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FEs06sol - MATH 127 Big eas have little eas Upon their...

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MATH 127 Solutions to Final Exam 1. 10 points According to the poem by Ogden Nash, Big fleas have little fleas, Upon their backs to bite ’em, And little fleas have lesser fleas, And so, ad infinitum. Assume each flea has exactly one flea which bites it. If the largest flea weighs 0 . 03 grams, and each flea is 1 / 10 the weight of the flea it bites, what is the total weight of all the fleas? image adapted from http://biodidac.bio.uottawa.ca The weight of the first flea is w 1 = . 03 , the weight of the second is w 2 = . 03 · 1 10 , the weight of the third is w 3 = . 03 · 1 100 , and so on. So the total weight of all the fleas is w = summationdisplay n =1 w n = w 1 + w 2 + w 3 + ... = . 03 + . 03 · 1 10 + . 03 · 1 100 + ... = summationdisplay n =0 . 03 1 10 n . This is a geometric series with | r | =1 / 10 < 1 and first term .03. So it converges to w = summationdisplay n =0 . 03 1 10 n = . 03 1 1 10 = 3 / 100 9 / 10 = 1 30 grams MATH 127 Solutions to Final Exam Page 1 of 13 May 15, 2006
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2. 12 points Find the Maclaurin series of each of the given functions. I suggest you use a familiar power series as your starting point. (a) e 3 x 2 Since e x = summationdisplay n =0 x n n ! , e 3 x 2 = summationdisplay n =0 (3 x 2 ) n n ! = summationdisplay n =0 3 n ( x 2 ) n n ! = summationdisplay n =0 3 n n ! x 2 n Since the first series converges for all x , so does the second series; i.e. the radius and interval of convergence of the series for e 3 x 2 are and ( −∞ , ) , respec- tively. (b) 1 1 + 4 x Since 1 1 x = summationdisplay n =0 x n whenever | x | < 1 , 1 1 + 4 x = 1 1 ( 4 x ) = summationdisplay n =0 ( 4 x ) n = summationdisplay n =0 ( 4) n x n Since the first series converges whenever x lies in the interval ( 1 , 1) , the second series converges whenever 1 < 4 x < 1 ⇐⇒ 1 > 4 x > 1 ⇐⇒ 1 < 4 x < 1 ⇐⇒ 1 / 4 < x < 1 / 4 . Thus, the radius and interval of convergence of the series for 1 1 + 4 x are 1 / 4 and ( 1 / 4 , 1 / 4) , respectively. (c) sin2 x Since sin x = summationdisplay n =0 ( 1) n (2 n +1)! x 2 n +1 , sin2 x = summationdisplay n =0 ( 1) n (2 n +1)! (2 x ) 2 n +1 = summationdisplay n =0 ( 1) n (2 n +1)! 2 · 2 2 n x 2 n +1 = 2 summationdisplay n =0 ( 1) n 4 n (2 n +1)! x 2 n +1 Since the first series converges for all x , so does the second series; i.e. the radius and interval of convergence of the series for sin2 x are and ( −∞ , ) , respec- tively. MATH 127 Solutions to Final Exam Page 2 of 13 May 15, 2006
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3. 12 points For each of the series below, decide if it converges or diverges. You must justify your answer to receive full credit. You do not need to find the sum of convergent series. (a) summationdisplay n =1 2 n 5 n + 3 n = 1 4 + 2 17 + 1 19 + 8 353 + ... This series converges as 2 n / (5 n +3 n ) looks like 2 n / 5 n = (2 / 5) n and a geometric series with r =2 / 5 converges. To actually justify this, you can use: the Ratio Test (note the exponents of n ): | a n +1 | | a n | = 2 n +1 / (5 n +1 +3 n +1 ) 2 n / (5 n +3 n ) = 2 n +1 2 n · 5 n + 3 n 5 n +1 +3 n +1 = 2 · 5 n / 5 n + 3 n / 5 n 5 · 5 n / 5 n + 3 · 3 n / 5 n = 2 · 1 + (3 / 5) n 5 + 3 · (3 / 5) n −→ 2 5 Since 2 / 5 < 1 , the series converges.
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