cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-3 of 3 results.

A332525 Decimal expansion of the minimal distance between (0,0) and the branch of the graph of y = tan x that passes through (Pi, 0).

Original entry on oeis.org

2, 5, 5, 7, 0, 1, 5, 6, 1, 4, 2, 4, 1, 3, 5, 8, 5, 2, 6, 0, 1, 3, 6, 6, 3, 5, 4, 1, 9, 0, 6, 7, 7, 1, 3, 7, 9, 6, 9, 9, 9, 8, 9, 0, 8, 9, 7, 8, 1, 2, 2, 8, 7, 7, 1, 8, 6, 6, 8, 9, 0, 4, 7, 4, 9, 1, 3, 7, 0, 4, 0, 1, 1, 5, 5, 6, 7, 8, 6, 6, 2, 0, 0, 5, 1, 2
Offset: 1

Views

Author

Clark Kimberling, Jun 15 2020

Keywords

Comments

Let T be the branch of the graph of y = tan x that passes through (Pi,0). There is a unique point (u,v) on T that is closer to (0,0) than any other point on T. Let d = distance between (u,v) and (0,0).
The first code in the Mathematica section gives
u = 2.319805307509200010738867057136510870483647988277... ;
v = -1.07556133564118881053529612226074179471679754375... ;
d = 2.557015614241358526013663541906771379699989089781... .
The second code shows (u,v) as the intersection of T and the circle centered at (0,0) with radius d.
The third code shows minimal distance-to-origin points on 16 branches of the tangent function.

Examples

			2.557015614241358526013663541906771379699989089781...

Crossrefs

Cf. A332526, A332527.

Programs

  • Mathematica
    (* This code computes (x,y) coordinates and the minimal distance. *)
x = x /. FindRoot[FullSimplify[D[Sqrt[x^2 + Tan[x]^2], x]] == 0, {x, 2},
   WorkingPrecision -> 150]
y = Tan[x]
d = Sqrt[x^2 + Tan[x]^2]
RealDigits[x][[1]]
RealDigits[y][[1]]
RealDigits[d][[1]]
(* Peter J. C. Moses, May 04 2020 *)
(* This code shows the two points on the graph of y = tan x and on a circle whose radius is the minimal distance. *)
g1 = Plot[Tan[x], {x, -2 \[Pi], 2 \[Pi]}, AspectRatio -> 1];
g2 = Graphics[Circle[{0, 0}, Sqrt[Tan[#]^2 + #^2] &[x /. FindRoot[
       FullSimplify[D[Sqrt[x^2 + Tan[x]^2], x]] == 0, {x, 2},
       WorkingPrecision -> 30]]]];
Show[g1, g2]
(* Peter J. C. Moses, May 04 2020 *)
(* This code shows minimal distance points on 16 branches of the tangent function. *)
max = 25;
ptX = Table[x /. FindRoot[# == 0, {x, nn}, WorkingPrecision -> 10], {nn, 2,
      max, Pi}] &[FullSimplify[D[Sqrt[x^2 + Tan[x]^2], x]]];
Show[Plot[Tan[x], {x, -#, #}, PlotRange -> {-#, #}] &[max],
   Map[Graphics[{Red, Circle[{0, 0}, Sqrt[Tan[#]^2 + #^2]]}] &, #],
   Map[Graphics[{PointSize[Large], Point[-{#, Tan[#]}], Point[{0, 0}],
        Point[{#, Tan[#]}]}] &, #], AspectRatio -> Automatic,
        ImageSize -> 600] &[ptX]
(* Peter J. C. Moses, May 05 2020 *)

Formula

u = - sin u sec^3 u.
v = tan u.
d = sqrt(u^2 + v^2).

A332526 Decimal expansion of the minimal distance between distinct branches of the tangent function; see Comments.

Original entry on oeis.org

2, 3, 7, 5, 0, 6, 9, 1, 4, 6, 0, 4, 0, 1, 7, 6, 3, 4, 9, 4, 3, 9, 8, 5, 1, 5, 5, 8, 7, 7, 8, 9, 8, 2, 4, 8, 7, 8, 6, 6, 2, 6, 7, 8, 0, 6, 5, 0, 8, 8, 4, 1, 7, 9, 2, 9, 2, 6, 9, 8, 5, 6, 4, 5, 9, 7, 5, 4, 8, 6, 6, 7, 0, 2, 9, 6, 9, 1, 3, 1, 6, 3, 3, 4, 1, 1
Offset: 1

Views

Author

Clark Kimberling, Jun 15 2020

Keywords

Comments

Let T0 and T1 be the branches of the graph of y = tan x that passes through (0,0,) and (Pi,0), respectively. There exist points P = (p,q) on T0 and U = (u,v) on T1 such that the distance between P and U is the minimal distance, d, between points on T0 and T1.
u = 2.549082584017596768984130292562154758705824602711...
v = -0.67319711901285205370684801604861382107848678888...
p = Pi - u
q = - v
d = 2.375069146040176349439851558778982487866267806508...

