🧠:<think>Okay, so I have this problem about particle motion in three-dimensional space, and I need to find the velocity vector and the total distance traveled. Let me try to break this down step by step.First, the position of the particle is given by the parametric equations:x(t) = e^{kt} cos(ωt)y(t) = e^{kt} sin(ωt)z(t) = e^{kt}I remember that the velocity vector is the derivative of the position vector with respect to time. So, I need to find the derivatives of x(t), y(t), and z(t) individually.Starting with x(t). The function is e^{kt} multiplied by cos(ωt). This looks like a product of two functions, so I should use the product rule for differentiation. The product rule states that d/dt [u*v] = u’v + uv’. Here, u = e^{kt} and v = cos(ωt).Let me compute u’ first. The derivative of e^{kt} with respect to t is k*e^{kt} because the derivative of e^{at} is a*e^{at}.Now, v’ is the derivative of cos(ωt). The derivative of cos(at) is -a sin(at), so here it would be -ω sin(ωt).Putting it together, the derivative of x(t) is:dx/dt = k*e^{kt} cos(ωt) + e^{kt} (-ω sin(ωt)) = e^{kt} (k cos(ωt) - ω sin(ωt))Okay, that seems right. Let me check the signs: the second term is negative because of the derivative of cosine, so yes, it's minus ω sin(ωt).Now moving on to y(t). Similarly, y(t) is e^{kt} sin(ωt). Again, a product of two functions, so product rule applies.u = e^{kt}, so u’ = k e^{kt}v = sin(ωt), so v’ = ω cos(ωt)So, dy/dt = k e^{kt} sin(ωt) + e^{kt} ω cos(ωt) = e^{kt} (k sin(ωt) + ω cos(ωt))That looks good. The derivative of sine is positive cosine, so the second term is positive.Now z(t) is e^{kt}, so its derivative is straightforward. dz/dt = k e^{kt}So, putting it all together, the velocity vector v(t) is:v(t) = [dx/dt, dy/dt, dz/dt] = [e^{kt}(k cos ωt - ω sin ωt), e^{kt}(k sin ωt + ω cos ωt), k e^{kt}]Hmm, that seems correct. Let me just verify each component again.For x(t): derivative is k e^{kt} cos ωt - ω e^{kt} sin ωt. Yep, that's right.For y(t): derivative is k e^{kt} sin ωt + ω e^{kt} cos ωt. Correct.For z(t): derivative is k e^{kt}. Yep.So, velocity vector is done.Now, the second part is to calculate the total distance traveled from t=0 to t=T. The total distance is the integral of the magnitude of the velocity vector over time. So, I need to find |v(t)| and integrate it from 0 to T.First, let's find |v(t)|. The magnitude of the velocity vector is the square root of the sum of the squares of its components.So,|v(t)| = sqrt[(dx/dt)^2 + (dy/dt)^2 + (dz/dt)^2]Let me compute each component squared.First, (dx/dt)^2 = [e^{kt}(k cos ωt - ω sin ωt)]^2 = e^{2kt} (k cos ωt - ω sin ωt)^2Similarly, (dy/dt)^2 = [e^{kt}(k sin ωt + ω cos ωt)]^2 = e^{2kt} (k sin ωt + ω cos ωt)^2And (dz/dt)^2 = [k e^{kt}]^2 = k^2 e^{2kt}So, adding them up:|v(t)|^2 = e^{2kt} ( (k cos ωt - ω sin ωt)^2 + (k sin ωt + ω cos ωt)^2 ) + k^2 e^{2kt}Wait, actually, no. Wait, (dz/dt)^2 is already k^2 e^{2kt}, so when we add all three, it's e^{2kt} times the sum of the squares of the x and y components plus k^2 e^{2kt}.But let me compute the sum inside first.Let me compute (k cos ωt - ω sin ωt)^2 + (k sin ωt + ω cos ωt)^2.Expanding the first square:= k^2 cos^2 ωt - 2kω cos ωt sin ωt + ω^2 sin^2 ωtExpanding the second square:= k^2 sin^2 ωt + 2kω sin ωt cos ωt + ω^2 cos^2 ωtNow, adding these two together:k^2 cos^2 ωt - 2kω cos ωt sin ωt + ω^2 sin^2 ωt + k^2 sin^2 ωt + 2kω sin ωt cos ωt + ω^2 cos^2 ωtLet me combine like terms.The -2kω cos ωt sin ωt and +2kω sin ωt cos ωt cancel each other out.Then, we have:k^2 cos^2 ωt + k^2 sin^2 ωt + ω^2 sin^2 ωt + ω^2 cos^2 ωtFactor out k^2 and ω^2:= k^2 (cos^2 ωt + sin^2 ωt) + ω^2 (sin^2 ωt + cos^2 ωt)But cos^2 + sin^2 = 1, so this simplifies to:= k^2 * 1 + ω^2 * 1 = k^2 + ω^2Wow, that's nice! So, the sum of the squares of the x and y components is (k^2 + ω^2) e^{2kt}But wait, no. Wait, actually, the sum inside was (k cos ωt - ω sin ωt)^2 + (k sin ωt + ω cos ωt)^2 = k^2 + ω^2, so when multiplied by e^{2kt}, it's (k^2 + ω^2) e^{2kt}Then, adding the (dz/dt)^2 which is k^2 e^{2kt}, so total |v(t)|^2 is:(k^2 + ω^2) e^{2kt} + k^2 e^{2kt} = (k^2 + ω^2 + k^2) e^{2kt} = (2k^2 + ω^2) e^{2kt}Wait, hold on. Wait, no. Wait, the sum of the x and y components squared is (k^2 + ω^2) e^{2kt}, and then we have the z component squared which is k^2 e^{2kt}. So total |v(t)|^2 is (k^2 + ω^2 + k^2) e^{2kt} = (2k^2 + ω^2) e^{2kt}Wait, that seems a bit off. Let me double-check.Wait, no. Wait, the x and y components squared sum to (k^2 + ω^2) e^{2kt}, and the z component squared is k^2 e^{2kt}. So total |v(t)|^2 is (k^2 + ω^2) e^{2kt} + k^2 e^{2kt} = (2k^2 + ω^2) e^{2kt}Yes, that's correct.So, |v(t)| = sqrt( (2k^2 + ω^2) e^{2kt} ) = sqrt(2k^2 + ω^2) * e^{kt}Because sqrt(e^{2kt}) is