Math 06: Review for Final Exam, Part II - SOLUTIONS. Use a second-degree Taylor polynomial to estimate 8. We choose f(x) x and x 0 7 because 7 is the perfect cube closest to 8. f(x) x /3 f(7) 3 f (x) 3 x /3 f (7) 3x /3 3 7 /3 7 f (x) 9 x 5/3 f (7) 9x 5/3 9 7 5/3 87 Now plug in to the Taylor polynomial formula with x 0 7. P (x) f(x 0 ) + f (x 0 )(x x 0 ) + f (x 0 ) (x x 0 ) 3 + (x 7) (x 7)! 7 87 Finally, evaluate at x 8. 8 P (8) 3 + (8 7) 7 87 (8 7) 664 87 3.0365797. What is the largest possible error that could have occurred in your previous estimate? We now that f(x) P n (x) K n+ (n + )! x x 0 n+. In this case, n, x 0 7, and x 8. K 3 max of f (x) on [7, 8] max of 0 7x 8/3 on [7, 8] 0 0 7 78/3 7747 Putting this all together, we have f(x) P (x) 0 7747 8 7 3 5 5344 0.0000094. 3. Use a comparison to show whether each of the following converges or diverges. If an integral converges, give a good upper bound for its value. (a) (b) 7 + 5 sin x x For all x, we have 0 7 + 5 sinx 7 + 5() because the maximum of sin x is. x x x t x t x t t x [ ] t t [0 ( )] Therefore, the original integral in question must converge to a value less than. + 3x + x 3 0x + 7x 0 For x, we have + 3x + x 3 0x + 7x 0 x 3 0x + 7x 0. (We ve made the numerator smaller and the denominator larger, so the new fraction is smaller.)
x 3 But 0x + 7x x3 x3 7x 3x 4 3 x and we now that 3 yourself or notice that p ). Therefore the original integral must also diverge. x diverges (compute for 4. Decide if each of the following sequences {a } converges or diverges. If a sequence converges, compute its it. (a) a 3 + 0 Terms are 3., 3.0, 3.00,3.000,... ( 3 + ) 3, so the sequence converges to 3. 0 (b) a ( ) Terms are,,,,... ( ) doesn t exist, so the sequence diverges. (c) a 3 + 5 Terms are 8/9, 3/, 8/3, 3/5,... 7 + 3 + 5 7 + 5 (by L Hopital s Rule or by inspection), so the sequence converges to 5. 5. Decide if each of the following series converges or diverges. If a series converges, find its value. (a) 3. + 3.0 + 3.00 + 3.000 +... a 3 0, so the series diverges by the nth Term Test. (We eep adding 3 s forever.) [Compare this with the first sequence of the previous problem.] (b) + / + /3 + /4 +... This is the famous Harmonic Series, which diverges even though the terms do approach 0. We can use the Integral Test: (c) 5 5/3 + 5/9 5/7 +... x diverges, which means that This is a geometric series with r a, so it converges to 3 must diverge too. r 5 ( /3) 5 4. 6. Decide if each of the following series converges or diverges. If a series converges, find upper and lower bounds for its value. [The :0 section need not find the bounds.] (a) ( ) + The terms of this series alternate in sign. And,... 0. 3 3 4 And, + 0. [Alternating Series Test] Therefore, by the Alternating Series Test, the series must converge. We now that any two consecutive partial sums will provide upper and lower bounds: lower bound S [To get better bounds, use later partial sums, such as S 5 and S 6.] upper bound S + 3 3
(b) ()! 3 (!) a + a [(+)]! 3 + [(+)!] ()! 3 (!) ( + )! 3 (!) ()! 3 + [( + )!] ( + )( + ) 3 ( + ) 4 + 6 + 3 + 6 + 3 [Ratio Test] Use L Hopital or divide each term by. (c) (d) 4 3 Since the it of the ratio is greater than, the series diverges. ( 00 + ) 5 5) [nth Term Test] ( 00 + 5) 00 98 + 5 6 0, so, by the nth Term Test, the series diverges. 5 + 3 98 + 5 For, we have 6 9 8 5 > + 3 5 + 3 5 > 0. 9 8 But, 5 + 3 34 5 5 5 5 and we now that p ). Therefore, the original series, which has larger terms, must diverge also. [Comparison Test] diverges (use Integral Test or note that (e) (ln()) [Integral Test] x(lnx) t t t t xt x(lnx) x u xt x t lnx t 0 ln ln t u du Substitute u lnx, so du x. [ lnt ln The integral converges, so the series must converge too. ]
