🧠:<think>Okay, so I have these two problems to solve about vectors and matrices related to a book series. Let me try to break them down step by step.Starting with the first problem: Each novel is represented by a vector ( mathbf{v}_i ) in ( mathbb{R}^n ). The reader thinks the themes follow a circular pattern, forming a closed loop. I need to prove that the sum of all these vectors is the zero vector.Hmm, a closed loop in vector space. I remember that in geometry, if you have vectors forming a closed polygon, the sum of those vectors is zero because you end up where you started. So, if you imagine walking along each vector ( mathbf{v}_1, mathbf{v}_2, ldots, mathbf{v}_k ) in sequence, you'd end up back at the starting point. That means the resultant vector is zero.In mathematical terms, if you have vectors ( mathbf{v}_1, mathbf{v}_2, ldots, mathbf{v}_k ) arranged head-to-tail, the resultant vector is ( mathbf{v}_1 + mathbf{v}_2 + cdots + mathbf{v}_k ). Since it's a closed loop, the starting and ending points are the same, so the resultant must be zero. That makes sense.So, to formalize this, if the vectors form a closed loop, then the sum must be zero. Therefore, ( mathbf{v}_1 + mathbf{v}_2 + cdots + mathbf{v}_k = mathbf{0} ). I think that's the proof.Moving on to the second problem: The reader wants to analyze thematic diversity by calculating the determinant of a matrix ( A ), where each row is a vector of a novel's themes. The matrix ( A ) is ( k times n ) with ( k < n ). I need to prove that the determinant is zero and interpret this.Wait, determinant is only defined for square matrices. But ( A ) is ( k times n ), which is not square unless ( k = n ). But here ( k < n ), so ( A ) is a rectangular matrix. Hmm, maybe the determinant is being considered for a square matrix somehow?Perhaps the reader is considering the matrix ( A^T A ), which would be ( n times n ), since ( A ) is ( k times n ). The determinant of ( A^T A ) would be zero if the columns of ( A ) are linearly dependent. Since ( k < n ), the rank of ( A ) is at most ( k ), which is less than ( n ). Therefore, ( A^T A ) is rank-deficient and its determinant is zero.Alternatively, if we think about the determinant of ( A ) itself, but since ( A ) isn't square, it doesn't have a determinant. Maybe the question is referring to the determinant of a square matrix formed by these vectors? Or perhaps it's a misinterpretation.Wait, the problem says "the determinant of a matrix ( A )", but ( A ) is ( k times n ). Maybe the determinant is zero because the matrix doesn't have full rank? But determinant is only for square matrices. Hmm, perhaps the question is implying that if you take the determinant of any ( k times k ) minor of ( A ), but that might not necessarily be zero.Wait, no. If ( k < n ), then any ( k times k ) minor could potentially be non-zero, but the determinant of the entire matrix isn't defined. Maybe the question is referring to the determinant of ( A A^T ), which would be ( k times k ). Since ( A ) has rank at most ( k ), ( A A^T ) is a square matrix of rank at most ( k ). But since ( k < n ), does that imply the determinant is zero? Wait, no, because ( A A^T ) is ( k times k ), and if ( A ) has full rank ( k ), then ( A A^T ) is invertible and determinant is non-zero. Hmm, that doesn't fit.Wait, maybe I'm overcomplicating. The problem says "the determinant of a matrix ( A )", but ( A ) is ( k times n ). Maybe the question is incorrectly assuming ( A ) is square? Or perhaps it's referring to the determinant of a square submatrix.Wait, another thought: If ( A ) is ( k times n ) with ( k < n ), then the rows are in ( mathbb{R}^n ). If we consider the determinant of a matrix formed by these