e^{kt}.So, |v(t)| = sqrt(2k^2 + ω^2) * e^{kt}Therefore, the total distance traveled is the integral from t=0 to t=T of |v(t)| dt, which is:Integral from 0 to T of sqrt(2k^2 + ω^2) * e^{kt} dtSince sqrt(2k^2 + ω^2) is a constant with respect to t, we can factor it out:sqrt(2k^2 + ω^2) * Integral from 0 to T of e^{kt} dtThe integral of e^{kt} dt is (1/k) e^{kt} + CSo, evaluating from 0 to T:sqrt(2k^2 + ω^2) * [ (1/k) e^{kT} - (1/k) e^{0} ] = sqrt(2k^2 + ω^2) * (1/k) (e^{kT} - 1)Therefore, the total distance traveled is (sqrt(2k^2 + ω^2)/k) (e^{kT} - 1)Wait, let me make sure I didn't make a mistake in the integration.Yes, integral of e^{kt} is (1/k) e^{kt}, so evaluated from 0 to T gives (1/k)(e^{kT} - 1). Then multiplied by sqrt(2k^2 + ω^2). So yes, that seems correct.So, to recap:1. Velocity vector v(t) is [e^{kt}(k cos ωt - ω sin ωt), e^{kt}(k sin ωt + ω cos ωt), k e^{kt}]2. The magnitude |v(t)| simplifies to sqrt(2k^2 + ω^2) e^{kt}3. The total distance is integral of |v(t)| from 0 to T, which is (sqrt(2k^2 + ω^2)/k)(e^{kT} - 1)I think that's it. Let me just verify the steps again.Differentiated each component correctly, used product rule, got the derivatives right. Then, when computing |v(t)|, expanded the squares, noticed that cross terms cancel, ended up with k^2 + ω^2 for x and y components, added z component squared which is k^2 e^{2kt}, so total inside the sqrt was (2k^2 + ω^2) e^{2kt}, so sqrt of that is sqrt(2k^2 + ω^2) e^{kt}. Then integrated that from 0 to T, got the expression.Yes, seems solid.Final Answer1. The velocity vector is (boxed{mathbf{v}(t) = e^{kt} left( (k cos(omega t) - omega sin(omega t)) mathbf{i} + (k sin(omega t) + omega cos(omega t)) mathbf{j} + k mathbf{k} right)}).2. The total distance traveled is (boxed{frac{sqrt{2k^2 + omega^2}}{k} left( e^{kT} - 1 right)}).</think>

🧠:<think>Okay, so I need to figure out the expected number of shutdown days per year and the expected financial loss because of a critical security breach. Let me break this down step by step.First, the risk assessment part. The probability of a critical security breach in any given month is 3%. Hmm, so that's 3% each month. Since there are 12 months in a year, I guess I need to calculate the probability of at least one breach occurring in a year. But wait, the question is about the expected number of shutdown days, not just the probability of a breach. So maybe I don't need to calculate the probability of at least one breach, but rather the expected number of breaches and then the expected shutdown days per breach.Let me think. The expected number of breaches per year would be the probability per month times the number of months. So that's 0.03 * 12 = 0.36 breaches per year on average. Okay, so about 0.36 breaches expected each year.Now, if a breach occurs, there's a 60% chance the shutdown lasts more than 3 days. But the question mentions a potential 5-day shutdown. Wait, does that mean that if the shutdown is more than 3 days, it's exactly 5 days? Or is 5 days the maximum? The wording says "a potential 5-day shutdown," so maybe if it's more than 3 days, it's 5 days. So perhaps if a breach happens, there's a 60% chance of a 5-day shutdown and a 40% chance of a shutdown of 3 days or less. But the question is about the expected number of days, so I need to consider both possibilities.Wait, actually, the problem says "the probability of a shutdown lasting more than 3 days is 60%." So if a breach occurs, 60% chance it's more than 3 days, which I think we can assume is 5 days as per the question's mention of a 5-day shutdown. So maybe the shutdown duration is either 5 days with 60% probability or less than or equal to 3 days with 40% probability.But the question is about the expected number of shutdown days per year. So perhaps I should model it as, for each breach, the expected shutdown duration is 0.6*5 + 0.4*3? Wait, no, because if it's more than 3 days, it's 5 days, but if it's less than or equal to 3 days, what's the exact duration? The problem doesn't specify, so maybe we can assume that if it's more than 3 days, it's exactly 5 days, and if it's less, it's 3 days? Or maybe it's 0 days? Hmm, no, that doesn't make sense because a shutdown would have to occur if there's a breach.Wait, actually, the problem says "the probability of a shutdown lasting more than 3 days is 60%." So perhaps the shutdown duration is either 5 days with 60% probability or 3 days with 40% probability. Because if it's more than 3 days, it's 5 days, and if it's not, it's 3 days. That seems reasonable.So for each breach, the expected shutdown duration is 0.6*5 + 0.4*3 = 3 + 1.2 = 4.2 days.But wait, the problem says "a potential 5-day shutdown," so maybe the shutdown is exactly 5 days if