Wikipedia:Reference desk/Archives/Mathematics/2016 May 3

= May 3 =

Im puzzled
if a minor league plays 123 games in a season won 15 more games than 3 times as many as they loss how many did they win and lose — Preceding unsigned comment added by 2602:30A:C04A:9870:EC77:4200:1746:8A13 (talk) 09:16, 3 May 2016 (UTC)
 * You may find our article on elementary algebra to be useful.--Jasper Deng (talk) 09:26, 3 May 2016 (UTC)
 * You may find our article on elementary algebra to be useful.--Jasper Deng (talk) 09:26, 3 May 2016 (UTC)

Max size Octahedron inscribed in a cube?
For a cube 1 unit in each direction, what is the maximum size of a regular Octahedron which can be inscribed in it?Naraht (talk) 12:31, 3 May 2016 (UTC)


 * The largest possible square inside a cube (with side length s) must have a pair of opposite sides on two opposite faces, and its vertices must be on the cube's edges, x units away from the vertices. Making use of the fact that all the sides have the same length, we obtain x = $1/4$ and s = $3√2/4$. From this we can immediately construct the largest possible octahedron, as a regular octahedron can be made from three mutually perpendicular squares. Thus the octahedron has side length $3√2/4$ and thus volume $9/16$. (Assuming I haven't messed up my literally back-of-envelope calculations.) Double sharp (talk) 14:03, 3 May 2016 (UTC)


 * What will we do when there are no more envelopes? This is equivalent to Prince Rupert's cube! —Tamfang (talk) 21:10, 4 May 2016 (UTC)

Reverse problem? (Max size Cube in an Octahedron?)
Cool! Is the reverse problem equally back of envelope? I fully expect some of the largest X in a Y where X and Y are regular polyhedra to be hideously ugly (Largest octahedron in a a unit dodecahedron? - urg), I wonder whether for the Dodec/Icos pairing whether the inscribed polyhedron can be "twisted" from simply being the one where its vertices touch the center of the original polyhedron's faces.Naraht (talk) 14:26, 3 May 2016 (UTC)


 * The problem has been solved for all twenty pairs of Platonic solids in 2014. They vary widely in difficulty: some are indeed back-of-envelope, and some look hideous. (The octahedron-in-dodecahedron one isn't too bad: it's $τ^{2}/√2$. The true horror story is the largest tetrahedron in an icosahedron, which is the root of a 32nd-degree equation.)
 * Cube-in-octahedron is not too bad. It is similar in that you start by noting that the cube's vertices have to all be on each of the six edges and positioned symmetrically (each cube vertex x units away from an octahedron vertex). You will find x = √2 − 1 and hence s = 1 − x = 2 − √2 (because the faces of the octahedron are equilateral triangles) to maximise s3 = 2(10 − 7√2). Similarly for dodecahedron-icosahedron (and vice versa), it is not an exact twist because the rotation allows the inscribed polyhedron a little more space to expand. Double sharp (talk) 14:44, 3 May 2016 (UTC)


 * Thanx! Looks like I've got my bedtime reading! Hmm. Wonder how extendible this work is. Max size 24-cell in a 120-cell? (ooh, and whether formulae can be generated for the 6 combinations of n-cube-, n-cross and n-simplex)Naraht (talk) 15:35, 3 May 2016 (UTC)


 * The case of a 16-cell in a tesseract would be the exact 4D analogue, and seems like a good place to start. It looks like you can start similarly (four mutually perpendicular octahedra define a 16-cell), but it gets harder to visualise what parameter x you need to control to maximise the volume. The middle cross-section (from the vertex first) of a tesseract is an octahedron, and one would expect the true maximum octahedron to be at a slightly (but not too different) angle. Double sharp (talk) 13:44, 5 May 2016 (UTC)

A few of the other 3D cases, by the way, are even easier than the two you asked about: the maximum tetrahedron in an octahedron shares one of the octahedron's faces and hence has edge-length 1; the maximum tetrahedron in a cube has half the cube's vertices and hence has edge-length √2; the maximum cube in a dodecahedron has eight of the vertices and hence has edge-length τ (and hence the maximum tetrahedron in a dodecahedron has edge-length τ√2). Double sharp (talk) 13:50, 5 May 2016 (UTC)
 * The max T in O makes sense, T in C definitely. C in D doesn't to me since the D has 30 verticies, I'll have to look at distribution. T in D also seems odd for the same reason.Naraht (talk) 00:19, 6 May 2016 (UTC)
 * Compound of five cubes and compound of five tetrahedra show the cases C in D and T in D. (BTW, a regular dodecahedron has 20 vertices and 30 edges.) Double sharp (talk) 07:13, 7 May 2016 (UTC)
 * Duh, for some reason, I was thinking the inscribed had to go to the edges. Duh.Naraht (talk) 12:30, 7 May 2016 (UTC)

Calc 2 simple question
Hello,

So I know that a Maclaurin series is equal to the function it approximates if and only if the limit of the difference between f(x) and p-sub-n(x) tends to zero as n tends to infinity. My question is: why? — Preceding unsigned comment added by 140.233.173.233 (talk) 17:42, 3 May 2016 (UTC)
 * By the rules of limits, $$\lim_{n\to\infty}p_n(x) = f(x) \iff \lim_{n\to\infty}(p_n(x)-f(x)) = 0$$. -- Meni Rosenfeld (talk) 17:57, 3 May 2016 (UTC)

Vertex of a cone
The article vertex describes a vertex as a point where two or more curves, lines, or edges meet, but cone refers to the point at the tip of a cone as a vertex. Since the vertex of a cone (as in a right circular cone, to avoid ambiguity) is not at the intersection of two edges, does it actually constitute a vertex? — crh 23   &thinsp;(Talk) 19:25, 3 May 2016 (UTC)
 * Sure it is. There are an infinite number of lines that meet at that vertex.  Those infinite lines define the surface of the cone.  Infinite is "two or more".  -- Jayron 32 19:28, 3 May 2016 (UTC)


 * See Cone. Note that mathematically a cone often refers to the double infinite cone shown in the pic. StuRat (talk) 20:14, 3 May 2016 (UTC)