rows, but since it's not square, the determinant isn't defined. Maybe the problem is referring to the determinant of a matrix where ( A ) is considered as a linear transformation, but again, determinant is for square matrices.Wait, maybe the question is actually about the rank of the matrix. If ( k < n ), then the rank of ( A ) is at most ( k ), which is less than ( n ). So, the matrix ( A ) does not have full column rank, meaning its columns are linearly dependent. But determinant is about square matrices.Wait, perhaps the question is referring to the determinant of a matrix formed by selecting ( n ) rows from ( A ), but since ( k < n ), that's impossible. Hmm, this is confusing.Wait, maybe the problem is incorrectly stated. If ( A ) is ( k times n ) with ( k < n ), then ( A ) cannot have a determinant because it's not square. So, perhaps the intended meaning is that the determinant of any ( n times n ) matrix formed by these vectors is zero because there are only ( k < n ) vectors, hence they are linearly dependent. So, if you try to form an ( n times n ) matrix with these vectors as rows or columns, since there are fewer than ( n ) vectors, the determinant would be zero.Wait, but if you have ( k < n ) vectors in ( mathbb{R}^n ), they can't span the space, but they can still be linearly independent if ( k leq n ). So, if ( k < n ), the set of vectors can be linearly independent, but they don't form a basis. However, if you try to form a square matrix with these vectors, say as rows, you would have an ( n times n ) matrix with only ( k < n ) non-zero rows (assuming each vector is a row), which would make the determinant zero because there are linearly dependent rows.Alternatively, if you have ( k < n ) vectors, and you try to make an ( n times n ) matrix by repeating them or something, but that might not be the case.Wait, perhaps the question is referring to the determinant of ( A ) when ( A ) is considered as a linear transformation from ( mathbb{R}^n ) to ( mathbb{R}^k ). But determinant is only defined for linear transformations from a space to itself, so ( mathbb{R}^n ) to ( mathbb{R}^n ). So, if ( A ) is ( k times n ), it's a linear transformation from ( mathbb{R}^n ) to ( mathbb{R}^k ), and determinant isn't defined here.Wait, maybe the problem is referring to the determinant of ( A^T A ), which is ( n times n ). Since ( A ) has rank at most ( k ), ( A^T A ) also has rank at most ( k ). If ( k < n ), then ( A^T A ) is rank-deficient, so its determinant is zero. That makes sense.So, perhaps the determinant of ( A^T A ) is zero because the rank is less than ( n ). Therefore, the determinant is zero, indicating that the columns of ( A ) are linearly dependent, meaning there's redundancy in the thematic elements across the novels.So, putting it together: Since ( A ) is ( k times n ) with ( k < n ), the matrix ( A^T A ) is ( n times n ) with rank at most ( k ). Therefore, the determinant of ( A^T A ) is zero. This implies that the columns of ( A ) are linearly dependent, meaning the themes are not all unique and there's redundancy among the novels.Alternatively, if we think about the determinant of a square matrix formed by these vectors, since we have fewer vectors than the dimension, any square matrix formed by them will have linearly dependent columns, hence determinant zero.I think that's the way to go. So, the determinant is zero because the matrix doesn't have full rank, indicating linear dependence among the theme vectors, which means some themes are redundant across the novels.Final Answer1. The sum of the vectors is the zero vector: boxed{mathbf{0}}.2. The determinant of the matrix ( A ) is zero, indicating thematic redundancy: boxed{0}.</think>

