Series #4 - Q&A

Question 1 What is the difference between Fourier transformation and Fourier series? Answer 1 A periodic function is represented as a discontinuous sum of complex exponentials in the Fourier series. However, the Fourier transform is applied to a continuous superposition or integral of complex exponentials to describe a broad, nonperiodic function. The limits of integration shift from one period to the next because the Fourier transform may be seen as the limit of the Fourier series of a function as the period approaches infinity. Answer 2 Periodic and non-periodic signals are converted from the time domain to the frequency domain using the Fourier transform. Given that Fourier series are a condensed version of time-domain periodic functions, it can also convert Fourier series into the frequency domain. Question 2 Find the Taylor expansion of ln ⁡ ( 1 − 𝑥 ) ln(1−x) about 𝑥 = 0 x=0. Answer 1 Differentiate ln ⁡ ( 1 − 𝑥 ) ln(1−x): 𝑑 𝑑 𝑥 ( ln ⁡ ( 1 − 𝑥 ) ) = − 1 1 − 𝑥 ,    ∣ 𝑥 ∣ < 1. dx d ​ (ln(1−x))= 1−x −1 ​ ,∣x∣<1. Express − 1 1 − 𝑥 1−x −1 ​ using the binomial expansion: − 1 1 − 𝑥 = − ( 1 + 𝑥 + 𝑥 2 + 𝑥 3 + …   ) . 1−x −1 ​ =−(1+x+x 2 +x 3 +…). Integrate the above: ln ⁡ ( 1 − 𝑥 ) = − 𝑥 − 𝑥 2 2 − 𝑥 3 3 − 𝑥 4 4 − … ln(1−x)=−x− 2 x 2 ​ − 3 x 3 ​ − 4 x 4 ​ −… Answer 2 Calculate derivatives using Taylor's formula: 𝑓 ( 𝑥 ) = ln ⁡ ( 1 − 𝑥 ) ⇒ 𝑓 ′ ( 𝑥 ) = − 1 1 − 𝑥 ,    𝑓 ( 2 ) ( 𝑥 ) = − 1 ( 1 − 𝑥 ) 2 ,    etc. f(x)=ln(1−x)⇒f ′ (x)=− 1−x 1 ​ ,f (2) (x)=− (1−x) 2 1 ​ ,etc. Expanding: ln ⁡ ( 1 − 𝑥 ) = − 𝑥 − 𝑥 2 2 − 𝑥 3 3 − … ln(1−x)=−x− 2 x 2 ​ − 3 x 3 ​ −… Question 3 Consider the telescoping series ∑ 𝑘 = 3 ∞ ( 𝑘 − 𝑘 − 2 ) ∑ k=3 ∞ ​ ( k ​ − k−2 ​ ): (a) Apply the Divergence Test to the series. What conclusions, if any, can you make about the series? (b) Write out the partial sums 𝑠 3 , 𝑠 4 , 𝑠 5 s 3 ​ ,s 4 ​ ,s 5 ​ , and 𝑠 6 s 6 ​ . (c) Compute the 𝑛 n-th partial sum 𝑠 𝑛 s n ​ , and put it in closed form. Answer 1 (a) The general term of the series is: 𝑎 𝑘 = 𝑘 − 𝑘 − 2 . a k ​ = k ​ − k−2 ​ . Take the limit as 𝑘 → ∞ k→∞: lim ⁡ 𝑘 → ∞ 𝑎 𝑘 = lim ⁡ 𝑘 → ∞ ( 𝑘 − 𝑘 − 2 ) . k→∞ lim ​ a k ​ = k→∞ lim ​ ( k ​ − k−2 ​ ). Multiply numerator and denominator by 𝑘 + 𝑘 − 2 k ​ + k−2 ​ : lim ⁡ 𝑘 → ∞ 𝑎 𝑘 = lim ⁡ 𝑘 → ∞ 2 𝑘 + 𝑘 − 2 = 0. k→∞ lim ​ a k ​ = k→∞ lim ​ k ​ + k−2 ​ 2 ​ =0. The divergence test states that if the 𝑛 n-th term of a series does not converge to zero, the series diverges. Here 𝑎 𝑘 → 0 a k ​ →0, so the series may or may not converge. (b) The 𝑛 n-th partial sum of series ∑ 𝑎 𝑛 ∑a n ​ is the sum of the first 𝑛 n terms of the series. For 𝑠 3 s 3 ​ : 𝑠 3 = ( 3 − 1 ) + ( 4 − 2 ) + ( 5 − 3 ) = 5 − 1 . s 3 ​ =( 3 ​ − 1 ​ )+( 4 ​ − 2 ​ )+( 5 ​ − 3 ​ )= 5 ​ − 1 ​ . For 𝑠 4 s 4 ​ : 𝑠 4 = 𝑠 3 + ( 6 − 4 ) = 6 − 1 . s 4 ​ =s 3 ​ +( 6 ​ − 4 ​ )= 6 ​ − 1 ​ . Similarly, 𝑠 5 = 7 − 1 ,    𝑠 6 = 8 − 1 . s 5 ​ = 7 ​ − 1 ​ ,s 6 ​ = 8 ​ − 1 ​ . (c) The 𝑛 n-th partial sum: 𝑠 𝑛 = 𝑛 + 2 − 1 = 𝑛 + 2 − 1. s n ​ = n+2 ​ − 1 ​ = n+2 ​ −1. Question 4 Find the sum of the following series. Round to the nearest hundredth if necessary: 6 + 12 + 24 + ⋯ + 6144. 6+12+24+⋯+6144. Answer 1 The sum of a finite geometric series: 𝑆 𝑛 = 𝑎 1 ( 1 − 𝑟 𝑛 ) 1 − 𝑟 . S n ​ = 1−r a 1 ​ (1−r n ) ​ . Here 𝑎 1 = 6 ,    𝑟 = 2 ,    𝐿 = 6144 a 1 ​ =6,r=2,L=6144. Solve for 𝑛 n: 𝑎 𝑛 = 𝑎 1 𝑟 𝑛 − 1 ⇒ 6144 = 6 ( 2 𝑛 − 1 ) . a n ​ =a 1 ​ r n−1 ⇒6144=6(2 n−1 ). 1024 = 2 𝑛 − 1 ⇒ 𝑛 − 1 = 10 ⇒ 𝑛 = 11. 1024=2 n−1 ⇒n−1=10⇒n=11. Substitute into the sum formula: 𝑆 𝑛 = 6 ( 1 − 2 11 ) 1 − 2 = 6 ( − 2047 ) − 1 = 12282. S n ​ = 1−2 6(1−2 11 ) ​ = −1 6(−2047) ​ =12282. Question 5 Confirm that the Integral Test can be applied to the series. Then use it to determine the convergence or divergence of the series: ∑ 𝑛 = 1 ∞ 1 𝑛 + 3 . n=1 ∑ ∞ ​ n+3 1 ​ . Answer 1 Integral Test: If 𝑓 ( 𝑥 ) f(x) is continuous, positive, and decreasing on [ 𝑘 , ∞ ) [k,∞) and 𝑓 ( 𝑛 ) = 𝑎 𝑛 f(n)=a n ​ , then: If ∫ 𝑓 ( 𝑥 )   𝑑 𝑥 ∫f(x)dx converges, ∑ 𝑎 𝑛 ∑a n ​ converges. If ∫ 𝑓 ( 𝑥 )   𝑑 𝑥 ∫f(x)dx diverges, ∑ 𝑎 𝑛 ∑a n ​ diverges. Here, 𝑓 ( 𝑥 ) = 1 𝑥 + 3 f(x)= x+3 1 ​ . Evaluate the integral: ∫ 1 ∞ 1 𝑥 + 3 𝑑 𝑥 = lim ⁡ 𝑏 → ∞ ln ⁡ ∣ 𝑥 + 3 ∣ 1 𝑏 = ln ⁡ ( ∞ ) − ln ⁡ ( 4 ) = ∞ . ∫ 1 ∞ ​ x+3 1 ​ dx= b→∞ lim ​ ln∣x+3∣ 1 b ​ =ln(∞)−ln(4)=∞. Since the integral diverges, the series ∑ 𝑛 = 1 ∞ 1 𝑛 + 3 ∑ n=1 ∞ ​ n+3 1 ​ also diverges.