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Conversely, given a 3-partition of B, the sum of each 3-set is a multiple of 4, so it must contain either two regular items and one pairing item, or two pairing items and one filler item: If a 3-set contains two pairing items u ij , u kl and one filler item, then the sum of the two pairing items must be 44T+4 = 4*(5T+6T)+2+2, so they must have ...
There is a method to construct all Pythagorean triples that contain a given positive integer x as one of the legs of the right-angled triangle associated with the triple. It means finding all right triangles whose sides have integer measures, with one leg predetermined as a given cathetus. [13] The formulas read as follows.
The Pythagorean triples thus lie on curves given by = | / |, that is, parabolas reflected at the a-axis, and the corresponding curves with a and b interchanged. If a is varied for a given n (i.e. on a given parabola), integer values of b occur relatively frequently if n is a square or a small multiple of a square. If several such values happen ...
To find the primitive Pythagorean triple associated with any such value t, compute (1 − t 2, 2t, 1 + t 2) and multiply all three values by the least common multiple of their denominators. (Alternatively, write t = n / m as a fraction in lowest terms and use the formulas from the previous section.)
The function q(n) gives the number of these strict partitions of the given sum n. For example, q(3) = 2 because the partitions 3 and 1 + 2 are strict, while the third partition 1 + 1 + 1 of 3 has repeated parts. The number q(n) is also equal to the number of partitions of n in which only odd summands are permitted. [20]
For example, if the summands x i are uncorrelated random numbers with zero mean, the sum is a random walk and the condition number will grow proportional to . On the other hand, for random inputs with nonzero mean the condition number asymptotes to a finite constant as n → ∞ {\displaystyle n\to \infty } .
It is used to prove Kronecker's lemma, which in turn, is used to prove a version of the strong law of large numbers under variance constraints. It may be used to prove Nicomachus's theorem that the sum of the first n {\displaystyle n} cubes equals the square of the sum of the first n {\displaystyle n} positive integers.
In essence, the suspension can be seen as the cartesian product of with the unit interval, modulo an equivalence relation to turn the interval into a loop. The curried form then maps the space X {\displaystyle X} to the space of functions from loops into Z {\displaystyle Z} , that is, from X {\displaystyle X} into Ω Z {\displaystyle \Omega Z ...