<p>MAT 271 (FIPE) Test 3 Review Chapters 9, 10</p><p>Sample Problems (not necessarily all inclusive)</p><p>A. Taylor Series</p><p>List the first 5 non-zero terms of the Taylor series for the functions in problems 1––4 about the given points. Then use the ratio test to determine the interval of convergence.</p><p>1 1. f x  ln1  x x  0 ans 2. f x  x  0 ans 1  x</p><p>1 3. f x  x 2 cos x x  0 ans 4. f x  x  2 ans x 2 1</p><p>5. Find the Taylor series of order 7 for f x  x about x  1 and use it to find a 10 digit approximation for 1.5 . ans</p><p>B. Geometric Series</p><p>1. A patient is given a 20 mg injection of a therapeutic drug. Each day, the patient’s body metabolizes 40% of the drug present. The patient is given a 20 mg injection of the drug each day. Determine the drug level in the patient’s body after 6 days. If the procedure is continued indefinitely, determine the amount of the drug level in the patient’s body. ans</p><p>12  9  6 3  5 1  2. Find the sum of the first 50 terms of the series 4 16 ans 3. Find the sum of the first 20 terms of the series 1 4  9  16 ans</p><p>4. If 47.90625  24  12  6    an , determine the value of n. ans</p><p>C. Convergence of Series, Intervals of Convergence Determine if the following series converge or diverge. State the test used and explain your reasoning. answers    n   2 1 2 1 1 2n  7 en 1.    2.  2 3.  n 4.  2 3 5.  3 n1 n2 nln n n1 3 1 n1 n n1 n</p><p>Determine the intervals of convergence</p><p> n  x 1 n 6.  7.  nx  2 n1 n n1 D. Fourier Series</p><p>1. Find the Fourier Series of orders ans 2. Find the Fourier Series of orders ans 4 and 8 for the function shown below. 2 and 4 for the function shown below.</p><p>Use F4 x to approximate f 1.5.</p><p>Possible Answers</p><p>A. 1 1 1 1 1. f := x x2 x3 x4 x5 1  x  1 2 3 4 5 , </p><p>1 3 5 35 63 g := 1 x x2 x3 x4 x5 1  x  1 2. 2 8 16 128 256 , </p><p>1 1 1 1 h := x2 x4 x6 x8 x10 3. 2 24 720 40320 , Converges for all x</p><p>1 2 1 2 11 3 227 4 , h1 := 3  3 ( x2 ) 3 ( x2 )  3 ( x2 )  3 ( x2 ) 1  x  3 4. 3 9 6 81 1944 </p><p>5.</p><p>1 1 1 1 5 7 h2 := x  x ( x1 )2 ( x1 )3 ( x1 )4 ( x1 )5 2 2 8 16 128 256 21 33  ( x1 )6 ( x1 )7 1024 2048 > h2(1.5); 1.224781036</p><p>B.</p><p>1. 28.6 mg, 30 mg 2. 6.857 3. 2870 4. 9</p><p>C. 1. Converges, Ratio test 2. Converges, Integral test</p><p> n   2  2 3. Converges, compare to   , geometric, r  4. Diverges, p-series, p  1 n1  3  3 5. Converges, p-series, p  1 6. 0  x  2 7. 1 x  3</p><p>D. 1  8 3  8 cos  x  cos  x  2  9 2  1. f := 1     4 2 2 ,</p><p>1  8 3  8 5  8 7  8 cos  x  cos  x  cos  x  cos  x  2  9 2  25 2  49 2  .4905251840 f := 1         8 2 2 2 2 , </p><p>1 sin(  x ) 1 sin( 2  x ) 1 sin(  x ) 1 sin( 2  x ) 1 sin( 3  x ) 1 sin( 4  x ) f :=   f :=     2. 2 2  2  , 4 2  2  3  4 </p>