it's more than 3 days. So the expected duration per breach is 0.6*5 + 0.4*0? No, that can't be because a shutdown would still occur even if it's less than 3 days. Maybe the shutdown is at least 1 day, but the problem doesn't specify. Hmm, this is a bit confusing.Wait, let's read the problem again. It says, "the probability of a shutdown lasting more than 3 days is 60%." So if a breach occurs, the shutdown could be either more than 3 days (which is 5 days as per the question) or less than or equal to 3 days. But the problem doesn't specify the exact duration for the less than or equal case. Maybe we can assume that if it's more than 3 days, it's 5 days, and if it's less, it's 3 days? Or maybe it's 1 day? Hmm, the problem doesn't specify, so perhaps we need to make an assumption.Alternatively, maybe the shutdown duration is 5 days with 60% probability and 0 days with 40% probability, but that doesn't make sense because a shutdown would occur regardless. So perhaps the shutdown is at least 1 day, but the problem doesn't specify. This is a bit unclear.Wait, maybe the problem is simplifying it by saying that if a breach occurs, there's a 60% chance of a 5-day shutdown and a 40% chance of no shutdown? But that contradicts the idea that a breach would cause a shutdown. So perhaps the shutdown is certain, but the duration is variable.Given that, perhaps the expected shutdown duration per breach is 0.6*5 + 0.4*3 = 4.2 days. Because if it's more than 3 days, it's 5 days, and if it's less, it's 3 days. That seems reasonable.So, the expected number of breaches per year is 0.36, and each breach causes an expected shutdown of 4.2 days. Therefore, the expected number of shutdown days per year is 0.36 * 4.2.Let me calculate that: 0.36 * 4.2. 0.36 * 4 = 1.44, and 0.36 * 0.2 = 0.072, so total is 1.44 + 0.072 = 1.512 days per year.Wait, but let me double-check. The expected number of breaches is 0.36 per year, and each breach leads to an expected shutdown of 4.2 days. So 0.36 * 4.2 = 1.512 days. So approximately 1.51 days per year.Okay, that seems reasonable.Now, moving on to the financial impact. The cost is 50,000 per day of shutdown. Additionally, there's a 70% chance that for each day of shutdown, the company loses an additional 10,000 in potential contracts due to reputational damage.So, for each day of shutdown, the expected cost is 50,000 plus 0.7*10,000 = 50,000 + 7,000 = 57,000 per day.Therefore, the expected financial loss per year is the expected number of shutdown days multiplied by 57,000.We already have the expected shutdown days as 1.512, so 1.512 * 57,000.Let me calculate that. 1 * 57,000 = 57,000. 0.512 * 57,000. Let's compute 0.5 * 57,000 = 28,500. 0.012 * 57,000 = 684. So total is 28,500 + 684 = 29,184. Therefore, total expected loss is 57,000 + 29,184 = 86,184.Wait, but hold on. Is that correct? Because the expected number of shutdown days is 1.512, so multiplying by 57,000 gives 1.512 * 57,000.Alternatively, perhaps it's better to compute it as 1.512 * (50,000 + 0.7*10,000) = 1.512 * 57,000.Yes, that's the same as above.So 1.512 * 57,000. Let me compute it more accurately.First, 1 * 57,000 = 57,000.0.5 * 57,000 = 28,500.0.012 * 57,000 = 684.So adding them together: 57,000 + 28,500 = 85,500 + 684 = 86,184.So the expected financial loss per year is 86,184.Wait, but let me think again. The expected number of shutdown days is 1.512, and each day has an expected cost of 57,000. So yes, 1.512 * 57,000 = 86,184.Alternatively, perhaps we should calculate it as:Expected shutdown days: 1.512Expected cost per shutdown day: 50,000 + 0.7*10,000 = 57,000.Total expected cost: 1.512 * 57,000 = 86,184.Yes, that seems correct.So, summarizing:1. Expected shutdown days per year: 1.512 days.2. Expected financial loss per year: 86,184.But let me check if I interpreted the shutdown duration correctly. The problem says "a potential 5-day shutdown due to a critical security breach." So if a breach occurs, the shutdown could be 5 days with 60% probability, and perhaps less than or equal to 3 days with 40% probability. But the problem doesn't specify the exact duration for the less than or equal case. So maybe I should assume that if it's more than 3 days, it's 5 days, and if it's less, it's 3 days? Or maybe it's 1 day? Hmm.Wait, actually, the problem says "the probability of a shutdown lasting more than 3 days is 60%." So if it's more than 3 days, it's 5 days, and if it's less than or equal to 3 days, it's 3 days. So the expected duration per breach is 0.6*5 + 0.4*3 = 4.2 days, as I calculated earlier.So that part seems correct.Alternatively, if the shutdown is either 5 days with 60% probability or 0 days with 40% probability, but that doesn't make sense because a shutdown would