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Lecture Note Sketches Hermann Riecke - ESAM Home Page

Lecture Note Sketches Hermann Riecke - ESAM Home Page

• the integral term

• the integral term gives exactly the remainder of the Taylor expansion. If that integral exists for all N and for small x f(x) = ∞� n=0 x n n! f(n) (0) Example 2: I(x) = � ∞ x e−t4 dt for x → ∞ Motivated by case ii) above we would like to integrate the exponential. Rewrite I(x) there- fore as Estimate the integral term � ∞ x I(x) = − 1 � ∞ 4 x 1 t 4e−t4 = − 1 4 1 t3 d dt 1 t3e−t4 � �∞ � � x � e −t4� dt + 1 � ∞ 4 x −3 e−t4dt t4 dt < 1 x4I(x) ≪ I(x) for x → ∞ Therefore we get I(x) ∼ 1 e 4 −x4 x3 x → ∞ Higher-order terms are obtained in the same way by repeated integration by parts. Example 3: I(x) = � x 0 t−1 2e−tdt for x → ∞ Try straight-forward integration by parts I(x) = −t −1 2e −t � � � x 0 − 1 2 � x This will not work: both terms diverge at the lower limit. 0 t −3 2e −t dt In this case one can take care of the lower limit by noting � ∞ t −1 2e −t dt = Γ( 1 2 ) = √ π Thus Note: I(x) = 0 � ∞ 0 = √ π + t −1 t −1 2e −t dt − � � 2e −t � ∞ x � ∞ x + 1 2 I(x) ∼ √ π − e−x √ x 48 t −1 2e −t dt � ∞ x t −3 2e −t dt

• if the integration by parts leads to a contribution that diverges or is even only much larger than I(x) itself the integration by parts will not work, because that large contribution needs to cancel between boundary terms and the resulting integral, i.e. the resulting integral will also be large. Example 4: Stieltjes integral I(x) = � ∞ 0 e−t dt for small and large x 1+xt For small x we want to integrate by parts repeatedly to generate terms x n , n > 0 Since we write I(x) = ���� i.b.p. = ���� i.b.p. d 1 1 = −x dt 1 + xt (1 + xt) 2 − 1 1 + xt e−t �∞ � � ∞ � 1 � + −x 0 0 (1 + xt) 2e−tdt 1 1 + x (1 + xt) 2e−t � �∞ � � + x 0 2 � ∞ 2 0 (1 + xt) 3e−tdt � ∞ 1 (1 + xt) n+1e−tdt = 1 − x + . . .(−1) n−1 (n − 1)!x n−1 + (−1) n The integrals generated by the integration by parts exist for all n. Therefore we get for small x the series ∞� I(x) = (−1) n n!x n x → 0 n=0 For large x we would like to generate terms x−n , n > 0. Try therefore integration by parts the other way around. � ∞ −t 1 d I(x) = e ln (1 + xt) dt 0 x dt = 1 � �∞ ln (1 + xt) e−t� x � + 0 1 � ∞ ln (1 + xt) e x 0 −t dt = 1 x2 [(1 + xt) ln (1 + xt) − (1 + xt)] e−t�� ∞ 1 + 0 x2 � ∞ [(1 + xt) ln (1 + xt) − (1 + xt)] e 0 −t dt � ∞ [(1 + xt) ln (1 + xt) − (1 + xt)]e −t dt = 1 1 + x2 x2 0 So it seems, that I(x) ∼ x−2 . But: the integral in the remainder diverges for x → ∞: � ∞ [(1 + xt) ln(1 + xt) − (1 + xt)] e 0 −t � ∞ dt > [xt ln (xt) − (1 + xt)]e 0 −t dt � ∞ = x ln x te −t �� ∞ dt + x Thus, the remainder goes like x −1 ln x ≫ x −2 . 49 0 0 0 t lnte −t − te −t dt � � ∞ − e −t dt 0

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