🧠:<think>Okay, so I have this problem about a statistics professor analyzing network traffic logs using Gaussian Processes for anomaly detection. There are two parts: first, deriving the posterior predictive distribution for a new observation, and second, calculating the anomaly detection threshold and the probability of a false positive. Let me try to work through each part step by step.Starting with part 1: Deriving the posterior predictive distribution for ( X_{N+1} ). I remember that Gaussian Processes are non-parametric models that define a distribution over functions, and they're particularly useful for regression tasks. The posterior predictive distribution is essentially the distribution of a new observation given the training data and the prior assumptions.Given that the prior is a Gaussian Process ( mathcal{GP}(mu(t), k(t, t')) ), and the covariance function is the squared exponential kernel ( k(t, t') = sigma^2 expleft(-frac{(t-t')^2}{2ell^2}right) ). So, the mean function is ( mu(t) ), which could be a constant or some function, but I think in many cases, especially for simplicity, it's often assumed to be zero. But since it's given as ( mu(t) ), I should keep it general.To find the posterior predictive distribution, I need to consider the joint distribution of the observed data ( X_1, X_2, ldots, X_N ) and the new observation ( X_{N+1} ). Since the Gaussian Process is a collection of Gaussian random variables, any finite set of them has a multivariate normal distribution.So, the joint distribution of ( mathbf{X} = [X_1, X_2, ldots, X_N]^T ) and ( X_{N+1} ) is:[begin{pmatrix}mathbf{X} X_{N+1}end{pmatrix}sim mathcal{N}left(begin{pmatrix}mu(mathbf{t}) mu(t_{N+1})end{pmatrix},begin{pmatrix}K(mathbf{t}, mathbf{t}) & K(mathbf{t}, t_{N+1}) K(t_{N+1}, mathbf{t}) & K(t_{N+1}, t_{N+1})end{pmatrix}right)]Where ( mu(mathbf{t}) ) is the vector of mean functions evaluated at each time point ( t_1, t_2, ldots, t_N ), and ( K(mathbf{t}, mathbf{t}) ) is the covariance matrix between all pairs of observed times, ( K(mathbf{t}, t_{N+1}) ) is the covariance vector between the observed times and the new time ( t_{N+1} ), and ( K(t_{N+1}, t_{N+1}) ) is the variance at the new time point.Given this joint distribution, the posterior predictive distribution of ( X_{N+1} ) given ( mathbf{X} ) can be found using the properties of the multivariate normal distribution. Specifically, the conditional distribution of a subset of variables given the others is also normal, with mean and covariance given by:[mathbb{E}[X_{N+1} | mathbf{X}] = mu(t_{N+1}) + K(t_{N+1}, mathbf{t}) K(mathbf{t}, mathbf{t})^{-1} (mathbf{X} - mu(mathbf{t}))]and[text{Var}(X_{N+1} | mathbf{X}) = K(t_{N+1}, t_{N+1}) - K(t_{N+1}, mathbf{t}) K(mathbf{t}, mathbf{t})^{-1} K(mathbf{t}, t_{N+1})]So, putting it all together, the posterior predictive distribution is:[X_{N+1} | mathbf{X} sim mathcal{N}left( mu(t_{N+1}) + K(t_{N+1}, mathbf{t}) K(mathbf{t}, mathbf{t})^{-1} (mathbf{X} - mu(mathbf{t})), K(t_{N+1}, t_{N+1}) - K(t_{N+1}, mathbf{t}) K(mathbf{t}, mathbf{t})^{-1} K(mathbf{t}, t_{N+1}) right)]This is the standard result for Gaussian Processes regression. So, I think that's the answer for part 1.Moving on to part 2: Calculating the anomaly detection threshold and determining the probability of a false positive. An anomaly is defined as any observation outside the 99.9% credible interval of the posterior predictive distribution. So, first, I need to find the threshold