Further, we now that Therefore, our lower bound is And our upper bound is a + x(lnx) x(lnx) (ln()) a + ln. x(lnx) (ln) + ln. x(lnx). 7. Does the first series from the previous problem converge absolutely or conditionally? ( ) +, which diverges by the Integral Test (chec for yourself). + Therefore, the first series from the previous problem converges conditionally. 8. Compute the radius and interval (including endpoints) of convergence for (x+3) + (+) 5 + (x+3) 5 (x + 3) + (x + 3) + (x + 3) 5 x + 3 5 So, we are guaranteed convergence when x + 3 5 5 5 + (x + 3) 5. Use L Hopital on the middle fraction. <. But this is equivalent to the following. < x + 3 < 5 5 < x + 3 < 5 8 < x < To chec convergence at the endpoints (where the Ratio Test is inconclusive), we plug in to the series itself. ( + 3) 5 At x, we have 5 5, which is the Harmonic Series and thus diverges. At x 8, we have Series Test. ( 8 + 3) 5 ( 5) 5 ( ), which converges by the Alternating Thus, the interval of convergence is 8 x < and the radius of convergence is 5. 9. Find the complete Taylor series (in summation notation) for f(x) ln( x) about x 0 and determine its interval of convergence. f(x) ln( x) f(0) 0 f (x) x f (0) f (x) ( x) f (0) f (x) ( x) 3 f (0) f (4) (x) 6 ( x) 4 f (0) 6
Now plug in to the Taylor series formula with x 0 0. f(x 0 ) + f (x 0 )x + f (x 0 )! (x x 0 ) + f (x 0 ) (x x 0 ) 3 +... 0 (x) + x + 6 x3 +... We now find the interval of convergence as in the previous problem. x x x3 3 x4 4... x x + (+) x x x + x + Use L Hopital on the second fraction. So, we are guaranteed convergence for x <, which is equivalent to < x <. Now chec the endpoints. At x, we have, which is the negative of the Harmonic Series and thus diverges. ( ) ( ) + At x, we have, which converges by the Alternating Series Test. So, the interval of convergence is x <. 0. (a) Write the complete series equal to 0 e x and show that it converges. We now e w + w + w! + w3 +..., so by substitution we obtain the following. Thus, 0 e x + ( x ) + ( x ) e x! + ( x ) 3 ] [ x + x4 0! x6 +... ] [x x3 3 + x5 5! x7 7 +... 3 3 + 5 5! 7 7 +... ( ) ( + )! 0 The terms of this series alternate in sign. And, 3 3 5 5! 7 7 ( + )! 0.... 0. 0 +... x + x4! x6 +... And, Therefore, by the Alternating Series Test, the series must converge.
(b) If f(x) e x, what is f (400) (0)? What is f (40) (0)? In the previous part, we found the following. e x + ( x ) + ( x )! + ( x ) 3 +... x + x4! x6 +... We now that f (400) (0) will appear in the Taylor series for f(x) in the coefficient of the x 400 term, which is f(400) (0) x 400. 400! From above, we see that term is x400 00!. Setting them equal gives f(400) (0) x 400 x400 400! 00!, which means that f(400) (0) 400! 00!. [Note that this does not simplify to 00!. Rather, 400! 400 399... 0 0.] 00! We now that f (40) (0) will appear in the Taylor series for f(x) in the coefficient of the x 40 term, which is f(40) (0) x 40. 40! However, the Taylor series above for f(x) has no terms with odd powers of x, meaning that the coefficients of all those terms must be 0. Therefore, we now that f (40) (0) 0.