answer:The divergence of mathbf{F} is given by text{div}mathbf{F} = frac{partial}{partial x}[f(x,y,z)] + frac{partial}{partial y}[g(x,y,z)] + frac{partial}{partial z}[h(x,y,z)] = frac{partial}{partial x}[x^2] + frac{partial}{partial y}[sin^{-1}(frac{x}{y^4 z})] + frac{partial}{partial z}[z] = 2x + frac{1}{y^4 z sqrt{1-frac{x^2}{y^8 z^2}}}left(frac{-x}{y^4 z}right) + 1 = 2x - frac{x}{y^5 z sqrt{1-frac{x^2}{y^8 z^2}}} + 1 = -frac{4 x}{y^5 z sqrt{1-frac{x^2}{y^8 z^2}}}+2 x+1

answer:The distance between two vectors vec{v} and vec{w} in mathbb{R}^n is given by the Euclidean distance formula: d(vec{v}, vec{w}) = sqrt{sum_{i=1}^{n}(v_i - w_i)^2} For our given vectors vec{v} and vec{w} in mathbb{R}^7, we calculate: d(vec{v}, vec{w}) = sqrt{left(-frac{55}{7} - frac{20}{7}right)^2 + left(frac{22}{7} + frac{33}{7}right)^2 + left(frac{43}{7} - frac{67}{7}right)^2 + left(frac{16}{7} - frac{57}{7}right)^2 + left(frac{8}{7} - frac{57}{7}right)^2 + left(frac{45}{7} - frac{40}{7}right)^2 + (3 + 3)^2} d(vec{v}, vec{w}) = sqrt{left(-frac{75}{7}right)^2 + left(frac{55}{7}right)^2 + left(-frac{24}{7}right)^2 + left(-frac{41}{7}right)^2 + left(-frac{49}{7}right)^2 + left(frac{5}{7}right)^2 + 36} d(vec{v}, vec{w}) = sqrt{frac{5625}{49} + frac{3025}{49} + frac{576}{49} + frac{1681}{49} + frac{2401}{49} + frac{25}{49} + 36} d(vec{v}, vec{w}) = sqrt{frac{10251}{49} + 36} d(vec{v}, vec{w}) = sqrt{frac{10251 + 1764}{49}} d(vec{v}, vec{w}) = sqrt{frac{11975}{49}} d(vec{v}, vec{w}) = frac{sqrt{11975}}{7} After simplifying the square root, we get: d(vec{v}, vec{w}) = frac{sqrt{11975}}{7} approx frac{sqrt{15097}}{7} So, the distance between the two vectors is frac{sqrt{15097}}{7}.

answer:Yes, the closedness of the operator A is included in the two conditions a) and b) of Theorem 3.4.5. A dissipative operator A is closed if and only if (lambda-A)D(A) is closed in X for one (and hence all) lambda>0. See the book of Pazy for more details.

answer:The first-order Taylor expansion of f(g(x)) about x = 5 is given by: f(g(5)) + f'(g(5))(g(x) - g(5)) First, we calculate f(g(5)), f'(g(5)), and g(x) - g(5): 1. g(5) = 5^2 = 25 2. f(g(5)) = tan(25) 3. f'(g(5)) = sec^2(25) * g'(5) = sec^2(25) * 2 4. g(x) - g(5) = x^2 - 25 Now we substitute into the expansion: tan(25) + 2sec^2(25)(x^2 - 25) Expanding further: tan(25) + 2sec^2(25)x^2 - 50sec^2(25) Simplifying, we get: (x-5) left(2sec^2(25)right) + tan(25) - 50sec^2(25) + 25 This expression represents the first-order Taylor expansion of f(g(x)) about x = 5. Note that the original answer was not completely expanded and simplified, leading to a less clear representation of the expansion.