MATH 4 (R) Winter 8 Intermediate Calculus I Solutions to Problem Set #4 Completion Date: Monday February, 8 Department of Mathematical and Statistical Sciences University of Alberta Question. [Sec..9, # 6] Find a power series representation for the function and determine the interval of convergence. + 9x Solution: Start with the geometric series /( x) x n, x <, then ( 9x ) ( 9x ) n ( ) n 9 n x n ( ) n (x) n. The series converges for 9x <, that is, for x < /9, or x < /, and the interval of convergence is ( /, /). Question [Sec..9, # 6] Find a power series representation for the function and determine the radius of convergence. x ( x) Solution: Start with the geometric series /( x) x n, x <, then x (x) n n x n converges for x <, that is, x <. Differentiating we obtain ( x) n nx n, n that is, ( x) x ( x) x n nx n n nx n, n n n n nx n n nx n+, n for x <, and the radius of convergence is R /.
Question. [Sec..9, # 8] Find a power series representation for the function and determine the radius of convergence. Solution: Recall that Also, 9 + x 9( + ( x ) ) 9 arctan(x/) 9 + x dx tan x + C for x /9 <, that is, x/ <, or x <. Therefore tan x 9 If we let x, then C, ( ( ) n x ) n dx ( ( x ) ) n 9 ( ) n tan x for x < and the radius of convergence is R. n xn+ ( ) n x n+ n+ (n + ), ( ( ) n x n + + C ) n ( ) n x n+ n+ (n + ) + C. Question 4. [Sec..9, # ] Use a power series representation to approximate the integral to six decimal places. / dx + x 6 Solution: We use the geometric series again, / dx / + x 6 dx / ( x 6 ) 6n + ( ) n x6n+ / ( x 6 ) n dx 7 7 +. ( ) n ( )6n+ 6n + / From the Alternating Series Estimation Theorem, we have R n b n+, ( ) n x 6n dx ().94 and 9 (9)., and we use the first terms of the alternating series: / dx + x 6 7 7 +.49889. ( ) n 6n+ (6n + )
Question 5. [Sec..9, # ] Show that the function is a solution to the differential equation ( ) n x n (n)! f (x) +. Solution: Differentiating, f (x) + f (x) f (x) n k n n ( ) n n x n (n)! n ( ) n (n ) x n (n )! ( ) n (n )! xn + ( ) k+ x k + (k)! k ( ) k x k (k)! + k k ( ) n (n )! xn, n ( ) n (n)! xn ( ) k x k (k)! ( ) k x k (k)! ( ) n (n )! xn, (k n in the st sum). Question 6. [Sec.., # 4] Find the Maclaurin series for sin x using the definition of a Maclaurin series. [Assume that f has a power series expansion. Do not show that R n (x).] Also find the associated radius of convergence. Solution: We need f() + f ()x + f () x +.! sin x, f() f (x) cos x, f () f (x) 4 sin x, f () f (x) 8 cos x, f () 8 f (4) (x) 6 sin x, f (4) () f (5) (x) cos x, f (5) () 5, etc, therefore f (n) () if n is even and f (n+) () ( ) n n+, and Also, ( ) n n+ x n+ x 8 (n + )!! x + 5! x5. a n+ a n ( ) n+ (n+)+ x (n+)+ (n + )! ((n + ) + )! ( ) n n+ x n+ x (n + )(n + ), lim a n+ n a n lim for all x R, and by the Ratio Test R. n x (n + )(n + ) <
Question 7. [Sec.., # 4] Find the Taylor series for ln x centered at the value a. [Assume that f has a power series expansion. Do not show that R n (x).] Solution: We have ln x f() ln f (x) x f () f (x) x f () and f (n) n (n )! () ( ) n f (n) () f (x) x f () f (4) (x) x 4 f (4) () 4, etc. for n, (x ) n ln + n From the Ratio Test ( ) n (x ) n+ (n + ) n+ and lim a n+ a n lim n ( ) n (n )! n n n ( ) n (x ) n n (x ) n ln + n x (n + ), n x (n + ) x. ( ) n n n (x ) n. Therefore the series converges absolutely if x <, that is, x <, and the radius of convergence is R. Question 8. [Sec.., # ] Prove that the Maclaurin series for represents cosh x for all x. Solution: First we find the Maclaurin series. so the Maclaurin series is f (n) () x n cosh x cosh x f() f (x) sinh x f () f (x) cosh x f () f (x) sinh x f (n) () x n (n)! f (), etc. Now, f (n+) (x) cosh x or sinh x and sinh x < cosh x for all x, n x n (n)! + x! + x4 4! + f (n+) (x) cosh x cosh d if x d.