🧠:<think>Okay, so I have this problem about calculating the Analytical Depth Index (ADI) for a legal article. The ADI is given by an integral involving two functions, f(x, D) and g(x, L). Let me try to break this down step by step.First, the ADI formula is:[ text{ADI}(R, D, L) = int_1^R left( f(x, D) cdot g(x, L) right) , dx ]where ( f(x, D) = x^{frac{1}{D}} ) and ( g(x, L) = e^{-frac{x}{L}} ).So, for part 1, I need to compute this integral when R = 15, D = 3, and L = 10. That means I'll substitute these values into the functions f and g and then integrate from 1 to 15.Let me write out the specific functions with the given parameters:( f(x, 3) = x^{frac{1}{3}} ) because D = 3.( g(x, 10) = e^{-frac{x}{10}} ) because L = 10.So, the integrand becomes:( x^{frac{1}{3}} cdot e^{-frac{x}{10}} )Therefore, the ADI is:[ text{ADI} = int_1^{15} x^{frac{1}{3}} e^{-frac{x}{10}} , dx ]Hmm, integrating this function might not be straightforward. Let me think about how to approach this. It seems like an integral that might not have an elementary antiderivative, so I might need to use integration techniques like substitution or perhaps look for a special function that can represent it.Wait, let me recall that integrals of the form ( int x^k e^{-ax} dx ) can be expressed in terms of the incomplete gamma function. The incomplete gamma function is defined as:[ Gamma(s, x) = int_x^infty t^{s-1} e^{-t} dt ]But in our case, the integral is from 1 to 15, and the integrand is ( x^{1/3} e^{-x/10} ). So, perhaps I can manipulate this into a form that resembles the incomplete gamma function.Let me make a substitution to adjust the limits and the exponent. Let me set ( t = frac{x}{10} ), so that ( x = 10t ) and ( dx = 10 dt ). Then, when x = 1, t = 1/10, and when x = 15, t = 15/10 = 1.5.Substituting into the integral:[ int_{1/10}^{1.5} (10t)^{1/3} e^{-t} cdot 10 dt ]Simplify the terms:First, ( (10t)^{1/3} = 10^{1/3} t^{1/3} ).Then, the integral becomes:[ 10^{1/3} cdot 10 int_{1/10}^{1.5} t^{1/3} e^{-t} dt ]Which simplifies to:[ 10^{4/3} int_{1/10}^{1.5} t^{1/3} e^{-t} dt ]Now, the integral ( int t^{1/3} e^{-t} dt ) is similar to the lower incomplete gamma function, which is:[ gamma(s, x) = int_0^x t^{s-1} e^{-t} dt ]In our case, s - 1 = 1/3, so s = 4/3. Therefore, the integral can be expressed as:[ gammaleft(frac{4}{3}, 1.5right) - gammaleft(frac{4}{3}, frac{1}{10}right) ]So, putting it all together, the ADI is:[ 10^{4/3} left[ gammaleft(frac{4}{3}, 1.5right) - gammaleft(frac{4}{3}, frac{1}{10}right) right] ]Hmm, but I don't have the exact values for the incomplete gamma functions at these points. Maybe I can approximate them numerically. Alternatively, I can use the relationship between the incomplete gamma function and the gamma function for certain values.Wait, another thought: perhaps I can express this in terms of the gamma function and the exponential integral. Alternatively, maybe I can use a series expansion for the exponential function and integrate term by term.Let me try expanding ( e^{-t} ) as a power series:[ e^{-t} = sum_{n=0}^{infty} frac{(-1)^n t^n}{n!} ]Then, the integral becomes:[ int t^{1/3} sum_{n=0}^{infty} frac{(-1)^n t^n}{n!} dt = sum_{n=0}^{infty} frac{(-1)^n}{n!} int t^{n + 1/3} dt ]Integrating term by term:[ sum_{n=0}^{infty} frac{(-1)^n}{n!