occur regardless. So I think my initial approach is correct.Therefore, the expected shutdown days per year are 1.512, and the expected financial loss is 86,184.Wait, but let me think about the financial impact again. The cost is 50,000 per day, and there's a 70% chance of losing an additional 10,000 per day. So for each day, the expected loss is 50,000 + 0.7*10,000 = 57,000. So that's correct.Therefore, multiplying 1.512 days by 57,000 gives the total expected loss.Yes, that seems right.So, to recap:1. Expected shutdown days = (Probability of breach per month * 12 months) * Expected shutdown duration per breach.Which is (0.03 * 12) * (0.6*5 + 0.4*3) = 0.36 * 4.2 = 1.512 days.2. Expected financial loss = Expected shutdown days * (50,000 + 0.7*10,000) = 1.512 * 57,000 = 86,184.So, the answers are approximately 1.51 days and 86,184.But let me check the calculations again to be sure.First, expected number of breaches per year: 0.03 per month * 12 = 0.36.Expected shutdown duration per breach: 0.6*5 + 0.4*3 = 3 + 1.2 = 4.2 days.So expected shutdown days per year: 0.36 * 4.2 = 1.512 days.Financial loss: 1.512 * (50,000 + 0.7*10,000) = 1.512 * 57,000.Calculating 1.512 * 57,000:1 * 57,000 = 57,0000.5 * 57,000 = 28,5000.012 * 57,000 = 684Adding them: 57,000 + 28,500 = 85,500 + 684 = 86,184.Yes, that's correct.So, the expected number of shutdown days is 1.512, and the expected financial loss is 86,184.I think that's the solution.</think>

🧠:<think>Okay, so I need to calculate the weighted similarity score between two articles, A and B, using cosine similarity combined with a time decay function. Hmm, let me break this down step by step because I'm still getting the hang of these concepts.First, I remember that cosine similarity measures how similar two vectors are regardless of their magnitude. The formula is the dot product of the vectors divided by the product of their magnitudes. So, I need to compute the dot product of A and B, then find the magnitudes of both vectors.Let me write down the vectors again to make sure I have them right:- Article A: (1, 3, 4, 0, 2)- Article B: (2, 1, 2, 3, 1)Alright, starting with the dot product. The dot product is calculated by multiplying corresponding components and then summing them up.So, let's compute each component:1*2 = 23*1 = 34*2 = 80*3 = 02*1 = 2Now, adding these up: 2 + 3 + 8 + 0 + 2 = 15. So, the dot product is 15.Next, I need the magnitudes of both vectors. The magnitude is the square root of the sum of the squares of each component.For vector A:1² = 13² = 94² = 160² = 02² = 4Adding these: 1 + 9 + 16 + 0 + 4 = 30So, the magnitude of A is sqrt(30). Let me calculate that. Sqrt(30) is approximately 5.477, but I'll keep it as sqrt(30) for now to be precise.For vector B:2² = 41² = 12² = 43² = 91² = 1Adding these: 4 + 1 + 4 + 9 + 1 = 19So, the magnitude of B is sqrt(19). That's approximately 4.358, but again, I'll keep it as sqrt(19).Now, the cosine similarity is the dot product divided by the product of the magnitudes. So:cosine_similarity = 15 / (sqrt(30) * sqrt(19))Let me compute the denominator first. sqrt(30)*sqrt(19) is sqrt(30*19) = sqrt(570). Hmm, sqrt(570) is approximately 23.874. So, 15 divided by 23.874 is roughly 0.628. But I should keep it exact for now.So, cosine_similarity = 15 / sqrt(570). Maybe I can simplify sqrt(570). Let me see: 570 factors into 5*114, which is 5*2*57, which is 5*2*3*19. So, no perfect squares except 1, so sqrt(570) remains as is.Alright, moving on to the time decay function. The formula is f(t) = e^(-λt). We have λ = 0.1, t_A = 2 days, and t_B = 3 days.So, f(t_A) = e^(-0.1*2) = e^(-0.2)Similarly, f(t_B) = e^(-0.1*3) = e^(-0.3)I need to compute these exponential values. Let me recall that e^(-0.2) is approximately 0.8187 and e^(-0.3) is approximately 0.7408. I can use these approximate values for calculation.So, f(t_A) ≈ 0.8187f(t_B) ≈ 0.7408Now, the weighted similarity score W(A, B) is the cosine similarity multiplied by f(t_A) and f(t_B). So:W(A, B) = (15 / sqrt(570)) * 0.8187 * 0.7408Let me compute this step by step.First, compute 15 / sqrt(570). Let's calculate sqrt(570):sqrt(570) ≈ 23.874So, 15 / 23.874 ≈ 0.628Now, multiply this by 0.8187:0.628 * 0.8187 ≈ 0.514Then, multiply by 0.7408:0.514 * 0.7408 ≈ 0.380So, approximately, the weighted similarity score is 0.38.Wait, let me double-check my calculations to make sure I didn't make a mistake.First, the dot product: 1*2 + 3*1 + 4*2 + 0*3 + 2*1 = 2 + 3 + 8 + 0 + 2 = 15. That's correct.Magnitudes:A: sqrt(1 + 9 + 16 + 0 + 4) = sqrt(30) ≈5.477B: sqrt(4 + 1 + 4 + 9 + 1) = sqrt(19) ≈4.358Cosine similarity: 15 / (5.477 * 4.358) ≈15 / 23.874 ≈0.628. That's correct.Time decay:f(t_A) = e^(-0.2) ≈0.8187f(t_B) = e^(-0.3) ≈0.7408Multiplying all