You’re right—using 1/distance makes all the weight come from the immediate neighborhood. I don’t think it’s a fatal problem. That’s not why the effect is strongest at the edges. The addition of s(w) makes it work out that way.
I’m not following the bit about maximum product of radius^3*opinion density. Is that something that came out after integrating? I’ve thrown my work out by now, so can’t check without redoing it.
The simplest w(v) is w(v) = v. The s(w) proposed is |w|. a(w(pi),w(pj))(1 + s(w(vj))(vi·vj)) when pj = is a(p_i, )(1 + |v_j|)(v_i · ) = (1+x)xx_i / [(x_i—x)^2 + y_i^2]. The problem with x_i being close to x is the same everywhere in the plane.
You’re right—using 1/distance makes all the weight come from the immediate neighborhood. I don’t think it’s a fatal problem. That’s not why the effect is strongest at the edges. The addition of s(w) makes it work out that way.
Right, that alone doesn’t make it strongest at the edges. It just means that you’re measuring how strongly a tiny few people agree, not how believable it is to any significant fraction of people, and it didn’t come across to me as clearly marked as such.
I’m not following the bit about maximum product of radius^3*opinion density.
Err, I misread parenthesis.
The simplest w(v) is w(v) = v. The s(w) proposed is |w|.
Right, so a(w(pi),w(pj))(1 + s(w(vj))(vi·vj)) simplifies to 1/distance(pi,pj)(1+|vj|)(vi·vj). Since we’re only interested in a tiny neighborhood, we can throw away 1/distance as having the same effect everywhere, and we can reduce the rest to (1+|vi|)(vi)^2 (which, if you misread parenthesis, is (1+|vi|vi^2) or 1+r^3 and we can ignore the constant 1)
So if the expected agreement’s position dependency is r^2+r^3, you get the highest values at the edges. But its not clear to me why s(w) should be |w| instead of some constant, and it’s not clear to me why the (vi·vj) shouldn’t be renormalized as (vi·vj)/(|vi|*|vj|), or why we shouldn’t have a Gaussian distribution of opinions.
You’re right—using 1/distance makes all the weight come from the immediate neighborhood. I don’t think it’s a fatal problem. That’s not why the effect is strongest at the edges. The addition of s(w) makes it work out that way.
I’m not following the bit about maximum product of radius^3*opinion density. Is that something that came out after integrating? I’ve thrown my work out by now, so can’t check without redoing it.
The simplest w(v) is w(v) = v. The s(w) proposed is |w|. a(w(pi),w(pj))(1 + s(w(vj))(vi·vj)) when pj = is a(p_i, )(1 + |v_j|)(v_i · ) = (1+x)xx_i / [(x_i—x)^2 + y_i^2]. The problem with x_i being close to x is the same everywhere in the plane.
Right, that alone doesn’t make it strongest at the edges. It just means that you’re measuring how strongly a tiny few people agree, not how believable it is to any significant fraction of people, and it didn’t come across to me as clearly marked as such.
Err, I misread parenthesis.
Right, so a(w(pi),w(pj))(1 + s(w(vj))(vi·vj)) simplifies to 1/distance(pi,pj)(1+|vj|)(vi·vj). Since we’re only interested in a tiny neighborhood, we can throw away 1/distance as having the same effect everywhere, and we can reduce the rest to (1+|vi|)(vi)^2 (which, if you misread parenthesis, is (1+|vi|vi^2) or 1+r^3 and we can ignore the constant 1)
So if the expected agreement’s position dependency is r^2+r^3, you get the highest values at the edges. But its not clear to me why s(w) should be |w| instead of some constant, and it’s not clear to me why the (vi·vj) shouldn’t be renormalized as (vi·vj)/(|vi|*|vj|), or why we shouldn’t have a Gaussian distribution of opinions.