🧠:<think>Okay, so I have this problem about Markov chains and democratic transitions. Let me try to figure it out step by step. First, part 1 is about verifying if the given matrix P is a valid stochastic matrix. I remember that a stochastic matrix is one where each row sums to 1. So, I need to check each row of P and make sure that the probabilities add up to 1.Looking at the matrix P:First row: 0.7 + 0.2 + 0.1. Let me add those up. 0.7 + 0.2 is 0.9, plus 0.1 is 1.0. Okay, that's good.Second row: 0.4 + 0.4 + 0.2. Adding those: 0.4 + 0.4 is 0.8, plus 0.2 is 1.0. That works too.Third row: 0.3 + 0.3 + 0.4. Hmm, 0.3 + 0.3 is 0.6, plus 0.4 is 1.0. Perfect.So, all rows sum to 1, which means P is indeed a valid stochastic matrix. Cool, that wasn't too bad.Now, the next part is to find the steady-state distribution vector π such that πP = π. I remember that the steady-state vector is a row vector where each element represents the long-term probability of being in that state. To find π, I need to solve the equation πP = π, and also make sure that the sum of the elements in π is 1.So, let's denote π as [π₁, π₂, π₃]. Then, multiplying π by P should give me back π. Let's write out the equations.From the first row of P:π₁ = π₁ * 0.7 + π₂ * 0.4 + π₃ * 0.3From the second row:π₂ = π₁ * 0.2 + π₂ * 0.4 + π₃ * 0.3From the third row:π₃ = π₁ * 0.1 + π₂ * 0.2 + π₃ * 0.4Also, we have the constraint that π₁ + π₂ + π₃ = 1.So, now I have four equations:1. π₁ = 0.7π₁ + 0.4π₂ + 0.3π₃2. π₂ = 0.2π₁ + 0.4π₂ + 0.3π₃3. π₃ = 0.1π₁ + 0.2π₂ + 0.4π₃4. π₁ + π₂ + π₃ = 1Let me rearrange equations 1, 2, and 3 to bring all terms to one side.From equation 1:π₁ - 0.7π₁ - 0.4π₂ - 0.3π₃ = 0Which simplifies to:0.3π₁ - 0.4π₂ - 0.3π₃ = 0Equation 2:π₂ - 0.2π₁ - 0.4π₂ - 0.3π₃ = 0Simplify:-0.2π₁ + 0.6π₂ - 0.3π₃ = 0Equation 3:π₃ - 0.1π₁ - 0.2π₂ - 0.4π₃ = 0Simplify:-0.1π₁ - 0.2π₂ + 0.6π₃ = 0So now, the system of equations is:1. 0.3π₁ - 0.4π₂ - 0.3π₃ = 02. -0.2π₁ + 0.6π₂ - 0.3π₃ = 03. -0.1π₁ - 0.2π₂ + 0.6π₃ = 04. π₁ + π₂ + π₃ = 1Hmm, this looks a bit complicated, but maybe I can express some variables in terms of others.Looking at equation 1: 0.3π₁ = 0.4π₂ + 0.3π₃. Let me write that as:π₁ = (0.4π₂ + 0.3π₃) / 0.3Simplify: π₁ = (4/3)π₂ + π₃Similarly, equation 2: -0.2π₁ + 0.6π₂ - 0.3π₃ = 0Let me plug in π₁ from equation 1 into equation 2.So, equation 2 becomes:-0.2*( (4/3)π₂ + π₃ ) + 0.6π₂ - 0.3π₃ = 0Let me compute each term:-0.2*(4/3 π₂) = - (8/30) π₂ = - (4/15) π₂-0.2*π₃ = -0.2π₃So, putting it all together:- (4/15) π₂ - 0.2π₃ + 0.6π₂ - 0.3π₃ = 0Combine like terms:For π₂: (-4/15 + 0.6) π₂Convert 0.6 to fifteenths: 0.6 = 9/15So, (-4/15 + 9/15) = 5/15 = 1/3For π₃: (-0.2 - 0.3) = -0.5So, equation 2 becomes:(1/3) π₂ - 0.5 π₃ = 0Multiply both sides by 6 to eliminate denominators:2π₂ - 3π₃ = 0So, 2π₂ = 3π₃ => π₂ = (3/2) π₃Okay, so π₂ is 1.5 times π₃.Now, let's go back to equation 1: π₁ = (4/3)π₂ + π₃But since π₂ = (3/2) π₃, substitute that in:π₁ = (4/3)*(3/2 π₃) + π₃ = (4/3)*(3/2) π₃ + π₃Simplify:(4/3)*(3/2) = (4/2) = 2So, π₁ = 2 π₃ + π₃ = 3 π₃So, π₁ = 3 π₃So, now we have π₁ = 3 π₃ and π₂ = 1.5 π₃Now, let's use equation 4: π₁ + π₂ + π₃ = 1Substitute:3 π₃ + 1.5 π₃ + π₃ = 1Add them up:3 + 1.5 + 1 = 5.5So, 5.5 π₃ = 1 => π₃ = 1 / 5.5Convert 5.5 to fraction: 5.5 = 11/2, so π₃ = 2/11Therefore, π₃ = 2/11Then, π₂ = 1.5 * (2/11) = 3/11And π₁ = 3 * (2/11) = 6/11So, the steady-state distribution vector π is [6/11, 3/11, 2/11]Let me check if this satisfies equation 3:Equation 3: -0.1π₁ - 0.2π₂ + 0.6π₃ = 0Plugging in:-0.1*(6/11) - 0.2*(3/11) + 0.6*(2/11)Compute each term:-0.1*(6/11) = -6/110 = -3/55-0.2*(3/11) = -6/110 = -3/550.6*(2/11) = 12/110 = 6/55Adding them up:-3/55 -3/55 + 6/55 = 0/55 = 0Perfect, it works.So, π = [6/11, 3/11, 2/11]Alright, that was part 1. Now, moving on to part 2.Part 2: The publisher wants to introduce a new metric called "resilience," which is the expected number of steps for a country to return to a stable democratic state (state 1) once it leaves. They want to use the steady-state distribution vector π found in part 1 to solve this.I remember that in Markov chains, the expected return time to a state is the reciprocal of its stationary probability. So, if π₁ is the stationary probability of being in state 1, then the expected return time is 1/π₁.Wait, let me recall. The expected return time to state i is indeed 1/π_i, where π_i is the stationary probability. So, if π₁ = 6/11, then the expected return time is 11/6.But let me make sure I'm not making a mistake here. Is this always the case? I think so, as long as the chain is irreducible and aperiodic, which I believe it is here because all the transition probabilities are positive, so it's irreducible and aperiodic.So, since the chain is finite, irreducible, and aperiodic, it's ergodic, and the stationary distribution is unique, and the expected return time is 1/π_i.So, in this case, the expected return time to state 1 is 11/6, which is approximately 1.8333 steps.But let me think again. Is resilience defined as the expected number of steps to return to state 1 once it leaves? So, starting from state 1, the expected number of steps until it returns is 1/π₁. Alternatively, if it leaves state 1, then it's the expected return time.But actually, in Markov chain theory, the expected return time to a state is the expected number of steps to return to that state starting from that state. So, if you start at state 1, the expected time to come back is 1/π₁.But the problem says "once it leaves," so maybe starting from state 1, the expected time until it returns after leaving. But actually, in Markov chains, the expected return time is defined as starting from the state, so it includes the time to leave and come back.Wait, but if you leave state 1, you're in another state, so the expected return time is the expected time to go from state 1 to state 1, which is 1/π₁.Alternatively, if you are in state 1, the expected time until you return is 1/π₁.So, perhaps the resilience is 1/π₁, which is 11/6.Let me double-check this formula.Yes, in Markov chains, the expected return time to state i is indeed 1/π_i, where π_i is the stationary probability of state i. So, in this case, π₁ is 6/11, so the expected return time is 11/6.So, the resilience is 11/6 steps.But just to be thorough, let me think about another way to compute this, maybe using first-step analysis.Suppose we define m_i as the expected return time to state 1 starting from state i.We need to find m_1, but actually, m_1 is the expected return time starting from state 1, which is 1/π₁.But let's see if we can compute it using equations.Define m_i as the expected number of steps to reach state 1 starting from state i.We want to find m_1, but actually, once you leave state 1, you have to come back. Wait, no. If you start at state 1, the expected time to return is m_1.But if you leave state 1, then you're in another state, so the expected time to return is m_j where j ≠ 1.But the problem says "once it leaves," so maybe it's the expected return time after leaving, meaning starting from the next state.Wait, the problem says: "the expected number of steps for a country to return to a stable democratic state (represented by state 1) once it leaves."So, once it leaves state 1, meaning starting from the state it leaves to, which could be state 2 or 3.But the problem doesn't specify which state it leaves to, so maybe we need to compute the expected return time starting from the distribution after leaving state 1.Alternatively, perhaps it's the expected return time starting from the steady-state distribution excluding state 1.Wait, I'm getting confused.Wait, maybe the resilience is the expected return time to state 1 starting from the distribution conditioned on not being in state 1. But that might complicate things.Alternatively, perhaps it's the expected return time starting from state 1, which is 1/π₁.But let me think again.In the problem statement: "the expected number of steps for a country to return to a stable democratic state (represented by state 1) once it leaves."So, once it leaves, meaning starting from the state it leaves to, which is either state 2 or 3.But the problem doesn't specify which state it leaves to, so