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## 2.3.3 Higher-order Terms

2.3.3 Higher-order Terms with Laplace’s Method Consider the case of a maximum of φ(t) at c ∈ (a, b) with φ ′ (c) = 0, φ ′′ (c) < 0, and f(c) �= 0. To get higher-order terms expand φ(t) and f(t) to higher orders. How far do we need to go? Without loss of generality assume c = 0 to simplify the algebra. � +ǫ � I(x) ∼ Notes: −ǫ � f(c) + f ′ (0)t + 1 2 f ′′ (0)t 2 + . . . e x[φ(0)+1 2 φ′′ (0)t 2 + 1 6 φ′′′ (0)t 3 + 1 4! φ(4) t 4 +...] dt • integral extends only over small |t|: expand the exponential in t: – in order to extend the integral to (−∞, +∞) we need to keep the quadratic term up in the exponential and cannot include it in the expansion of the exponential – the cubic and quartic terms should not be kept in the exponential: ∗ without expanding the exponential in the cubic and quartic term the extension to (−∞, +∞) would not be possible for φ (iv) (c) > 0, it would lead to a diverging integral ∗ the polynomial expansion in the exponent is likely to have maxima outside the interval (−ǫ, ǫ). These maxima are only a result of the approximation of φ(t) ⇒ if we were not to expand the exponential these spurious maxima may contribute to the integral once the domain of integration is extended to (−∞, +∞) although we know that the contributions from outside (−ǫ, ǫ) should be exponentially smaller. – if we did not extend the domain of integration we may be tempted to expand also the quadratic term in the exponential. As illustrated in the example above (cf. (27)), the resulting integrals would, however, depend on ǫ. Figure 10: Possible divergence and spurious maxima of Taylor expansion of φ(t). φ (t) −ε 58 +ε

Extend integration to (−∞, +∞) and rescale integration variable to extract the x-dependence from the integrals, s = √ xt I(x) ∼ 1 √ x � +∞ −∞ � ′ s f(0) + f √ + x 1 ′′s2 f 2 x + 1 6 f ′′′ s3 x 3 2 where all derivatives are evaluated at s = c = 0. In the integration odd terms in s cancel � » xφ(0)+ + . . . e 1 2φ′′ s2 + 1 s3 φ′′′ 6 x 1 2 + 1 4! φ(4) s4 x 1 s5 + φ(5) 5! x 3 – ... 2 ds ⇒ the first non-trivial term beyond the leading-order term will involve the terms of O( 1 x ) ⇒ we need to keep two additional terms in the expansion of φ(t) and f(t) to get one additional order in the expansion for I(x) I(x) ∼ 1 √ x e xφ(0) Using � +∞ −∞ � +∞ −∞ we get (reinstating again c) � I(x) ∼ Note: 2π −xφ ′′ (c) exφ(c) e 1 2φ′′ s2 � f(0) + 1 � � 1 f(0) x 4! φ(4) s 4 + 1 2 s 2n e −αs2 ds = (−1) n dn dαn � +∞ −∞ 1 1 6 6 φ′′′2 s 6 � e −αs2 ds = (−1) n dn dαn � π α + 1 6 f ′ φ ′′′ s 4 + 1 2 f ′′ s 2 � � 1 + O x2 �� d � f(c) + 1 � − x f ′′ (c) 2φ ′′ (c) + f(c)φ(iv) (c) 8φ ′′2 (c) + f ′ (c)φ ′′′ (c) 2φ ′′2 (c) − 5φ′′′2 (c)f(c) 24φ ′′3 � � 1 + O (c) x2 �� • since odd powers in s cancel the higher-order corrections are an expansion in 1 x rather than in 1 √ x . Example: Again I(x) = � π 2 0 e−x sin2 t dt To get a higher-order approximation expand � ǫ I(x) ∼ e 0 −x(t−1 6t3 +...) 2 � ǫ dt = 0 � ∞ = e 0 −xt2 � 1 + 1 3 xt4 � + . . . dt = 1 � � π 1 + 2 x 1 � + . . . x → ∞ 4x Note: e −x(t2 − 1 3 t4 +...) dt • if the exponential had not been expanded in the quartic term the integral over (0, ∞) would have (erroneously) diverged. 59

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