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1.
Solutions are obtained for the boundary value problem, y
(n) + f(x,y) = 0, y
(i)(0) = y(1) = 0, 0 i n – 2, where f(x,y) is singular at y = 0. An application is made of a fixed point theorem for operators that are decreasing with respect to a cone. 相似文献
2.
It is known that for all monotone functions f : {0, 1}n → {0, 1}, if x ∈ {0, 1}n is chosen uniformly at random and y is obtained from x by flipping each of the bits of x independently with probability ? = n?α, then P[f(x) ≠ f(y)] < cn?α+1/2, for some c > 0. Previously, the best construction of monotone functions satisfying P[fn(x) ≠ fn(y)] ≥ δ, where 0 < δ < 1/2, required ? ≥ c(δ)n?α, where α = 1 ? ln 2/ln 3 = 0.36907 …, and c(δ) > 0. We improve this result by achieving for every 0 < δ < 1/2, P[fn(x) ≠ fn(y)] ≥ δ, with:
- ? = c(δ)n?α for any α < 1/2, using the recursive majority function with arity k = k(α);
- ? = c(δ)n?1/2logtn for t = log2 = .3257 …, using an explicit recursive majority function with increasing arities; and
- ? = c(δ)n?1/2, nonconstructively, following a probabilistic CNF construction due to Talagrand.
3.
Another logarithmic functional equation 总被引:1,自引:0,他引:1
K. J. Heuvers 《Aequationes Mathematicae》1999,58(3):260-264
Summary. Let f : ]0,¥[? \Bbb R f :\,]0,\infty[\to \Bbb R be a real valued function on the set of positive reals. The functional equations¶¶f(x + y) - f(x) - f(y) = f(x-1 + y-1) f(x + y) - f(x) - f(y) = f(x^{-1} + y^{-1}) ¶and¶f(xy) = f(x) + f(y) f(xy) = f(x) + f(y) ¶are equivalent to each other. 相似文献
4.
Nicholas Tzanakis 《Journal of Number Theory》1984,19(2):203-208
Some general remarks are made concerning the equation f(x, y) = qn in the integral unknowns x, y, n, where f is an integral form and q > 1 is a given integer. It is proved that the only integral triads (x, y, n) satisfying x3 + 3y3 = 2n are (x, y, n) = (?1, 1, 1), (1, 1, 2), (?7, 5, 5,), (5, 1, 7). 相似文献
5.
6.
In this paper, the direct method and the fixed point alternative method are implemented to give Hyers-Ulam-Rassias stability
of the functional equation
6f(x + y) - 6f(x - y) + 4f(3y) = 3f(x + 2y) - 3f(x - 2y) + 9f(2y)6f(x + y) - 6f(x - y) + 4f(3y) = 3f(x + 2y) - 3f(x - 2y) + 9f(2y) 相似文献
7.
Margherita Fochi 《Aequationes Mathematicae》2009,78(3):309-320
Let X be a real inner product space of dimension greater than 2 and f be a real functional defined on X. Applying some ideas from the recent studies made on the alternative-conditional functional equation
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