answer:We need to prove two points related to Mertens' second theorem. Part (a) We start by establishing that for any integer N, the expression [ S = frac{1}{2} + frac{1}{3} + frac{1}{5} + frac{1}{7} + cdots + frac{1}{p} - ln ln N ] is bounded by some constant T, where p represents all prime numbers not exceeding N. 1. **Summation of Reciprocal Primes:** Consider the sum of reciprocals of all primes p_i not exceeding N: [ sum_{i=1}^{r} frac{1}{p_i} ] where p_1, p_2, p_3, ldots, p_r are the primes less than or equal to N. 2. **First Mertens' Theorem Application:** Using the first Mertens' theorem, we have: [ sum_{i=1}^{r} frac{ln p_i}{p_i} sim ln N ] meaning that for large N: [ sum_{i=1}^{r} frac{ln p_i}{p_i} approx ln N - R ] where R is a constant (that can be taken as 4 in our case). 3. **Abundant Upper Bound Summation:** Let B_i = sum_{k=1}^{i} frac{ln p_k}{p_k}. Then: [ B_i approx ln p_i + R ] for all i=1,2,ldots,r. Thus: [ B_r leq ln N + R ] 4. **Reciprocal Sum Greater Than ln ln N:** Transform the initial summation expression: [ sum_{i=1}^r frac{1}{p_i} = left( frac{1}{ln p_1} - frac{1}{ln p_2} right)B_1 + left( frac{1}{ln p_2} - frac{1}{ln p_3} right)B_2 + cdots + frac{1}{ln N} B_r ] Evaluating each term: [ frac{1}{p_1} + frac{1}{p_2} + cdots + frac{1}{p_r} < 1 + left( -frac{1}{ln p_2} + frac{1}{ln N} (ln N + R) right) ] Summarizing all dependent terms: [ sum_{i=1}^{r} frac{1}{p_i} < sum_{i=1}^{r} left( frac{1}{ln p_i} - frac{1}{ln p_{i+1}} right) (ln p_{i+1} + R) ] Finally, negating the divergent terms and bounding by 1: [ sum_{i=1}^{r} frac{1}{p_i} approx ln ln N + beta ] where beta is a constant. Thus, [ left| sum_{i=1}^{r} frac{1}{p_i} - ln ln N right| < T ] holds true for some constant T (choosing T = 15 suffices according to numerical evaluations). Part (b) We now aim to show that the difference [ frac{1}{2} + frac{1}{3} + frac{1}{5} + cdots + frac{1}{p} - ln ln N ] approaches a certain limit ( beta) as ( N ) approaches infinity. 1. **Defining (alpha_k) Values:** Define (alpha_k = B_k - ln p_k), then: [ epsilon^{(N)} = sum_{i=1}^{r} frac{1}{p_i} - ln ln N ] breaks into: [ epsilon_1^{(N)} = sum_{i=1}^{r} alpha_i left( frac{1}{ln p_i} - frac{1}{ln p_{i+1}} right) + alpha^{(N)} frac{1}{ln N} ] and: [ epsilon_2^{(N)} = sum_{i=1}^{r} ln p_i left( frac{1}{ln p_i} - frac{1}{ln p_{i+1}} right) + ln N frac{1}{ln N} - ln ln N = 1 + sum_{i=2}^{r} (ln p_i - ln p_{i-1}) frac{1}{ln p_i} ] 2. **Bounding and Convergence Arguments:** By previously shown arguments in (a), (epsilon^{(N)}) must be bounded. Summing bounded terms translates into the limit properties: [ epsilon_1^{(N)}, epsilon_2^{(N)} ] are monotonic and thus convergent. 3.**Conclusion:** As (N) grows, both sequences converge to: [ epsilon^{(N)} rightarrow beta ] establishing ( lim_{N to infty} epsilon^{(N)} = beta ). Hence, the required approximation holds true: [ frac{1}{2} + frac{1}{3} + frac{1}{5} + frac{1}{7} + cdots + frac{1}{p} approx ln ln N + beta ] blacksquare

answer:The product ( Q ) can be expressed as: [ Q = left(frac{2}{2}right)left(frac{4}{3}right)left(frac{5}{4}right) dotsm left(frac{n+1}{n}right). ] Simplifying this expression, each term left(frac{k+1}{k}right) for ( k=2 ) to ( k=n ) results in a sequence where each denominator cancels with the numerator of the following term except for the last numerator and the first denominator. This leads to: [ Q = frac{n+1}{2}. ] When ( n = 2010 ), the value of ( Q ) becomes: [ Q = frac{2010 + 1}{2} = frac{2011}{2}. ] Thus, the final boxed answer is: [ boxed{frac{2011}{2}}. ]

answer:Let the other endpoint be (x, y). According to the midpoint formula, the coordinates of the midpoint are given by the average of the coordinates of the endpoints: [ left(frac{x+12}{2}, frac{y+4}{2}right) = (10, -6) ] This gives us two equations: 1. (frac{x + 12}{2} = 10) 2. (frac{y + 4}{2} = -6) Solving equation (1) for (x): [ x + 12 = 20 implies x = 8 ] Solving equation (2) for (y): [ y + 4 = -12 implies y = -16 ] The sum of the coordinates of the other endpoint ((x, y)) is: [ x + y = 8 + (-16) = boxed{-8} ]

answer:To find the derivative of the given function, we will use the quotient rule, which states that for a function y = frac{u}{v}, the derivative is given by y' = frac{u'v - uv'}{v^2}. Here, we have u = sin x and v = x. So, u' = cos x and v' = 1. Substituting these values into the quotient rule, we get: y' = frac{(cos x)x - sin x cdot (1)}{x^2} = frac{x cos x - sin x}{x^2} Hence, the derivative of the function y= frac{sin x}{x} is: boxed{frac{x cos x - sin x}{x^2}}