20d_04s_fs

20d_04s_fs - Math 20D Final Exam Solutions June 2004 1(a...

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 DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Math 20D Final Exam Solutions June 2004 1. (a) Since | sin x | ≤ 1, converges by comparison with 2 ∑ 1 /n 2 (the p test). (b) We have ∞ summationdisplay n =1 2 | sin n | n = sin 1 1 + ∞ summationdisplay n =2 | sin( n- 1) | n- 1 + | sin n | n ≥ ∞ summationdisplay n =2 | sin( n- 1) | + | sin n | n > 1 2 ∞ summationdisplay n =2 1 n , by | sin( x- 1) | + | sin x | > 1 / 2. Since (1 / 2) ∑ 1 /n diverges to + ∞ and (b) is larger, it diverges. 2. The radius of convergence is 1. It converges conditionally for x = 2 and absolutely for 0 < x < 2. 3. (a) Separate variables: e − x dx = e t dt . Thus- e − x = e t + C and we get C =- 2. (b) The characteristic equation has roots 1 ± i , so the general solution is x = C 1 e t cos t + C 2 e t sin t. Since x (0) = C 1 , we have C 1 = 0. Then x ′ = C 2 e t (sin t + cos t ) and, since x ′ (0) = 2, C 2 = 2. In summary, x = 2 e t sin t ....
View Full Document

This note was uploaded on 06/18/2009 for the course MATH 20D taught by Professor Mohanty during the Fall '06 term at UCSD.

Page1 / 2

20d_04s_fs - Math 20D Final Exam Solutions June 2004 1(a...

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