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By R. Wong

I'm learning "Asyptotic Expantions" in graduate college. to review this box, this ebook is particularly very important reference certainly one of them.

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By Lemma 1 (ii) and (iii), we conclude that the same is true for B2m — B2m(x) in [0,1]. 20). The result in (ii) follows from the so-called error test, which states that if consecutive error terms associated with a series expansion have opposite signs, then each error term is numerically less than, and has the same sign as, the first neglected term of the series. This result is easy to prove, and is given in Steffensen (1950, p. 4) and Olver (1974a, p. 68). Since f(2m\x) and f(2m +2\x) have the same sign in (0, n), and since B2m — B2m(x) is of the same sign as B2m in [0,1], the remainders Rm(ri) and Rm +a(n) obviously have opposite signs.

Watson's Lemma 19 where Thus, one might naturally conclude that as x -* + oo, which is obviously false, since the integral S(x) is positive, whereas the terms in its series expansion are all negative. Also, since t < 1 + t <; x + t for x > 1, we have In view of the well-known identity (Olver 1974a, p. 33) can be explicitly evaluated to be T(N - |) r(|)/r(A/)*1/3. 30), violating the definition of an asymptotic expansion. The correct asymptotic expansion of S(x) is given in Chapter VI. For a similar example involving generalized asymptotic expansions, see Soni (1978).

46) at v = \ and let It is easily shown that By writing with 28 I Fundamental Concepts of Asymptotics we also have where and F(a, x) being the complementary incomplete Gamma function defined by On making the substitution t = x(l + T), the last integral can be written as Since it follows that if a > 1 and x > a — 1. Therefore if x > 2N - 1. 53). 51) then yields 5. 57). 45), there still remains the question of how good are these bounds. 59) shows immediately that the latter result is definitely superior to the former, even without numerical computation.

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