The function f has a Taylor series about x=1 that converges to f(x) for all x in the interval of convergence. It is known that f(1)=1,f^(')(1)=-(1)/(2), and the nth derivative of f at x=1 is given by f^((n))(1)=(-1)^(n)((n-1)!)/(2^(n)) for n = 2. (a) Write the first four nonzero terms and the general term of the Taylor series for f about x=1. (b) The Taylor series for f about x=1 has a radius of convergence of 2 . Find the interval of convergence. Show the work that leads to your answer. (c) The Taylor series for f about x=1 can be used to represent f(1.2) as an alternating series. Use the first three nonzero terms of the alternating series to approximate f(1.2). (d) Show that the approximation found in part (c) is within 0.001 of the exact value of f(1.2).

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The function  f has a Taylor series about  x=1 that converges to  f(x) for all  x in the interval of convergence. It is known that  f(1)=1,f^(')(1)=-(1)/(2), and the  nth derivative of  f at  x=1 is given by  f^((n))(1)=(-1)^(n)((n-1)!)/(2^(n)) for  n >= 2. (a) Write the first four nonzero terms and the general term of the Taylor series for  f about  x=1. (b) The Taylor series for  f about  x=1 has a radius of convergence of 2 . Find the interval of convergence. Show the work that leads to your answer. (c) The Taylor series for  f about  x=1 can be used to represent  f(1.2) as an alternating series. Use the first three nonzero terms of the alternating series to approximate  f(1.2). (d) Show that the approximation found in part (c) is within 0.001 of the exact value of  f(1.2).

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Question Prompt: Question prompt: Determine the interval of convergence for the Taylor series of the function f about x=1, given that the radius of convergence is 2.Consider Radius and Center: To find the interval of convergence, we need to consider the radius of convergence and the center of the Taylor series, which is x=1.Calculate Interval: Since the radius of convergence is 2, the interval of convergence will extend 2 units to the left and right of the center x=1.Calculate Endpoints: Calculating the endpoints of the interval, we get 1 - 2 = -1 on the left and 1 + 2 = 3 on the right.Check Endpoints: Therefore, the interval of convergence is (-1, 3), but we need to check the endpoints to see if they are included in the interval.Test x = -1: To check the endpoints, we need to test the series at x = -1 and x = 3 by substituting these values into the Taylor series and checking for convergence.Simplify nth term for x = -1: For x = -1, we substitute x = -1 into the nth term of the series: f^(n)(1) * ((-1 - 1)^(n)) / n! = (-1)^(n) * ((n-1)!) / (2^n) * (-2^n) / n!.Divergence at x = -1: Simplifying the nth term for x = -1, we get (-1)^(n) * ((n-1)!) / (2^n) * (-2^n) / n! = (-1)^(n) * (-1)^(n) * (n-1)! / n! = (-1)^(2n) * (1/n).Test x = 3: Since (-1)^(2n) is always 1, the series becomes a harmonic series 1/n, which diverges. Therefore, x = -1 is not included in the interval of convergence.Simplify nth term for x = 3: For x = 3, we substitute x = 3 into the nth term of the series: f^(n)(1) * ((3 - 1)^(n)) / n! = (-1)^(n) * ((n-1)!) / (2^n) * (2^n) / n!.Divergence at x = 3: Simplifying the nth term for x = 3, we get (-1)^(n) * ((n-1)!) / (2^n) * (2^n) / n! = (-1)^(n) * (n-1)! / n! = (-1)^(n) * (1/n).Final Interval: The series for x = 3 is the same as for x = -1, which is the harmonic series 1/n, and it also diverges. Therefore, x = 3 is not included in the interval of convergence.The interval of convergence for the Taylor series of f about x=1 is (-1, 3), excluding the endpoints.

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