values that define this interval.Assuming the posterior predictive distribution is normal, as derived in part 1, the 99.9% credible interval corresponds to the mean plus or minus approximately 3.29 standard deviations. Because for a normal distribution, about 99.9% of the data lies within 3.29 standard deviations from the mean.So, the anomaly detection thresholds would be:[mu_{text{pred}} pm 3.29 times sigma_{text{pred}}]Where ( mu_{text{pred}} ) is the mean of the posterior predictive distribution and ( sigma_{text{pred}} ) is its standard deviation.Now, the probability of a false positive is the probability that a normal observation (i.e., one generated by the Gaussian Process model) is classified as an anomaly. Since the model assumes that the data is generated by the GP, the probability that an observation lies outside the 99.9% credible interval is 0.1%. So, the probability of a false positive is 0.1%.Wait, but hold on. Is it exactly 0.1%? Because the credible interval is 99.9%, so the probability outside is 0.1%, which is 0.001. So, the false positive rate is 0.1%.But let me think again. The 99.9% credible interval for a normal distribution is indeed approximately 3.29 standard deviations. So, the probability that a new observation falls outside this interval is 0.1%, which is 0.001. Therefore, the probability of a false positive is 0.1%.But wait, is this under the assumption that the model is correct? Yes, because the professor is assuming the true underlying process is generated by the GP model. So, if the model is correct, then the probability of a false positive is 0.1%.So, summarizing:1. The posterior predictive distribution is a normal distribution with the mean and variance as derived above.2. The anomaly detection threshold is set at 3.29 standard deviations from the mean, and the probability of a false positive is 0.1%.I think that's it. Let me just check if I missed anything.For part 1, I used the standard GP regression formula, which is correct. For part 2, I used the properties of the normal distribution to find the credible interval and the false positive rate. Since the model is assumed correct, the false positive rate is exactly the probability outside the credible interval, which is 0.1%.Yeah, I think that's solid.Final Answer1. The posterior predictive distribution for ( X_{N+1} ) is ( mathcal{N}left( mu(t_{N+1}) + K(t_{N+1}, mathbf{t}) K(mathbf{t}, mathbf{t})^{-1} (mathbf{X} - mu(mathbf{t})), K(t_{N+1}, t_{N+1}) - K(t_{N+1}, mathbf{t}) K(mathbf{t}, mathbf{t})^{-1} K(mathbf{t}, t_{N+1}) right) ).2. The anomaly detection threshold is ( mu_{text{pred}} pm 3.29 times sigma_{text{pred}} ), and the probability of a false positive is ( boxed{0.1%} ).</think>

🧠:<think>Alright, so I'm trying to solve this problem about World War I propaganda posters using the logistic growth model. Let me break it down step by step.First, the logistic growth equation is given as:[ P(t) = frac{K}{1 + e^{-r(t - t_0)}} ]where:- ( P(t) ) is the number of posters at time ( t ),- ( K ) is the carrying capacity,- ( r ) is the growth rate,- ( t_0 ) is the inflection point.The problem states that in 1915, the poster reached 30% of ( K ), and in 1916, it reached 60% of ( K ). I need to find ( r ) and ( t_0 ).Let me denote 1914 as ( t = 0 ), so 1915 is ( t = 1 ) and 1916 is ( t = 2 ).So, at ( t = 1 ), ( P(1) = 0.3K ), and at ( t = 2 ), ( P(2) = 0.6K ).Plugging these into the logistic equation:For ( t = 1 ):[ 0.3K = frac{K}{1 + e^{-r(1 - t_0)}} ]Divide both sides by ( K ):[ 0.3 = frac{1}{1 + e^{-r(1 - t_0)}} ]Take reciprocal:[ frac{1}{0.3} = 1 + e^{-r(1 - t_0)} ][ frac{10}{3} = 1 + e^{-r(1 - t_0)} ]Subtract 1:[ frac{7}{3} = e^{-r(1 - t_0)} ]Take natural log:[ lnleft(frac{7}{3}right) = -r(1 - t_0) ]Let me compute ( ln(7/3) ):[ ln(7) ≈ 1.9459, ln(3) ≈ 1.0986 ]So, ( ln(7/3) ≈ 1.9459 - 1.0986 ≈ 0.8473 )Thus:[ 0.8473 = -r(1 - t_0) ]Equation 1: ( 0.8473 = -r(1 - t_0) )Similarly, for ( t = 2 ):[ 0.6K = frac{K}{1 + e^{-r(2 - t_0)}} ]Divide by ( K ):[ 0.6 = frac{1}{1 + e^{-r(2 - t_0)}} ]Take reciprocal:[ frac{1}{0.6} = 1 + e^{-r(2 - t_0)} ][ frac{5}{3} ≈ 1.6667 = 1 + e^{-r(2 - t_0)} ]Subtract 1:[ 0.6667 = e^{-r(2 - t_0)} ]Take natural log:[ ln(2/3) ≈ -0.4055 = -r(2 - t_0) ]Equation 2: ( -0.4055 = -r(2 - t_0) )Now, I have two equations:1. ( 0.8473 = -r(1 - t_0) )2. ( -0.4055 = -r(2 - t_0) )Let me simplify both equations:From Equation 1:[ 0.8473 = -r + r t_0 ]From Equation 2:[ -0.4055 = -2r + r t_0 ]Let me write them as:1. ( r t_0 - r = 0.8473 )2. ( r t_0 - 2r = -0.4055 )Let me subtract Equation 2 from Equation 1:[ (r t_0 - r) - (r t_0 - 2r) = 0.8473 - (-0.4055) ]Simplify:[ r t_0 - r - r t_0 + 2r = 0.8473 + 0.4055 ][ ( -r + 2r ) = 1.2528 ][ r = 1.2528 ]So, the growth rate ( r ≈ 1.2528 ) per year.Now, plug ( r ) back into Equation 1:[ 0.8473 = -1.2528(1 - t_0) ]Divide both sides by -1.2528:[ frac{0.8473}{-1.2528} = 1 - t_0 ]Calculate:[ ≈ -0.676 = 1 - t_0 ]Thus:[ t_0 = 1 + 0.676 ≈ 1.676 ]So, the inflection point ( t_0 ≈ 1.676 ) years after 1914, which is approximately mid-1916.Wait, let me check the calculations again because the signs might be tricky.From Equation 1:[ 0.8473 = -r(1 - t_0) ]So, ( 0.8473 = -r + r t_0 )Similarly, Equation 2:[ -0.4055 = -2r + r t_0 ]So, subtracting Equation 2 from Equation 1:[ (0.8473) - (-0.4055) = (-r + r t_0) - (-2r + r t_0) ][ 1.2528 = (-r + r t_0 + 2r - r t_0) ][ 1.2528 = r ]So, ( r ≈ 1.2528 )Then, plug back into Equation 1:[ 0.8473 = -1.2528(1 - t_0) ]Divide both sides by -1.2528:[ frac{0.8473}{-1.2528} ≈ -0.676 = 1 - t_0 ]Thus:[ t_0 = 1 + 0.676 ≈ 1.676 ]Yes, that seems correct. So, ( t_0 ≈ 1.676 ), which is about 1.676 years after 1914, so mid-1916.Now, for part 2, I need to calculate the cumulative impact, which is the integral of ( P(t) ) from 1914 to 1918. Let me denote the integral as ( C ):[ C = int_{0}^{4} P(t) dt = int_{0}^{4} frac{K}{1 + e^{-r(t - t_0)}} dt ]This integral can be solved using substitution. Let me set ( u = r(t - t_0) ), so ( du = r dt ), ( dt = du/r ). When ( t = 0 ), ( u = -r t_0 ), and when ( t = 4 ), ( u = r(4 - t_0) ).But this might get complicated. Alternatively, recall that the integral of ( frac{1}{1 + e^{-u}} du ) is ( u - ln(1 + e^{-u}) + C ).So, let me rewrite the integral:[ C = frac{K}{r} int_{-r t_0}^{r(4 - t_0)} frac{1}{1 + e^{-u}} du ]The integral of ( frac{1}{1 + e^{-u}} du ) is ( u - ln(1 + e^{-u}) ).Thus,[ C = frac{K}{r} left[ u - ln(1 + e^{-u}) right]_{-r t_0}^{r(4 - t_0)} ]Compute at upper limit ( u = r(4 - t_0) ):[ r(4 - t_0) - ln(1 + e^{-r(4 - t_0)}) ]Compute at lower limit ( u = -r t_0 ):[ -r t_0 - ln(1 + e^{r t_0}) ]So, the integral becomes:[ C = frac{K}{r} left[ r(4 - t_0) - ln(1 + e^{-r(4 - t_0)}) - (-r t_0 - ln(1 + e^{r t_0})) right] ]Simplify:[ C = frac{K}{r} left[ r(4 - t_0) + r t_0 - ln(1 + e^{-r(4 - t_0)}) + ln(1 + e^{r t_0}) right] ][ C = frac{K}{r} left[ 4r - lnleft(frac{1 + e^{-r(4 - t_0)}}{1 + e^{r t_0}}right) right] ][ C = K left[ 4 - frac{1}{r} lnleft(frac{1 + e^{-r(4 - t_0)}}{1 + e^{r t_0}}right) right] ]This expression can be simplified further. Let me compute the logarithm term:[ lnleft(frac{1 + e^{-r(4 - t_0)}}{1 + e^{r t_0}}right) = ln(1 + e^{-r(4 - t_0)}) - ln(1 + e^{r t_0}) ]But this might not lead to a significant simplification. Alternatively, perhaps we can express it in terms of the logistic function.Alternatively, recall that the integral of the logistic function from ( t = a ) to ( t = b ) is:[ int_{a}^{b} frac{K}{1 + e^{-r(t - t_0)}} dt = frac{K}{r} left[ (b - t_0) - (a - t_0) - lnleft(frac{1 + e^{-r(b - t_0)}}{1 + e^{-r(a - t_0)}}right) right] ]Wait, let me check:Let me make substitution ( u = r(t - t_0) ), so ( du = r dt ), ( dt = du/r ). Then:[ int frac{K}{1 + e^{-u}} cdot frac{du}{r} = frac{K}{r} int frac{1}{1 + e^{-u}} du ]Which is ( frac{K}{r} [u - ln(1 + e^{-u})] + C ).So, the definite integral from ( t = a ) to ( t = b ) is:[ frac{K}{r} left[ u - ln(1 + e^{-u}) right]_{u = r(a - t_0)}^{u = r(b - t_0)} ]Which is:[ frac{K}{r} left[ r(b - t_0) - ln(1 + e^{-r(b - t_0)}) - (r(a - t_0) - ln(1 + e^{-r(a - t_0)})) right] ][ = frac{K}{r} left[ r(b - a) - lnleft(frac{1 + e^{-r(b - t_0)}}{1 + e^{-r(a - t_0)}}right) right] ][ = K(b - a) - frac{K}{r} lnleft(frac{1 + e^{-r(b - t_0)}}{1 + e^{-r(a - t_0)}}right) ]So, in our case, ( a = 0 ), ( b = 4 ), so:[ C = K(4 - 0) - frac{K}{r} lnleft(frac{1 + e^{-r(4 - t_0)}}{1 + e^{-r(0 - t_0)}}right) ][ = 4K - frac{K}{r} lnleft(frac{1 + e^{-r(4 - t_0)}}{1 + e^{-r(-t_0)}}right) ]Simplify the denominator in the log:[ 1 + e^{-r(-t_0)} = 1 + e^{r t_0} ]So,[ C = 4K - frac{K}{r} lnleft(frac{1 + e^{-r(4 - t_0)}}{1 + e^{r t_0}}right) ]This is the same expression as before.Now, let me compute this with the values we have:We have ( r ≈ 1.2528 ), ( t_0 ≈ 1.676 ).First, compute ( r(4 - t_0) ):( 4 - t_0 ≈ 4 - 1.676 ≈ 2.324 )So, ( r(4 - t_0) ≈ 1.2528 * 2.324 ≈ 2.907 )Similarly, ( r t_0 ≈ 1.2528 * 1.676 ≈ 2.100 )Now, compute ( e^{-r(4 - t_0)} ≈ e^{-2.907} ≈ 0.054 )And ( e^{r t_0} ≈ e^{2.100} ≈ 8.166 )So, the fraction inside the log is:[ frac{1 + 0.054}{1 + 8.166} ≈ frac{1.054}{9.166} ≈ 0.1148 ]Thus, the log term is:[ ln(0.1148) ≈ -2.157 ]So, the cumulative impact:[ C ≈ 4K - frac{K}{1.2528} * (-2.157) ][ ≈ 4K + frac{K * 2.157}{1.2528} ]Calculate ( 2.157 / 1.2528 ≈ 1.722 )Thus,[ C ≈ 4K + 1.722K ≈ 5.722K ]So, the cumulative impact from 1914 to 1918 is approximately ( 5.722K ).Now, if the carrying capacity ( K ) is doubled in 1917, let's see how this affects the cumulative impact.Wait, the problem says "hypothetically doubled due to increased production capabilities in 1917". So, does this mean that from 1917 onwards, ( K ) becomes ( 2K )?Assuming that, we need to adjust the integral accordingly. The original integral was from 1914 to 1918, so if ( K ) doubles in 1917 (which is ( t = 3 )), we need to split the integral into two parts: from 1914 to 1917 (t=0 to t=3) with ( K ), and from 1917 to 1918 (t=3 to t=4) with ( 2K ).So, the new cumulative impact ( C' ) would be:[ C' = int_{0}^{3} frac{K}{1 + e^{-r(t - t_0)}} dt + int_{3}^{4} frac{2K}{1 + e^{-r(t - t_0)}} dt ]Compute each integral separately.First integral from 0 to 3:Using the same method as before:[ C1 = K(3 - 0) - frac{K}{r} lnleft(frac{1 + e^{-r(3 - t_0)}}{1 + e^{r t_0}}right) ]Compute ( r(3 - t_0) ≈ 1.2528*(3 - 1.676) ≈ 1.2528*1.324 ≈ 1.656 )So, ( e^{-1.656} ≈ 0.191 )Thus, the fraction:[ frac{1 + 0.191}{1 + 8.166} ≈ frac{1.191}{9.166} ≈ 0.1299 ][ ln(0.1299) ≈ -2.028 ]So,[ C1 ≈ 3K - frac{K}{1.2528}*(-2.028) ≈ 3K + frac{K * 2.028}{1.2528} ≈ 3K + 1.619K ≈ 4.619K ]Second integral from 3 to 4 with ( 2K ):[ C2 = 2K(4 - 3) - frac{2K}{r} lnleft(frac{1 + e^{-r(4 - t_0)}}{1 + e^{-r(3 - t_0)}}right) ][ = 2K - frac{2K}{1.2528} lnleft(frac{1 + e^{-2.907}}{1 + e^{-1.656}}right) ]Compute the fraction:Numerator: ( 1 + e^{-2.907} ≈ 1 + 0.054 ≈ 1.054 )Denominator: ( 1 + e^{-1.656} ≈ 1 + 0.191 ≈ 1.191 )So, fraction ≈ 1.054 / 1.191 ≈ 0.885Thus, ( ln(0.885) ≈ -0.123 )So,[ C2 ≈ 2K - frac{2K}{1.2528}*(-0.123) ≈ 2K + frac{2K * 0.123}{1.2528} ≈ 2K + 0.196K ≈ 2.196K ]Thus, total cumulative impact ( C' = C1 + C2 ≈ 4.619K + 2.196K ≈ 6.815K )Comparing to the original ( C ≈ 5.722K ), the cumulative impact increases by approximately ( 6.815K - 5.722K = 1.093K ), which is about a 19% increase.So, doubling the carrying capacity in 1917 leads to a significant increase in the cumulative impact, as the posters can be distributed more widely in the later years, amplifying the cultural impact.I think that's the gist of it. Let me just recap:1. Found ( r ≈ 1.2528 ) and ( t_0 ≈ 1.676 ) (mid-1916).2. Calculated the cumulative impact from 1914 to 1918 as ≈5.722K.3. If K doubles in 1917, the cumulative impact becomes ≈6.815K, a notable increase.</think>

🧠:<think>Okay, so I'm trying to help Emily figure out if this painting is a forgery or not. There are two sub-problems here, and I need to solve both to determine the authenticity. Let me take it step by step.Starting with Sub-problem 1: The chemical analysis of the pigment decay. The formula given is ( P(t) = P_0 e^{-lambda t} ). I know that ( P(t) ) is the current proportion of pigment, which is 35% or 0.35. ( P_0 ) is the initial amount, so that would be 100% or 1. The decay constant ( lambda ) is 0.005 per year. I need to find the age ( t ) of the painting.So, plugging the known values into the equation:( 0.35 = 1 times e^{-0.005 t} )Simplifying that:( 0.35 = e^{-0.005 t} )To solve for ( t ), I need to take the natural logarithm of both sides. Remember, ( ln(e^x) = x ), so that should help.Taking ln on both sides:( ln(0.35) = ln(e^{-0.005 t}) )Simplifying the right side:( ln(0.35) = -0.005 t )Now, solving for ( t ):( t = frac{ln(0.35)}{-0.005} )Let me calculate ( ln(0.35) ). I remember that ( ln(1) = 0 ), and ( ln(0.5) ) is about -0.6931. Since 0.35 is less than 0.5, the ln should be more negative. Let me use a calculator for precision.Calculating ( ln(0.35) ):Using a calculator, ( ln(0.35) ) is approximately -1.0498.So, plugging that in:( t = frac{-1.0498}{-0.005} )Dividing two negatives gives a positive:( t = 209.96 ) years.So, approximately 210 years old. Hmm, that seems quite old. I wonder if that's plausible for the artist in question. If the artist was known to have lived, say, 200 years ago, this would fit. But if the painting is supposed to be from a more recent period, this could indicate a forgery. I'll keep that in mind.Moving on to Sub-problem 2: The brushstroke pattern follows a Poisson distribution. The average rate ( lambda_b ) is 4 brushstrokes per square meter. The section of the painting is 2 square meters, so I need to adjust the rate accordingly.In Poisson distribution, the average rate ( lambda ) is the expected number of occurrences in the given interval. Since the section is 2 square meters, the new ( lambda ) would be ( 4 times 2 = 8 ) brushstrokes.We need to find the probability that there are exactly 10 brushstrokes in this section. The formula is:( P(X=k) = frac{lambda^k e^{-lambda}}{k!