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A216891
Decimal expansion of the minimal zero x(1) of the function F(x) = f(f(x)) - x, where f(x) = cos(x) - sin(x).
7
8, 3, 0, 1, 9, 8, 5, 1, 7, 0, 6, 7, 8, 2, 3, 9, 3, 4, 5, 5, 2, 2, 5, 6, 2, 7, 1, 9, 5, 5, 2, 7, 1, 0, 6, 5, 7, 7, 8, 2, 0, 6, 3, 0, 8, 4, 3, 9, 4, 5, 4, 3, 7, 3, 1, 9, 3, 9, 5, 5, 2, 4, 1, 2, 2, 1, 6, 0, 8, 4, 8, 3, 2, 0, 4, 5, 6, 1, 8, 8, 9, 6, 2, 2, 6, 4, 1, 6, 3, 8, 6, 9, 7, 2, 6, 2, 9, 1, 2, 1, 5, 9, 1, 2, 3
OFFSET
0,1
COMMENTS
Let x(2) and x(3) denote the remaining zeros of F(x), x(2) < x(3). Then it could be proved that f(x(1)) = x(3), f(x(3)) = x(1), and f(x(2)) = x(2).
The decimal expansions of x(2) and x(3) in A206291 and A216863 respectively are presented.
We note that the plot of the restriction of F(x) to the interval [-2,2] "is very similar" to the plot of the polynomial (x-x(1))*(x-x(2))*(x-x(3)) for x in [-2,2].
Let A = {x in R: f^n(x) = x(2) for some nonnegative integer n, where f^n denotes the n-th iteration of f}. Then if z is a real number, which does not belong to A, and z(0):= z, z(n+1) = f(z(n)) = sqrt(2)*sin(Pi/4 - z(n)), n in N, then one of the subsequences either {z(2*n-1)} or {z(2*n)} is convergent to x(1) and the second one is convergent to x(3).
LINKS
R. Witula, D. Slota and Szeged Problem Group "Fejentalaltuka", An Iteration Convergence: 11318[2007,745], Amer. Math. Monthly, 116 No 7 (2009), 648-649.
EXAMPLE
We have x(1) = -0.830198517...
KEYWORD
nonn,cons
AUTHOR
Roman Witula, Sep 19 2012
STATUS
approved