perhaps we have to compute the expected return time starting from either state 2 or 3, weighted by the probability of leaving to each.Wait, but when you leave state 1, you can go to state 2 with probability 0.2 or state 3 with probability 0.1.So, the expected return time would be the expected number of steps starting from state 2 with probability 0.2/(0.2+0.1) and state 3 with probability 0.1/(0.2+0.1). Wait, no, actually, when you leave state 1, you go to state 2 with probability 0.2 and state 3 with probability 0.1. So, the expected return time is 0.2 * m_2 + 0.1 * m_3, where m_i is the expected return time to state 1 starting from state i.But I think that might not be the case. Alternatively, maybe it's the expected return time starting from state 1, which is 1/π₁.Wait, perhaps I should set up the equations for m_i.Let me define m_i as the expected number of steps to reach state 1 starting from state i.We need to find m_1, m_2, m_3.But actually, m_1 is the expected return time to state 1 starting from state 1, which is 1/π₁.But let's see if we can compute it.We have:For state 1: m_1 = 1 + 0.2 m_2 + 0.1 m_3Because from state 1, you take 1 step, and then with probability 0.2 go to state 2, and with probability 0.1 go to state 3.For state 2: m_2 = 1 + 0.4 m_2 + 0.3 m_3Because from state 2, you take 1 step, and with probability 0.4 stay in state 2, and with probability 0.3 go to state 3.For state 3: m_3 = 1 + 0.3 m_2 + 0.4 m_3Because from state 3, you take 1 step, and with probability 0.3 go to state 2, and with probability 0.4 stay in state 3.Wait, but actually, once you reach state 1, the process stops, so m_1 = 0? Wait, no. Wait, m_i is the expected number of steps to reach state 1 starting from state i. So, if you're already in state 1, m_1 = 0. But if you're in state 1 and you take a step, then m_1 = 1 + ... So, actually, m_1 is the expected return time, which is different.Wait, I think I need to clarify.If m_i is the expected number of steps to reach state 1 starting from state i, then m_1 = 0, because you're already there. But if you start at state 1 and take a step, then m_1 would be 1 + expected time from the next state.But in our case, the problem says "once it leaves," so starting from state 1, you leave, so you have to go to state 2 or 3, and then compute the expected time to return.So, perhaps the resilience is m_1', where m_1' is the expected return time starting from state 1, which is 1 + 0.2 m_2 + 0.1 m_3.But in that case, m_1' is different from m_1.Wait, maybe I need to define m_i as the expected return time to state 1 starting from state i, including the step that leaves state 1.So, m_1 = 1 + 0.2 m_2 + 0.1 m_3Similarly, m_2 = 1 + 0.4 m_2 + 0.3 m_3m_3 = 1 + 0.3 m_2 + 0.4 m_3And m_1 is the expected return time starting from state 1, which is what we need.So, let's write these equations:1. m_1 = 1 + 0.2 m_2 + 0.1 m_32. m_2 = 1 + 0.4 m_2 + 0.3 m_33. m_3 = 1 + 0.3 m_2 + 0.4 m_3We can solve this system of equations.Let me rearrange equations 2 and 3.From equation 2:m_2 - 0.4 m_2 - 0.3 m_3 = 1Simplify:0.6 m_2 - 0.3 m_3 = 1Multiply both sides by 10 to eliminate decimals:6 m_2 - 3 m_3 = 10Equation 2a: 6 m_2 - 3 m_3 = 10From equation 3:m_3 - 0.3 m_2 - 0.4 m_3 = 1Simplify:0.6 m_3 - 0.3 m_2 = 1Multiply both sides by 10:6 m_3 - 3 m_2 = 10Equation 3a: -3 m_2 + 6 m_3 = 10Now, we have equations 2a and 3a:2a: 6 m_2 - 3 m_3 = 103a: -3 m_2 + 6 m_3 = 10Let me write them together:6 m_2 - 3 m_3 = 10-3 m_2 + 6 m_3 = 10Let me add these two equations:(6 m_2 - 3 m_3) + (-3 m_2 + 6 m_3) = 10 + 10Simplify:3 m_2 + 3 m_3 = 20Divide both sides by 3:m_2 + m_3 = 20/3 ≈ 6.6667Let me call this equation 4.Now, let's solve equations 2a and 3a.From equation 2a: 6 m_2 - 3 m_3 = 10From equation 3a: -3 m_2 + 6 m_3 = 10Let me solve for one variable. Let's solve equation 2a for m_2:6 m_2 = 10 + 3 m_3m_2 = (10 + 3 m_3)/6Now, plug this into equation 3a:-3*( (10 + 3 m_3)/6 ) + 6 m_3 = 10Simplify:- (30 + 9 m_3)/6 + 6 m_3 = 10Simplify the fraction:-5 - 1.5 m_3 + 6 m_3 = 10Combine like terms:-5 + 4.5 m_3 = 10Add 5 to both sides:4.5 m_3 = 15Multiply both sides by 2/9:m_3 = 15 * (2/9) = 30/9 = 10/3 ≈ 3.3333Now, plug m_3 = 10/3 into equation 4:m_2 + 10/3 = 20/3So, m_2 = 20/3 - 10/3 = 10/3 ≈ 3.3333Now, we have m_2 = 10/3 and m_3 = 10/3Now, go back to equation 1:m_1 = 1 + 0.2 m_2 + 0.1 m_3Substitute m_2 and m_3:m_1 = 1 + 0.2*(10/3) + 0.1*(10/3)Compute each term:0.2*(10/3) = 2/3 ≈ 0.66670.1*(10/3) = 1/3 ≈ 0.3333So, m_1 = 1 + 2/3 + 1/3 = 1 + 1 = 2Wait, that's interesting. So, m_1 = 2.But earlier, I thought it was 11/6 ≈ 1.8333.Hmm, so which one is correct?Wait, let's check the calculations again.From equation 1:m_1 = 1 + 0.2 m_2 + 0.1 m_3We found m_2 = 10/3 and m_3 = 10/3So, 0.2*(10/3) = 2/30.1*(10/3) = 1/3So, 2/3 + 1/3 = 1Thus, m_1 = 1 + 1 = 2So, according to this, the expected return time is 2 steps.But earlier, using the stationary distribution, I got 11/6 ≈ 1.8333.Hmm, so which one is correct?Wait, perhaps I made a mistake in the setup.Wait, in the first approach, I considered m_i as the expected return time to state 1 starting from state i, including the step that leaves state 1.But in the second approach, using the stationary distribution, the expected return time is 1/π₁ = 11/6 ≈ 1.8333.But according to the equations, it's 2.So, which one is correct?Wait, maybe I messed up the definition of m_i.In the first approach, I defined m_i as the expected number of steps to reach state 1 starting from state i, which would mean m_1 = 0, because you're already there. But in the problem, the resilience is the expected number of steps to return to state 1 once it leaves, so starting from state 1, you leave, so you have to take at least one step.Wait, perhaps the correct way is to define m_i as the expected return time to state 1 starting from state i, which includes the step that leaves state 1.So, in that case, m_1 = 1 + 0.2 m_2 + 0.1 m_3And m_2 and m_3 are defined as the expected return times starting from state 2 and 3, respectively.So, in that case, solving the equations gives m_1 = 2.But according to the stationary distribution, the expected return time is 1/π₁ = 11/6 ≈ 1.8333.So, which one is correct?Wait, maybe I need to reconcile these two results.Wait, the stationary distribution π gives the long-term proportion of time spent in each state. So, the expected return time is 1/π₁, which is 11/6 ≈ 1.8333.But according to the equations, it's 2.So, perhaps I made a mistake in setting up the equations.Wait, let me think again.If m_i is the expected return time to state 1 starting from state i, then for state 1, m_1 = 1 + 0.2 m_2 + 0.1 m_3Because from state 1, you take one step, and then you're in state 2 with probability 0.2 or state 3 with probability 0.1.For state 2, m_2 = 1 + 0.4 m_2 + 0.3 m_3Because from state 2, you take one step, and with probability 0.4 stay in state 2, and with probability 0.3 go to state 3.Similarly, for state 3, m_3 = 1 + 0.3 m_2 + 0.4 m_3Because from state 3, you take one step, and with probability 0.3 go to state 2, and with probability 0.4 stay in state 3.Wait, but in this setup, we have m_1 = 1 + 0.2 m_2 + 0.1 m_3But when solving, we got m_1 = 2.But according to the stationary distribution, it's 11/6.So, which one is correct?Wait, perhaps the issue is that the expected return time to state 1 is indeed 1/π₁, which is 11/6, but the way I set up the equations, I might have included an extra step.Wait, let me think about the definition.In Markov chain theory, the expected return time to state i is the expected number of steps to return to i starting from i.So, in that case, m_1 is the expected return time starting from state 1, which is 1/π₁.But in our equations, we have m_1 = 1 + 0.2 m_2 + 0.1 m_3Which is the expected return time starting from state 1, including the step that leaves state 1.So, that should be equal to 1/π₁.But according to our calculations, m_1 = 2, which is not equal to 11/6.So, there must be a mistake in the setup.Wait, let me check the equations again.From state 1: m_1 = 1 + 0.2 m_2 + 0.1 m_3From state 2: m_2 = 1 + 0.4 m_2 + 0.3 m_3From state 3: m_3 = 1 + 0.3 m_2 + 0.4 m_3Wait, in state 2, the transition probabilities are 0.4 to stay, 0.4 to go to state 