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
Unformatted text preview: suleimenov (bs26835) – test2review – rusin – (55565) 1 This printout should have 38 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. Again this review is optional and non graded. Thursday’s 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 you’re trying to say. 001 0.0 points Determine the interval of convergence of the series ∞ summationdisplay k =0 k 4 3 k (4 x − 5) k . 1. interval convergence = parenleftBig 1 2 , 2 parenrightBig correct 2. interval convergence = parenleftBig − 1 2 , 1 2 parenrightBig 3. series converges only at x = 5 4 4. interval convergence = ( −∞ , ∞ ) 5. interval convergence = parenleftBig − 2 , − 1 2 parenrightBig Explanation: The given series has the form ∞ summationdisplay k = 0 c k ( x − a ) k with c k = k 4 parenleftBig 4 3 parenrightBig k , a = 5 4 . But then, lim k →∞ c k +1 c k = lim k →∞ 4 3 parenleftBig k + 1 k parenrightBig 4 = 4 3 . By the Ratio Test, the series thus (i) converges when  x − a  < 3 4 , (ii) diverges when  x − a  > 3 4 . Now at the point x − a = 3 4 the series reduces to ∞ summationdisplay k = 0 k 4 , while at x − a = − 3 4 it reduces to ∞ summationdisplay k = 0 ( − 1) k k 4 . But in both cases these series diverge by the Divergent Test. Consequently, the interval of convergence of the given series is ( a − 3 4 , a + 3 4 ) = parenleftBig 1 2 , 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 = 7 , c 2 n +1 = 2 ( n ≥ 0) . (i) Find the interval of the convergence of the series. 1. interval of convergence = ( − 1 , 1) cor rect 2. interval of convergence = ( − 2 , 2) 3. interval of convergence = [ − 7 , 7) 4. interval of convergence = [ − 1 , 1) 5. interval of convergence = [ − 2 , 2) 6. interval of convergence = ( − 7 , 7) Explanation: From the definition of the coefficients c n we see that f ( x ) = 7 + 2 x + 7 x 2 + 2 x 3 + . . . . suleimenov (bs26835) – 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 ( ∗ ) 7 + 7 x 2 + 7 x 4 + . . . = ∞ summationdisplay n = 0 7 x 2 n and ( ‡ ) 2 x + 2 x 3 + 2 x 5 + . . . = ∞ summationdisplay n = 0 2 x 2 n +1 separately. But ∞ summationdisplay n = 0 7 x 2 n = ∞ summationdisplay n = 0 ar n is an infinite geometric series with a = 7 , r = x 2 ....
View
Full
Document
This note was uploaded on 09/15/2011 for the course M 55565 taught by Professor Rusin during the Spring '11 term at University of Texas.
 Spring '11
 RUSIN

Click to edit the document details