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A335444 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 states that F[n]^2 <= G[n] = (x^2 + 1)^2*(x^2 + 2)^(n-3) for all n >= 3 and all real x. The sequence gives a triangle of the coefficients of the even exponents of G[n] - F[n]^2 read by rows. 0

%I #27 Jun 12 2020 04:19:38

%S 0,2,1,3,6,2,8,19,14,3,15,52,58,26,4,32,128,192,132,42,5,63,300,558,

%T 518,253,62,6,128,679,1496,1742,1152,433,86,7,255,1506,3801,5294,4413,

%U 2248,684,114,8,512,3292,9308,14999,15040,9680,3992,1018,146,9

%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 states that F[n]^2 <= G[n] = (x^2 + 1)^2*(x^2 + 2)^(n-3) for all n >= 3 and all real x. The sequence gives a triangle of the coefficients of the even exponents of G[n] - F[n]^2 read by rows.

%C Swamy's (1966) inequality states that F[n]^2 <= G[n] for all real x and all integers n >= 3.

%C Row n >= 3 of this irregular table gives the coefficients of the even powers of the polynomial G[n] - F[n]^2 (with exponents in increasing order). The coefficients of the odd powers are zero, and they are thus omitted. The degree of G[n] - F[n]^2 is 2*n - 6, so row n >= 3 contains n - 2 terms.

%C To prove that the degree of G[n] - F[n]^2 is 2*n - 6, note that the first few terms of G[n] are x^(2*n-2) + 2*(n-2)*x^(2*n-4) + (2*n^2 - 10*n + 13)*x^(2*n-6) + ... while the first few terms of F[n]^2 are x^(2*n-2) + 2*(n-2)*x^(2*n-4) + (2*n^2 - 11*n + 16)*x^(2*n-6) + ..., so the leading term of the polynomial G[n] - F[n]^2 is (n-3)*x^(2*n-6).

%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.

%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; see 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 T(n,0) = 2^(n-3) - (1 - (-1)^n)/2 = A166920(n-3) for n >= 3.

%F T(n,1) = 2^(n-4)*(n + 1) - floor(n/2)*ceiling(n/2) = A045623(n-2) - A002620(n) for n >= 4.

%F T(n, n-4) = 2*(n^2 - 7*n + 13) = A051890(n-3) for n >= 4.

%F T(n, n-3) = n - 3 for n >= 3.

%e Triangle T(n,k) (with rows n >= 3 and columns k = 0..n-3) begins:

%e 0;

%e 2, 1;

%e 3, 6, 2;

%e 8, 19, 14, 3;

%e 15, 52, 58, 26, 4;

%e 32, 128, 192, 132, 42, 5;

%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]^2); for(n=3, nn, for(k=0, n-3, print1(polcoef(h[n], 2*k, x), ", ")); print(); ); }

%Y Cf. A002620, A045623, A049310, A051890, A053119, A128932 (similar), A166920.

%K nonn,tabl

%O 3,2

%A _Petros Hadjicostas_, Jun 10 2020

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