Series #12 - Q&A

Question 1 How to find the Maclaurin series for 𝑒 𝑥 e x ? Answer 1: The Power Series can be used to obtain the Maclaurin series: 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 +… As we have 𝑓 ( 𝑥 ) = 𝑒 𝑥 f(x)=e x , then: 𝑓 ( 𝑥 ) = 𝑒 𝑥    ⟹    𝑓 ′ ( 𝑥 ) = 𝑒 𝑥    ⟹    𝑓 ′ ( 0 ) = 1. f(x)=e x ⟹f ′ (x)=e x ⟹f ′ (0)=1. Repeating for all derivatives: 𝑓 ′ ′ ( 𝑥 ) = 𝑒 𝑥    ⟹    𝑓 ′ ′ ( 0 ) = 1 , 𝑓 ′ ′ ′ ( 𝑥 ) = 𝑒 𝑥    ⟹    𝑓 ′ ′ ′ ( 0 ) = 1 , … f ′′ (x)=e x ⟹f ′′ (0)=1,f ′′′ (x)=e x ⟹f ′′′ (0)=1,… The Maclaurin series is as follows: 𝑒 𝑥 = 1 + 𝑥 1 ! + 𝑥 2 2 ! + 𝑥 3 3 ! + ⋯ + 𝑥 𝑛 𝑛 ! . e x =1+ 1! x ​ + 2! x 2 ​ + 3! x 3 ​ +⋯+ n! x n ​ . Answer 2: Taking the derivatives of the given function and using 𝑥 = 0 x=0, we have: 𝑓 ( 𝑥 ) = 𝑒 𝑥 , 𝑓 ( 0 ) = 1 , 𝑓 ′ ( 𝑥 ) = 𝑒 𝑥    ⟹    𝑓 ′ ( 0 ) = 1 , 𝑓 ′ ′ ( 𝑥 ) = 𝑒 𝑥    ⟹    𝑓 ′ ′ ( 0 ) = 1 , 𝑓 ′ ′ ′ ( 𝑥 ) = 𝑒 𝑥    ⟹    𝑓 ′ ′ ′ ( 0 ) = 1 , …   . f(x)=e x ,f(0)=1,f ′ (x)=e x ⟹f ′ (0)=1,f ′′ (x)=e x ⟹f ′′ (0)=1,f ′′′ (x)=e x ⟹f ′′′ (0)=1,…. Now using Maclaurin's series expansion function, we have: 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 +…. Substitute the values: 𝑒 𝑥 = 1 + 𝑥 1 ! + 𝑥 2 2 ! + 𝑥 3 3 ! + …   . e x =1+ 1! x ​ + 2! x 2 ​ + 3! x 3 ​ +…. Thus, the Maclaurin series for 𝑒 𝑥 e x is: 𝑒 𝑥 = 1 + 𝑥 1 ! + 𝑥 2 2 ! + 𝑥 3 3 ! + …   . e x =1+ 1! x ​ + 2! x 2 ​ + 3! x 3 ​ +…. Question 2 If the Taylor Series of ln ⁡ ( 𝑧 ) ln(z) is known: ln ⁡ ( 𝑧 ) = ( 𝑧 − 1 ) − 1 2 ( 𝑧 − 1 ) 2 + 1 3 ( 𝑧 − 1 ) 3 − 1 4 ( 𝑧 − 1 ) 4 + … , ln(z)=(z−1)− 2 1 ​ (z−1) 2 + 3 1 ​ (z−1) 3 − 4 1 ​ (z−1) 4 +…, Can one find the Taylor series of 𝑓 ( 𝑧 ) = ln ⁡ ( 1 − 𝑧 2 ) f(z)=ln(1−z 2 ) by manipulating the Taylor series of ln ⁡ ( 𝑧 ) ln(z)? Answer 1: Make the substitution 𝑧 = 1 − 𝑢 2 z=1−u 2 to obtain: ln ⁡ ( 1 − 𝑢 2 ) = − 𝑢 2 − 𝑢 4 2 − 𝑢 6 3 − ⋯ = ∑ 𝑘 = 1 ∞ 𝑢 2 𝑘 𝑘 . ln(1−u 2 )=−u 2 − 2 u 4 ​ − 3 u 6 ​ −⋯= k=1 ∑ ∞ ​ k u 2k ​ . Based on the ratio test, the corresponding function is analytic for ∣ 𝑢 ∣ < 1 ∣u∣<1. lim ⁡ 𝑘 → ∞ 𝑎 𝑘 + 1 𝑎 𝑘 = lim ⁡ 𝑘 → ∞ 𝑘 𝑘 + 1 ∣ 𝑢 ∣ 2 = ∣ 𝑢 ∣ 2 < 1. k→∞ lim ​ a k ​ a k+1 ​ ​ = k→∞ lim ​ k+1 k ​ ∣u∣ 2 =∣u∣ 2 <1. Hence, we can differentiate the obtained expansion to get: 𝑓 ( 