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Please send the answers within Half hour. Thanks.

Please Send The Answers Within Half Hour Thanks Follow The Steps 1 (19.54 KiB) Viewed 26 times

Follow the Steps:

Please Send The Answers Within Half Hour Thanks Follow The Steps 2 (67.36 KiB) Viewed 26 times

Q3 (7 points) (20-²(x-2)x) Let f(x) = cos(x) + 8(x - 1) sin(x) 276 given that: 3(x-2) 3(x-2)² (x-2)³ (1) P₂ (T) = 10 76 is the cubic 274 672 Taylor polynomial of f(x) at x = 2. (2) The Taylor series of f(x) at x = 2 is NOT alternating. (3) ƒ(¹) (x) = (2x − x²) cos(TT) Find the maximum possible error in using P3(z) to approximate f(x) in the interval 0 ≤ ≤ 2. You are
4. Find the Taylor series about x = -3 for the function 2 f(x) = (²+3) ². x+5 Write your final answer in sigma notation, simplified as much as possible. You must use a method that guarantees that the series converges to f(x). What is the interval of convergence of the series? Hint: Do not divide x +3 by x + 5. x+3 We begin by writing the function in the form f(x)= (x+3)² (x + 5)² Now, consider 1 n = 2+6 (2+3)+2-1+4= (-²+³)". Σ(13) Σ = x+5 valid for - (x+3) 2 < 1 x + 3 < 2 -2<x+3 < 2 -5 < x < -1. With a positive radius of convergence, we can differentiate the series ∞ -1 = (-1)"n 2n+1 (x + 5)² (x+3)-¹. n=0 Multiplication by -(x+3)2 now yields 80 ∞ (-1) n+¹n f(x) = (x+3)² (x + 5)² (-1)(n-1), Σ ²(x+3) ²+¹ = [₁ 2n+1 -(x + 3)". 2n n=0 n=2 Since differentiation never picks up an end point of the open interval interval of convergence, we can say that the interval of convergence is -5 < x < -1. = (-1) ¹ 2n+1(x+3)",