together: 0.628 * 0.8187 * 0.7408Let me compute 0.628 * 0.8187 first:0.628 * 0.8 = 0.50240.628 * 0.0187 ≈0.01173Adding together: ≈0.5024 + 0.01173 ≈0.5141Then, 0.5141 * 0.7408 ≈0.5 * 0.7408 = 0.37040.0141 * 0.7408 ≈0.01045Adding together: ≈0.3704 + 0.01045 ≈0.38085So, approximately 0.3809. Rounding to three decimal places, that's 0.381.Wait, but let me check if I should carry more decimal places for accuracy.Alternatively, maybe I can compute it more precisely using exact values.Alternatively, perhaps I can compute it step by step with more precise intermediate steps.Alternatively, maybe I can compute it using logarithms or something, but that might complicate.Alternatively, perhaps I can compute the exact value symbolically first.So, W(A,B) = (15 / sqrt(570)) * e^(-0.2) * e^(-0.3) = (15 / sqrt(570)) * e^(-0.5)Because -0.2 + (-0.3) = -0.5.So, e^(-0.5) is approximately 0.6065.So, then W(A,B) = (15 / sqrt(570)) * 0.6065We already have 15 / sqrt(570) ≈0.628So, 0.628 * 0.6065 ≈0.6 * 0.6065 = 0.36390.028 * 0.6065 ≈0.01698Adding together: ≈0.3639 + 0.01698 ≈0.3809Same result as before. So, approximately 0.381.Therefore, the weighted similarity score is approximately 0.381.But let me see if I can compute it more accurately.Compute e^(-0.5):e^(-0.5) ≈0.60653066Compute 15 / sqrt(570):sqrt(570) ≈23.874281515 / 23.8742815 ≈0.6280313Multiply by e^(-0.5):0.6280313 * 0.60653066 ≈Let me compute 0.6280313 * 0.6 = 0.376818780.6280313 * 0.00653066 ≈0.004096Adding together: ≈0.37681878 + 0.004096 ≈0.38091478So, approximately 0.3809, which is about 0.381.So, rounding to three decimal places, 0.381.Alternatively, if I use more precise values:Compute 15 / sqrt(570):sqrt(570) = sqrt(570) ≈23.874281515 / 23.8742815 ≈0.6280313e^(-0.5) ≈0.60653066Multiplying these:0.6280313 * 0.60653066Let me compute this more accurately:0.6280313 * 0.6 = 0.376818780.6280313 * 0.00653066 ≈First, 0.6280313 * 0.006 = 0.00376818780.6280313 * 0.00053066 ≈0.0003333Adding these: ≈0.0037681878 + 0.0003333 ≈0.0041015So, total ≈0.37681878 + 0.0041015 ≈0.38092028So, approximately 0.38092, which is about 0.381.Therefore, the weighted similarity score is approximately 0.381.I think that's as precise as I can get without a calculator. So, the final answer is approximately 0.381.</think>

🧠:<think>Okay, so I have this problem about Manoel Victor Cavalcante and his collection of rare books. Each book has a unique ID from 1 to n, and he has a special subset called "The Prime Collection," which includes all the books with prime numbers as their IDs. The first part asks me to prove that the number of primes less than or equal to n, denoted as π(n), is greater than n divided by the natural logarithm of n for sufficiently large n. Hmm, I remember that π(n) is approximately n / ln(n) from the Prime Number Theorem, but I need to prove that π(n) is actually greater than n / ln(n) for large n. I think the Prime Number Theorem gives an asymptotic estimate, meaning that as n becomes very large, π(n) approaches n / ln(n). But the problem is asking to show that π(n) is greater than n / ln(n) for sufficiently large n. Maybe I can use some bounds on π(n). I recall that there are explicit bounds for π(n). For example, I think Rosser's theorem states that π(n) > n / ln(n) for all n ≥ 1. Is that correct? Let me check. Yes, Rosser's theorem indeed proves that π(n) > n / ln(n) for all n ≥ 1. So, that should suffice for the first part. I just need to reference Rosser's theorem or maybe provide a sketch of the proof. But since the problem says "prove," I might need to outline the main ideas. Rosser's proof involves showing that the nth prime is greater than n ln(n), which then implies that π(n) is greater than n / ln(n). Alternatively, maybe I can use the integral test or some properties of the prime-counting function. Wait, another approach is to use the fact that the density of primes decreases, but the exact count is still more than n / ln(n). Alternatively, I can consider the integral of 1 / ln(t) from 2 to n, which approximates π(n). Maybe I can compare it to n / ln(n). Let me think. The integral of 1 / ln(t) dt from 2 to n is approximately Li(n), which is a better approximation for π(n). But I need to show that π(n) > n / ln(n). Maybe I can use the fact that Li(n) is greater than n / ln(n) for sufficiently large n. Since Li(n) is a better approximation for π(n), and it's known that π(n) is asymptotic to Li(n), which is greater than n / ln(n). But I need a more concrete argument. Alternatively, perhaps I can use the fact that the sum of the reciprocals of primes diverges. But that might not directly help. Maybe I can use some inequalities involving the prime-counting function. Wait, I think I can use the fact that π(n) is bounded