1, and 0.2 to go to state 3? Wait, no, looking back at the transition matrix P:P is:0.7 0.2 0.10.4 0.4 0.20.3 0.3 0.4So, from state 2, the transitions are:to state 1: 0.4to state 2: 0.4to state 3: 0.2Wait, in my previous setup, I had from state 2: 0.4 to stay, 0.3 to go to state 3, but that's incorrect.Wait, no, in the transition matrix P, the rows are the current state, and columns are the next state.So, row 2 is [0.4, 0.4, 0.2], meaning from state 2, you can go to state 1 with 0.4, state 2 with 0.4, and state 3 with 0.2.Similarly, row 3 is [0.3, 0.3, 0.4], so from state 3, you can go to state 1 with 0.3, state 2 with 0.3, and state 3 with 0.4.Wait, so in my previous setup, I incorrectly wrote the transitions for state 2 and 3.I think that's where the mistake was.So, let me correct the equations.From state 1: m_1 = 1 + 0.2 m_2 + 0.1 m_3From state 2: m_2 = 1 + 0.4 m_1 + 0.4 m_2 + 0.2 m_3From state 3: m_3 = 1 + 0.3 m_1 + 0.3 m_2 + 0.4 m_3Wait, no, that's not correct either.Wait, m_i is the expected return time to state 1 starting from state i.So, from state 2, you can go to state 1 with probability 0.4, which would mean you've reached state 1 in one step, so m_2 = 1 + 0.4*0 + 0.4 m_2 + 0.2 m_3Wait, no, that's not correct.Wait, let me think carefully.If m_i is the expected number of steps to reach state 1 starting from state i, then:From state 1: m_1 = 0, because you're already there.But if you start at state 1 and take a step, then m_1' = 1 + 0.2 m_2 + 0.1 m_3But in our case, the problem is about the expected return time once it leaves, so starting from state 1, you leave, so you have to compute m_1' = 1 + 0.2 m_2 + 0.1 m_3But in the standard definition, the expected return time to state 1 is m_1 = 1 + 0.2 m_2 + 0.1 m_3But in that case, m_1 is the expected return time starting from state 1, which is 1/π₁.But according to the equations, we have:From state 1: m_1 = 1 + 0.2 m_2 + 0.1 m_3From state 2: m_2 = 1 + 0.4 m_1 + 0.4 m_2 + 0.2 m_3Wait, no, from state 2, you can go to state 1 with probability 0.4, which would mean you've reached state 1 in one step, so m_2 = 1 + 0.4*0 + 0.4 m_2 + 0.2 m_3Similarly, from state 3: m_3 = 1 + 0.3*0 + 0.3 m_2 + 0.4 m_3Wait, that's correct.So, let me write the corrected equations:1. m_1 = 1 + 0.2 m_2 + 0.1 m_32. m_2 = 1 + 0.4*0 + 0.4 m_2 + 0.2 m_3 = 1 + 0.4 m_2 + 0.2 m_33. m_3 = 1 + 0.3*0 + 0.3 m_2 + 0.4 m_3 = 1 + 0.3 m_2 + 0.4 m_3So, now, let's rewrite these equations:Equation 1: m_1 = 1 + 0.2 m_2 + 0.1 m_3Equation 2: m_2 = 1 + 0.4 m_2 + 0.2 m_3Equation 3: m_3 = 1 + 0.3 m_2 + 0.4 m_3Now, let's solve these equations.First, equation 2:m_2 = 1 + 0.4 m_2 + 0.2 m_3Subtract 0.4 m_2 from both sides:0.6 m_2 = 1 + 0.2 m_3Equation 2a: 0.6 m_2 - 0.2 m_3 = 1Equation 3:m_3 = 1 + 0.3 m_2 + 0.4 m_3Subtract 0.4 m_3 from both sides:0.6 m_3 = 1 + 0.3 m_2Equation 3a: -0.3 m_2 + 0.6 m_3 = 1Now, we have:Equation 2a: 0.6 m_2 - 0.2 m_3 = 1Equation 3a: -0.3 m_2 + 0.6 m_3 = 1Let me write them as:2a: 0.6 m_2 - 0.2 m_3 = 13a: -0.3 m_2 + 0.6 m_3 = 1To eliminate decimals, multiply equation 2a by 10:6 m_2 - 2 m_3 = 10Multiply equation 3a by 10:-3 m_2 + 6 m_3 = 10Now, we have:Equation 2b: 6 m_2 - 2 m_3 = 10Equation 3b: -3 m_2 + 6 m_3 = 10Let me solve these two equations.Let me multiply equation 2b by 3:18 m_2 - 6 m_3 = 30Now, add equation 3b:18 m_2 - 6 m_3 -3 m_2 + 6 m_3 = 30 + 10Simplify:15 m_2 = 40So, m_2 = 40 / 15 = 8/3 ≈ 2.6667Now, plug m_2 = 8/3 into equation 2b:6*(8/3) - 2 m_3 = 10Compute:16 - 2 m_3 = 10Subtract 16:-2 m_3 = -6Divide by -2:m_3 = 3Now, we have m_2 = 8/3 and m_3 = 3Now, go back to equation 1:m_1 = 1 + 0.2 m_2 + 0.1 m_3Substitute m_2 and m_3:m_1 = 1 + 0.2*(8/3) + 0.1*3Compute each term:0.2*(8/3) = 1.6/3 ≈ 0.53330.1*3 = 0.3So, m_1 = 1 + 0.5333 + 0.3 ≈ 1.8333Which is 11/6 ≈ 1.8333So, m_1 = 11/6Therefore, the expected return time to state 1 is 11/6.So, earlier, I made a mistake in setting up the equations because I incorrectly wrote the transitions for state 2 and 3. After correcting that, the result matches the stationary distribution method.So, the resilience is 11/6 steps.Therefore, the answer is 11/6.Final Answer1. The steady-state distribution vector is (boxed{left[dfrac{6}{11}, dfrac{3}{11}, dfrac{2}{11}right]}).2. The expected return time to state 1 is (boxed{dfrac{11}{6}}).</think>

🧠:<think>Alright, so I've got these two problems from Dr. Evelyn, the veterinarian who uses math to help turtles. Let me try to tackle them one by one. Starting with the first problem: It's a differential equation modeling the growth of a turtle's shell. The equation given is:[ frac{dS(t)}{dt} = k cdot S(t) cdot left(1 - frac{S(t)}{L}right) ]Hmm, this looks familiar. I think it's the logistic growth model. Yeah, logistic equation is used to model population growth where there's a carrying capacity, right? In this case, the shell size can't grow beyond L, which makes sense because there's a maximum size for the shell.So, the equation is a first-order ordinary differential equation. To solve it, I remember that we can use separation of variables. Let me try to rewrite the equation so that all terms involving S(t) are on one side and the terms involving t are on the other.Starting with:[ frac{dS}{dt} = k S left(1 - frac{S}{L}right) ]I can rewrite this as:[ frac{dS}{S left(1 - frac{S}{L}right)} = k , dt ]Now, I need to integrate both sides. The left side looks a bit tricky because of the denominator. Maybe I can use partial fractions to simplify it. Let's set up the partial fraction decomposition.Let me denote:[ frac{1}{S left(1 - frac{S}{L}right)} = frac{A}{S} + frac{B}{1 - frac{S}{L}} ]Multiplying both sides by ( S left(1 - frac{S}{L}right) ) gives:[ 1 = A left(1 - frac{S}{L}right) + B S ]Expanding the right side:[ 1 = A - frac{A S}{L} + B S ]Now, let's collect like terms:The constant term: AThe terms with S: ( left(-frac{A}{L} + Bright) S )Since the left side is 1, which has no S term, the coefficient of S must be zero, and the constant term must be 1. So, we have two equations:1. ( A = 1 )2. ( -frac{A}{L} + B = 0 )From the first equation, A is 1. Plugging into the second equation:[ -frac{1}{L} + B = 0 implies B = frac{1}{L} ]So, the partial fractions decomposition is:[ frac{1}{S left(1 - frac{S}{L}right)} = frac{1}{S} + frac{1}{L left(1 - frac{S}{L}right)} ]Wait, hold on. Let me check that again. If B is 1/L, then:[ frac{1}{S left(1 - frac{S}{L}right)} = frac{1}{S} + frac{1}{L} cdot frac{1}{1 - frac{S}{L}} ]Yes, that's correct.So, going back to the integral:[ int left( frac{1}{S} + frac{1}{L} cdot frac{1}{1 - frac{S}{L}} right) dS = int k , dt ]Let me compute the left integral term by term.First term: ( int frac{1}{S} dS = ln |S| + C )Second term: ( int frac{1}{L} cdot frac{1}{1 - frac{S}{L}} dS )Let me make a substitution for the second integral. Let ( u = 1 - frac{S}{L} ), then ( du = -frac{1}{L} dS ), which implies ( -L du = dS ).So, substituting:[ int frac{1}{L} cdot frac{1}{u} (-L du) = - int frac{1}{u} du = -ln |u| + C = -ln left| 1 - frac{S}{L} right| + C ]Putting it all together, the left integral becomes:[ ln |S| - ln left| 1 - frac{S}{L} right| + C ]Which can be written as:[ ln left| frac{S}{1 - frac{S}{L}} right| + C ]And the right integral is:[ int k , dt = k t + C ]So, combining both sides:[ ln left( frac{S}{1 - frac{S}{L}} right) = k t + C ]I can exponentiate both sides to eliminate the natural log:[ frac{S}{1 - frac{S}{L}} = e^{k t + C} = e^{C} e^{k t} ]Let me denote ( e^{C} ) as another constant, say ( C' ), since it's just an arbitrary constant.So,[ frac{S}{1 - frac{S}{L}} = C' e^{k t} ]Now, let's solve for S. Multiply both sides by the denominator:[ S = C' e^{k t} left( 1 - frac{S}{L} right) ]Expanding the right side:[ S = C' e^{k t} - frac{C' e^{k t} S}{L} ]Bring the term involving S to the left side:[ S + frac{C' e^{k t} S}{L} = C' e^{k t} ]Factor out S:[ S left( 1 + frac{C' e^{k t}}{L} right) = C' e^{k t} ]Then,[ S = frac{C' e^{k t}}{1 + frac{C' e^{k t}}{L}} ]Multiply numerator and denominator by L to simplify:[ S = frac{C' L e^{k t}}{L + C' e^{k t}} ]Now, let's apply the initial condition to find C'. At t=0, S=S0.So,[ S_0 = frac{C' L e^{0}}{L + C' e^{0}} = frac{C' L}{L + C'} ]Solving for C':Multiply both sides by denominator:[ S_0 (L + C') = C' L ]Expand:[ S_0 L + S_0 C' = C' L ]Bring terms with C' to one side:[ S_0 L = C' L - S_0 C' ]Factor C':[ S_0 L = C' (L - S_0) ]Thus,[ C' = frac{S_0 L}{L - S_0} ]So, plugging back into the expression for S(t):[ S(t) = frac{ left( frac{S_0 L}{L - S_0} right) L e^{k t} }{ L + left( frac{S_0 L}{L - S_0} right) e^{k t} } ]Simplify numerator and denominator:Numerator:[ frac{S_0 L^2 e^{k t}}{L - S_0} ]Denominator:[ L + frac{S_0 L e^{k t}}{L - S_0} = frac{L (L - S_0) + S_0 L e^{k t}}{L - S_0} ]Simplify denominator:[ frac{L^2 - L S_0 + S_0 L e^{k t}}{L - S_0} ]So, putting numerator over denominator:[ S(t) = frac{ frac{S_0 L^2 e^{k t}}{L - S_0} }{ frac{L^2 - L S_0 + S_0 L e^{k t}}{L - S_0} } ]The ( L - S_0 ) cancels out:[ S(t) = frac{S_0 L^2 e^{k t}}{L^2 - L S_0 + S_0 L e^{k t}} ]Factor L in the denominator:[ S(t) = frac{S_0 L^2 e^{k t}}{L (L - S_0) + S_0 L e^{k t}} ]Factor L from both terms in denominator:[ S(t) = frac{S_0 L^2 e^{k t}}{L [ (L - S_0) + S_0 e^{k t} ] } ]Cancel one L from numerator and denominator:[ S(t) = frac{S_0 L e^{k t}}{ (L - S_0) + S_0 e^{k t} } ]We can factor S0 in the denominator:Wait, actually, let me write it as:[ S(t) = frac{S_0 L e^{k t}}{ L - S_0 + S_0 e^{k t} } ]Alternatively, we can factor S0 from the denominator:[ S(t) = frac{S_0 L e^{k t}}{ L - S_0 + S_0 e^{k t} } = frac{S_0 L e^{k t}}{ L + S_0 (e^{k t} - 1) } ]But maybe it's better to write it as:[ S(t) = frac{L}{1 + frac{(L - S_0)}{S_0} e^{-k t}} ]Wait, let me check that. Let's see:Starting from:[ S(t) = frac{S_0 L e^{k t}}{ L - S_0 + S_0 e^{k t} } ]Divide numerator and denominator by S0 e^{k t}:[ S(t) = frac{L}{ frac{L - S_0}{S_0 e^{k t}} + 1 } ]Which is:[ S(t) = frac{L}{1 + frac{L - S_0}{S_0} e^{-k t}} ]Yes, that's a common form of the logistic equation solution. So, that's another way to write it.But either form is correct. So, depending on which form is preferred, we can present the solution.So, summarizing, the solution to the differential equation is:[ S(t) = frac{L}{1 + frac{L - S_0}{S_0} e^{-k t}} ]Alternatively,[ S(t) = frac{S_0 L e^{k t}}{ L - S_0 + S_0 e^{k t} } ]Either is acceptable, but the first form might be more elegant because it shows the carrying capacity L clearly.So, that's the solution for the first problem.Moving on to the second problem: Dr. Evelyn is tracking bacterial population, modeled by:[ B(t) = B_0 cdot e^{rt} ]She wants to know when the bacterial count exceeds a threshold ( B_{max} ). So, we need to find the time t when ( B(t) = B_{max} ).So, set up the equation:[ B_{max} = B_0 e^{rt} ]We need to solve for t.Starting with:[ B_{max} = B_0 e^{rt} ]Divide both sides by B0:[ frac{B_{max}}{B_0} = e^{rt} ]Take the natural logarithm of both sides:[ ln left( frac{B_{max}}{B_0} right) = rt ]Solve for t:[ t = frac{1}{r} ln left( frac{B_{max}}{B_0} right) ]So, that's the time when the bacterial count reaches B_max.But wait, the question says "when the bacterial count exceeds B_max". So, technically, the count exceeds B_max at any time t > t_critical, where t_critical is the time when B(t) = B_max. So, the time when it exceeds is t > t_critical, but if they want the exact time when it surpasses, it's at t_critical. So, depending on interpretation, but I think they just want t_critical.So, the answer is:[ t = frac{1}{r} ln left( frac{B_{max}}{B_0} right) ]Alternatively, using logarithm properties, it can be written as:[ t = frac{ln B_{max} - ln B_0}{r} ]But the first form is probably more straightforward.So, to recap, for the first problem, we solved the logistic differential equation and found the expression for S(t). For the second problem, we solved for t when the bacterial population reaches a certain threshold.I think that's it. Let me just double-check my steps.For the first problem, separation of variables, partial fractions, integrating, applying initial condition—all steps seem correct. The solution matches the standard logistic growth solution, so that's reassuring.For the second problem, it's a straightforward exponential growth model. Solving for t when B(t) = B_max is just taking logarithms, which is a standard approach. So, that seems solid.Yeah, I think I'm confident with these solutions.Final Answer1. The shell size at time ( t ) is given by (boxed{S(t) = dfrac{L}{1 + dfrac{L - S_0}{S_0} e^{-kt}}}).2. The time at which the bacterial count exceeds ( B_{text{max}} ) is (boxed{t = dfrac{1}{r} lnleft(dfrac{B_{text{max}}}{B_0}right)}).</think>

🧠:<think>Alright, so I have these two sub-problems to solve related to Alex and his dog Max. Let me take them one at a time.Starting with Sub-problem 1: Alex uses a Fibonacci sequence for command intervals. Each interval ( t_n ) is defined as ( t_n = t_{n-1} + t_{n-2} ), with ( t_1 = 1 ) second and ( t_2 = 1 ) second. I need to calculate the total time Alex spends giving commands over the first 15 intervals. Okay, so Fibonacci sequence. I remember that each term is the sum of the two preceding ones. The first few terms are 1, 1, 2, 3, 5, 8, and so on. So, for the first 15 intervals, I need to list out the Fibonacci numbers up to the 15th term and then sum them all together.Let me write out the sequence up to the 15th term:1. ( t_1 = 1 )2. ( t_2 = 1 )3. ( t_3 = t_2 + t_1 = 1 + 1 = 2 )4. ( t_4 = t_3 + t_2 = 2 + 1 = 3 )5. ( t_5 = t_4 + t_3 = 3 + 2 = 5 )6. ( t_6 = t_5 + t_4 = 5 + 3 = 8 )7. ( t_7 = t_6 + t_5 = 8 + 5 = 13 )8. ( t_8 = t_7 + t_6 = 13 + 8 = 21 )9. ( t_9 = t_8 + t_7 = 21 + 13 = 34 )10. ( t_{10} = t_9 + t_8 = 34 + 21 = 55 )11. ( t_{11} = t_{10} + t_9 = 55 + 34 = 89 )12. ( t_{12} = t_{11} + t_{10} = 89 + 55 = 144 )13. ( t_{13} = t_{12} + t_{11} = 144 + 89 = 233 )14. ( t_{14} = t_{13} + t_{12} = 233 + 144 = 377 )15. ( t_{15} = t_{14} + t_{13} = 377 + 233 = 610 )So, the 15th term is 610 seconds. Now, to find the total time, I need to sum all these terms from ( t_1 ) to ( t_{15} ).Let me list them again for clarity:1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610.I can add them step by step:Start with 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 = 143.143 + 89 = 232.232 + 144 = 376.376 + 233 = 609.609 + 377 = 986.986 + 610 = 1596.Wait, let me verify that addition step by step because it's easy to make a mistake.Starting from the beginning:1 (t1) = 11 + 1 = 2 (t2)2 + 2 = 4 (t3)4 + 3 = 7 (t4)7 + 5 = 12 (t5)12 + 8 = 20 (t6)20 + 13 = 33 (t7)33 + 21 = 54 (t8)54 + 34 = 88 (t9)88 + 55 = 143 (t10)143 + 89 = 232 (t11)232 + 144 = 376 (t12)376 + 233 = 609 (t13)609 + 377 = 986 (t14)986 + 610 = 1596 (t15)So, the total time is 1596 seconds.But wait, I remember that the sum of the first n Fibonacci numbers has a formula. Let me recall. I think it's ( S_n = t_{n+2} - 1 ). Let me check.For example, the sum of the first 1 term is 1, and ( t_3 - 1 = 2 - 1 = 1 ). Correct.Sum of first 2 terms: 1 + 1 = 2, ( t_4 - 1 = 3 - 1 = 2 ). Correct.Sum of first 3 terms: 1 + 1 + 2 = 4, ( t_5 - 1 = 5 - 1 = 4 ). Correct.So, yes, the formula seems to hold. Therefore, the sum of the first 15 Fibonacci numbers is ( t_{17} - 1 ).Wait, hold on. If ( S_n = t_{n+2} - 1 ), then for n=15, it's ( t_{17} - 1 ).But earlier, when I listed up to t15, the 17th term would be t16 and t17.Wait, let me compute t16 and t17.From t15 = 610,t16 = t15 + t14 = 610 + 377 = 987t17 = t16 + t15 = 987 + 610 = 1597So, ( S_{15} = t_{17} - 1 = 1597 - 1 = 1596 ). That matches my manual addition.So, the total time is 1596 seconds.Okay, that seems solid.Moving on to Sub-problem 2: Alex gives treats in a geometric progression after each Fibonacci interval. The number of treats after each interval is 1, 3, 6, 12, etc., doubling each time. Wait, hold on, the problem says: "1 treat after the first interval, 3 treats after the second interval, and continues doubling the number of treats after each subsequent interval."Wait, let me parse that again.After the first interval: 1 treat.After the second interval: 3 treats.Then, continues doubling the number of treats after each subsequent interval.So, after the third interval: 6 treats (double of 3).After the fourth interval: 12 treats (double of 6).And so on, up to the 15th interval.So, the number of treats after each interval is: 1, 3, 6, 12, 24, ..., up to the 15th term.Wait, let me confirm: first term is 1, second