problemset5

Where l is the lipshitz constant 1 st 1 x2 s 2x 4

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Unformatted text preview: 0 �t ≤ |x1 | |t − t0 | + M |t − s| ds t0 = |x1 | |t − t0 | + M |t − t0 |2 2 ≤ B |t − t0 | �t (2) |xn (t) − xn−1 (t)| ≤ (t − s)|f (s, xn−1 (s)) − f (s, xn−2 (s))| ds t0 �t ≤ L (t − s)|xn−1 − xn−2 | ds t0 |t − t0 |2n (2n)! where L is the Lipshitz constant. � � � ∞ √ � ∞ 1 −st 1 −x2 s 2x 4. (a) L √ = e √ dt = e dx, where x2 = st. x s t � t 0 0 ∞ � 2 2 =√ e−x dx = π /s s0 � � ∞ √ √ √ e−st ∞ � ∞ e−st 1 −st √ dt by parts. (b...
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