} cdot frac{t^{n + 4/3}}{n + 4/3} ]Evaluated from 1/10 to 1.5:[ sum_{n=0}^{infty} frac{(-1)^n}{n! (n + 4/3)} left[ (1.5)^{n + 4/3} - (1/10)^{n + 4/3} right] ]This seems complicated, but maybe I can compute a few terms numerically to approximate the integral.Alternatively, perhaps using numerical integration would be more straightforward. Since I don't have access to computational tools right now, maybe I can estimate the integral using Simpson's rule or another numerical method.But given that this is a problem-solving scenario, perhaps the integral is expected to be expressed in terms of the incomplete gamma function, and then evaluated numerically.Alternatively, maybe I can use substitution to express the integral in terms of the gamma function.Wait, let me recall that:[ int x^{k} e^{-ax} dx = frac{1}{a^{k+1}} Gamma(k+1, ax) + C ]So, in our case, k = 1/3 and a = 1/10.Thus, the integral from 1 to 15 is:[ int_1^{15} x^{1/3} e^{-x/10} dx = left[ frac{1}{(1/10)^{4/3}} Gammaleft( frac{4}{3}, frac{x}{10} right) right]_1^{15} ]Simplify:Since ( frac{1}{(1/10)^{4/3}} = 10^{4/3} ), so:[ 10^{4/3} left[ Gammaleft( frac{4}{3}, frac{15}{10} right) - Gammaleft( frac{4}{3}, frac{1}{10} right) right] ]Which is the same as:[ 10^{4/3} left[ Gammaleft( frac{4}{3}, 1.5 right) - Gammaleft( frac{4}{3}, 0.1 right) right] ]Now, the incomplete gamma function ( Gamma(s, x) ) can be related to the gamma function ( Gamma(s) ) and the lower incomplete gamma function ( gamma(s, x) ):[ Gamma(s, x) = Gamma(s) - gamma(s, x) ]So, we can write:[ 10^{4/3} left[ (Gamma(frac{4}{3}) - gamma(frac{4}{3}, 1.5)) - (Gamma(frac{4}{3}) - gamma(frac{4}{3}, 0.1)) right] ]Simplify:[ 10^{4/3} left[ -gamma(frac{4}{3}, 1.5) + gamma(frac{4}{3}, 0.1) right] ]Which is:[ 10^{4/3} left[ gamma(frac{4}{3}, 0.1) - gamma(frac{4}{3}, 1.5) right] ]But I still need numerical values for these incomplete gamma functions. I know that ( gamma(s, x) ) can be expressed as:[ gamma(s, x) = x^s Gamma(s) sum_{k=0}^{infty} frac{(-x)^k}{k! Gamma(s + k + 1)} ]But this might not be helpful for exact computation. Alternatively, perhaps I can use known approximations or look up tables for the incomplete gamma function.Alternatively, I can use the relationship with the error function for specific cases, but I don't think that applies here since s = 4/3 is not an integer.Alternatively, perhaps I can use the series expansion for the lower incomplete gamma function:[ gamma(s, x) = frac{x^s}{s} sum_{n=0}^{infty} frac{(-1)^n x^n}{n! (s + n)} ]So, for s = 4/3, x = 0.1 and x = 1.5.Let me compute ( gamma(4/3, 0.1) ) first.Using the series:[ gamma(4/3, 0.1) = frac{(0.1)^{4/3}}{4/3} sum_{n=0}^{infty} frac{(-1)^n (0.1)^n}{n! (4/3 + n)} ]Simplify:[ gamma(4/3, 0.1) = frac{3}{4} (0.1)^{4/3} sum_{n=0}^{infty} frac{(-1)^n (0.1)^n}{n! (4/3 + n)} ]Similarly, for ( gamma(4/3, 1.5) ):[ gamma(4/3, 1.5) = frac{(1.5)^{4/3}}{4/3} sum_{n=0}^{infty} frac{(-1)^n (1.5)^n}{n! (4/3 + n)} ]This seems quite involved, but maybe I can compute a few terms to approximate the sum.Let me start with ( gamma(4/3, 0.1) ):First, compute ( (0.1)^{4/3} ). Since 0.1 = 10^{-1}, so (0.1)^{4/3} = 10^{-4/3} ≈ 0.04641588834.Then, multiply by 3/4: 0.04641588834 * 0.75 ≈ 0.03481191625.Now, the sum:[ sum_{n=0}^{infty} frac{(-1)^n (0.1)^n}{n! (4/3 + n)} ]Let me compute the first few terms:n=0:[ frac{1}{1 * (4/3)} = 3/4 = 0.75 ]n=1:[ frac{-0.1}{1! * (4/3 +1)} = frac{-0.1}{(7/3)} ≈ -0.1 * 3/7 ≈ -0.04285714286 ]n=2:[ frac{(0.1)^2}{2! * (4/3 +2)} = frac{0.01}{2 * (10/3)} = 0.01 / (20/3) ≈ 0.01 * 3/20 ≈ 0.0015 ]n=3:[ frac{-(0.1)^3}{3! * (4/3 +3)} = frac{-0.001}{6 * (13/3)} ≈ -0.001 / 26 ≈ -0.0000384615 ]n=4:[ frac{(0.1)^4}{4! * (4/3 +4)} = frac{0.0001}{24 * (16/3)} ≈ 0.0001 / 128 ≈ 0.00000078125 ]So, adding these up:0.75 - 0.04285714286 + 0.0015 - 0.0000384615 + 0.00000078125 ≈0.75 - 0.042857 ≈ 0.7071430.707143 + 0.0015 ≈ 0.7086430.708643 - 0.000038 ≈ 0.7086050.708605 + 0.00000078 ≈ 0.70860578So, the sum is approximately 0.70860578.Therefore, ( gamma(4/3, 0.1) ≈ 0.03481191625 * 0.70860578 ≈ 0.02464 ).Now, let's compute ( gamma(4/3, 1.5) ):First, compute ( (1.5)^{4/3} ). 