term is 3, then each subsequent term is double the previous. So, it's a geometric sequence starting from the second term with a common ratio of 2.But the first term is 1, the second term is 3, which is not double of 1. So, the sequence is 1, 3, 6, 12, 24, ..., up to 15 terms.So, the first term is 1, the second term is 3, and from the third term onwards, each term is double the previous.So, the sequence is: 1, 3, 6, 12, 24, 48, 96, 192, 384, 768, 1536, 3072, 6144, 12288, 24576.Wait, let me list them:Term 1: 1Term 2: 3Term 3: 6 (3*2)Term 4: 12 (6*2)Term 5: 24 (12*2)Term 6: 48Term 7: 96Term 8: 192Term 9: 384Term 10: 768Term 11: 1536Term 12: 3072Term 13: 6144Term 14: 12288Term 15: 24576So, that's 15 terms. Now, I need to find the total number of treats, which is the sum of this sequence.This is a bit tricky because the first two terms don't follow the same pattern. The first term is 1, the second is 3, and then starting from the third term, it's a geometric progression with ratio 2.So, perhaps I can split the sum into two parts: the first two terms and then the geometric series from term 3 to term 15.Sum = Term1 + Term2 + Sum from Term3 to Term15.Term1 = 1Term2 = 3Sum from Term3 to Term15: This is a geometric series with first term a = 6, common ratio r = 2, and number of terms n = 13 (since from term3 to term15 is 13 terms).Wait, term3 is 6, term4 is 12, ..., term15 is 24576.Number of terms from term3 to term15: 15 - 2 = 13 terms.The formula for the sum of a geometric series is ( S_n = a times frac{r^n - 1}{r - 1} ).So, plugging in the values:( S_{13} = 6 times frac{2^{13} - 1}{2 - 1} = 6 times (8192 - 1) = 6 times 8191 = 49146 ).Wait, let me compute that:2^13 is 8192.8192 - 1 = 8191.6 * 8191: Let's compute 6*8000 = 48000, 6*191 = 1146. So, 48000 + 1146 = 49146.So, the sum from term3 to term15 is 49146.Adding the first two terms: 1 + 3 = 4.Total treats = 4 + 49146 = 49150.Wait, let me verify:Sum from term3 to term15: 6 + 12 + 24 + ... + 24576.Number of terms: 13.First term a = 6, ratio r = 2.Sum = a*(r^n - 1)/(r - 1) = 6*(2^13 - 1)/1 = 6*(8192 - 1) = 6*8191 = 49146.Yes, that's correct.Adding term1 and term2: 1 + 3 = 4.Total: 49146 + 4 = 49150.So, the total number of treats is 49,150.Wait, let me think again. Is the sequence correctly identified?Term1: 1Term2: 3Term3: 6Term4: 12...Term15: 24576Yes, that's correct. So, the sum is 49,150.Alternatively, another way to think about it is that starting from term2, it's 3, 6, 12, ..., which is a geometric series with first term 3, ratio 2, and 14 terms (from term2 to term15). But wait, term2 is 3, term3 is 6, ..., term15 is 24576. So, that's 14 terms.Wait, hold on, maybe I made a mistake earlier.Wait, if I consider term2 as the start of the geometric series, then term2 is 3, term3 is 6, term4 is 12, ..., term15 is 24576.So, number of terms from term2 to term15 is 14 terms.So, sum from term2 to term15: a = 3, r = 2, n =14.Sum = 3*(2^14 - 1)/(2 - 1) = 3*(16384 - 1) = 3*16383 = 49149.Then, adding term1: 1, total treats = 49149 +1 = 49150.Yes, same result. So, that's consistent.Therefore, the total number of treats is 49,150.So, summarizing:Sub-problem 1: Total command time is 1596 seconds.Sub-problem 2: Total treats given is 49,150.Final AnswerSub-problem 1: The total time is boxed{1596} seconds.Sub-problem 2: The total number of treats is boxed{49150}.</think>

🧠:<think>Okay, so I'm trying to solve this problem about the influence of Bollywood movies on societal trends using network theory. It's divided into two parts, and I need to tackle them one by one. Let me start with the first part.Problem 1: Expected Proportion Adopting the BehaviorWe have a society of ( n ) individuals, each represented as a node in a network. The network is an Erdős-Rényi random graph ( G(n, p) ), which means each pair of nodes is connected with probability ( p ). Initially, 10% of the individuals adopt a new behavior randomly. The adoption probability for each individual is proportional to the number of direct connections (edges) they have with individuals who have already adopted the behavior.I need to find the expected proportion of the entire society that will eventually adopt the behavior.Hmm, okay. So this sounds like a type of epidemic or information spreading model on a network. In such models, nodes can be in different states, like susceptible, infected, etc. Here, the behavior adoption is similar to an infection spreading through the network.In the Erdős-Rényi model, each node has a degree that follows a Poisson distribution with mean ( lambda = p(n-1) approx np ) for large ( n ). So, the degree distribution is approximately Poisson.The adoption process is such that each individual adopts the behavior with a probability proportional to the number of adopters they are connected to. So, if a person has ( k ) neighbors who have adopted the behavior, their probability of adopting is proportional to ( k ).But wait, the problem says "proportional to the number of direct connections." So, does that mean the probability is ( k times gamma ), where ( gamma ) is some constant? Or is it normalized in some way?Wait, the problem says "proportional," but doesn't specify the constant. Maybe it's just ( k times gamma ), but without knowing ( gamma ), we might need to express the result in terms of ( p ) and the initial condition.Alternatively, perhaps the probability is ( frac{k}{n-1} ), but that seems like a normalized version. Wait, in part 2, they introduce a parameter ( beta ) such that the probability becomes ( min(1, beta times frac{k}{n-1}) ). So maybe in part 1, the probability is just proportional to ( k ), without the normalization by ( n-1 ). But since it's proportional, maybe it's ( gamma k ), where ( gamma ) is a constant.But since the problem doesn't specify ( gamma ), I think the key is to model this as a threshold model or a linear threshold model. In the linear threshold model, each node has a threshold, and if the number of adopters among its neighbors exceeds the threshold, it adopts the behavior.But in this case, the adoption probability is proportional to the number of adopters. So, it's more like a probabilistic adoption where the probability increases with the number of adopters.Wait, maybe it's similar to the SIS (Susceptible-Infected-Susceptible) model, but with a probability proportional to the number of infected neighbors.Alternatively, perhaps it's a type of bootstrap percolation, where nodes activate if a certain number of their neighbors are active.But in this case, it's probabilistic rather than deterministic.Given that the network is large and behaves like an Erdős-Rényi graph, we can use some mean-field approximations.In mean-field theory for such processes, we can model the expected fraction of adopters over time.Let me denote ( q(t) ) as the fraction of adopters at time ( t ). Initially, ( q(0) = 0.1 ).Each node has degree ( k ), which is Poisson distributed with mean ( lambda = np ). The probability that a node adopts at time ( t+1 ) is proportional to the number of adopters among its neighbors.But since the network is large, we can approximate the expected number of adopters among the neighbors as ( k times q(t) ). So, the probability that a susceptible node adopts is proportional to ( k times q(t) ).But we need to normalize this probability so that it doesn't exceed 1. So, perhaps the probability is ( gamma k q(t) ), where ( gamma ) is a constant such that ( gamma k q(t) leq 1 ).But without knowing ( gamma ), maybe we can set ( gamma = frac{1}{lambda} ), so that the expected probability is ( frac{k q(t)}{lambda} ). Since ( lambda = np ), and for large ( n ), ( k approx lambda ), so this would make the probability ( frac{lambda q(t)}{lambda} = q(t) ). But that seems too simplistic.Wait, maybe the probability is ( 1 - e^{-gamma k q(t)} ), which is a common form in percolation models. But I'm not sure.Alternatively, perhaps the probability is ( gamma k q(t) ), but since ( k ) is random, we can take the expectation over the degree distribution.Wait, let's think in terms of the expected change in ( q(t) ). The rate of adoption would be proportional to the number of susceptible nodes times the probability of adoption.So, the