Date: Dec 27, 2012 6:02 PM Author: kirby urner Subject: Re: A Point on Understanding On Thu, Dec 27, 2012 at 2:09 PM, Robert Hansen <bob@rsccore.com> wrote:

>

> On Dec 27, 2012, at 2:16 PM, kirby urner <kirby.urner@gmail.com> wrote:

>

> Judge Hansen: Debater A wins because you can never have "infinity

> triangles" i.e. 1/n approaches 0 as n->infinity but at no point is 1/n

> actually 0, just as 0.999... is never 1 if you stop with the 9s at

> some point. [Editor: "..." is one of the most magical of all symbols

> in math, given what its allowed to do for us in our imaginations]

>

>

> If you didn't understand then why didn't you just say so?

>

It was you who didn't understand with your pizza analogy. That was

getting in the way for my readers, who are reading such syllabus items

as:

http://www.amazon.com/Divided-Spheres-Geodesics-Orderly-Subdivision/dp/1466504293

I need my readers to understand about only 5-spoke and 6-spoke wheels,

not pizzas with arbitrarily thin slices.

Where your pizza comes in is if you take out the tiniest wedge and

close the perimeter, the center pops up or down (if the crust is crisp

enough) making a cone, which is a better model for concavity /

convexity.

> Do you understand that your contradiction is essentially between limit(n) *

> limit(720/n) as n->infinity and limit(n*720/n) as n->infinity?

>

n is the number of vertexes and goes to infinity, so there is no lim

n, unless maybe you mean something different by that notation.

lim 720/n as n-> infinity is 0, as is any limit C/n where C is

constant (lets no get into negatives -- say positive Reals).

> Do you understand that the left side of that comparison is the part that is

> not valid, that limit(n) as n->infinity does not exist?

>

I wasn't suggesting any limit to n. The lim | 360 - v | where v is

degrees around each n (lets say on average) is what goes to 0. n is

not limited.

n * (360 - v) = 720 where v = F(n) i.e. where v is a function of how

many vertexes there are in total. n-> infinity, so (360 - F(n)) goes

to 0 as close as we like. Give me an epsilon, I'll give you a delta.

The 720 is there for all finite n, i.e. it's a constant.

Also, given the described algorithm, n is not just 1, 2, 3... that'd

be a measure of 'frequency', which might symbolized as lowercase f.

So n = V(f) might be better, where f = 1, 2, 3...

V(f) * (360 - F(V(f))) = 720 as f --> infinity. Rewritten: lim (n *

~(720/n)) = Descartes' Deficit for all n = V(f), where f = 1,2,3...

(360 - v) might be called the "tax" or "tithe" each vertex pays to the

sum total 720 so...

Sigma [ V(f) * tax ] = 720 where f = frequency and V(f) = number

of vertexes and tax = ~720/n (average tax per vertex).

f->1, 2, 3...

This may make it look like we're getting closer and closer to 720 as f

- -> infinity but actually the 720 is a constant.

> You may have meant something else, but you never stated the contradiction

> explicitly. You just said, how can these somethings go to zero yet if you

> add them all up they add up to 720. Translated to something explicit, that

> becomes limit(n) * limit(720/n) as n->infinity.

>

> Bob Hansen

>

>

I think it's "explicit" before it gets mangled by a lot of nonsense notation.

That's called a "straw man" when you put words in my mouth only to

show they're nonsense.

If you were on a plane, you *could* say |360 - v| = 0, which is

certainly how it appears locally (triangular bathroom tiles), once f

and therefore n = V(f) is high enough.

Once you add "there's curvature, you are on a ball" to the list of

facts (things dip below the horizon as they increase in distance),

then we know |360 - v| > 0, so this added fact of curvature (convex /

concave) tells us something new: that we don't have perfect flatness.

What is V(f) by the way? V(f) = 10 * f * f + 2, i.e. V(f) is 12, 42, 92, 162...

If we drop two vertexes then the ratio of N:F:E == 1:2:3 (N = V - 2)

(this is for the omnitriangulated sphere, not the hexapent).

Also (of course) V + F == E + 2 (V = number of vertexes, F = number

of faces, E = number of Edges)

Kirby