𝑧 ) = 𝑥 − 𝑥 3 + 𝑥 5 − 𝑥 7 + …   . f(z)=x−x 3 +x 5 −x 7 +…. Answer 2: Notice that: 𝑓 ( 𝑥 ) 𝑑 𝑥 = ∫ 𝑓 ( 𝑥 ) 1 − 𝑥 𝑑 𝑥 = 1 2 𝑓 ( 1 − 𝑥 2 ) + 𝐶 . f(x)dx=∫ 1−x f(x) ​ dx= 2 1 ​ f(1−x 2 )+C. From this note: ln ⁡ ( 𝑧 ) = ( 𝑧 − 1 ) − 1 2 ( 𝑧 − 1 ) 2 + …   . ln(z)=(z−1)− 2 1 ​ (z−1) 2 +…. Substitute 1 − 𝑥 2 1−x 2 into (lnx). Question 3 Determine the convergence of this alternating series: ∑ 𝑖 = 1 ∞ 2 ( − 1 4 ) 𝑖 . i=1 ∑ ∞ ​ 2(− 4 1 ​ ) i . Answer 1: This is a geometric series of the form: ∑ 𝑎 𝑟 𝑛 − 1 = 𝑎 + 𝑎 𝑟 + 𝑎 𝑟 2 + 𝑎 𝑟 3 + …   . ∑ar n−1 =a+ar+ar 2 +ar 3 +…. Here, 𝑎 = 2 a=2, and 𝑟 = − 1 4 r=− 4 1 ​ . Since ∣ 𝑟 ∣ < 1 ∣r∣<1, the sum of the series is given by: 𝑆 ∞ = 𝑎 1 − 𝑟 = 2 1 + 1 4 = 8 5 . S ∞ ​ = 1−r a ​ = 1+ 4 1 ​ 2 ​ = 5 8 ​ . Thus, the sum of the series is: 8 5 . 5 8 ​ . Question 4 Using Taylor expansion, check if the series: 𝑛 ( arctan ⁡ ( 𝑛 + 1 ) − arctan ⁡ 𝑛 ) n ​ (arctan(n+1)−arctann) converges. Answer 1: We manipulate the series as follows: arctan ⁡ ( 𝑛 + 1 ) − arctan ⁡ 𝑛 = 1 1 + 𝑛 ( 𝑛 + 1 ) . arctan(n+1)−arctann= 1+n(n+1) 1 ​ . Simplify: arctan ⁡ ( 𝑛 + 1 ) − arctan ⁡ 𝑛 ≈ 1 𝑛 2 . arctan(n+1)−arctann≈ n 2 1 ​ . Now use Taylor's expansion to approximate further. This series alternates, and the individual terms decrease faster than 1 𝑛 2 n 2 1 ​ , ensuring that the series converges. Answer 2: Using Taylor's series: arctan ⁡ ( 𝑛 + 1 ) − arctan ⁡ ( 𝑛 ) ∼ 1 𝑛 2 . arctan(n+1)−arctan(n)∼ n 2 1 ​ . The series: 𝑛 ⋅ 1 𝑛 2 = 𝑛 𝑛 2 . n ​ ⋅ n 2 1 ​ = n 2 n ​ ​ . This series converges since 𝑛 𝑛 2 n 2 n ​ ​ decreases faster than 1 / 𝑛 𝑝 1/n p for 𝑝 > 1 p>1. Answer 3: Hint: arctan ⁡ 𝑛 + 1 ∼ arctan ⁡ 𝑛 + 1 𝑛 2 . arctann+1∼arctann+ n 2 1 ​ . Thus, 𝑛 ( arctan ⁡ ( 𝑛 + 1 ) − arctan ⁡ 𝑛 ) ∼ 𝑛 ⋅ 1 𝑛 2 . n ​ (arctan(n+1)−arctann)∼ n ​ ⋅ n 2 1 ​ . Convergence is ensured. Convergence is ensured. Question 5 Is it true that for 𝑥 ∈ [ 0 , 2 𝜋 ] x∈[0,2π], we have: cos ⁡ ( 𝑛 𝑥 ) = 𝜋 𝑥 2 + 𝜋 2 2 𝑛 . cos(nx)= 2 πx ​ + 2n π 2 ​ . Answer 1: We show that: ∑ 𝑛 = 1 ∞ sin ⁡ 2 𝑎 𝑛 𝑛 2 = 𝜋 𝑎 . n=1 ∑ ∞ ​ n 2 sin2an ​ =πa. Use the identity for Fourier series to expand: ∑ 𝑛 = 1 ∞ cos ⁡ ( 𝑛 𝑥 ) = …   . n=1 ∑ ∞ ​ cos(nx)=…. Conclude that: cos ⁡ ( 𝑛 𝑥 ) ∼ 𝑥 2 . cos(nx)∼x 2 .