below by n / (ln(n) + 2) for n ≥ 41, which is a result from Rosser and Schoenfeld. Since n / (ln(n) + 2) is greater than n / ln(n) for sufficiently large n, this would imply π(n) > n / ln(n). So, if I can show that n / (ln(n) + 2) > n / ln(n) for sufficiently large n, then π(n) > n / ln(n). Let's see: n / (ln(n) + 2) > n / ln(n) simplifies to 1 / (ln(n) + 2) > 1 / ln(n), which is true because ln(n) + 2 < ln(n) for large n? Wait, no, that's not true. ln(n) + 2 is greater than ln(n), so 1 / (ln(n) + 2) is less than 1 / ln(n). Hmm, that seems contradictory. Maybe I made a mistake here. Wait, actually, if π(n) > n / (ln(n) + 2), and n / (ln(n) + 2) is less than n / ln(n), then π(n) > n / (ln(n) + 2) doesn't directly imply π(n) > n / ln(n). So maybe that approach isn't helpful. Perhaps I need to go back to Rosser's theorem. Rosser proved that for n ≥ 2, π(n) > n / ln(n). So, actually, it's already proven for all n ≥ 2, so for sufficiently large n, it's definitely true. So, maybe I just need to state that according to Rosser's theorem, π(n) > n / ln(n) for all n ≥ 2, which certainly holds for sufficiently large n. Alright, that seems acceptable for the first part. Moving on to the second part. Manoel wants to arrange his "Prime Collection" books in a special sequence such that the sum of the identification numbers of every consecutive pair is also a prime number. I need to determine the conditions under which such an arrangement is possible and provide a method to construct the sequence if possible. This sounds like constructing a Hamiltonian path in a graph where each node is a prime number, and edges connect primes whose sum is also prime. So, essentially, we're looking for a permutation of the primes up to n where each adjacent pair sums to a prime. First, let's think about the properties required. For such a sequence to exist, the graph must have a Hamiltonian path. The graph is defined with primes as vertices, and edges between primes p and q if p + q is also prime. I need to figure out under what conditions on n such a path exists. Let's consider small cases first. Let's take n = 2. The primes are [2]. There's only one book, so trivially, the sequence is just [2]. n = 3. Primes are [2, 3]. The sum is 2 + 3 = 5, which is prime. So the sequence can be [2, 3] or [3, 2]. n = 5. Primes are [2, 3, 5]. Let's see possible sequences. Start with 2: Next can be 3 (2+3=5, prime) or 5 (2+5=7, prime). If we go 2, 3: Then next can be 5? 3 + 5 = 8, which is not prime. So that doesn't work. If we go 2, 5: Then next can be 3? 5 + 3 = 8, not prime. So that doesn't work either. Alternatively, start with 3: Next can be 2 or 5. 3, 2: Then next can be 5? 2 + 5 = 7, prime. So 3, 2, 5. That works. Similarly, 5, 2, 3: 5 + 2 = 7, 2 + 3 = 5, both primes. So, for n=5, it's possible. n=7. Primes are [2, 3, 5, 7]. Let's try to arrange them. Let me try starting with 2. 2 can go to 3 (sum 5) or 5 (sum 7) or 7 (sum 9, not prime). So 2 can go to 3 or 5. If I go 2, 3: Next, 3 can go to 2 (already used) or 4 (not prime) or 8 (not prime). Wait, no, the next prime after 3 is 5 or 7. Wait, 3 + 5 = 8 (not prime), 3 + 7 = 10 (not prime). So 3 can't go anywhere else. So that path is stuck. If I go 2, 5: Then 5 can go to 2 (used), 3 (sum 8, not prime), or 7 (sum 12, not prime). So stuck again. Alternatively, start with 3. 3 can go to 2 or 4 (not prime) or 8 (not prime). So 3 can go to 2. 3, 2: Then 2 can go to 5 or 7. 3, 2, 5: Then 5 can go to 2 (used) or 3 (used) or 7 (sum 12, not prime). Stuck. 3, 2, 7: 2 + 7 = 9, not prime. So that's invalid. Alternatively, start with 5. 5 can go to 2 or 3 or 7. 5, 2: Then 2 can go to 3 or 5 (used). 2, 3: 5, 2, 3. Then 3 can go to 2 (used) or 5 (used) or 7 (sum 10, not prime). Stuck. 5, 3: 5 + 3 = 8, not prime. So invalid. 5, 7: 5 + 7 = 12, not prime. Invalid. Start with 7. 7 can go to 2 (sum 9, not prime) or 4 (not prime) or 10 (not prime). So no edges from 7. Wait, so for n=7, is it impossible? Because starting from any prime, we can't form a sequence that includes all primes. But wait, maybe I missed something. Let me try a different approach. Is there a way to arrange [2, 3, 5, 7] such that each consecutive sum is prime? Let's list all possible edges: - 2 can connect to 3 (5), 5 (7), but not 7 (9). - 3 can connect to 2 (5), but 3 + 5 = 8 (not prime), 3 + 7 = 10 (not prime). - 5 can connect to 2 (7), but 5 + 3 = 8 (not prime), 5 + 7 = 12 (not prime). - 7 can't connect to anyone because 7 + 2 = 9, 7 + 3 = 10, 7 + 5 = 12. So, the graph has edges: 2-3, 2-5. That's it. So the graph is two disconnected edges: 2-3 and 2-5, and 7 is isolated. Therefore, it's impossible to have a Hamiltonian path because 7 is isolated and the other primes form a disconnected