1.5 = 3/2, so (3/2)^{4/3} ≈ e^{(4/3) ln(3/2)} ≈ e^{(4/3)(0.4055)} ≈ e^{0.5406} ≈ 1.716.Then, multiply by 3/4: 1.716 * 0.75 ≈ 1.287.Now, the sum:[ sum_{n=0}^{infty} frac{(-1)^n (1.5)^n}{n! (4/3 + n)} ]This will converge slower because 1.5 is larger. Let's compute a few terms:n=0:[ frac{1}{1 * (4/3)} = 0.75 ]n=1:[ frac{-1.5}{1! * (7/3)} = frac{-1.5}{7/3} = -1.5 * 3/7 ≈ -0.642857 ]n=2:[ frac{(1.5)^2}{2! * (10/3)} = frac{2.25}{2 * (10/3)} = 2.25 / (20/3) ≈ 2.25 * 3/20 ≈ 0.3375 ]n=3:[ frac{-(1.5)^3}{3! * (13/3)} = frac{-3.375}{6 * (13/3)} = -3.375 / 26 ≈ -0.129807 ]n=4:[ frac{(1.5)^4}{4! * (16/3)} = frac{5.0625}{24 * (16/3)} = 5.0625 / 128 ≈ 0.03955078125 ]n=5:[ frac{-(1.5)^5}{5! * (19/3)} = frac{-7.59375}{120 * (19/3)} = -7.59375 / 760 ≈ -0.0099917 ]n=6:[ frac{(1.5)^6}{6! * (22/3)} = frac{11.390625}{720 * (22/3)} = 11.390625 / 5280 ≈ 0.002157 ]n=7:[ frac{-(1.5)^7}{7! * (25/3)} = frac{-17.0859375}{5040 * (25/3)} = -17.0859375 / 42000 ≈ -0.0004068 ]n=8:[ frac{(1.5)^8}{8! * (28/3)} = frac{25.62890625}{40320 * (28/3)} = 25.62890625 / 376320 ≈ 0.0000681 ]Adding these up:0.75 - 0.642857 ≈ 0.1071430.107143 + 0.3375 ≈ 0.4446430.444643 - 0.129807 ≈ 0.3148360.314836 + 0.0395508 ≈ 0.3543870.354387 - 0.0099917 ≈ 0.3443950.344395 + 0.002157 ≈ 0.3465520.346552 - 0.0004068 ≈ 0.3461450.346145 + 0.0000681 ≈ 0.346213So, the sum is approximately 0.346213.Therefore, ( gamma(4/3, 1.5) ≈ 1.287 * 0.346213 ≈ 0.4456 ).Now, putting it all together:[ text{ADI} = 10^{4/3} left[ 0.02464 - 0.4456 right] ]Wait, that would be negative, which doesn't make sense because the integrand is positive over [1,15]. I must have made a mistake in the signs.Wait, no, because the integral is from 1 to 15, and we expressed it as:[ 10^{4/3} left[ gamma(4/3, 0.1) - gamma(4/3, 1.5) right] ]But since 0.1 < 1.5, ( gamma(4/3, 0.1) < gamma(4/3, 1.5) ), so the difference is negative. But the integral from 1 to 15 should be positive. Therefore, I must have made a mistake in the substitution.Wait, let's go back to the substitution step.We had:[ int_1^{15} x^{1/3} e^{-x/10} dx = 10^{4/3} int_{1/10}^{1.5} t^{1/3} e^{-t} dt ]Which is correct.Then, ( int t^{1/3} e^{-t} dt = gamma(4/3, t) ) evaluated from 1/10 to 1.5.So, it's ( gamma(4/3, 1.5) - gamma(4/3, 1/10) ).Therefore, the ADI is:[ 10^{4/3} left[ gamma(4/3, 1.5) - gamma(4/3, 0.1) right] ]Which is positive because ( gamma(4/3, 1.5) > gamma(4/3, 0.1) ).So, substituting the approximate values:[ 10^{4/3} (0.4456 - 0.02464) = 10^{4/3} (0.42096) ]Now, compute 10^{4/3}. 10^{1/3} ≈ 2.15443469, so 10^{4/3} = 10 * 10^{1/3} ≈ 10 * 2.15443469 ≈ 21.5443469.Therefore, ADI ≈ 21.5443469 * 0.42096 ≈Let me compute 21.5443469 * 0.42096:First, 21.5443469 * 0.4 = 8.6177387621.5443469 * 0.02 = 0.43088693821.5443469 * 0.00096 ≈ 0.020708Adding these up:8.61773876 + 0.430886938 ≈ 9.04862579.0486257 + 0.020708 ≈ 9.0693337So, approximately 9.0693.Therefore, the ADI is approximately 9.07.But let me check my approximations because I might have made errors in the series expansions.Alternatively, perhaps I can use a calculator or computational tool to compute the integral numerically.But since I don't have access to that, I'll proceed with the approximation.So, for part 1, the ADI is approximately 9.07.Now, moving on to part 2: verifying the hypothesis that the impact of the number of references on the ADI diminishes as R increases. To do this, I need to find the limit of the partial derivative of ADI with respect to R as R approaches infinity.First, let's find the partial derivative ( frac{partial text{ADI}}{partial R} ).Given:[ text{ADI}(R, D, L) = int_1^R x^{frac{1}{D}} e^{-frac{x}{L}} dx ]By the Fundamental Theorem of Calculus, the derivative with respect to R is:[ frac{partial text{ADI}}{partial R} = R^{frac{1}{D}} e^{-frac{R}{L}} ]Now, we need to find the limit