rate equation would be:( frac{dq}{dt} = gamma (1 - q(t)) cdot E[k] cdot q(t) )But ( E[k] = lambda = np ). So,( frac{dq}{dt} = gamma np (1 - q) q )This is a differential equation that can be solved. The solution would give us the expected proportion ( q(t) ) over time.But we need to find the expected proportion that will eventually adopt the behavior, i.e., the steady-state value as ( t to infty ).Solving the differential equation:( frac{dq}{dt} = gamma np q (1 - q) )This is a logistic equation. The solution is:( q(t) = frac{q(0)}{q(0) + (1 - q(0)) e^{-gamma np t}} )As ( t to infty ), ( q(t) ) approaches 1 if ( gamma np > 0 ). But that can't be right because if ( gamma np ) is too small, the adoption might not spread.Wait, actually, in the logistic model, the steady-state is 1 if the growth rate is positive. But in our case, the adoption process might have a threshold. If the initial adoption is too small, it might die out, otherwise, it can spread to the entire network.But in the problem, the initial adoption is 10%, which is non-zero. So, depending on ( gamma np ), it might either reach a certain proportion or spread to everyone.Wait, but in the logistic model, it always approaches 1 as ( t to infty ), regardless of the initial condition, as long as ( gamma np > 0 ). But that doesn't seem right in the context of network adoption, because in some cases, the adoption might not spread beyond the initial seeds.Hmm, maybe I need to consider a different approach. Perhaps using the concept of the giant component in the Erdős-Rényi graph.In an Erdős-Rényi graph ( G(n, p) ), there is a phase transition at ( p = frac{ln n}{n} ). Below this, the graph consists mostly of small components, and above this, there is a giant component of size proportional to ( n ).But in our case, the adoption process is probabilistic and depends on the number of adopters in the neighborhood.Wait, maybe we can model this as a branching process. Each adopter can influence their neighbors, and each neighbor has a probability of adopting based on the number of adopters they have.But since the network is random, we can approximate the expected number of new adopters each step.Let me think. The initial number of adopters is ( 0.1n ). Each adopter has ( k ) neighbors, where ( k ) is Poisson with mean ( lambda = np ).The probability that a neighbor adopts is proportional to the number of adopters they have. But since each neighbor can have multiple adopters, the probability might be a function of the number of adopters they have.Wait, maybe it's better to think in terms of the expected number of new adopters in the next step.Let ( q(t) ) be the fraction of adopters at time ( t ). The expected number of adopters at time ( t+1 ) would be ( q(t) + (1 - q(t)) cdot E[text{probability of adopting}] ).The probability of a susceptible node adopting is proportional to the number of adopters among its neighbors. Since the network is random, the number of adopters among its neighbors is approximately Poisson with mean ( lambda q(t) ).So, if the probability is proportional to the number of adopters, say ( gamma k ), but since ( k ) is the number of adopters, which is a random variable, we need to take the expectation.Wait, perhaps the probability that a susceptible node adopts is ( 1 - e^{-gamma k} ), but I'm not sure.Alternatively, if the probability is linear, like ( gamma k ), but since ( k ) is the number of adopters, which is a random variable, the expected probability would be ( gamma E[k | text{node is susceptible}] ).But since the node is susceptible, its neighbors are a mix of adopters and non-adopters. The expected number of adopters among its neighbors is ( k times q(t) ), where ( k ) is the degree.Wait, maybe the probability is ( gamma k q(t) ), but since ( k ) is random, we can take the expectation over the degree distribution.So, the expected probability for a susceptible node to adopt is ( gamma E[k] q(t) ). Since ( E[k] = lambda = np ), this becomes ( gamma np q(t) ).But this probability must be less than or equal to 1. So, ( gamma np q(t) leq 1 ).Therefore, the rate of change of ( q(t) ) is:( frac{dq}{dt} = gamma np (1 - q(t)) q(t) )This is the same logistic equation as before.Solving this, we have:( q(t) = frac{q(0)}{q(0) + (1 - q(0)) e^{-gamma np t}} )As ( t to infty ), ( q(t) to 1 ) if ( gamma np > 0 ). But this suggests that eventually, everyone adopts, which might not be the case.Wait, perhaps the model is different. Maybe the adoption is not cumulative, but rather, each node has a threshold, and once the number of adopters exceeds the threshold, the node adopts.But in the problem, it's stated that the probability is proportional to the number of adopters. So, it's more like a probabilistic adoption where the chance increases with more adopters.Alternatively, maybe it's a type of epidemic where each adopter can infect their neighbors with a certain probability.Wait, in the standard SIR model, each infected node infects each susceptible neighbor with probability ( beta ) per time step. The expected number of new infections is ( beta k I(t) ), where ( k ) is the degree.But in our case, the probability is proportional to the number of adopters, so it's similar but the probability is ( gamma k ), where ( gamma ) is a constant.Wait, perhaps the probability is ( gamma times frac{k}{n-1} ), similar to part 2, but without the min function. But in part 1, it's just proportional to ( k ).But without knowing the constant, maybe we can express the result in terms of ( p ) and the initial condition.Wait, perhaps the expected proportion is 1, meaning everyone adopts, because the process is supercritical. But that doesn't seem right because in reality, the adoption might not spread to everyone.Wait, maybe I need to consider the critical threshold for the adoption to spread. In percolation theory, there's a critical probability above which a giant component emerges. Similarly, in epidemic models, there's a critical threshold for the epidemic to take off.In our case, the adoption process might have a critical value of ( gamma np ). If ( gamma np > 1 ), the adoption can spread to a significant portion of the network, otherwise, it dies out.But in our problem, the initial adoption is 10%, which is non-zero. So, if ( gamma np > 1 ), the adoption will spread to a significant fraction, otherwise, it might not.But the problem asks for the expected proportion that will eventually adopt, assuming the network is large.So, perhaps the expected proportion is 1 if ( gamma np > 1 ), otherwise, it's the initial proportion.But wait, in the logistic model, even if ( gamma np > 1 ), the adoption grows to 1. But in reality, in networks, the adoption might reach a certain fraction depending on the structure.Wait, maybe I need to use the concept of the giant component. If the network has a giant component, then the adoption can spread through it. The size of the giant component in ( G(n, p) ) is known to be ( theta(n) ) when ( p > frac{ln n}{n} ).But in our case, the adoption is probabilistic, so even if there's a giant component, the adoption might not reach everyone.Alternatively, perhaps the expected proportion is the initial proportion plus the expected number of new adopters influenced by the initial seeds.Wait, maybe I can model this as a branching process. Each adopter can influence their neighbors, and each neighbor has a probability of adopting based on the number of adopters they have.But since each adopter can influence multiple neighbors, and each neighbor can be influenced by multiple adopters, it's a bit complex.Alternatively, perhaps the expected proportion can be found using the formula for the expected size of the infected set in an epidemic model.In the SIR model on a random graph, the expected proportion of infected nodes can be found using the equation:( q = 1 - e^{-tau q} )where ( tau = beta lambda ), ( beta ) is the transmission probability, and ( lambda ) is the average degree.But in our case, the "transmission probability" is proportional to the number of adopters, so maybe it's similar.Wait, let's think about it. If each adopter can influence their neighbors, and the probability that a neighbor