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Extra resources for Acta Applicandae Mathematicae: An International Survey Journal on Applying Mathematics and Mathematical Applications ~ Volume 106, Number 3 June, 2009 pp325-499
It is simple to check that the function f (x) = 2 x − x 2 has a global maximum of 2 when x = , so that S[ψ] ≤ . Since t ∈ I , this contradicts the assumption that St > . Corollary 1 If S is a symmetric operator with finite deficiency indices such that S = > 0, then S is simple, regular, the deficiency indices (n, n) of S are equal, and the spectrum of any self-adjoint extension of S is purely discrete and consists of eigenvalues of finite multiplicity at most n with no finite accumulation point.
Alternatively, we may write the above equation as ∞ gk (y)t k = exp −y(et − 1)/t + y . (39) k=0 By putting y = 0 we see immediately that gk (0) = 0 for k ≥ 1 and g0 (0) = 1. Introducing the gk (y) into the Equivalence (38) yields ∞ ∞ ak t k = k=0 dy e−y 1 − 0 = 1− y y2 y ty 2 y2 y3 + t2 − − + − t3 + ··· 2! 8 3! 8 · 3! 2 · 3! 3! (2) (2) t 1 1 (4) + t2 − − + − t3 + ··· . 2 4 6 8 · 3! 2 · 3! 4! e. B0 = 1, B1 = −1/2, B2 = 1/6 and B3 = 0. Thus, we can obtain specific values by continuing the expansion.
2 (2k)! k=1 (3) The major difference between the above results and those appearing in the mathematical literature is that equivalence symbols have replaced equals signs. This is because the rhs’s of both equivalences can become divergent. According to p. 138 of Press et al. , Equivalence (3) and hence, Equivalence (1) are not convergent because the Bernoulli numbers diverge. This statement, however, is incorrect because although the Bk diverge as k → ∞, B2k /(2k)! converge. To observe this, from No.