} )Plugging in ( k = 10 ) and ( lambda = 8 ):( P(X=10) = frac{8^{10} e^{-8}}{10!} )First, let me compute each part step by step.Calculating ( 8^{10} ):( 8^1 = 8 )( 8^2 = 64 )( 8^3 = 512 )( 8^4 = 4096 )( 8^5 = 32768 )( 8^6 = 262144 )( 8^7 = 2097152 )( 8^8 = 16777216 )( 8^9 = 134217728 )( 8^{10} = 1073741824 )Wait, that seems too large. Maybe I made a mistake here. Wait, 8^10 is 8 multiplied by itself 10 times. Let me verify:Yes, 8^10 is 1073741824. That's correct.Next, ( e^{-8} ). The value of ( e ) is approximately 2.71828. So, ( e^{-8} ) is 1 divided by ( e^8 ).Calculating ( e^8 ):( e^1 = 2.71828 )( e^2 ≈ 7.38906 )( e^3 ≈ 20.0855 )( e^4 ≈ 54.5981 )( e^5 ≈ 148.413 )( e^6 ≈ 403.4288 )( e^7 ≈ 1096.633 )( e^8 ≈ 2980.911 )So, ( e^{-8} ≈ 1 / 2980.911 ≈ 0.00033546 )Now, ( 10! ) is 10 factorial, which is 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1.Calculating 10!:10 × 9 = 9090 × 8 = 720720 × 7 = 50405040 × 6 = 3024030240 × 5 = 151200151200 × 4 = 604800604800 × 3 = 18144001814400 × 2 = 36288003628800 × 1 = 3628800So, 10! = 3,628,800.Putting it all together:( P(X=10) = frac{1,073,741,824 × 0.00033546}{3,628,800} )First, multiply numerator:1,073,741,824 × 0.00033546Let me compute that:First, 1,073,741,824 × 0.0001 = 107,374.1824So, 0.00033546 is approximately 3.3546 × 0.0001.So, 107,374.1824 × 3.3546 ≈ ?Calculating 107,374.1824 × 3 = 322,122.5472107,374.1824 × 0.3546 ≈ Let's compute 107,374.1824 × 0.3 = 32,212.25472107,374.1824 × 0.0546 ≈ Approximately 107,374.1824 × 0.05 = 5,368.70912Plus 107,374.1824 × 0.0046 ≈ 493.90124So, adding up:32,212.25472 + 5,368.70912 = 37,580.9638437,580.96384 + 493.90124 ≈ 38,074.86508So, total numerator ≈ 322,122.5472 + 38,074.86508 ≈ 360,197.4123Now, divide by 3,628,800:360,197.4123 / 3,628,800 ≈ Let's compute this.First, 3,628,800 × 0.1 = 362,880So, 360,197.4123 is slightly less than 362,880, which is 0.1.So, approximately 0.0992.Wait, let me compute it more accurately.360,197.4123 ÷ 3,628,800Divide numerator and denominator by 1000: 360.1974123 / 3628.8Now, 3628.8 × 0.1 = 362.88So, 360.1974123 is about 0.992 times 362.88.Wait, that might not be the right approach. Alternatively, let me use decimal division.360,197.4123 ÷ 3,628,800Let me write it as:360,197.4123 ÷ 3,628,800 ≈ 0.0992Yes, approximately 0.0992, or 9.92%.So, the probability is roughly 9.92%.Hmm, that's a decent probability. It's not super high, but it's not extremely low either. So, getting exactly 10 brushstrokes in a 2 square meter section isn't that uncommon if the average is 8.Now, putting both results together.From Sub-problem 1, the painting is about 210 years old. If the artist was active around that time, this would support authenticity. However, if the painting is supposed to be from a more recent period, this would indicate a forgery.From Sub-problem 2, the probability of exactly 10 brushstrokes is about 9.92%, which is not too low. So, if the painting has exactly 10 brushstrokes, it's within the realm of possibility for a genuine piece.But wait, maybe I should consider whether 10 is a typical number. The average is 8, so 10 is two more than the mean. The Poisson distribution is skewed, but 10 isn't extremely unlikely. The probability is about 10%, which is not negligible.However, if the painting is supposed to be from a time when the artist was known for more vigorous brushstrokes, maybe 10 is plausible. If it's from a later period when the