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Unformatted text preview: 642:527 SOLUTIONS: ASSIGNMENT 1 FALL 2007 Some of these solutions were written by Prof. Dan Ocone. 4.2 Rather than using (7a) or (7b) it is usually better use the ratio test directly. 1. (a) Since lim n →∞ vextendsingle vextendsingle vextendsingle vextendsingle ( n + 1) x n +1 nx n vextendsingle vextendsingle vextendsingle vextendsingle =  x  lim n →∞ parenleftbigg n + 1 n parenrightbigg =  x  , the series ∑ ∞ nx n converges if  x  < 1 and diverges if  x  > 1. Thus the radius of convergence is 1. (c) The power series ∑ ∞ 5 e n x n is a geometric series in powers of ( ex ) n , and so will converge only if  ex  < 1 or, equivalently  x  < 1 /e . You can also use the ratio test to get this, but this argument is simpler. The geometric series is frequently of use; you should recognize it readily and know when it converges. (d) Since the upper limit on the sum is finite, this is just a polynomial. It is its own Taylor series and the radius of convergence is infinite. If the upper limit of the summation had been ∞ , however, the ratio test would give a radius of convergence 0. (g) lim n →∞ vextendsingle vextendsingle vextendsingle vextendsingle ( n + 1) 50 ( x + 7) n +1 / ( n + 1)! n 50 ( x + 7) n /n ! vextendsingle vextendsingle vextendsingle vextendsingle =  x + 7  lim n →∞ 1 n + 1 parenleftbigg n + 1 n parenrightbigg 50 =  x + 7  (0)(1) = 0 . The series converges for all x , i.e., R = ∞ . (h) lim n →∞ vextendsingle vextendsingle vextendsingle vextendsingle (ln( n + 1)) n +2 ( x − 2) n +1 (ln n ) n +1 ( x − 2) n vextendsingle vextendsingle vextendsingle vextendsingle =  x − 2  lim n →∞ ln( n + 1) parenleftbigg ln( n + 1) ln n parenrightbigg n +1 = braceleftbigg , if x = 2 ∞ , if x negationslash = 2. The series converges only if x = 2, i.e., R = 0. The last limit looks tricky, but notice that ln( n + 1) → ∞ and (ln( n + 1) / ln n ) n +1 > 1, so the limit must be infinite. 2. In all these problems the radius of convergence is the distance from the center, x , to the nearest zero of the denominator. (b) The denominator has zeros at ± 3; x = 3 is the zero closest to x = 2 and so R = 1, (d) The only zero in the denominator is at − 2, so R = 24. (g) The denominator factors as ( x − 1)( x 2 + 4), and the numerator as ( x − 1)( x + 2), so we can cancel the factor x − 1 and study ( x + 2) / ( x 2 + 4), for which the denominator has singularities at ± 2 i . These are a distance R = 2 √ 2 from the center x = 2. 6. (a) For the calculation using (16) we have f ( x ) = 1 x − 1 , f ′ ( x ) = − 1 ( x − 1) 2 , f ′′ ( x ) = 2 ( x − 1) 3 , . . ., f ( n ) ( x ) = ( − 1) n n ! ( x − 1) n +1 , and so f ( n ) (4) = ( − 1) n n ! / 3 n +1 and the Taylor series is ∞ summationdisplay n f ( n ) ( x ) n !...
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This note was uploaded on 09/29/2009 for the course 642 527 taught by Professor Speer during the Fall '07 term at Rutgers.
 Fall '07
 SPEER

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