3.4.M M M8 M85. Σ6.n=11. I =2. I =3. I =(-1)n+122n32nn!(-1)n+122n-132n(2η)!5. I =(−1)n+122n-132nn!(-1)n+122n32nn!(part 2 of 4)Use the Taylor series for ethe integralk = 0I=nΣk = 0Σk = 0004. I = Σk = 0- Σk = 0an(−1)kk!= 12erdr1k!(2k + 1)1(-1) kk!(2k + 1)2n+1χηx²n+1-2.32k-2.32kk!dx.(-1) k-2.32k+12k + 1to evaluate2.32k+1-2.3²k+1(part 1 of 4)Find the Maclaurin series for the functionf(x) sin23x .1. ΣM8 M82.n=1=(-1)n+122n-132n(2η)!(−1)n+122n32n(2n)!η 2η22η

Answered: Image /qna-images/answer/404fd00c-85d6-4a0c-979f-e81f9a8115b8.jpg

3. 4. M M M8 M8 5. Σ 6. n=1 1. I = 2. I = 3. I = (-1)n+122n32n n! 5. I = (-1)n+122n-132n (2η)! (-1)n+122n-132n n! (-1)n+122n32n n! (part 2 of 4) Use the Taylor series for el the integral I= k = 0 n Σ k = 0 Σ k = 0 00 4. I = Σ k = 0 k = 0 χη = 12erdr dx. 1 k!(2k + 1) (-1) k -2.32k k! (-1) k 2k + 1 (-1) k k!(2k + 1) 1 k! 2n+1 χη x²n+1 -2.32k to evaluate 2.32k+1 -2.32k+1 -2.3²k+1 (part 1 of 4) Find the Maclaurin series for the function sin23x . 1. Σ M8 M8 2. n=1 f(x) = (-1)n+122n-132n (2η)! (−1)n+122n32n (2η)! η 2η 22η

3. 4. M M M8 M8 5. Σ 6. n=1 1. I = 2. I = 3. I = (-1)n+122n32n n! 5. I = (-1)n+122n-132n (2η)! (-1)n+122n-132n n! (-1)n+122n32n n! (part 2 of 4) Use the Taylor series for el the integral I= k = 0 n Σ k = 0 Σ k = 0 00 4. I = Σ k = 0 k = 0 χη = 12erdr dx. 1 k!(2k + 1) (-1) k -2.32k k! (-1) k 2k + 1 (-1) k k!(2k + 1) 1 k! 2n+1 χη x²n+1 -2.32k to evaluate 2.32k+1 -2.32k+1 -2.3²k+1 (part 1 of 4) Find the Maclaurin series for the function sin23x . 1. Σ M8 M8 2. n=1 f(x) = (-1)n+122n-132n (2η)! (−1)n+122n32n (2η)! η 2η 22η

Algebra & Trigonometry with Analytic Geometry

13th Edition

ISBN:9781133382119

Author:Swokowski

Publisher:Swokowski

Chapter10: Sequences, Series, And Probability

Section10.3: Geometric Sequences

Problem 49E

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