ without their all being $0$.

that'd mean the $u_k$'s are linearly dependent

Well, with independence it's all super easy.

Well. If you have $\sum_k c_k=0$ without them being zero, then $\sum_k |c_k|>0$. Hence you definitely would have a larger $\ell^1$ norm than when $a=1$ and the rest are zero.

Yeah, right, this is what I commented earlier.

ah.

If $\sum \lambda_i = 1$ and we want to minimize $\sum |\lambda_1|$, then by the triangle inequality we must take all the $\lambda_i\ge 0$ and we get $1$.

shrug

$$1 = \left|\sum_i \lambda_i\right| \leq \sum_i |a_i| = \|a\|_1$$

Yup, just what I said :P

Makes sense

I was wondering about this in relation to my question math.stackexchange.com/questions/3442479/…

Specifically whether replacing the minimization of the $\ell_2$ norm in the linked paper with the minimization of the $\ell_1$ norm could yield exact interpolants.

Using the term "basis point" is misleading unless these points are affinely independent.

This is something that numerical analysts should have thought about, I would imagine.

When you use the language of coordinate systems, there's linear independence and no redundancy.

Yeah I wasn’t sure what to call the points, hence the quotes.

But barycentric coordinates are linearly dependent.

The sentence obtained by substituting "The sentence obtained by substituting $x$ into itself is false" into itself is false.

ugh, recursion

@user76284: I don't think so. You use affinely independent points to give barycentric coordinates on a simplex.

smacks DogAteMy

Non-unique coordinates make for a troublesome coordinate system (e.g., polar coordinates) ...

@TedShifrin Yeah that’s the right way to put it. I phrased it very poorly.

Can someone give some light about calculating normal in [Walton-Meek Gregory Patch]?(computergraphics.stackexchange.com/questions/9317/…)

But generalized barycentric coordinates use non-affinely independent points.

e.g. more than $d+1$ points in $\mathbb{R}^d$.

But then you really don't have a coordinate system. That doesn't fit my world view ...

@YardenJ2R: This is all foreign to me.

I suppose additional constraints make them injective if that’s what you mean.

Ok, thanks Ted!

Right ... On an appropriate set, I want a bijection between points and sets of coordinates.

Like in the paper the constraint is one of minimizing the L2 norm which I think ensures uniqueness.

Fair enough. As I said, this is clearly the purview of numerical analysts :)

Yeah. I guess what I’m looking for is a generalized barycentric coordinate system that assigns the “standard basis” to the given points.

By “standard basis” in this context I just mean the one-hot coefficient vectors.

Probably you can only do that with affine independence.

But I'm honestly not thinking about it enough.

Hey guys! Could somebody please give me a hand with the following question, please? If $f$ is periodic with a period of $2a$ for some $a > 0$, then $f(x) = f(x + 2a)$ for all $x ∈ R$. Show that if $f$ is continous, there exists some $c ∈ [0, a]$ such that $f(c) = f(c+a)$.

I thought about using the Intermediate Value Theorem, but I'm not sure if it's applicable in this case..

Yes, @Abwatts, that's precisely the right thing to use.

You just have to rephrase the problem slightly.

@TedShifrin I see! Would that be necessary to break down the proof into 3 separated cases in this instance?

Hello!

No cases needed if you do it right. What are you thinking?

heya @Stan

So I'm confused about something. Let me show you the specific sentence involved