adopts is proportional to the number of adopters they have, then the expected number of new adopters from a single adopter is ( lambda times gamma ), where ( gamma ) is the proportionality constant.But if ( gamma lambda > 1 ), the process is supercritical and can lead to a large outbreak, otherwise, it dies out.But in our case, the initial number of adopters is 10%, which is non-zero. So, if ( gamma lambda > 1 ), the adoption will spread to a significant fraction, otherwise, it might not.But the problem doesn't specify ( gamma ), so maybe we need to express the result in terms of ( p ) and the initial condition.Wait, perhaps the expected proportion is 1, but that seems too simplistic.Alternatively, maybe the expected proportion is the initial proportion plus the expected number influenced by the initial seeds.But I'm getting stuck here. Maybe I should look for similar models.Wait, in the linear threshold model, each node has a threshold, and if the fraction of adopters among its neighbors exceeds the threshold, it adopts. The expected proportion of adopters can be found using mean-field approximations.In our case, the adoption probability is proportional to the number of adopters, which is similar to a linear threshold model with a threshold of 0, but with a probabilistic adoption.Wait, maybe it's better to model it as a continuous-time Markov process.Each susceptible node has a hazard rate of adoption equal to ( gamma k q(t) ), where ( k ) is the degree and ( q(t) ) is the fraction of adopters.But since the network is random, we can approximate ( k ) as ( lambda ), so the hazard rate is ( gamma lambda q(t) ).The expected time until adoption is ( frac{1}{gamma lambda q(t)} ).But integrating this over time to find the expected proportion is tricky.Alternatively, using the approximation that the process can be described by the differential equation:( frac{dq}{dt} = gamma lambda (1 - q) q )This is the same logistic equation as before.Solving this, we get:( q(t) = frac{q(0)}{q(0) + (1 - q(0)) e^{-gamma lambda t}} )As ( t to infty ), ( q(t) to 1 ) if ( gamma lambda > 0 ). But this suggests that eventually, everyone adopts, which might not be the case.Wait, perhaps the model is different. Maybe the adoption is not cumulative, but rather, each node can adopt only once, and the probability is based on the current number of adopters.But in that case, the process would still lead to everyone adopting eventually, which doesn't seem right.Wait, maybe the adoption is not permanent. Maybe adopters can revert, but the problem doesn't mention that.Alternatively, perhaps the adoption is a one-time event, and once adopted, the node remains adopted.Given that, and the process being probabilistic, the expected proportion would depend on the initial condition and the parameters.But without knowing the exact form of the probability function, it's hard to derive the exact expression.Wait, maybe the problem is expecting a result from known models. In the Erdős-Rényi graph, the expected proportion of adopters can be found using the formula for the giant component.But the giant component size is given by ( q = 1 - e^{-lambda q} ), where ( lambda = np ).Wait, that's the equation for the size of the giant component in the configuration model, which is similar to Erdős-Rényi for large ( n ).So, if we set ( q = 1 - e^{-lambda q} ), solving for ( q ) gives the expected proportion in the giant component.But in our case, the adoption process is different. It's not just percolation, but a probabilistic adoption based on the number of adopters.Wait, maybe the expected proportion is similar to the giant component, but scaled by the initial condition.Alternatively, perhaps the expected proportion is the initial proportion plus the expected number influenced by the initial seeds.But I'm not sure.Wait, maybe I can use the concept of the expected number of adopters influenced by each initial adopter.Each initial adopter has ( k ) neighbors, and each neighbor has a probability ( gamma k ) of adopting. But since ( k ) is random, we can take the expectation.The expected number of new adopters influenced by one initial adopter is ( E[k] times gamma E[k] ), but this seems recursive.Alternatively, maybe the expected number of adopters is the initial number plus the expected number influenced by them, and so on.This sounds like a branching process where each adopter can cause new adopters, and so on.In a branching process, the expected number of adopters is the initial number multiplied by the expected number of new adopters per adopter.If the expected number of new adopters per adopter is ( gamma lambda ), then if ( gamma lambda > 1 ), the process is supercritical and can lead to a large number of adopters.But the problem is about the proportion, not the absolute number.Wait, maybe the expected proportion is given by the equation ( q = 0.1 + (1 - 0.1) times gamma lambda q ).But this is a linear equation:( q = 0.1 + 0.9 gamma lambda q )Solving for ( q ):( q (1 - 0.9 gamma lambda) = 0.1 )( q = frac{0.1}{1 - 0.9 gamma lambda} )But this assumes that the influence is linear and doesn't account for saturation.Alternatively, maybe it's a quadratic equation because each adopter can influence others, leading to a multiplicative effect.Wait, perhaps it's better to model it as a differential equation where the rate of adoption is proportional to the number of adopters times the number of non-adopters.So,( frac{dq}{dt} = gamma lambda q (1 - q) )This is the logistic equation again, whose solution is:( q(t) = frac{q(0)}{q(0) + (1 - q(0)) e^{-gamma lambda t}} )As ( t to infty ), ( q(t) to 1 ) if ( gamma lambda > 0 ). But this suggests that everyone adopts, which might not be the case.Wait, maybe the problem is expecting the result from the giant component. The expected proportion of the giant component in ( G(n, p) ) is ( q = 1 - e^{-lambda q} ), where ( lambda = np ).But how does this relate to our problem?In our case, the adoption process is influenced by the number of adopters, so maybe the expected proportion is similar to the giant component.But the initial adoption is 10%, so maybe the expected proportion is the solution to ( q = 0.1 + (1 - 0.1) (1 - e^{-lambda q}) ).Wait, that might make sense. Because each non-adopter has a probability of being connected to the giant component, which is ( 1 - e^{-lambda q} ), and thus the expected proportion would be the initial 10% plus the expected number influenced by the giant component.But I'm not sure. Let me think.In the giant component, the size ( q ) satisfies ( q = 1 - e^{-lambda q} ). But in our case, the adoption is influenced by the number of adopters, so maybe the expected proportion is the solution to ( q = 0.1 + (1 - 0.1) (1 - e^{-lambda q}) ).Let me write that down:( q = 0.1 + 0.9 (1 - e^{-lambda q}) )Simplify:( q = 0.1 + 0.9 - 0.9 e^{-lambda q} )( q = 1 - 0.9 e^{-lambda q} )But this is similar to the giant component equation ( q = 1 - e^{-lambda q} ), except scaled by 0.9.Wait, maybe the expected proportion is the solution to ( q = 0.1 + (1 - 0.1) (1 - e^{-lambda q}) ).Yes, that seems plausible. Because each non-adopter (which is 90% initially) has a probability ( 1 - e^{-lambda q} ) of being connected to the giant component of adopters, thus adopting themselves.Therefore, the expected proportion ( q ) satisfies:( q = 0.1 + 0.9 (1 - e^{-lambda q}) )Solving this equation for ( q ) would give the expected proportion.But since this is a transcendental equation, it might not have a closed-form solution, but we can express it implicitly.So, the expected proportion ( q ) is the solution to:( q = 0.1 + 0.9 (1 - e^{-lambda q}) )Where ( lambda = np ).Therefore, the expression is:( q = 0.1 + 0.9 (1 - e^{-np q}) )This is the implicit equation for ( q ).But the problem asks for an expression, so maybe this is acceptable.Alternatively, if we rearrange:( e^{-np q} = 1 - frac{q - 0.1}{0.9} )But I don't think this helps much.So, perhaps the answer is that the expected proportion ( q ) satisfies the equation:( q = 0.1 + 0.9 (1 - e^{-np q}) )Which can be written as:( q = 1 - 0.9 e^{-np q} )But let me check the logic again.In the Erdős-Rényi graph, the size of the giant component