Wikipedia:Reference desk/Archives/Mathematics/2024 June 7

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June 7

Reduced Dedekind Phi function

Euler totient function $\phi (n)$ (sequence A000010 in the OEIS) has a reduced function: Carmichael lambda function $\lambda (n)$ (sequence A002322 in the OEIS), and $\lambda (xy)=lcm(\lambda (x),\lambda (y))$ instead of $\lambda (x)\lambda (y)$ for coprime x, y, while $\lambda (p^{r})=\phi (p^{r})=(p-1)p^{r-1}$ if p is prime and r>=1, and a composite number n is Carmichael number if and only if $\lambda (n)$ divides $n-1$, and Dedekind psi function $\psi (n)$ (sequence A001615 in the OEIS) should also have a reduced function: $\eta (n)$, and $\eta (xy)=lcm(\eta (x),\eta (y))$ instead of $\eta (x)\eta (y)$ for coprime x, y, while $\eta (p^{r})=\psi (p^{r})=(p+1)p^{r-1}$ if p is prime and r>=1, and a composite number n is Lucas-Carmichael number if and only if $\eta (n)$ divides $n+1$, but I cannot even find the function $\eta (n)$ in OEIS (it should start with (start from n=1) 1, 3, 4, 6, 6, 12, 8, 12, 12, 6, 12, 12, 14, 24, 12, 24, 18, 12, 20, 6, 8, 12, 24, 12, …). 125.230.9.88 (talk) 02:38, 7 June 2024 (UTC)

$\eta (399)={\text{lcm}}(\eta (3),\eta (7),\eta (19))={\text{lcm}}(4,8,20)=40,$ yet $40\not \mid 398.$ Did you mean to write "if and only if $\eta (n)$ divides $n+1$"? But for any prime $p,$ $\eta (p)=p+1.$ The modified criterion requires $n$ to be composite.  --Lambiam 06:55, 7 June 2024 (UTC)

Sorry, typo, $n+1$ instead of $n-1$, fixed. 2402:7500:92C:A1B7:8906:5B5F:EA02:8E54 (talk) 15:02, 7 June 2024 (UTC)

It is also true for Carmichael numbers, n is a number n is Carmichael number if and only if $\lambda (n)$ divides $n-1$, but for any prime p, $\lambda (p)=p-1$, thus both require that n is composite … I fixed it. 2402:7500:92C:A1B7:8906:5B5F:EA02:8E54 (talk) 15:06, 7 June 2024 (UTC)

Also, you state above that $\lambda (p^{r})=\phi (p^{r}),$ yet, according to both Carmichael function and A002322, $\lambda (2^{3})=2$ while $\phi (2^{3})=4.$ "Your" $\lambda$ function is given as OEIS sequence A011773 (Variant of Carmichael's lambda function).  --Lambiam 13:44, 7 June 2024 (UTC)

Well, I forgot that it is only worked for odd prime p, and not worked for p=2 and r>=3, thus this function is in fact (sequence A011773 in the OEIS). 2402:7500:92C:A1B7:8906:5B5F:EA02:8E54 (talk) 15:04, 7 June 2024 (UTC)

Choosing a point inside a quadrilateral

Is it possible to choose a point strictly inside a simple quadrilateral such that all four sides (and hence the whole interior) are visible from the point? The chosen point is required to be a smooth function of the the vertices, reasonably middling and free from any nastiness (such as depending on the labeling of the vertices). Thanks, catslash (talk) 16:58, 7 June 2024 (UTC)

All four sides would obviously be visible from any random point inside a quadrilateral that isn't concave. I suspect your question is more complex that that but you haven't expressed it optimally. -- Jack of Oz ^{[pleasantries]} 19:25, 7 June 2024 (UTC)

If you take each of the four edges and consider the half-planes on the interior sides of each edge, then the intersection of all half-planes is the precise set of points which can "see" all sides. Although I haven't formalized a proof, I'm confident that this set is always nonempty (and also a convex quadrilateral) if the original quadrilateral is non-trivial.

As for a smooth function, I'm sure that there is some definition of the center of a convex quadrilateral that, applied to this set of points, would suffice. GalacticShoe (talk) 20:02, 7 June 2024 (UTC)

On reflection, since convex quadrilaterals can only take one specific form, with one concave vertex, two "wingtips", and a "head", it's very easy to describe this shape. It's just the quadrilateral formed from the concave vertex, the head, and the two intersections of opposite sides. As for the center, you can use the intersection of diagonals of this internal quadrilateral. GalacticShoe (talk) 22:34, 7 June 2024 (UTC)

When just one angle of the original quadrilateral is very close to 180°, then so is one angle of the convex inner quadrilateral, at the same vertex, and the intersection of its diagonals is very close to that vertex. In the degenerate case that the angle is equal to 180°, as when a concave quadrilateral is continuously transformed into a convex quadrilateral, the point of intersection coincides with the vertex. Using the vertex centroid or area centroid of the convex inner quadrilateral as the chosen point avoids this.  --Lambiam 07:01, 8 June 2024 (UTC)

In general, for any area A enclosed by a Jordan curve, the set P(A) of panoptic points is convex. It is the same as A iff A is convex. If the boundary is a polygon, P(A) is empty, a single point, a line segment, or the area enclosed by a polygon.  --Lambiam 15:43, 8 June 2024 (UTC)

I think I thought – but I was wrong – that the midpoint of the shortest diagonal satisfies the visibility criterion, but it is not a continuous function of the locations of the vertices. I conjecture that there is a continuous function w from these locations, ordered cyclically, to the unit interval, where w(v₂, v₃, v₄, v₁) = 1 − w(v₁, v₂, v₃, v₄), invariant under similarity transformations, such that the weighted mean of the diagonal midpoints (v₁ + v₃) / 2 and (v₂ + v₄) / 2, with weights w and 1 − w, satisfies both requirements. For any configuration, there is an interval [w_lo, w_hi] of w-values that guarantee visibility. If we can prove that both endpoints vary continuously with the vertices, a solution is provided by w = (w_lo + w_hi) / 2.  --Lambiam 21:15, 7 June 2024 (UTC)

The midpoint of the shortest diagonal does not work. Consider (0, 0), (1, 4), (0, 3), (-1, 4) taken in that order. The diagonal (1, 4), (-1, 4) is shorter than (0, 0), (0, 3), but its midpoint lies outside the quadrilateral. Of course if both diagonals lie inside the quadrilateral then it's convex and any interior point will do. --RDBury (talk) 02:06, 8 June 2024 (UTC)

The vertex centroid of the (convex) panoptic quadrilateral seems to do the job. Many thanks. If there is a practical way of choosing a point on the Newton line of the original quadrilateral, that would also be very interesting. catslash (talk) 12:57, 8 June 2024 (UTC)

At least one of the midpoints of the two diagonals is contained in the panoptic quad, so the panoptic part of the Newton line is nonempty. It should be easy to determine the endpoints of the (possibly degenerate) panoptic segment of this line and take its midpoint. If the original quadrilateral is convex, this is a trivial exercise. Otherwise, you have to determine where the extended sides adjacent to the concave vertex cut the segment connecting the midpoints short. As with many problems in computational geometry, the hardest part is not the computations themselves but to get the case distinctions correct and complete.

I think though the result will in extreme cases be markedly less "middling" than the vertex centroid.  --Lambiam 15:26, 8 June 2024 (UTC)

I hoped to find a relevant theorem in Star-shaped polygon, but no. —Tamfang (talk) 18:35, 19 June 2024 (UTC)

Kernel is less descriptive and perhaps duller than panoptic region. catslash (talk) 09:09, 20 June 2024 (UTC)