as R approaches infinity:[ lim_{R to infty} R^{frac{1}{D}} e^{-frac{R}{L}} ]We can analyze this limit. As R grows large, the exponential term ( e^{-R/L} ) decays to zero much faster than the polynomial term ( R^{1/D} ) grows. Therefore, the product will approach zero.Thus, the limit is zero, which means that as R increases, the marginal impact on ADI (the derivative) diminishes to zero. This supports the hypothesis.So, summarizing:1. The ADI for the given article is approximately 9.07.2. The limit of the partial derivative as R approaches infinity is zero, confirming that the impact of increasing references diminishes.Final Answer1. The ADI for the article is approximately boxed{9.07}.2. The limit of the partial derivative as ( R to infty ) is boxed{0}.</think>

🧠:<think>Okay, so I have this problem about Kabir, an early Hindi literary pioneer, and his poem-writing pattern. The number of poems he wrote in the nth year is given by P(n) = a·F(n) + b·L(n), where F(n) is the nth Fibonacci number and L(n) is the nth Lucas number. I need to find the constants a and b using the information that in the 5th year, he wrote 22 poems, and in the 7th year, he wrote 47 poems. Then, using those a and b, calculate the total number of poems in the first 10 years.Alright, let's start by recalling what Fibonacci and Lucas numbers are. The Fibonacci sequence starts with F(1) = 1, F(2) = 1, and each subsequent term is the sum of the two preceding ones. So, F(3) = 2, F(4) = 3, F(5) = 5, F(6) = 8, F(7) = 13, and so on.The Lucas sequence is similar but starts with different initial terms. L(1) = 1, L(2) = 3, and each subsequent term is the sum of the two preceding ones. So, L(3) = 4, L(4) = 7, L(5) = 11, L(6) = 18, L(7) = 29, etc.Given that, let's note down the values for n=5 and n=7.For n=5:F(5) = 5L(5) = 11So, P(5) = a·5 + b·11 = 22For n=7:F(7) = 13L(7) = 29So, P(7) = a·13 + b·29 = 47Now, we have a system of two equations:1. 5a + 11b = 222. 13a + 29b = 47I need to solve this system for a and b. Let's write it out:Equation 1: 5a + 11b = 22 Equation 2: 13a + 29b = 47I can use the method of elimination or substitution. Let's try elimination. Maybe multiply Equation 1 by 13 and Equation 2 by 5 to make the coefficients of a the same.Multiplying Equation 1 by 13:65a + 143b = 286Multiplying Equation 2 by 5:65a + 145b = 235Now, subtract the first new equation from the second new equation:(65a + 145b) - (65a + 143b) = 235 - 286 65a - 65a + 145b - 143b = -51 0a + 2b = -51 So, 2b = -51 Therefore, b = -51 / 2 b = -25.5Wait, that seems odd. b is negative? Let me check my calculations.Wait, 235 - 286 is indeed -51, and 145b - 143b is 2b. So 2b = -51, so b = -25.5. Hmm, that's a fraction, but maybe it's correct.Let me plug b back into Equation 1 to find a.Equation 1: 5a + 11b = 22 So, 5a + 11*(-25.5) = 22 Calculate 11*(-25.5): 11*25 = 275, 11*0.5=5.5, so total is -275 -5.5 = -280.5So, 5a - 280.5 = 22 Add 280.5 to both sides: 5a = 22 + 280.5 = 302.5 Therefore, a = 302.5 / 5 = 60.5So, a = 60.5 and b = -25.5Wait, that seems a bit strange because a and b are constants, but they are fractional. Is that possible? The problem doesn't specify that a and b have to be integers, so maybe it's okay.Let me verify these values with the original equations.First, Equation 1: 5a + 11b 5*60.5 = 302.5 11*(-25.5) = -280.5 302.5 - 280.5 = 22, which matches P(5)=22.Equation 2: 13a + 29b 13*60.5 = 786.5 29*(-25.5) = -739.5 786.5 - 739.5 = 47, which matches P(7)=47.Okay, so the values of a and b are correct, even though they are fractional. So, a = 60.5 and b = -25.5.Now, moving on to part 2: Calculate the total number of poems written by Kabir in the first 10 years.So, I need to compute the sum S = P(1) + P(2) + ... + P(10).Given that P(n) = a·F(n) + b·L(n), so S = a*(F(1)+F(2)+...