artist's style changed, maybe 10 is too high or too low.But without more context, it's hard to say definitively. However, the age of 210 years is quite specific. If the artist's work is typically from that era, it's consistent. If not, it's a red flag.So, considering both factors, if the age aligns with the artist's known period and the brushstroke count is within an expected range, the painting might be authentic. If either the age or the brushstroke count is inconsistent with the artist's known work, it could be a forgery.But based solely on these calculations, the age suggests it's quite old, and the brushstroke probability isn't extremely low. So, unless there's other evidence, it's hard to conclude it's a forgery. Maybe more analysis is needed.Wait, but the age is 210 years, which is a long time. If the pigment decay model is accurate, that's a strong indicator. But forgeries can sometimes use old pigments or techniques to mimic age. So, it's not foolproof.Similarly, the brushstroke pattern could be replicated by a skilled forger. So, while these are indicators, they aren't absolute proof.In conclusion, based on these two analyses, the painting appears to be about 210 years old with a reasonable brushstroke count. If the artist's work is known to be from that time and the brushstroke style matches, it's likely authentic. If not, it might be a forgery. But without additional information, it's a bit of a toss-up.But since the problem asks to determine whether the painting is likely to be authentic based on these calculations, I think the age is a strong point. If the artist is known to have painted around that time, it's more likely authentic. The brushstroke probability isn't high, but it's not low enough to rule it out.So, tentatively, I would say the painting is likely authentic, but more evidence might be needed for a definitive conclusion.Final AnswerThe painting is estimated to be boxed{210} years old, and the probability of exactly 10 brushstrokes is approximately 9.92%. Based on these calculations, the painting is likely authentic.boxed{210}boxed{0.0992}</think>

🧠:<think>Okay, so I have these two math problems related to the Sultanate of Malacca. Let me try to figure them out one by one.Starting with the first problem: It says that the Sultanate imposed a tax on passing ships proportional to the square of the ship's cargo weight. The total tax collected in a year is modeled by the function ( T(x) = k int_{0}^{L} x^2 , dx ), where ( k ) is a constant and ( L ) is the total cargo weight. They give me ( k = 5 ) and ( L = 100 ) units. I need to find the total tax collected.Hmm, okay. So, this is an integral problem. The function ( T(x) ) is defined as the integral from 0 to L of ( x^2 ) dx, multiplied by a constant k. So, I need to compute that integral first and then multiply by k.Let me recall how to compute the integral of ( x^2 ). The integral of ( x^n ) is ( frac{x^{n+1}}{n+1} ), right? So, for ( x^2 ), the integral should be ( frac{x^3}{3} ). Let me write that down.So, ( int_{0}^{L} x^2 , dx = left[ frac{x^3}{3} right]_0^L = frac{L^3}{3} - frac{0^3}{3} = frac{L^3}{3} ).Okay, so the integral simplifies to ( frac{L^3}{3} ). Then, the total tax ( T ) is ( k ) times that. So, ( T = k times frac{L^3}{3} ).Given that ( k = 5 ) and ( L = 100 ), let's plug those values in.First, compute ( L^3 ). 