05 - ASSIGNMENT 5 SOLUTIONS MAT 572 A FALL 2007 Problem...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: ASSIGNMENT 5 · SOLUTIONS MAT 572 A · FALL 2007 Problem 1 ( cf. [1, Exercise IV.1 #7]) . Show that γ : [0 , 1] → C defined by γ ( t ) = t + i t sin(1 /t ) t 6 = 0 t = 0 is a path, but is not rectifiable. Proof. Since lim t → + t sin(1 /t ) = 0, γ is continuous on [0 , 1], so γ is a path. See Figure 1 for the sketch. For the rest, consider partitions of the form P n = , 2 (2 n- 1) π , . . . , 2 5 π , 2 3 π , 2 π , 1 . (There are n + 2 elements of P n , which we label as usual as t = 0 < t 1 < ··· < t n +1 = 1.) Then for 2 ≤ k ≤ n , we have sin(1 /t k ) = (- 1) n- k , so the successive differences become sums: v ( γ ; P n ) = n +1 X k =1 | γ ( t k )- γ ( t k- 1 ) | ≥ n X k =2 | Im γ ( t k )- Im γ ( t k- 1 ) | = n X k =2 | t k sin(1 /t k )- t k- 1 sin(1 /t k- 1 ) | = n X k =2 ( t k- 1 + t k ) = 2 π 1 2 n- 1 + 1 2 n- 3 + ··· + 1 5 + 1 3 + 1 3 + 1 1 ≥ 2 π 1 2 n- 1 + 1 2 n- 2 + ··· + 1 5 + 1 4 + 1 3 + 1 2 = 2 n- 1 X k =2 1 k ....
View Full Document

{[ snackBarMessage ]}

Page1 / 3

05 - ASSIGNMENT 5 SOLUTIONS MAT 572 A FALL 2007 Problem...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online