# REVIEW2 - hussain (tsh476) test2review rusin (55565) 1 This...

This preview shows pages 1–3. Sign up to view the full content.

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

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

Unformatted text preview: hussain (tsh476) test2review rusin (55565) 1 This print-out should have 38 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. Again this review is optional and non- graded. Thursdays exam will focus on mate- rial since the last test, hence these problems are limited to those topics. Please remember that the exam format is not like the Quest for- mat: give yourself some practice writing out your solutions in a way that makes it clear what youre trying to say. 001 0.0 points Determine the interval of convergence of the series summationdisplay k =0 k 3 2 k (3 x 4) k . 1. interval convergence = parenleftBig 2 3 , 2 3 parenrightBig 2. interval convergence = parenleftBig 2 , 2 3 parenrightBig 3. series converges only at x = 4 3 4. interval convergence = ( , ) 5. interval convergence = parenleftBig 2 3 , 2 parenrightBig correct Explanation: The given series has the form summationdisplay k = 0 c k ( x a ) k with c k = k 3 parenleftBig 3 2 parenrightBig k , a = 4 3 . But then, lim k c k +1 c k = lim k 3 2 parenleftBig k + 1 k parenrightBig 3 = 3 2 . By the Ratio Test, the series thus (i) converges when | x a | < 2 3 , (ii) diverges when | x a | > 2 3 . Now at the point x a = 2 3 the series reduces to summationdisplay k = 0 k 3 , while at x a = 2 3 it reduces to summationdisplay k = 0 ( 1) k k 3 . But in both cases these series diverge by the Divergent Test. Consequently, the interval of convergence of the given series is ( a 2 3 , a + 2 3 ) = parenleftBig 2 3 , 2 parenrightBig 002 (part 1 of 2) 0.0 points A function f is defined by the series f ( x ) = summationdisplay n = 0 c n x n in which the coefficients c n are specified by c 2 n = 5 , c 2 n +1 = 3 ( n 0) . (i) Find the interval of the convergence of the series. 1. interval of convergence = [ 1 , 1) 2. interval of convergence = [ 3 , 3) 3. interval of convergence = ( 1 , 1) cor- rect 4. interval of convergence = [ 5 , 5) 5. interval of convergence = ( 3 , 3) 6. interval of convergence = ( 5 , 5) Explanation: From the definition of the coefficients c n we see that f ( x ) = 5 + 3 x + 5 x 2 + 3 x 3 + . . . . hussain (tsh476) test2review rusin (55565) 2 Now the sum summationdisplay n = 0 ( a n + b n ) of two convergent series is again convergent, so consider the series ( ) 5 + 5 x 2 + 5 x 4 + . . . = summationdisplay n = 0 5 x 2 n and ( ) 3 x + 3 x 3 + 3 x 5 + . . . = summationdisplay n = 0 3 x 2 n +1 separately. But summationdisplay n = 0 5 x 2 n = summationdisplay n = 0 ar n is an infinite geometric series with a = 5 , r = x 2 ....
View Full Document

## This note was uploaded on 04/03/2011 for the course MATH 408D taught by Professor Chu during the Spring '09 term at University of Texas at Austin.

### Page1 / 22

REVIEW2 - hussain (tsh476) test2review rusin (55565) 1 This...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online