+F(10)) + b*(L(1)+L(2)+...+L(10))So, I need to compute the sum of the first 10 Fibonacci numbers and the sum of the first 10 Lucas numbers, then multiply each by a and b respectively, and add them together.First, let's recall the Fibonacci numbers up to F(10):F(1) = 1 F(2) = 1 F(3) = 2 F(4) = 3 F(5) = 5 F(6) = 8 F(7) = 13 F(8) = 21 F(9) = 34 F(10) = 55Sum of F(1) to F(10): Let's add them up.1 + 1 = 2 2 + 2 = 4 4 + 3 = 7 7 + 5 = 12 12 + 8 = 20 20 + 13 = 33 33 + 21 = 54 54 + 34 = 88 88 + 55 = 143So, sum of Fibonacci numbers from 1 to 10 is 143.Now, Lucas numbers up to L(10):L(1) = 1 L(2) = 3 L(3) = 4 L(4) = 7 L(5) = 11 L(6) = 18 L(7) = 29 L(8) = 47 L(9) = 76 L(10) = 123Sum of L(1) to L(10):1 + 3 = 4 4 + 4 = 8 8 + 7 = 15 15 + 11 = 26 26 + 18 = 44 44 + 29 = 73 73 + 47 = 120 120 + 76 = 196 196 + 123 = 319So, sum of Lucas numbers from 1 to 10 is 319.Therefore, total poems S = a*143 + b*319We have a = 60.5 and b = -25.5Compute 60.5*143:First, 60*143 = 8580 0.5*143 = 71.5 So, total is 8580 + 71.5 = 8651.5Compute (-25.5)*319:First, 25*319 = 7975 0.5*319 = 159.5 So, 25.5*319 = 7975 + 159.5 = 8134.5 But since it's -25.5, it's -8134.5Now, add these two results:8651.5 + (-8134.5) = 8651.5 - 8134.5 = 517So, the total number of poems written in the first 10 years is 517.Wait, let me double-check the calculations because 60.5*143 and 25.5*319.Alternatively, maybe I can compute 60.5*143:60.5 * 143 = (60 + 0.5) * 143 = 60*143 + 0.5*143 = 8580 + 71.5 = 8651.5Similarly, 25.5*319:25.5 * 319 = (25 + 0.5) * 319 = 25*319 + 0.5*319 = 7975 + 159.5 = 8134.5So, 8651.5 - 8134.5 = 517. That seems correct.Alternatively, I can compute 60.5*143 -25.5*319:But 60.5 is 121/2 and 25.5 is 51/2.So, 121/2 * 143 - 51/2 * 319 = (121*143 - 51*319)/2Compute numerator:121*143: 121*100=12100 121*40=4840 121*3=363 Total: 12100 + 4840 = 16940 + 363 = 1730351*319: 50*319=15950 1*319=319 Total: 15950 + 319 = 16269So, numerator: 17303 - 16269 = 1034Then, 1034 / 2 = 517Yes, same result. So, total is 517.Therefore, the total number of poems in the first 10 years is 517.Final Answer1. The values of (a) and (b) are (boxed{60.5}) and (boxed{-25.5}) respectively.2. The total number of poems written in the first 10 years is (boxed{517}).</think>

🧠:<think>Alright, so I have this problem about a city employee monitoring water leaks. There are two parts to it. Let me try to break them down one by one.Problem 1: Expected Total LeakageOkay, so the city is divided into n sectors, each with m sensors. Each sensor has a leakage rate L_ij, which follows a Gaussian distribution with mean μ_ij and standard deviation σ_ij. The employee needs to find the expected total leakage across all sectors.Hmm, expected value. I remember that for independent random variables, the expected value of the sum is the sum of the expected values. So, since each L_ij is independent, the total leakage would just be the sum of all individual expected leakages.So, the expected total leakage E[Total] should be the sum from i=1 to n, and for each i, sum from j=1 to m of E[L_ij]. Since each L_ij is Gaussian with mean μ_ij, E[L_ij] is just μ_ij. Therefore, the expected total leakage is the sum of all μ_ij across all sectors and sensors.Let me write that down:E[Total] = Σ_{i=1 to n} Σ_{j=1 to m} μ_ijThat seems straightforward. I don't think I need to worry about variances or anything else because the question is only about the expected total, not the distribution of the total.Problem 2: Sector with Highest Probability of Critical LeakNow, a critical leak is defined as L_ij exceeding a threshold T. The employee needs to find the sector with the highest probability of at least one sensor exceeding T.So, for