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May 1

Extension for ternary plot with 4 and more variables

How can the ternary plot be extended to represent a case with 4 (and more) variables? Thanks!--82.137.15.31 (talk) 00:41, 1 May 2017 (UTC)[reply]

You could use a tetrahedron for four variables. The problem would be representing it, as in displaying it in a way which was easy for people to understand, as it is three-dimensional. It’s one thing drawing a tetrahedron, but the plot would still look like a 2D plot inside it. Doing it in 3D with a 3D engine would be one way. Displaying it stereoscopically would be another.--JohnBlackburne^words_deeds 02:28, 1 May 2017 (UTC)[reply]

How are the relations from the triangle of ternary plot extended in a tetrahedron? For instance does a paralel to an edge of the triangle correspond to a plane paralel to a face of the tetrahedron? How about a line passing through a vertex of the triangle, what is its correspondent in the tetrahedron?--82.137.15.31 (talk) 12:53, 1 May 2017 (UTC)[reply]

The answer to your second question is "yes." The answer to your third question is "a line passing through a vertex of the tetrahedron." --JBL (talk) 13:32, 1 May 2017 (UTC)[reply]

Interesting and/or somewhat surprising! What is then the significance of the plane containing the line passing through a vertex of the tetrahedron?--82.137.15.31 (talk) 20:02, 1 May 2017 (UTC)[reply]

It is known that in a ternary plot a line passing through a vertex of the triangle represents mixtures with constant ratio of compositions of components located at the other vertices. How is the situation with an interior plane passing through a vertex of the tetrahedron?--82.137.15.31 (talk) 20:22, 1 May 2017 (UTC)[reply]

Bound on the totient function

Euler's totient function has a good lower bound on phi(n). Is there a good upper bound on phi(n) when n is composite, i.e. better than n − 1? Bubba73 ^{You talkin' to me?} 13:04, 1 May 2017 (UTC)[reply]

I googled and found phi(n) <= n - sqrt(n) for composite n. Is that the best upper bound? Bubba73 ^{You talkin' to me?} 13:09, 1 May 2017 (UTC)[reply]

It is best in the sense that equality is achieved for infinitely many values of n, namely, the squares of prime numbers. (In general, requiring more prime factors pushes φ(n) away from n.) --JBL (talk) 13:35, 1 May 2017 (UTC)[reply]

Thank you. Bubba73 ^{You talkin' to me?} 13:55, 1 May 2017 (UTC) [reply]

Resolved

You are welcome. To add a tiny bit of detail: if n has k prime factors counting multiplicities, then the smallest of these factors cannot be larger than

{\sqrt[{k}]{n}}

. Therefore, this factor will have at least

{\frac {n}{\sqrt[{k}]{n}}}=n^{\frac {k-1}{k}}

multiples less than or equal to n, and so

\varphi (n)\leq n-n^{\frac {k-1}{k}}

. Equality is achieved exactly for the kth powers of primes. --JBL (talk) 14:11, 1 May 2017 (UTC)[reply]

That is good to know. For my present purpose I have to consider all composite n, though. Bubba73 ^{You talkin' to me?} 15:11, 1 May 2017 (UTC)[reply]

Polyiamond problem

What's the smallest polyiamond with fewer vertexes than faces?? The familiar hexagon divided into 6 triangles has 6 faces but 7 vertexes. Georgia guy (talk) 18:22, 1 May 2017 (UTC)[reply]

If I understand the question correctly, I'm pretty sure it would be

  --------
 /\  /\  /\
/  \/  \/  \
------------
\  /\  /\  /
 \/  \/  \/
  --------
  \  /\  /
   \/  \/ 
    ----

In other words a triangle completely surrounded by a moat of other triangles. In order to get more faces than vertices you would need at least three vertices not on the boundary, or equivalently, three vertices completely surrounded by triangles within the figure. (You could prove this from the hexagonal lattice version of Pick's theorem.) In order to minimize the number of faces, these three vertices would have to be packed as close together as possible, which would seem to mean they form a single face. That gives you the figure above with 12 vertices and 13 faces. Lot's of hand waving in that argument but I, at least, am persuaded. --RDBury (talk) 21:19, 2 May 2017 (UTC)[reply]

Chaos game

The first paragraph of Chaos game says that if you use a factor of 1/2, you get the Sierpinski triangle. What happens if you use a different factor, say 1/3 or 3/4? Bubba73 ^{You talkin' to me?} 23:56, 1 May 2017 (UTC)[reply]

If I remember correctly, you get something similar, just with the original figure made smaller, so with gaps in between the parts (for > 1/2), or made larger, so with overlapping parts (for < 1/2). But don't take my word for it -- try it out yourself and report back! --Deacon Vorbis (talk) 00:05, 2 May 2017 (UTC)[reply]

Thanks, I think I did that about 1990. Bubba73 ^{You talkin' to me?} 17:49, 2 May 2017 (UTC)[reply]