4 3 x 1 x 4 summationdisplay k 1 1 k x k correct Explanation Using the hint we

# 4 3 x 1 x 4 summationdisplay k 1 1 k x k correct

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Unformatted text preview: 4 + 3 x 1 + x = 4 + ∞ summationdisplay k =1 (- 1) k x k correct Explanation: Using the hint we get 4 + 3 x 1 + x = 4 1 + x + 3 x 1 + x , and 1 1 + x = 1- x + x 2 + . . . = ∞ summationdisplay k = 0 (- 1) k x k . Thus 4 + 3 x 1 + x = 4 ∞ summationdisplay k =0 (- 1) k x k + 3 x ∞ summationdisplay k =0 (- 1) k x k . But 4 ∞ summationdisplay k =0 (- 1) k x k = ∞ summationdisplay k =0 (- 1) k 4 x k , while 3 x ∞ summationdisplay k = 0 (- 1) k x k = ∞ summationdisplay k = 0 (- 1) k 3 x k +1 . To combine the infinite sums we need to ex- press the last one as a sum of powers of x k : ∞ summationdisplay k =0 (- 1) k 3 x k +1 = 3 x- 3 x 2 + 3 x 3- . . . =- ∞ summationdisplay k = 1 (- 1) k 3 x k . Since the last sum now goes from k = 1 to k = ∞ , we next write: ∞ summationdisplay k = 0 (- 1) k 4 x k = 4 + ∞ summationdisplay k =1 (- 1) k 4 x k , for then we can add the two series: ∞ summationdisplay k =0 (- 1) k 4 x k + ∞ summationdisplay k =0 (- 1) k 3 x k +1 = 4 + ∞ summationdisplay k =1 (- 1) k 4 x k + parenleftBig- ∞ summationdisplay k = 1 (- 1) k 3 x k parenrightBig . Consequently, 4 + 3 x 1 + x = 4 + ∞ summationdisplay k =1 (- 1) k x k . 017 10.0 points Find a power series representation for the function f ( x ) = x 4 tan- 1 ( x ) on (- 1 , 1). 1. f ( x ) = ∞ summationdisplay n =0 (- 1) n n + 1 x n +5 2. f ( x ) = ∞ summationdisplay n =0 1 n + 1 x n +5 3. f ( x ) = ∞ summationdisplay n =0 (- 1) n 2 n + 1 x 2 n +5 correct 4. f ( x ) = ∞ summationdisplay n =0 1 2 n + 1 x 2 n +5 5. f ( x ) = ∞ summationdisplay n =0 (- 1) n ( n + 1)! x n +5 6. f ( x ) = ∞ summationdisplay n =0 (- 1) n (2 n + 1)! x 2 n +5 Explanation: cadena (jc59484) – HW14 – lawn – (55930) 11 The interval of convergence of the geomet- ric series 1 1- x = 1 + x + x 2 + . . . is (- 1 , 1). Thus on (- 1 , 1) 1 1 + x 2 = 1- x 2 + x 4- . . . = ∞ summationdisplay n =0 (- 1) n x 2 n . On the other hand, tan- 1 ( x ) = integraldisplay x 1 1 + t 2 dt . Thus on (- 1 , 1) tan- 1 ( x ) = integraldisplay x ∞ summationdisplay n =0 (- 1) n t 2 n dt = ∞ summationdisplay n = 0 parenleftbiggintegraldisplay x (- 1) n t 2 n dt parenrightbigg = ∞ summationdisplay n =0 (- 1) n 2 n + 1 x 2 n +1 . Consequently, on (- 1 , 1) f ( x ) = ∞ summationdisplay n =0 (- 1) n 2 n + 1 x 2 n +5 ....
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