Examples

			minimal distance = 2.375069146040176349439851558778982487866267806508...

Crossrefs

Cf. A332500, A332525, A332527.

Programs

  • Mathematica
    min = Quiet[FindMinimum[Sqrt[(#[[1]][[1]] - #[[2]][[1]])^2 + (#[[1]][[2]] - \
#[[2]][[2]])^2] &[{{#, Tan[#]} &[x /. FindRoot[# Cos[#]^2 - x Cos[#]^2 + Tan[#] == Tan[x], {x, 0}, WorkingPrecision -> 500]], {#, Tan[#]} &[#]} &[y]], {y, 2}, WorkingPrecision -> 100]]
Show[Plot[{Tan[x], (-# Sec[#]^2) + x Sec[#]^2 + Tan[#], {(# Cos[#]^2) - x Cos[#]^2 + Tan[#]}}, {x, 0, Pi}, AspectRatio -> Automatic, ImageSize -> 300, PlotRange -> {-2, 2}], Graphics[{PointSize[Large], Point[{Pi/2, 0}], Point[{#, Tan[#]}], Point[{Pi - #, -Tan[#]}]}]] &[y /. min[[2]][[1]]]
(* Peter J. C. Moses, May 06 2020 *)

A332528 Decimal expansion of the maximal curvature of the secant function.

Original entry on oeis.org

1, 1, 1, 5, 3, 9, 8, 6, 1, 6, 3, 6, 7, 0, 8, 4, 7, 8, 8, 5, 2, 3, 6, 7, 2, 0, 2, 2, 8, 1, 3, 2, 6, 6, 5, 7, 6, 5, 4, 7, 5, 1, 2, 0, 8, 1, 8, 4, 5, 3, 1, 7, 7, 3, 7, 4, 0, 4, 4, 4, 1, 0, 3, 3, 5, 8, 5, 6, 3, 4, 6, 7, 6, 8, 6, 4, 7, 3, 9, 2, 1, 7, 4, 3, 2, 3
Offset: 0

Views

Author

Clark Kimberling, Jun 21 2020

Keywords

Comments

The maximal curvature of the graph of y = sec x occurs at two points (x,y) on every branch. One of the points has y > 0. Let T be the branch passes through (0,1) and lies in the first quadrant. The maximal curvature, K, occurs at a point (u,v):
u = 0.469952511643664772466732023628843853062603014858623133147...
v = 1.121592022152314185447110000884194699579672726085862403985...
K = 1.11539861636708478852367202281326657654751208184531773740...
The osculating circle at (u,v) has
center = (x,y) = (0.02618081309772817465,,,, 1.900598757881329358432040976889617...).
radius = 1/K = 0.896540470219566446984489512566284091376257661443...
maximal curvature: K = 1.11539861636708478852367202281326657654751208184531773740...

Crossrefs

Cf. A332527.

Programs

  • Mathematica
    maxC = ArcCos[   Sqrt[(2*(2 - Sqrt[31]*Cos[(Pi + ArcTan[(9*Sqrt[302])/73])/3]))/3]];
centerMaxC = {(-3*x + x*Cos[2*x] + Sin[2*x] + 2*Tan[x]^3)/(-3 +
       Cos[2*x]), -((Cos[x]*Cos[2*x])/(-3 + Cos[2*x])) + (3*Sec[x])/
      2} /. x -> maxC;
rMin = (Sqrt[(7 + Cos[4 x])^3] Sec[x]^3)/(Sqrt[2] 8 (3 - Cos[2 x])) /. x -> maxC;
Show[Plot[Sec[x], {x, 0 - 3/2, 2}],
Graphics[{PointSize[Large], Red, Point[centerMaxC],
   Point[{maxC, Sec[maxC]}], Circle[centerMaxC, rMin],
   Line[{centerMaxC, {maxC, Sec[maxC]}}]}], AspectRatio -> Automatic,
PlotRange -> {0, 2}]
x = ArcCos[Sqrt[(2*(2 - Sqrt[31]*Cos[(Pi + ArcTan[(9*Sqrt[302])/73])/3]))/
    3]];  (* maximal curvature occurs at (x, sec x) *)
{N[x, 150], N[Sec[x], 150]}
{cx, cy} = {(-3*x + x*Cos[2*x] + Sin[2*x] + 2*Tan[x]^3)/(-3 + Cos[2*x]), -((Cos[x]*Cos[2*x])/(-3 + Cos[2*x])) + (3*Sec[x])/2}; (* center of osculating circle *)
{N[cx, 150], N[cy, 150]}
r = N[Sqrt[(x - cx)^2 + (Sec[x] - cy)^2], 150] (* radius of curvature *)
1/r  (* curvature *)
kr = RealDigits[1/r][[1]]
(* Peter J. C. Moses, May 07 2020 *)
Showing 1-3 of 3 results.

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