Series #3 - Q&A

Question 1 Find a power series for the function, centered at 𝑐 c, and determine the interval of convergence: 𝑔 ( 𝑥 ) = 3 2 𝑥 − 1 ;    𝑐 = 2 g(x)= 2x−1 3 ​ ;c=2 Answer 1 Geometric Power Series Centered at 𝑐 c: A geometric power series centered at 𝑐 c is a series of the form: 𝑎 1 − ( 𝑟 − 𝑐 ) = ∑ 𝑛 = 0 ∞ 𝑎 ( 𝑟 − 𝑐 ) 𝑛 ,    ∣ 𝑟 − 𝑐 ∣ < 1 1−(r−c) a ​ = n=0 ∑ ∞ ​ a(r−c) n ,∣r−c∣<1 where 𝑎 a is the first term and 𝑟 − 𝑐 r−c is the common ratio. The function is given by: 𝑓 ( 𝑥 ) = 3 2 𝑥 − 1 ;    𝑐 = 2 f(x)= 2x−1 3 ​ ;c=2 Writing 𝑓 ( 𝑥 ) f(x) in the form 𝑎 1 − 𝑟 1−r a ​ produces: 3 2 𝑥 − 1 = 2 1 − ( 2 𝑥 − 2 ) + 3 2x−1 3 ​ = 1−(2x−2) 2 ​ +3 = 3 ( 1 1 + 𝑥 − 2 2 ) =3( 1+ 2 x−2 ​ 1 ​ ) So the power series is: 2 3 ( 𝑥 − 2 ) 𝑛 3 2 ​ (x−2) n The convergence interval is ( (1.5, 2.5) ANSWER 1 Extracted Text Continued: Question 2 A pair of honest dice is rolled once. Find the expected value of the sum of the two numbers rolled. Answer 1 Below is the sample space for the sum of a pair of dice: 𝑆 = { 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 , 10 , 11 , 12 } S={2,3,4,5,6,7,8,9,10,11,12} Probability of each event: Sum Probability 2 1/36 3 2/36 4 3/36 5 4/36 6 5/36 7 6/36 Using the expected value formula: 𝐸 = ∑ 𝑝 𝑖 ⋅ 𝑥 𝑖 = 2 36 + . . . . = 4. E=∑p i ​ ⋅x i ​ = 36 2 ​ +....=4. Extracted Text Continued: Question 3 Use linear approximation (or differentials) to estimate: 99.2 99.2 ​ Answer 1 Use the Taylor series for 𝑥 x ​ about 𝑥 = 100 x=100. The reason is that 100 is the closest square to 99.2. 𝑓 ( 𝑥 ) = 𝑓 ( 100 ) + 𝑓 ′ ( 100 ) ( 𝑥 − 100 ) + higher order terms . f(x)=f(100)+f ′ (100)(x−100)+higher order terms. Hence, 99.2 ≈ 100 + 1 2 100 ( 99.2 − 100 ) . 99.2 ​ ≈ 100 ​ + 2 100 ​ 1 ​ (99.2−100). = 10 − 0.8 20 = 10 − 0.04 = 9.96 =10− 20 0.8 ​ =10−0.04=9.96 Question 4 Calculate the following: (a) Find the Maclaurin series of cos ⁡ ( 𝑥 ) cos(x) and find the radius of this series without using any power or Maclaurin series. Besides geometric. (a) The Maclaurin series of a function 𝑓 ( 𝑥 ) f(x) is given by: 𝑓 ( 𝑥 ) = 𝑓 ( 0 ) + 𝑓 ′ ( 0 ) 𝑥 + 𝑓 ′ ′ ( 0 ) 2 ! 