each sector, which has m sensors, we need to compute the probability that at least one sensor in that sector has L_ij > T. Then, compare these probabilities across all sectors and pick the one with the highest probability.First, for a single sensor, the probability that L_ij > T is P(L_ij > T). Since L_ij is Gaussian, this can be calculated using the standard normal distribution. Specifically, if we standardize L_ij, we get Z = (L_ij - μ_ij)/σ_ij, which follows a standard normal distribution. Then, P(L_ij > T) = P(Z > (T - μ_ij)/σ_ij) = 1 - Φ((T - μ_ij)/σ_ij), where Φ is the standard normal CDF.But since we're looking for the probability that at least one sensor in the sector exceeds T, and the sensors are independent, we can use the complement rule. The probability that at least one exceeds T is 1 minus the probability that all of them are less than or equal to T.So, for a sector, the probability is:P(at least one > T) = 1 - [P(L_1j <= T) * P(L_2j <= T) * ... * P(L_mj <= T)]Since each sensor is independent, we can multiply their individual probabilities.Therefore, for sector i, the probability is:1 - Π_{j=1 to m} [Φ((T - μ_ij)/σ_ij)]So, for each sector, compute this product, subtract it from 1, and that's the probability that at least one sensor in that sector exceeds T. Then, compare these probabilities across all sectors and pick the sector with the highest value.Wait, but is there a more efficient way to compute this? I mean, if m is large, computing the product might be computationally intensive, but I don't think that's the concern here. The problem is just asking for the formulation, not the computation.So, to recap:For each sector i:1. For each sensor j in sector i, compute P(L_ij <= T) = Φ((T - μ_ij)/σ_ij)2. Multiply all these probabilities together for the sector: Π_{j=1 to m} Φ((T - μ_ij)/σ_ij)3. Subtract this product from 1 to get the probability that at least one sensor exceeds T: 1 - Π_{j=1 to m} Φ((T - μ_ij)/σ_ij)Then, among all sectors, the one with the highest value from step 3 is the sector with the highest probability of a critical leak.I think that's the correct approach. It makes sense because if a sector has even one sensor with a high probability of exceeding T, the overall probability for the sector increases. But since we're taking the product of all sensors' probabilities of not exceeding T, the more sensors you have, the lower the product, hence the higher the probability that at least one exceeds T. But it depends on the individual μ_ij and σ_ij as well.Let me think if there's any edge cases or something I'm missing. For example, if all sensors in a sector have a very low probability of exceeding T, then the product would be close to 1, so 1 - product would be close to 0. Conversely, if even one sensor has a high probability, the product would be lower, making 1 - product higher.Yes, that seems correct. So, the sector with the highest probability is the one where the product of the probabilities of each sensor not exceeding T is the smallest, which translates to the highest 1 - product.I think that's solid. I don't see any mistakes in the reasoning.Final Answer1. The expected total leakage is the sum of all individual expected leakages. So, the expected total leakage is boxed{sum_{i=1}^{n} sum_{j=1}^{m} mu_{ij}}.2. The probability that at least one sensor in a sector exceeds the threshold ( T ) is ( 1 - prod_{j=1}^{m} Phileft(frac{T - mu_{ij}}{sigma_{ij}}right) ). The sector with the highest probability is the one with the maximum value of this expression. Therefore, the sector with the highest probability of a critical leak is the one where boxed{1 - prod_{j=1}^{m} Phileft(frac{T - mu_{ij}}{sigma_{ij}}right)} is maximized.</think>