100 cubed is 1,000,000. So, ( 100^3 = 1,000,000 ).Then, divide that by 3: ( frac{1,000,000}{3} ) is approximately 333,333.333...But since we're dealing with exact values, I should keep it as a fraction. So, ( frac{1,000,000}{3} ).Then, multiply by k, which is 5: ( 5 times frac{1,000,000}{3} = frac{5,000,000}{3} ).Hmm, so that's the total tax. Let me see if I can write that as a mixed number or something, but I think as an improper fraction, it's fine. Alternatively, if I convert it to a decimal, it's approximately 1,666,666.666..., but since the question doesn't specify, maybe I should leave it as ( frac{5,000,000}{3} ).Wait, let me double-check my steps. The integral of ( x^2 ) from 0 to L is indeed ( frac{L^3}{3} ). Then, multiplying by k gives ( frac{k L^3}{3} ). Plugging in k=5 and L=100, that's ( frac{5 times 100^3}{3} ). 100 cubed is 1,000,000, so 5 times that is 5,000,000. Divided by 3, that's correct. So, I think that's right.Alright, moving on to the second problem. It says that the population followed an exponential growth model: ( P(t) = P_0 e^{rt} ). They tell me that the population doubled over a period of 25 years, and I need to find the growth rate ( r ).Okay, so exponential growth. The formula is given, and I know that when the population doubles, ( P(t) = 2 P_0 ). So, at time t=25, the population is twice the initial population.So, plugging into the formula: ( 2 P_0 = P_0 e^{r times 25} ).I can divide both sides by ( P_0 ) to simplify, since ( P_0 ) is not zero. That gives me ( 2 = e^{25r} ).Now, to solve for ( r ), I need to take the natural logarithm of both sides. Remember, the natural logarithm is the inverse of the exponential function with base e.So, taking ln of both sides: ( ln(2) = ln(e^{25r}) ).Simplify the right side: ( ln(e^{25r}) = 25r ), because ( ln(e^x) = x ).So, now we have ( ln(2) = 25r ).To solve for ( r ), divide both sides by 25: ( r = frac{ln(2)}{25} ).Hmm, that seems straightforward. Let me compute the numerical value if needed, but since the question doesn't specify, I think expressing it in terms of natural logarithm is acceptable. However, sometimes people prefer to write it as a decimal.Calculating ( ln(2) ) is approximately 0.6931. So, ( r approx frac{0.6931}{25} approx 0.027724 ). So, approximately 0.0277 per year.But let me check if I did everything correctly. The exponential growth model is correct. When the population doubles, ( P(t) = 2 P_0 ). Plugging into the formula, dividing both sides by ( P_0 ), taking natural logs, solving for r. Yes, that seems right.Alternatively, sometimes people use the formula ( r = frac{ln(2)}{t_{double}} ), where ( t_{double} ) is the doubling time. So, in this case, ( t_{double} = 25 ), so ( r = frac{ln(2)}{25} ). Yep, that's consistent.So, I think that's the correct growth rate.Let me just recap both problems to make sure I didn't make any mistakes.First problem: Calculated the integral of ( x^2 ) from 0 to L, which is ( L^3 / 3 ). Then multiplied by k=5 and L=100, so ( 5 * (100^3)/3 = 5,000,000 / 3 ). That seems correct.Second problem: Used the exponential growth formula, set ( P(t) = 2 P_0 ) at t=25, solved for r by taking natural logs, resulting in ( r = ln(2)/25 ). That also seems correct.I don't think I made any calculation errors here. The first problem was straightforward integration, and the second was applying the exponential growth formula. Both steps make sense.So, I think I'm confident with these answers.Final Answer1. The total tax collected is boxed{dfrac{5000000}{3}}.2. The growth rate ( r ) is boxed{dfrac{ln 2}{25}}.</think>