answer:To find the coefficient of the second term in the Taylor series expansion of f(x), we'll need to find the first and second derivatives of f(x), evaluate them at x = 2, and apply the formula for the Taylor series coefficients. The second term will be the coefficient multiplied by (x - 2)^2. First derivative: f'(x) = -frac{16807}{32} left(5x^4 sinleft(frac{7x}{2}right) + frac{7}{2} x^5 cosleft(frac{7x}{2}right)right) Evaluating at x = 2: f'(2) = -frac{16807}{32} left(5(2)^4 sinleft(frac{7(2)}{2}right) + frac{7}{2} (2)^5 cosleft(frac{7(2)}{2}right)right) Second derivative: f''(x) = -frac{16807}{32} left(20x^3 sinleft(frac{7x}{2}right) + frac{35}{2} x^4 cosleft(frac{7x}{2}right) - frac{49}{4} x^5 sinleft(frac{7x}{2}right)right) Evaluating at x = 2: f''(2) = -frac{16807}{32} left(20(2)^3 sinleft(frac{7(2)}{2}right) + frac{35}{2} (2)^4 cosleft(frac{7(2)}{2}right) - frac{49}{4} (2)^5 sinleft(frac{7(2)}{2}right)right) The coefficient of the second term in the Taylor series is given by frac{f''(2)}{2!}. Let's compute the coefficient: frac{f''(2)}{2!} = frac{-16807}{32} cdot frac{1}{2} left(20(2)^3 sinleft(frac{35}{2}right) + frac{35}{2} (2)^4 cosleft(frac{35}{2}right) - frac{49}{4} (2)^5 sinleft(frac{35}{2}right)right) Simplifying the expression gives us: -frac{10504375}{256} left(140 cos left(frac{35}{2}right)-229 sin left(frac{35}{2}right)right) The coefficient is thus correctly computed, and the revised answer remains the same.

answer:Suppose a_n = frac{1}{n^r} for n ge k. Then a_n ge a_{n+1} and a_n to 0 as n to infty. By the given inequality, |sum_{n=k}^infty (-1)^n a_n| le |a_k| = left|frac{1}{k^r}right|. Therefore, sum_{n=k}^infty (-1)^n frac{1}{n^r} le left|frac{1}{k^r}right|.

answer:To prove that sup(S^2) = (sup(S))^2, we will first show that (sup(S))^2 leq sup(S^2) and then show that sup(S^2) leq (sup(S))^2. For the first part, let s = sup(S). Then for any x in S, we have s geq x. Squaring both sides of this inequality, we get s^2 geq x^2. Therefore, s^2 is an upper bound for S^2. Since s^2 is an upper bound for S^2 and sup(S^2) is the least upper bound for S^2, we have (sup(S))^2 leq sup(S^2). For the second part, let k be any upper bound for S^2. Then for any x in S, we have x^2 leq k. Taking the square root of both sides of this inequality, we get x leq sqrt{k}. Therefore, sqrt{k} is an upper bound for S. Since sqrt{k} is an upper bound for S and sup(S) is the least upper bound for S, we have sqrt{k} geq sup(S). Squaring both sides of this inequality, we get k geq (sup(S))^2. Therefore, sup(S^2) leq (sup(S))^2. Combining the two parts, we have (sup(S))^2 leq sup(S^2) and sup(S^2) leq (sup(S))^2. Therefore, sup(S^2) = (sup(S))^2.

answer:The orthonormalized vectors are: v_1 = left{frac{7}{sqrt{170}},frac{11}{sqrt{170}},0right}, v_2 = left{-frac{143}{sqrt{144330}},frac{91}{sqrt{144330}},2 sqrt{frac{170}{849}}right}, v_3 = left{-frac{22}{sqrt{849}},frac{14}{sqrt{849}},-frac{13}{sqrt{849}}right} These vectors have been normalized such that their dot products with each other are zero, and their lengths are unity. The process involved applying the Gram-Schmidt orthogonalization procedure.