answer:The correct notation for asymptotic analysis is f(n) in O(g(n)), which means that f(n) is an element of the set of functions that are asymptotically bounded by g(n). This notation is more precise and theoretically correct because it emphasizes that f(n) is a member of a class of functions, rather than being equal to a specific function. On the other hand, the notation f(n) = O(g(n)) is often used in practice because it is more convenient and concise. It is interpreted as "f(n) is asymptotically bounded by g(n)", which is equivalent to f(n) in O(g(n)). However, it is important to note that this notation is not strictly correct from a mathematical standpoint. The main difference between the two notations is that f(n) = O(g(n)) implies that f(n) is equal to a specific function in the class of functions that are asymptotically bounded by g(n), while f(n) in O(g(n)) simply states that f(n) is a member of that class. In general, it is preferable to use the notation f(n) in O(g(n)) when writing mathematical proofs or when discussing asymptotic analysis in a formal setting. However, the notation f(n) = O(g(n)) is acceptable in less formal contexts, such as when communicating ideas or results to a broader audience.

answer:To show this, we start by expressing the probability of X taking the value n as Pr(X=n) = a_n, where the probability generating function G(t) can be written as the infinite series: [ G(t) = a_0 + a_1t + a_2t^2 + a_3t^3 + a_4t^4 + a_5t^5 + cdots ] The probability of X taking on an even value is the sum of the probabilities for even n, which is a_0 + a_2 + a_4 + cdots. This can be expressed as: [ Pr(X text{ is even}) = a_0 + a_2t^2 + a_4t^4 + cdots ] Now, notice that this is the coefficient of the even powers in the expansion of (1+t)^2 G(t). Expanding (1+t)^2 gives 1 + 2t + t^2, so we have: [ (1+t)^2 G(t) = (1 + 2t + t^2)(a_0 + a_1t + a_2t^2 + cdots) ] When we collect the even powers of t from this product, we get: [ Pr(X text{ is even}) = frac{1}{2}(G(1) + G(-1)) ] Finally, since G(1) represents the sum of all the probabilities (which must equal 1 for a valid probability distribution), we have: [ G(1) = a_0 + a_1 + a_2 + a_3 + cdots = 1 ] Thus, the probability that the random variable X takes on an even value is: [ Pr(X text{ is even}) = frac{1}{2}(1 + G(-1)) ]

answer:Acute inflammation is a rapid, short-term response to injury or infection, characterized by redness, swelling, pain, and heat. Chronic inflammation, conversely, persists over a longer duration due to ongoing exposure to irritants, such as in autoimmune diseases or long-term infections. Both types involve the immune system's activation in response to harmful stimuli. They share the feature of being triggered by injury, infection, or irritants, but the key difference lies in their duration and the consequences they have on the body. Acute inflammation typically resolves on its own, while chronic inflammation can lead to tissue damage and contribute to various health issues.

answer:For a group of order 65=5times 13, the Sylow theorems imply that n_5=n_{13}=1, so both Sylow subgroups are normal. By a standard result, the group is isomorphic to the direct product of these subgroups, so Gcong mathbb{Z_5}timesmathbb{Z_{13}}cong mathbb{Z_{65}}. Thus, there is only one group of order 65 up to isomorphism. For a group of order 105=3times 5times 7, the Sylow theorems imply that n_7=1 and n_3=n_5=7. Since the 7-Sylow subgroup is normal, we have Gcong Htimes mathbb{Z_7} where H is a group of order 15. By the Sylow theorems, H has a normal 5-Sylow subgroup and a normal 3-Sylow subgroup, so Hcong mathbb{Z_3}timesmathbb{Z_5}cong mathbb{Z_{15}}. Therefore, Gcong mathbb{Z_{15}}times mathbb{Z_7}cong mathbb{Z_{105}}. Thus, there is also only one group of order 105 up to isomorphism.