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A128932
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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.
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1
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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, 0, 1, 15, 0, 58, 0, 99, 0, 87, 0, 41, 0, 10, 0, 1, 32, -4, 144, -10, 272, -6, 280, -1, 170, 0, 61, 0, 12, 0, 1, 63, 0, 310, 0, 673, 0, 825, 0, 619, 0, 292, 0, 85, 0, 14, 0, 1
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OFFSET
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3,6
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COMMENTS
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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.
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.
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.
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)
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REFERENCES
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Thomas Koshy, Fibonacci and Lucas numbers with Applications, Vol. 2, Wiley, 2019; see p. 33. [Vol. 1 was published in 2001.]
D. S. Mitrinovic, Analytic Inequalities, Springer-Verlag, 1970; p. 232, Sect. 3.3.38.
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LINKS
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M. N. S. Swamy and Joel Pitcain, Comment to Problem E1846, 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.]
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FORMULA
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T(n,0) = 2^(n-3) - (1 - (-1)^n)/2 = A166920(n-3) for n >= 3.
Sum_{k=0}^{2*n-2} T(n,k) = 4*3^(n-3) - Fib(n) = A003946(n-2) - A000045(n) for n >= 3. (End)
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EXAMPLE
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Triangle T(n,k) (with rows n >= 3 and columns k = 0..2*n-2) begins:
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, 0, 1;
15, 0, 58, 0, 99, 0, 87, 0, 41, 0, 10, 0, 1;
...
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PROG
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(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
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CROSSREFS
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KEYWORD
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sign,tabf
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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