University:
Morehouse College
Course:
HMTH 105 | College Algebra 1
Academic year:

2024
Views:

317

Pages:

3
Author:

Reynaldo Wong

Question 1 Representing functions by power series: Identify the functions represented by the following power series: ∑ 𝑘 = 1 ∞ 𝑥 2 𝑘 𝑘 k=1 ∑ ∞ k x 2k Answer 1 Given: A power series: ∑ 𝑘 = 1 ∞ 𝑥 2 𝑘 𝑘 k=1 ∑ ∞ k x 2k Known properties of power series: ln ⁡ ( 1 − 𝑦 ) = − ∑ 𝑛 = 1 ∞ 𝑦 𝑛 𝑛 ln(1−y)=− n=1 ∑ ∞ n y n As the power series of ln ⁡ ( 1 − 𝑦 ) ln(1−y): ln ⁡ ( 1 − 𝑦 ) = − ∑ 𝑛 = 1 ∞ 𝑦 𝑛 𝑛 ln(1−y)=− n=1 ∑ ∞ n y n So, when 𝑦 = 𝑥 2 y=x 2 : ∑ 𝑘 = 1 ∞ 𝑥 2 𝑘 𝑘 = − ln ⁡ ( 1 − 𝑥 2 ) k=1 ∑ ∞ k x 2k =−ln(1−x 2 ) Question 2 Find a formula for the nth partial sum of the series and use it to determine whether the series converges or diverges. If a series converges, find its sum: ∑ 𝑛 = 1 ∞ ( cos ⁡ − 1 ( 1 𝑛 + 1 ) − cos ⁡ − 1 ( 1 𝑛 + 2 ) ) n=1 ∑ ∞ (cos −1 ( n+1 1 )−cos −1 ( n+2 1 )) Answer 1 The series: ∑ 𝑛 = 1 ∞ ( cos ⁡ − 1 ( 1 𝑛 + 1 ) − cos ⁡ − 1 ( 1 𝑛 + 2 ) ) n=1 ∑ ∞ (cos −1 ( n+1 1 )−cos −1 ( n+2 1 )) Consider a general term: 𝑎 𝑛 = cos ⁡ − 1 ( 1 𝑛 + 1 ) − cos ⁡ − 1 ( 1 𝑛 + 2 ) a n =cos −1 ( n+1 1 )−cos −1 ( n+2 1 ) The nth partial sum is: 𝑆 𝑛 = cos ⁡ − 1 ( 1 2 ) − cos ⁡ − 1 ( 1 𝑛 + 2 ) S n =cos −1 ( 2 1 )−cos −1 ( n+2 1 ) As 𝑛 → ∞ n→∞: 𝑆 𝑛 = cos ⁡ − 1 ( 1 2 ) − cos ⁡ − 1 ( 0 ) = 𝜋 3 − 0 = 𝜋 3 S n =cos −1 ( 2 1 )−cos −1 (0)= 3 π −0= 3 π The series converges, and its sum is 𝜋 3 3 π . Question 3 Find the Maclaurin series for the function 𝑓 ( 𝑥 ) = cos ⁡ 4 𝑥 f(x)=cos4x. Answer 1 Using the known series for cos ⁡ 𝑥 cosx: cos ⁡ 𝑥 = ∑ 𝑛 = 0 ∞ ( − 1 ) 𝑛 𝑥 2 𝑛 ( 2 𝑛 ) ! cosx= n=0 ∑ ∞ (2n)! (−1) n x 2n Substitute 4 𝑥 4x for 𝑥 x: cos ⁡ 4 𝑥 = ∑ 𝑛 = 0 ∞ ( − 1 ) 𝑛 ( 4 𝑥 ) 2 𝑛 ( 2 𝑛 ) ! = ∑ 𝑛 = 0 ∞ ( − 1 ) 𝑛 4 2 𝑛 𝑥 2 𝑛 ( 2 𝑛 ) ! cos4x= n=0 ∑ ∞ (2n)! (−1) n (4x) 2n = n=0 ∑ ∞ (2n)! (−1) n 4 2n x 2n Question 4 Find the coefficient of 𝑥 5 𝑦 8 x 5 y 8 in ( 𝑥 + 𝑦 ) 13 (x+y) 13 . Answer 1 Using the binomial theorem: ( 𝑥 + 𝑦 ) 𝑛 = ∑ 𝑗 = 0 𝑛 ( 𝑛 𝑗 ) 𝑥 𝑗 𝑦 𝑛 − 𝑗 (x+y) n = j=0 ∑ n ( j n )x j y n−j We need the term where 𝑥 5 𝑦 8 x 5 y 8 , so 𝑗 = 5 j=5 and 𝑛 = 13 n=13. The coefficient is: ( 13 5 ) = 13 ! 5 ! ( 13 − 5 ) ! = 13 ! 5 ! ⋅ 8 ! = 1287 ( 5 13 )= 5!(13−5)! 13! = 5!⋅8! 13! =1287 Question 5 The Maclaurin series of 𝑓 ( 𝑥 ) f(x): Its first four terms should be written out. Given 𝑓 ( 0 ) = 2 f(0)=2, 𝑓 ′ ( 0 ) = 3 f ′ (0)=3, 𝑓 ′ ′ ( 0 ) = 4 f ′′ (0)=4, 𝑓 ( 3 ) ( 0 ) = 12 f (3) (0)=12. Answer 1 The Maclaurin series is: 𝑓 ( 𝑥 ) = 𝑓 ( 0 ) + 𝑓 ′ ( 0 ) 𝑥 + 𝑓 ′ ′ ( 0 ) 2 ! 𝑥 2 + 𝑓 ( 3 ) ( 0 ) 3 ! 𝑥 3 + … f(x)=f(0)+f ′ (0)x+ 2! f ′′ (0) x 2 + 3! f (3) (0) x 3 +… Substitute the given values: 𝑓 ( 𝑥 ) = 2 + 3 𝑥 + 4 2 𝑥 2 + 12 6 𝑥 3 f(x)=2+3x+ 2 4 x 2 + 6 12 x 3 𝑓 ( 𝑥 ) = 2 + 3 𝑥 + 2 𝑥 2 + 2 𝑥 3 f(x)=2+3x+2x 2 +2x 3