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Lecture Note Sketches Hermann Riecke - ESAM Home Page

Lecture Note Sketches Hermann Riecke - ESAM Home Page

Since sin t = t − 1 6

Since sin t = t − 1 6 t3 + 1 120 t5 . . . one might expect t − 1 6 t3 < sin t < t This can be shown to be true for all t > 0. Estimate now Thus, for t = x −α with α > 1 4 � � x sin 2 t − xt 2 � � = x |sin t + t| |sin t − t| < x 2t 1 6 t3 = 1 3 xt4 one has To leading order one then has also and also � x −α � �x sin 2 t − xt 2� � < 1 3 x1−4α → 0 x → ∞ 0 e −x sin2 t ∼ e −xt 2 e −xsin2 t dt ∼ � x −α 0 t ≤ x −α , x → ∞ e −xt2 dt x → ∞ What about the integrals from x−α to ǫ? They are both exponentially small compared to I(x): � ǫ x−α e −x sin2 t −xsin dt < e 2 x−α ǫ � ǫ x −α e −xt2 dt < e −x1−2α ǫ For the latter estimate to be meaningful we need to choose α < 1 2 . Example: Modified Bessel function In(x) = 1 π � π 0 excos t cosnt dt φ(t) has its maximum at t = 0. Try to approximate both cosines by lowest order terms and consider � ǫ e x 1dt = ǫe x (27) Why does this result depend on the artificial parameter ǫ? 0 The maximum at the boundary is at the same time a local maximum: need to keep the quadratic terms of φ(t) (cf. (24)). In(x; ǫ) = 1 π � ǫ 0 = 1 π ex1 2 e x−1 2 xt2 dt � 2π x Now there is no ǫ-dependence any more, as required. Example: I(x) = � ∞ 0 e−1t e−xtdt 56 = 1 √ 2πx e x

It looks like a case for Watson’s lemma with f(t) = e −1 t . However: e −1 t and all its derivatives vanish at t = 0 ⇒ Watson’s lemma would yield I(x) = 0 The dominant contribution to the integral comes from the maximum of the integrand. In this case the maximum depends on x, Note: tmax = 1 √ x • a maximum that depends on x is called a movable maximum. • to apply Laplace’s method to a movable maximum transform t into a coordinate in which the maximum is fixed, independent of x. We could introduce a shift of the coordinate s = t − tmax This would lead to complicated expressions in e −1 t . Instead, the maximum can be fixed by a rescaling s = √ xt smax = 1 I(x) = 1 � ∞ √ x 0 e −√ x( 1 s +s) ds Using (26) and noting that √ x plays the role of x in (26) we get and φ ′′ (1) = −2 Note: 3 I(x) ∼ 1 � √ x 2π 2 √ x e−2√ x = √ π x 3 4 e −2√ x • I(x) decreases faster than any power. Watson’s lemma can only generate results that depend on x as a power series. 3 Striling’s formula in Hinch p.33 is nice example for movable maximum and for important formula 57

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