From Taylor s Inequality R n (x) M x n+ (n + )! cosh d x n+, (n + )! and lim n x n+ (n + )!, cosh d x n+ lim n (n + )! By the Squeeze Theorem, lim n R n(x) for x d, and since d is any real number, then cosh x is equal to its Maclaurin series for all x. Question 9. [Sec.., # ] Use a known Maclaurin series to obtain the Maclaurin series for the function cos x. Hint: Use cos x ( + cos x).. Solution: Note that ( + cos x), cos x ( ) n x n (n)! (with R ) and this implies ( ) n (x) n cos x (n)! ( ) n n x n (n)! ( ) n n x n + (n)! n therefore for < x <. cos x + { + ( ) n n x n } ( ) n n x n +, (n)! (n)! n n Question. [Sec.., # 46] Use series to approximate the definite integral to within the accuracy error <.. / x e x dx Solution: Since e x x n + x + x! + x +, for < x <, then! ( x ) n ( ) n x n e x x e x x x 4 + x6! x8! + x + x4! x6! + ( ) n x n+
and.5 [ x x e x dx x5 5 + x7 7! x9 9! + (.5) (.5)5 5 + (.5)7 4 ].5 (.5)9 54 From the Alternating Series Estimate Theorem, R n b n+ and therefore (.5) 5.65 5 (.5) 7.56 <. 4 +..5 x e x dx (.5) (.5)5 5.54. Question. [Sec.., # 56] Find the sum of the series ( ) n π n 6 n (n)!. Solution: We have since cos x ( ) n x n. (n)! ( ) n π n 6 n (n)! ( ) n (n)! ( π 6 ) n cos π 6, Question. [Sec.., # 6a,b] Approximate cos x by a Taylor polynomial T n with degree n 4 at the number a π, and use Taylor s Inequality to estimate the accuracy of the approximation f(x) T n (x) when x lies in the interval x π/. Solution: (a) We have cos x, f (x) sin x, f (x) cos x, f (x) sin x, f (4) (x) cos x, f (5) (x) sin x, f ( ) π f ( ) π f ( ) π f ( ) π f (4)( ) π
(b) We have for f (n+) (x) M. T 4 (x) ( ) x π! R n (x) If x π/, then x π/ π/ and ( ) x π +! M x a n+, x a d (n + )! f (5) (x) sin x sin x M ( ) x π ( ) + x π 4. 4! and therefore R 4 (x) x π/ 5 5! (π/)5 5!.49. Question. [Sec.., # a,b] Approximate x ln x by a Taylor polynomial T n with degree n at the number a, and use Taylor s Inequality to estimate the accuracy of the approximation f(x) T n (x) when x lies in the interval / x /. solution: (a) We have x ln x, f() f (x) ln x +, f () f (x) x, f () f (x) x, f () f (4) (x) x. Therefore, T (x) f() + f ()(x ) + f ()! (x ) + (x )! (x ). (x ) + f () (x )! (b) We want and if.5 x.5, then x.5, and implies R (x) M x 4 /4!, (.5) x (.5) x (.5)
implies f (4) (x) x (.5) 6 M therefore R (x) 6(.5)4 4!.47. Question 4. [Sec.., # 6] How many terms of the Maclaurin series for ln( + x) do you need to use to estimate ln.4 to within.? Solution: We have so the Maclaurin series is ln( + x), f() f (x) + x, f () f (x) ( + x), f () f (x) ( + x), f () f (4) 6 (x) ( + x) 4, f (4) () 6, etc. ln( + x) (x )! (x ) +! (x ) therefore ln(.4) (.4) (.4) + (.4) 4 (.4)4 +. Since this is an alternating series, we use the Alternating Series Estimation Theorem, (.4) 4 (.4) 5 (.4) 6.64,., 4 5 6 and we need the first 5 (nonzero) terms of the Maclaurin series..7 <.,