8.6_Boerner_MAT_266_ONLINE_B_Spring_2020.lmwhitne.Section_8.6.pdf

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Larena Whitney-krug Boerner MAT 266 ONLINE B Spring 2020 Assignment Section 8.6 due 04/19/2020 at 11:59pm MST 1. (1 point) Consider the function 1 1 - x 4 Write a partial sum for the power series which represents this function consisting of the first 5 nonzero terms. For example, if the series were n = 0 3 n x 2 n , you would write 1 + 3 x 2 + 3 2 x 4 + 3 3 x 6 + 3 4 x 8 . Also indicate the radius of convergence. Partial Sum: Radius of Convergence: Solution: SOLUTION We know that for | x | < 1, the power series representation of 1 1 - x is 1 1 - x = n = 0 x n , | x | < 1 = 1 + x + x 2 + x 3 + x 4 + x 5 + ··· , | x | < 1 So the power series of 1 1 - x 4 for x 4 < 1 is 1 1 - x 4 = n = 0 ( x 4 ) n = n = 0 x 4 n , x 4 < 1 = 1 + x 4 + x 8 + x 12 + x 16 + x 20 + ··· , x 4 < 1 And so we see that the first five nonzero terms in the partial sum of the power series representation of 1 1 - x 4 for x 4 < 1 are 1 + x 4 + x 8 + x 12 + x 16 As noted above, the power series 1 1 - x 4 = n = 0 x 4 n converges for x 4 < 1 = ⇒ | x | 4 < 1 = ⇒ | x | < 1. This is equivalent to - 1 < x < 1, which is an interval centered at 0, with radius R = 1. Answer(s) submitted: 1+xˆ4+xˆ(2*4)+xˆ(3*4)+xˆ(4*4) 1 (correct) Correct Answers: 1 + x**4 + x**(2*4) + x**(3*4) + x**(4*4) 1 2. (1 point) Find the first five non-zero terms of power series representation centered at x = 0 for the function below. f ( x ) = x 1 + 5 x Answer: f ( x ) = + + + + + ··· What is the interval of convergence? Answer (in interval notation): Solution: SOLUTION We know that for | x | < 1, the power series representation of 1 1 - x is 1 1 - x = n = 0 x n , | x | < 1 = 1 + x + x 2 + x 3 + x 4 + ··· , | x | < 1 And so the power series of ( x ) 1 + 5 x for |- 5 x | < 1 is ( x ) 1 + 5 x = ( x ) 1 1 - ( - 5 x ) = ( x ) n = 0 ( - 5 x ) n = ( x ) 1 +( - 5 x )+( - 5 x ) 2 +( - 5 x ) 3 +( - 5 x ) 4 + ··· = ( x ) ( 1 - 5 x + 5 2 x 2 - 5 3 x 3 + 5 4 x 4 - + ··· ) = x 1 - 5 x 2 + 25 x 3 - 125 x 4 + 625 x 5 - + ··· With the first five nonzero terms of the power series listed above. This series converges absolutely for |- 5 x | < 1 = 5 | x | < 1 = ⇒ | x | < 1 5 This is equivalent to - 1 5 < x < 1 5 , or in other words, convergence occurs for x in the interval ( - 1 5 , 1 5 ) . Answer(s) submitted: x -5xˆ2 25xˆ3 -125xˆ4 625xˆ5 (-1/5,1/5) (correct) Correct Answers: x -5*xˆ2 25*xˆ3 -125*xˆ4 625*xˆ5 1
(-0.2,0.2) 3. (1 point) The function f ( x ) = 4 1 + 36 x 2 is represented as a power series f ( x ) = n = 0 c n x n . Find the first few coefficients in the power series. c 0 = c 1 = c 2 = c 3 = c 4 = Find the radius of convergence R of the series. R = . Solution: SOLUTION We know that for | x | < 1, the power series representation of 1 1 - x is 1 1 - x = n = 0 x n , | x | < 1 = 1 + x + x 2 + x 3 + x 4 + ··· , | x | < 1 And so the power series of 4 1 + 36 x 2 for ( - 36 x 2 ) < 1 is 4 1 + 36 x 2 = 4 1 1 - ( - 36 x 2 ) = 4 n = 0 ( - 36 x 2 ) n = 4 1 + ( - 36 x 2 ) + ( - 36 x 2 ) 2 + ( - 36 x 2 ) 3 + ··· = 4 1 - 36 x 2 + 36 2 x 4 - 36 3 x 6 + -··· = 4 - 144 x 2 + 4 ( 36 2 ) x 4 - 4 ( 36 3 ) x 6 + -··· Equating this to f ( x ) = n = 0 c n x n = c 0 + c 1 x + c 2 x 2 + c 3 x 3 + c 4 x 4 + ··· , we have c 0 = 4, c 1 = 0, c 2 = - 144, c 3 = 0, and c 4 = 4 ( 36 2 ) = 5184.

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