graph. So, for n=7, it's impossible. Wait, but 7 is a prime, so it must be included. Since 7 can't connect to any other prime in the collection, it's impossible to include it in the sequence without violating the sum condition. Therefore, the condition must be that all primes in the collection can be connected in such a way that each consecutive pair sums to a prime. But in the case of n=7, 7 can't be connected to any other prime, so the arrangement is impossible. So, perhaps the condition is that the graph is connected and has a Hamiltonian path. But how can we ensure that? Alternatively, maybe the only way this is possible is if the primes can be arranged in a sequence where each consecutive pair sums to a prime, which might require that the primes are arranged in a specific order, perhaps alternating between even and odd? Wait, but except for 2, all primes are odd. So, 2 is the only even prime. So, if we have 2 in the collection, it can only be adjacent to primes that are 3 mod 4 or something? Wait, no, 2 + p must be prime. So, 2 + p is prime. Since p is odd (except 2), 2 + p is odd + even = odd, which can be prime. But for p > 2, p is odd, so 2 + p is odd, which could be prime. But for two odd primes, p and q, p + q is even, so it can only be prime if p + q = 2, which is impossible since p and q are at least 2. So, the only way two odd primes can be adjacent is if their sum is 2, which is impossible. Therefore, two odd primes cannot be adjacent in the sequence unless their sum is 2, which is not possible. Wait, that can't be right because in the case of n=5, we had 3, 2, 5, which worked because 3 + 2 = 5 and 2 + 5 = 7, both primes. But if we have two odd primes, say 3 and 5, their sum is 8, which is not prime. So, they can't be adjacent. Therefore, in the sequence, after an odd prime, we must have 2, and after 2, we can have another odd prime. So, the sequence must alternate between odd primes and 2. But 2 can only be used once, right? Because each book is unique. Wait, no, in the sequence, each prime is used exactly once. So, if we have multiple odd primes, we need to interleave them with 2, but since 2 can only be placed once, we can only have one transition from an odd prime to 2 and then to another odd prime. Wait, let's think about it. Suppose we have primes: 2, 3, 5, 7, 11, etc. If we want to arrange them such that consecutive sums are prime, we have to consider that except for 2, all primes are odd. So, if we have two odd primes next to each other, their sum is even, which can only be prime if the sum is 2, which is impossible. Therefore, two odd primes cannot be adjacent in the sequence. Therefore, the only way to have two odd primes in the sequence is if they are separated by 2. But since 2 can only be used once, we can only have one such transition. Wait, but in the case of n=5, we had 3, 2, 5. So, 3 is odd, 2 is even, 5 is odd. That works because 3 + 2 = 5 and 2 + 5 = 7, both primes. Similarly, if we have more primes, say 2, 3, 5, 7, 11. Can we arrange them as 3, 2, 5, 2, 7, 2, 11? No, because 2 can only be used once. Alternatively, we need to find a way to arrange all the odd primes with 2 in between, but since 2 can only be used once, we can only have one transition from odd to even to odd. Wait, but if we have multiple odd primes, we need to have multiple 2s, which we don't have. Therefore, perhaps the only way such a sequence is possible is if there is only one odd prime, meaning n=2 or n=3. But wait, for n=5, we had 3, 2, 5, which worked. So, that's three primes: 2, 3, 5. Wait, but in that case, we have two odd primes and one even prime. So, the sequence alternates between odd, even, odd. But if we have more than two odd primes, say three odd primes, we would need two 2s to separate them, which we don't have. Therefore, perhaps the condition is that the number of odd primes is at most two, meaning that n must be such that the number of primes less than or equal to n is at most three. Wait, but for n=7, we have four primes: 2, 3, 5, 7. So, three odd primes. But as we saw earlier, it's impossible to arrange them because we can't have two odd primes adjacent, and we only have one 2 to separate them. Therefore, the condition is that the number of odd primes is at most two, which would mean that n must be such that π(n) ≤ 3. But π(n) ≤ 3 implies n ≤ 5, because π(5)=3 (primes 2, 3, 5). Wait, but for n=7, π(n)=4, which includes 2, 3, 5, 7. So, three odd primes. Therefore, the condition is that the number of odd primes is at most two, meaning that n must be ≤5. But wait, in the case of n=5, we have three primes: 2, 3, 5. So, two odd primes. Wait, actually, the number of odd primes is π(n) - 1, since 2 is the only even prime. So, if π(n) - 1 ≤ 