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%I #44 Jun 12 2020 09:27:09
%S 0,0,1,0,1,2,-2,5,-1,4,0,1,3,0,9,0,12,0,6,0,1,8,-3,28,-4,38,-1,25,0,8,
%T 0,1,15,0,58,0,99,0,87,0,41,0,10,0,1,32,-4,144,-10,272,-6,280,-1,170,
%U 0,61,0,12,0,1,63,0,310,0,673,0,825,0,619,0,292,0,85,0,14,0,1
%N Define the Fibonacci polynomials by F[1] = 1, F[2] = x; for n > 2, F[n] = x*F[n-1] + F[n-2] (cf. A049310, A053119). Swamy's inequality implies that F[n] <= F[n]^2 <= G[n] = (x^2 + 1)^2*(x^2 + 2)^(n-3) for n >= 3 and x >= 1. The sequence gives a triangle of coefficients of G[n] - F[n] read by rows.
%C From _Petros Hadjicostas_, Jun 10 2020: (Start)
%C Swamy's (1966) inequality states that F[n]^2 <= G[n] for all real x and all integers n >= 3. Because F[n] >= 1 for all real x >= 1, we get F[n] <= G[n] for all integers n >= 3 and all real x >= 1.
%C Row n >= 3 of this irregular table gives the coefficients of the polynomial G[n] - F[n] (with exponents in increasing order). The degree of G[n] - F[n] is 2*n - 2, so row n >= 3 contains 2*n - 1 terms.
%C Guilfoyle (1967) notes that F[n] = det(A_n), where A_n is the (n-1) X (n-1) matrix [[x, -1, 0, 0, ..., 0, 0, 0], [1, x, -1, 0, ..., 0, 0, 0], [0, 1, x, -1, ..., 0, 0, 0], ..., [0, 0, 0, 0, ..., 1, x, -1], [0, 0, 0, 0, ..., 0, 1, x]], and Swamy's original inequality follows from Hadamard's inequality.
%C Koshy (2019) writes Swamy's original inequality in the form x^(n-3)*F[n]^2 <= F[3]^2*F[4]^(n-3) for x >= 1, and gives a counterpart inequality for Lucas polynomials. Notice, however, that the original form of Swamy's inequality is true for all real x. (End)
%D Thomas Koshy, Fibonacci and Lucas numbers with Applications, Vol. 2, Wiley, 2019; see p. 33. [Vol. 1 was published in 2001.]
%D D. S. Mitrinovic, Analytic Inequalities, Springer-Verlag, 1970; p. 232, Sect. 3.3.38.
%H Richard Guilfoyle, <a href="https://www.jstor.org/stable/2314912">Comment to the solution of Problem E1846</a>, Amer. Math. Monthly, 74(5), 1967, 593. [It is pointed out that the inequality is a special case of Hadamard's inequality.]
%H M. N. S. Swamy, <a href="https://www.jstor.org/stable/2313936">Problem E1846 proposed for solution</a>, Amer. Math. Monthly, 73(1) (1966), 81.
%H M. N. S. Swamy and R. E. Giudici, <a href="https://www.jstor.org/stable/2314912">Solution to Problem E1846</a>, Amer. Math. Monthly, 74(5), 1967, 592-593.
%H M. N. S. Swamy and Joel Pitcain, <a href="https://www.jstor.org/stable/2314976">Comment to Problem E1846</a>, Amer. Math. Monthly, 75(3) (1968), 295. [It is pointed out that I^{n-1}*F[n](x) = U_{n-1}(I*x/2), where U_{n-1}(cos(t)) = sin(n*t)/sin(t) and I = sqrt(-1); Cf. A049310 and A053119, but with different notation.]
%H Wikipedia, <a href="https://en.wikipedia.org/wiki/Fibonacci_polynomials">Fibonacci polynomials</a>.
%H Wikipedia, <a href="https://en.wikipedia.org/wiki/Hadamard%27s_inequality">Hadamard's inequality</a>.
%F From _Petros Hadjicostas_, Jun 10 2020: (Start)
%F T(n,0) = 2^(n-3) - (1 - (-1)^n)/2 = A166920(n-3) for n >= 3.
%F Sum_{k=0}^{2*n-2} T(n,k) = 4*3^(n-3) - Fib(n) = A003946(n-2) - A000045(n) for n >= 3. (End)
%e Triangle T(n,k) (with rows n >= 3 and columns k = 0..2*n-2) begins:
%e 0, 0, 1, 0, 1;
%e 2, -2, 5, -1, 4, 0, 1;
%e 3, 0, 9, 0, 12, 0, 6, 0, 1;
%e 8, -3, 28, -4, 38, -1, 25, 0, 8, 0, 1;
%e 15, 0, 58, 0, 99, 0, 87, 0, 41, 0, 10, 0, 1;
%e ...
%o (PARI) lista(nn) = {my(f=vector(nn)); my(g=vector(nn)); my(h=vector(nn)); f[1]=1; f[2]=x; g[1]=0; g[2]=0; for(n=3, nn, g[n] = (x^2+1)^2*(x^2+2)^(n-3)); for(n=3, nn, f[n] = x*f[n-1]+f[n-2]); for(n=1, nn, h[n] = g[n]-f[n]); for(n=3, nn, for(k=0, 2*n-2, print1(polcoef(h[n], k, x), ",")); print(););} \\ _Petros Hadjicostas_, Jun 10 2020
%Y Cf. A000045, A003946, A049310, A053119, A166920.
%K sign,tabf
%O 3,6
%A _N. J. A. Sloane_, Apr 28 2007
%E Name edited by _Petros Hadjicostas_, Jun 10 2020
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