is ( q = 1 - e^{-lambda q} ). In our case, the adoption process is influenced by the giant component of adopters. The initial adoption is 10%, so the expected proportion is the initial 10% plus the expected number influenced by the giant component.Each non-adopter has a probability ( 1 - e^{-lambda q} ) of being connected to the giant component, thus adopting. Therefore, the total expected proportion is:( q = 0.1 + 0.9 (1 - e^{-lambda q}) )Yes, that makes sense.So, the expression is:( q = 0.1 + 0.9 (1 - e^{-np q}) )Therefore, the expected proportion is the solution to this equation.Problem 2: Influence Factor ( beta ) and Majority AdoptionNow, the anthropologist introduces a parameter ( beta ), modifying the adoption probability to ( min(1, beta times frac{k}{n-1}) ), where ( k ) is the number of neighbors who have adopted the behavior. We need to determine how the critical value of ( beta ) influences the behavior adoption threshold and find the minimum ( beta ) such that more than 50% adopt.So, in part 2, the adoption probability is ( min(1, beta times frac{k}{n-1}) ). This is similar to a threshold model where the adoption probability is a function of the fraction of adopters among neighbors.In the first part, the probability was proportional to ( k ), but now it's proportional to ( frac{k}{n-1} ), which is the fraction of adopters among neighbors, scaled by ( beta ).We need to find the critical ( beta ) such that the adoption can reach a majority, i.e., more than 50% adoption.In threshold models, there's a critical threshold above which the adoption can spread to a significant portion of the network. Similarly, here, the critical ( beta ) would determine whether the adoption can reach a majority.To find the minimum ( beta ), we can set up a similar equation as in part 1, but with the new adoption probability.Let me denote ( q ) as the expected proportion of adopters. The adoption probability for a susceptible node is ( min(1, beta times frac{k}{n-1}) ), where ( k ) is the number of adopters among its neighbors.Since the network is large, we can approximate ( frac{k}{n-1} ) as the fraction of adopters among neighbors, which is ( q ). Therefore, the adoption probability becomes ( min(1, beta q) ).But since ( q ) is the fraction of adopters, and ( beta q ) can be greater than 1, we take the minimum.Therefore, the probability that a susceptible node adopts is ( beta q ) if ( beta q leq 1 ), otherwise 1.But in the context of the expected proportion, we can model the rate of adoption as:( frac{dq}{dt} = (1 - q) times beta q )But this is only valid when ( beta q leq 1 ). Once ( beta q > 1 ), the adoption rate becomes ( 1 - q ), meaning everyone adopts.But to find the critical ( beta ), we need to consider the point where the adoption can sustain itself, i.e., the rate of adoption is positive.Wait, perhaps it's better to model this as a fixed point equation.At equilibrium, the rate of change is zero, so:( 0 = (1 - q) times beta q )But this gives ( q = 0 ) or ( q = 1 ). But this doesn't help.Alternatively, perhaps we need to consider the deterministic approximation where the expected proportion ( q ) satisfies:( q = 0.1 + (1 - 0.1) times min(1, beta q) )But this is similar to part 1, but with the adoption probability being ( min(1, beta q) ).Wait, let's think about it. Each non-adopter has a probability ( min(1, beta q) ) of adopting. So, the expected proportion is:( q = 0.1 + 0.9 times min(1, beta q) )We need to find the minimum ( beta ) such that ( q > 0.5 ).So, let's consider two cases:1. ( beta q leq 1 ): Then, ( q = 0.1 + 0.9 beta q )2. ( beta q > 1 ): Then, ( q = 0.1 + 0.9 times 1 = 1 )But we are looking for ( q > 0.5 ). So, let's see when ( q ) can reach 0.5.Case 1: ( beta q leq 1 )( q = 0.1 + 0.9 beta q )Rearranging:( q - 0.9 beta q = 0.1 )( q (1 - 0.9 beta) = 0.1 )( q = frac{0.1}{1 - 0.9 beta} )We want ( q > 0.5 ):( frac{0.1}{1 - 0.9 beta} > 0.5 )Multiply both sides by ( 1 - 0.9 beta ) (assuming ( 1 - 0.9 beta > 0 )):( 0.1 > 0.5 (1 - 0.9 beta) )( 0.1 > 0.5 - 0.45 beta )( -0.4 > -0.45 beta )Multiply both sides by -1 (inequality sign reverses):( 0.4 < 0.45 beta )( beta > frac{0.4}{0.45} approx 0.8889 )But we also need ( beta q leq 1 ):( beta times frac{0.1}{1 - 0.9 beta} leq 1 )Substitute ( beta = frac{0.4}{0.45} approx 0.8889 ):( 0.8889 times frac{0.1}{1 - 0.9 times 0.8889} leq 1 )Calculate denominator:( 1 - 0.9 times 0.8889 = 1 - 0.8 = 0.2 )So,( 0.8889 times frac{0.1}{0.2} = 0.8889 times 0.5 = 0.44445 leq 1 )Yes, so this is valid.Therefore, the critical ( beta ) is ( frac{0.4}{0.45} = frac{8}{9} approx 0.8889 ).But wait, this is the value where ( q = 0.5 ). To have ( q > 0.5 ), ( beta ) needs to be greater than ( frac{8}{9} ).But let's check if this is correct.If ( beta = frac{8}{9} ), then:( q = frac{0.1}{1 - 0.9 times frac{8}{9}} = frac{0.1}{1 - 0.8} = frac{0.1}{0.2} = 0.5 )So, at ( beta = frac{8}{9} ), ( q = 0.5 ). Therefore, to have ( q > 0.5 ), ( beta ) must be greater than ( frac{8}{9} ).But wait, this is under the assumption that ( beta q leq 1 ). If ( beta ) increases beyond ( frac{8}{9} ), ( q ) would increase beyond 0.5, but we need to check if ( beta q ) remains less than or equal to 1.At ( beta = frac{8}{9} ), ( q = 0.5 ), so ( beta q = frac{8}{9} times 0.5 = frac{4}{9} < 1 ). Therefore, the assumption ( beta q leq 1 ) holds.If ( beta ) increases further, ( q ) increases, but ( beta q ) could exceed 1.Wait, let's find the point where ( beta q = 1 ):( beta q = 1 )But ( q = frac{0.1}{1 - 0.9 beta} )So,( beta times frac{0.1}{1 - 0.9 beta} = 1 )Multiply both sides by ( 1 - 0.9 beta ):( 0.1 beta = 1 - 0.9 beta )( 0.1 beta + 0.9 beta = 1 )( beta = 1 )So, at ( beta = 1 ), ( q = frac{0.1}{1 - 0.9 times 1} = frac{0.1}{0.1} = 1 )Therefore, for ( beta > 1 ), the adoption probability becomes 1, meaning everyone adopts.But we are interested in the minimum ( beta ) such that ( q > 0.5 ). From earlier, ( beta > frac{8}{9} approx 0.8889 ).But let's verify this with another approach.Consider the fixed point equation:( q = 0.1 + 0.9 times min(1, beta q) )We want ( q > 0.5 ). Let's assume ( beta q leq 1 ), so:( q = 0.1 + 0.9 beta q )Rearranged:( q (1 - 0.9 beta) = 0.1 )( q = frac{0.1}{1 - 0.9 beta} )We set ( q = 0.5 ):( 0.5 = frac{0.1}{1 - 0.9 beta} )Solve for ( beta ):( 1 - 0.9 beta = frac{0.1}{0.5} = 0.2 )( 0.9 beta = 1 - 0.2 = 0.8 )( beta = frac{0.8}{0.9} = frac{8}{9} approx 0.8889 )So, this confirms our earlier result.Therefore, the minimum ( beta ) such that more than 50% adopt is ( beta > frac{8}{9} ).But the problem asks for the minimum value of ( beta ), so it's ( beta = frac{8}{9} ), but at this value, ( q = 0.5 ). To have more than 50%, ( beta ) must be greater than ( frac{8}{9} ).Therefore, the critical value is ( beta = frac{8}{9} ), and the minimum ( beta ) for majority adoption is just above ( frac{8}{9} ).But since the problem asks for the minimum value, we can express it as ( beta = frac{8}{9} ), but noting that it's the threshold.Alternatively, if we consider the exact point where ( q = 0.5 ), then ( beta = frac{8}{9} ).So, summarizing:1. The expected proportion ( q ) satisfies ( q = 0.1 + 0.9 (1 - e^{-np q}) ).2. The critical ( beta ) is ( frac{8}{9} ), and the minimum ( beta ) for majority adoption is ( beta > frac{8}{9} ).But the problem asks for the minimum value of ( beta ) such that more than 50% adopt, so the answer is ( beta = frac{8}{9} ), but strictly, it's the threshold, so the minimum ( beta ) is ( frac{8}{9} ).Wait, but in reality, at ( beta = frac{8}{9} ), ( q = 0.5 ). To have ( q > 0.5 ), ( beta ) must be greater than ( frac{8}{9} ). So, the minimum ( beta ) is ( frac{8}{9} ), but to achieve more than 50%, it needs to be slightly above.But since the problem asks for the minimum value, perhaps it's acceptable to state ( beta = frac{8}{9} ) as the critical value, beyond which majority adoption occurs.Therefore, the minimum ( beta ) is ( frac{8}{9} ).Final Answer1. The expected proportion is given by the solution to ( q = 0.1 + 0.9(1 - e^{-npq}) ). Thus, the expression is (boxed{q = 0.1 + 0.9(1 - e^{-npq})}).2. The minimum value of ( beta ) such that more than 50% of the network adopts the behavior is (boxed{dfrac{8}{9}}).</think>