2, then π(n) ≤ 3, which corresponds to n ≤5. Therefore, the condition is that π(n) ≤ 3, meaning n ≤5. But let's test n=7 again. π(7)=4, which is greater than 3, so it's impossible. Similarly, n=11: π(11)=5, which is greater than 3, so impossible. Wait, but what about n=2? π(2)=1, which is ≤3. Trivially possible. n=3: π(3)=2, which is ≤3. Possible. n=5: π(5)=3, which is ≤3. Possible. n=7: π(7)=4, which is >3. Impossible. So, the condition is that π(n) ≤3, meaning n ≤5. But wait, let me think again. Suppose we have more than three primes, but somehow arrange them in a way that doesn't require multiple 2s. Wait, but as we saw, two odd primes cannot be adjacent, so they must be separated by 2. But since 2 can only be used once, we can only have one such separation. Therefore, the maximum number of odd primes we can have is two, meaning π(n) -1 ≤2, so π(n) ≤3. Therefore, the condition is that π(n) ≤3, which corresponds to n ≤5. But wait, let's consider n=7 again. π(7)=4. If we try to arrange the primes [2, 3, 5, 7], we can't because we can't place 7 anywhere without violating the sum condition. Alternatively, maybe there's a different arrangement where 7 is at the end, but then the previous prime plus 7 must be prime. Let's try: 3, 2, 5, 7. 3 + 2 =5 (prime), 2 +5=7 (prime), 5 +7=12 (not prime). So, invalid. Alternatively, 5, 2, 3, 7. 5 +2=7 (prime), 2 +3=5 (prime), 3 +7=10 (not prime). Invalid. Alternatively, 7, 2, 3, 5. 7 +2=9 (not prime). Invalid. Alternatively, 2, 7, ... but 2 +7=9, not prime. So, no possible arrangement. Therefore, the conclusion is that such an arrangement is only possible if the number of primes is at most 3, i.e., π(n) ≤3, which corresponds to n ≤5. But wait, let me check n=7 again. Maybe there's a way to arrange the primes without using 2 in the middle. Wait, but 2 is the only even prime, so without 2, all other primes are odd, and their sums would be even, hence not prime (except 2). Therefore, without 2, it's impossible to have a sequence of odd primes where consecutive sums are prime. Therefore, 2 must be included, and it must be placed between two odd primes. But since we can only use 2 once, we can only have two odd primes in the sequence. Therefore, the maximum number of primes in the collection is 3: 2, p, q, where p and q are odd primes. Thus, the condition is that π(n) ≤3, which happens when n ≤5. Therefore, the arrangement is possible if and only if n ≤5. But wait, let me think about n=7 again. Suppose we don't include 7. But the problem states that it's the entire Prime Collection, meaning all primes up to n. So, if n=7, we must include 7. Therefore, for n ≥7, it's impossible. But wait, what about n=11? π(11)=5. So, primes are [2, 3, 5, 7, 11]. Is there a way to arrange them? Let's try. Start with 3: 3, 2, 5, 2, 7, 2, 11. But we can't use 2 multiple times. Alternatively, 3, 2, 5, 11. 3 +2=5, 2 +5=7, 5 +11=16 (not prime). Alternatively, 5, 2, 3, 11. 5 +2=7, 2 +3=5, 3 +11=14 (not prime). Alternatively, 7, 2, 5, 3, 11. 7 +2=9 (not prime). Alternatively, 11, 2, 3, 5, 7. 11 +2=13 (prime), 2 +3=5, 3 +5=8 (not prime). Alternatively, 5, 2, 11, 3, 7. 5 +2=7, 2 +11=13, 11 +3=14 (not prime). Alternatively, 3, 2, 11, 5, 7. 3 +2=5, 2 +11=13, 11 +5=16 (not prime). Alternatively, 7, 2, 11, 3, 5. 7 +2=9 (not prime). Alternatively, 2, 3, 5, 11, 7. 2 +3=5, 3 +5=8 (not prime). Alternatively, 2, 5, 3, 11, 7. 2 +5=7, 5 +3=8 (not prime). Alternatively, 2, 11, 3, 5, 7. 2 +11=13, 11 +3=14 (not prime). Alternatively, 2, 7, 3, 5, 11. 2 +7=9 (not prime). So, seems impossible. Therefore, the condition is that the number of primes is at most 3, i.e., n ≤5. But wait, let me think about n=7 again. If we exclude 7, but the problem says the entire Prime Collection, so we can't exclude any primes. Therefore, the conclusion is that such an arrangement is possible if and only if the number of primes is at most 3, which corresponds to n ≤5. Therefore, the conditions are that n must be such that π(n) ≤3, i.e., n ≤5. To construct the sequence, if possible, we can arrange the primes as follows: - If n=2: [2] - If n=3: [2, 3] or [3, 2] - If n=5: [3, 2, 5] or [5, 2, 3] These are the only possible arrangements because any more primes would require placing 2 multiple times or having two odd primes adjacent, which isn't allowed. Therefore, the method is to check if π(n) ≤3. If yes, arrange the primes with 2 in the middle, separating the two odd primes. If not, it's impossible. So, summarizing: 1. π(n) > n / ln(n) for sufficiently large n, proven by Rosser's theorem. 2. The arrangement is possible if and only if n ≤5. The sequence can be constructed by placing 2 between the two odd primes if π(n)=3, or just [2] or [2,3] if π(n)=1 or 2. I think that's the solution.</think>