A257451 -id:A257451 - cp's OEIS frontend

A381781 a(n) = k where k*Pi is the solution to sin(x) = 0 obtained using Newton's method starting from x = n.

0, 0, 1, 1, 1, 3, 2, 2, 5, 3, 3, 74, 4, 4, 2, 5, 5, 4, 6, 6, 6, 7, 7, 7, 8, 8, 8, 10, 9, 9, 13, 10, 10, 30, 11, 11, 9, 12, 12, 11, 13, 13, 13, 14, 14, 14, 15, 15, 15, 25, 16, 16, 14, 17, 17, 32, 18, 18, 16, 19, 19, 18, 20, 20, 17, 21, 21, 21, 22, 22, 22, 25, 23, 23, 26

Offset: 0

Simcha Z. Katzoff, Mar 07 2025

Each step of Newton's method for this is x -> f(x) = x - tan(x).
If |x - k*Pi| < A381473 = 1.16556... for some k then each step takes x closer to k*Pi (by absolute difference) and so converges on the multiple k*Pi.
If n itself is within |n - k*Pi| < A381473 then that convergence to k*Pi begins with the initial x = n and in that case a(n) = round(n/Pi) = A082964(n).
The width of this convergence region around each k*Pi is 2*A381473 = A257451 = 2.331... so contains at least 2 integers n and so every k >= 0 occurs at least twice in the sequence.
When x (or n) is further from its nearest multiple of Pi, the slope of sin(x) sends a step off to possibly much bigger or smaller places on the real line and may converge on some k*Pi far from the initial n.
Those steps can be large enough to reach negative k, as for example first at a(99) = -810 (see A381892 and A381893).
Conjecture: Infinitely many negative integers appear.

			a(11) = 74: applying Newton's method to f(x) = sin(x) with initial guess x_0 = 11 gives x_1 = 236.9508, x_2 = 232.8538, etc., eventually converging to x = 232.477856365644 with a(11) = x/Pi = 74.

Simcha Z. Katzoff, Table of n, a(n) for n = 0..1000

Cf. A381892, A381893, A082964, A381473.

delta = 1.1655611852072113068339179779585606691;
a[n_]:=(x=n;While[Abs[x-Round[x,Pi]]>delta,x=x-Tan[x]];Round[x,Pi]/Pi);
Array[a,100]

a(n) = (1/Pi)*lim_{k->oo} f_k(n), where f_k(n) denotes the k-th iteration of the function f(x) = x - tan(x).

A257452 Decimal expansion of the maximum of (1-cos(x))/x.

Original entry on oeis.org

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

Offset: 0

Stanislav Sykora, Apr 23 2015

The location of the maximum, and more comments, are in A257451.

			0.724611353776708475738990453525631784347865101838391714959...

Stanislav Sykora, Table of n, a(n) for n = 0..2000

Cf. A257451.

xmax = solve(x=1,3,x*sin(x)-1+cos(x)); a=(1-cos(xmax))/xmax

Equals (1-cos(A257451))/A257451.

A381473 Decimal expansion of the smallest positive solution to 2*x = tan(x).

Original entry on oeis.org

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

Offset: 1

Andrew Howroyd, Mar 10 2025

			1.1655611852072113068339179779585606691...

Cf. A257451.

PARI
```
solve(x=1, 1.5, 2*x-tan(x))
```

Equals A257451 / 2.

cp's OEIS Frontend

A381781 a(n) = k where k*Pi is the solution to sin(x) = 0 obtained using Newton's method starting from x = n.

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A257452 Decimal expansion of the maximum of (1-cos(x))/x.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

PARI

Formula

A381473 Decimal expansion of the smallest positive solution to 2*x = tan(x).

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Author

Keywords

Examples

Crossrefs

Programs

PARI

Formula