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# HW8 - ANSWERS TO SELECTED EXERCISES HW8 Exercise 8.7.7...

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ANSWERS TO SELECTED EXERCISES: HW8 Exercise 8.7.7 Suppose f ( x ) = k =0 a k ( x - c ) k has radius of convergence R > 0. For | x - c | < R , set F ( x ) = R x c f ( t ) dt . Prove that F ( x ) = X k =0 a k k + 1 ( x - c ) k +1 , | x - c | < R. Proof : Let ρ with 0 < ρ < R be arbitrary. Then by Theorem 8.7.3c, we know that the series f ( x ) = k =0 a k ( x - c ) k converges uniformly for all x with | x - c | 6 ρ . Applying Corollary 8.4.2 on the interval [ c, x ] (or [ x, c ]) we deduce that f is integrable on this interval and furthermore, we can integrate term-by-term, that is, F ( x ) = Z x c f ( t ) dt = X k =0 Z x c a k ( x - c ) k = X k =0 a k k + 1 ( x - c ) k +1 , for all x with | x - c | 6 ρ . Since ρ was arbitrary with 0 < ρ < R , we conclude that the desired equality holds for all x with | x - c | < R . Exercise 8.7.8 a. Use the previous exercise and the fact that d dx arctan x = 1 1 + x 2 to obtain the Taylor series expansion of arctan x about c = 0. b. Use part (a) to obtain a series expansion for π . Proof : a. The given equality yields arctan x = Z x 0 1 1 + t 2 dt, for all t R , by the fundamental theorem of calculus. The following equality

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