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Math 118 :: Winter 2009 :: Homework 4 Solution∗ March 6, 2009 1 If z ∈ R and −1 ≤ z ≤ 1, √ √ |z − z 2 − 1| |z − 1 − z 2 i| 1 φ(z) = log = log = log = − log 2 2 2 2 (1) 2 (a) N p(θ) = 1 X ikθ ˆ e fk 2π (2) k=1 Therefore all such functions are the linear space spanned by {eikθ , k = 1, · · · , N } (3) (b) By a similar argument as (a), all such functions are the linear space spanned by the basis, Tn (x), n = 0, 1, 2, · · · (4) ∗ contact Kaiyuan Zhang through [email protected] for any questions. 1 3 (a) ¯Z ¯ ¯ π/h ¯ ¯ ¯ eikxj (û(k) − v̂(k))dk ¯ ¯ ¯ −π/h ¯ Z π/h 1 ≤ |û(k) − v̂(k)|dk 2π −π/h Z π/h 1 ² 1dk ≤ 2π −π/h ² = h 1 |uj − vj | = 2π (5) (6) (7) (8) (b) By Theorem 3 on p.33, we have |û(k) − v̂(k)| ≤ M hp+1 (9) \ |(u(ν) − v (ν) )(k)| =|(ik)ν û(k) − (ik)ν v̂(k)| =|k|ν |û(k) − v̂(k)| π ≤( )ν M hp+1 h =M π ν hp−ν+1 (10) (11) (12) for some constant M . Then (13) (14) Let ² = M π ν hp−ν+1 , by part (a), |u(ν) (xj ) − wj | = |u(ν) (xj ) − v (ν) (xj )| ≤ ²/h ≤ M π ν hp−ν 2 (15) 4 Z 1 f (x)dx − h 0 = N −1 X ÃZ = ! f (x)dx − hf ((j + 1/2)h) (17) (f (x) − f (xj )) dx (18) jh N −1 Z (j+1)h X j=0 (16) (j+1)h N −1 Z (j+1)h X j=0 f ((j + 1/2)h) j=0 jh j=0 = N −1 X µ ¶ 1 1 f 0 (xj )(x − xj ) + f 00 (xj )(x − xj )2 + f (3) (xj )(x − xj )3 + o(h3 ) dx 2 6 µ ¶ 1 00 f (xj )(x − xj )2 + o(h3 ) dx 2 (20) ¶ 1 00 h2 3 f (xj ) + o(h ) dx 2 12 (21) jh (19) = N −1 Z (j+1)h X j=0 = N −1 Z (j+1)h X j=0 = jh N −1 µ X j=0 = jh µ ¶ 1 00 h3 4 f (xj ) + o(h ) dx 2 12 (22) N −1 h3 X 00 f (xj ) + o(h3 ) 24 j=0 Thus (23) ¯ ¯ ¯Z 1 ¯ N −1 X ¯ ¯ kf 00 k∞ 2 ¯ ¯≤ h + o(h3 ) f (x)dx − h f ((j + 1/2)h) ¯ ¯ 24 ¯ 0 ¯ j=0 5 (a) Since f ∈ L2 , let p = 0, then we know that the error is O(h). 3 (24) (b) The error equals the difference between red shadow part and the green shadow part, which does not exceed h = 2π/N . (c) We claim that the error is π/N now. If N is even, µ ¶ N/2−2 N −1 X X 1+ 1 2πj 2π(j + 1) h f( ) + f( ) =h 1+ 2 N N 2 j=0 j=0 1 2 + 1 2 +0 = hN = π 2 2 (25) If N is odd, h N −1 X j=0 1 2 µ ¶ (N −3)/2 X 2πj 2π(j + 1) 1 + 0 = h N = π (26) f( ) + f( ) = h 1+ N N 2 2 j=0 Therefore the error is π − π = 0. 6 f ∗ g = f tells fˆ(k)ĝ(k) = fˆ(k). Therefore ĝ(k) = 1, −B ≤ k ≤ B (27) We need to extend ĝ to the entire real axis with sufficient smoothness. ĝ(k) = exp(−(|k| − 1)4 ) 4 (28) for |k| > 1. We claim that ĝ ∈ C 2 and ĝ 00 ∈ L1 (R). In fact, at k = 1, exp(−(|k| − 1)4 ) = 1 − (k − 1)4 + o((k − 1)4 ) (29) is C 2 connected with ĝ|−1≤k≤1 = 1. Similarly, the connection at k = −1 is also C 2 , therefore ĝ ∈ C 2 . kĝ 00 kL1 (R) ¯ Z ∞¯ 2 ¯d ¯ 4 ¯ ¯ =2 exp(−(k − 1) ) ¯ dk 2 ¯ dk 1 ¯ Z ∞¯ 2 ¯ ¯d 4 ¯ ¯ exp(−k ) =2 ¯ dk 2 ¯ dk 0 Z ∞ =2 exp(−k 4 )|16k 6 − 12k 2 |dk (30) (31) (32) (33) 0 <∞ (34) We know that −x2 g(x) is the Fourier inverse transform of ĝ 00 , hence |x2 g(x)| ≤ kĝ 00 kL1 (R) ≤ M ⇒ |g(x)| ≤ M/|x|2 (35) Let C = 2 max{M, sup|x|≤1 |g(x)|} < ∞, we have when |x| ≤ 1, |g(x)| ≤ when |x| > 1, |g(x)| ≤ C 1 ≤C 2 1 + x2 (36) C 1 ≤C 2 2x 1 + x2 (37) 5