𝑥 2 + 𝑓 ′ ′ ′ ( 0 ) 3 ! 𝑥 3 + … f(x)=f(0)+f ′ (0)x+ 2! f ′′ (0) ​ x 2 + 3! f ′′′ (0) ​ x 3 +… For cos ⁡ ( 𝑥 ) cos(x): 𝑓 ( 𝑥 ) = cos ⁡ ( 𝑥 ) ⇒ 𝑓 ′ ( 𝑥 ) = − sin ⁡ ( 𝑥 ) ,    𝑓 ′ ′ ( 𝑥 ) = − cos ⁡ ( 𝑥 ) ,    𝑓 ′ ′ ′ ( 𝑥 ) = sin ⁡ ( 𝑥 ) ,    𝑓 ( 4 ) ( 𝑥 ) = cos ⁡ ( 𝑥 ) . f(x)=cos(x)⇒f ′ (x)=−sin(x),f ′′ (x)=−cos(x),f ′′′ (x)=sin(x),f (4) (x)=cos(x). At 𝑥 = 0 x=0: 𝑓 ( 0 ) = 1 ,    𝑓 ′ ( 0 ) = 0 ,    𝑓 ′ ′ ( 0 ) = − 1 ,    𝑓 ′ ′ ′ ( 0 ) = 0 ,    𝑓 ( 4 ) ( 0 ) = 1. f(0)=1,f ′ (0)=0,f ′′ (0)=−1,f ′′′ (0)=0,f (4) (0)=1. Thus, the Maclaurin series for cos ⁡ ( 𝑥 ) cos(x) is: cos ⁡ ( 𝑥 ) = 1 − 𝑥 2 2 ! + 𝑥 4 4 ! − 𝑥 6 6 ! + ⋯ = ∑ 𝑛 = 0 ∞ ( − 1 ) 𝑛 ( 2 𝑛 ) ! 𝑥 2 𝑛 . cos(x)=1− 2! x 2 ​ + 4! x 4 ​ − 6! x 6 ​ +⋯= n=0 ∑ ∞ ​ (2n)! (−1) n ​ x 2n . Radius of convergence: Using the ratio test: lim ⁡ 𝑛 → ∞ ∣ 𝑎 𝑛 + 1 𝑎 𝑛 ∣ = lim ⁡ 𝑛 → ∞ ∣ 𝑥 2 𝑛 + 2 ( 2 𝑛 + 2 ) ! ⋅ ( 2 𝑛 ) ! 𝑥 2 𝑛 ∣ = lim ⁡ 𝑛 → ∞ 𝑥 2 ( 2 𝑛 + 2 ) ( 2 𝑛 + 1 ) = 0 (for all  𝑥 ) . n→∞ lim ​ ​ a n ​ a n+1 ​ ​ ​ = n→∞ lim ​ ​ (2n+2)! x 2n+2 ​ ⋅ x 2n (2n)! ​ ​ = n→∞ lim ​ (2n+2)(2n+1) x 2 ​ =0(for all x). Thus, the radius of convergence is ∞ ∞. (b) For cos ⁡ ( − 2 𝑥 ) cos(−2x): 𝑓 ( 𝑥 ) = cos ⁡ ( − 2 𝑥 ) = 1 − ( − 2 𝑥 ) 2 2 ! + ( − 2 𝑥 ) 4 4 ! − ⋯ = ∑ 𝑛 = 0 ∞ ( − 1 ) 𝑛 ( 2 𝑛 ) ! ( 2 𝑥 ) 2 𝑛 . f(x)=cos(−2x)=1− 2! (−2x) 2 ​ + 4! (−2x) 4 ​ −⋯= n=0 ∑ ∞ ​ (2n)! (−1) n ​ (2x) 2n . = ∑ 𝑛 = 0 ∞ ( − 1 ) 𝑛 2 2 𝑛 𝑥 2 𝑛 ( 2 𝑛 ) ! . = n=0 ∑ ∞ ​ (2n)! (−1) n 2 2n x 2n ​ . Radius of convergence remains ∞ ∞, as it does not depend on the coefficient of 𝑥 x. Question 5 Find the values of 𝑥 x for which the given geometric series converges. Also, find the sum of the series (as a function of 𝑥 x) for those values of 𝑥 x: ∑ 𝑛 = 0 ∞ ( − 1 2 ) 𝑛 ( 𝑥 − 5 ) 𝑛 n=0 ∑ ∞ ​ (− 2 1 ​ ) n (x−5) n Answer 1 The given series is a geometric series of the form: ∑ 𝑛 = 0 ∞ 𝑎 𝑟 𝑛 ,    where  𝑟 = − 1 2 ( 𝑥 − 5 ) . n=0 ∑ ∞ ​ ar n ,where r=− 2 1 ​ (x−5). For convergence: ∣ 𝑟 ∣ < 1 ⇒ ∣ − 1 2 ( 𝑥 − 5 ) ∣ < 1. ∣r∣<1⇒ ​ − 2 1 ​ (x−5) ​ <1. 1 2 ∣ 𝑥 − 5 ∣ < 1 ⇒ ∣ 𝑥 − 5 ∣ < 2 ⇒ 3 < 𝑥 < 7. 2 1 ​ ∣x−5∣<1⇒∣x−5∣<2⇒3