answer：Alright, so I have this problem about Markov Decision Processes (MDPs) and Bellman equations. It's part of a study on how environmental factors influence decision-making. Hmm, okay, let's break this down.First, the problem is divided into two parts. The first part is about formulating the Bellman equation for the expected cumulative reward under a given policy π. The second part is about finding the optimal policy using value iteration. Let me tackle them one by one.Starting with part 1: Formulating the Bellman equation for V^π(s). I remember that in MDPs, the Bellman equation relates the value of a state to the values of subsequent states. The value function V^π(s) represents the expected cumulative reward starting from state s and following policy π thereafter.So, the Bellman equation for a policy π is given by:V^π(s) = E[R(s, a) + γ V^π(s') | s, π]Where R(s, a) is the immediate reward, γ is the discount factor, and s' is the next state. Since the policy π is a probability distribution over actions, we need to take the expectation over both the policy and the transition probabilities.Breaking it down, for each state s, we sum over all possible actions a, weighted by the probability π(a|s), and for each action, we sum over all possible next states s', weighted by the transition probability P(s'|s, a). Then, we multiply by the immediate reward R(s, a) plus the discounted future reward γ V^π(s').So, putting it all together, the Bellman equation becomes:V^π(s) = Σ_{a ∈ A} π(a|s) [ R(s, a) + γ Σ_{s' ∈ S} P(s'|s, a) V^π(s') ]That's the general form. Now, the question specifically asks to express V^π(s₁) in terms of π, P, and R for a single step decision process. Wait, single step? Hmm, does that mean we're only considering one step ahead, so γ might be 0? Or is it just the immediate reward?Wait, no, the discount factor γ is still present, but perhaps they just want the expression without considering multiple steps, so it's just the immediate reward plus the discounted next state value. So, for a single step, it's the same as the Bellman equation, but perhaps without the recursive aspect beyond one step.But actually, the Bellman equation inherently includes the recursive aspect. So, maybe they just want the expression for V^π(s₁) using the Bellman equation as is. Let me write that out.For state s₁, the value function is:V^π(s₁) = Σ_{a ∈ A} π(a|s₁) [ R(s₁, a) + γ Σ_{s' ∈ S} P(s'|s₁, a) V^π(s') ]So, that's the expression in terms of π, P, and R. Since it's a single step, we're not iterating multiple times, just expressing it once.Moving on to part 2: Determining the optimal policy π* that maximizes the expected cumulative reward. They want the expression for the optimal value function V*(s) after n iterations using value iteration.I recall that value iteration is an algorithm that iteratively updates the value function until it converges to the optimal value function V*. The recursive relation is based on the Bellman optimality equation, which for each state s is:V*(s) = max_{a ∈ A} [ R(s, a) + γ Σ_{s' ∈ S} P(s'|s, a) V*(s') ]But since we're doing value iteration, we update the value function iteratively. The general form of the recursive relation is:V_{k+1}(s) = max_{a ∈ A} [ R(s, a) + γ Σ_{s' ∈ S} P(s'|s, a) V_k(s') ]Where V_k(s) is the value function at iteration k. So, after n iterations, the optimal value function V^*(s) is approximated by V_n(s), which is obtained by applying this update n times.Therefore, the expression for V^*(s) after n iterations is the result of applying the Bellman optimality operator n times, starting from an initial value function V₀(s).So, in summary, the recursive relation is:V_{k+1}(s) = max_{a ∈ A} [ R(s, a) + γ Σ_{s' ∈ S} P(s'|s, a) V_k(s') ]And after n iterations, V^*(s) ≈ V_n(s).Wait, but the question says "derive the expression for the optimal value function V*(s) for any state s after n iterations." So, it's not just the recursive relation, but the expression after n steps. Hmm, that's a bit more involved.I think the optimal value function after n iterations can be expressed as the result of applying the Bellman optimality operator n times to some initial value function. If we start with V₀(s) = 0 for all s, then V₁(s) is the first iteration, V₂(s) the second, and so on up to V_n(s).But expressing V_n(s) explicitly would require expanding the recursive relation n times, which gets complicated. However, the general form is that each V_{k+1}(s) is the maximum over actions of the immediate reward plus the discounted expected future reward based on V_k.So, perhaps the answer is just the recursive relation, as that's the standard way value iteration is presented. The optimal value function after n iterations is obtained by applying this recursion n times.Alright, I think I have a handle on both parts now.

answer：Alright, so I have this math problem about Dr. Smith studying polyp growth. It's divided into two parts. Let me try to tackle them one by one.Starting with the first sub-problem: solving the differential equation given the initial conditions and the function for k. The differential equation is a logistic growth model, right? It's given by dP/dt = kP(1 - P/N). I remember that the logistic equation has a known solution, so maybe I can use that.First, let me write down the equation:dP/dt = kP(1 - P/N)This is a separable differential equation. So I can rewrite it as:dP / [P(1 - P/N)] = k dtTo integrate both sides, I might need partial fractions for the left side. Let me set up the partial fractions:1 / [P(1 - P/N)] = A/P + B/(1 - P/N)Multiplying both sides by P(1 - P/N):1 = A(1 - P/N) + BPLet me solve for A and B. Let's plug in P = 0:1 = A(1 - 0) + B(0) => A = 1Now, plug in P = N:1 = A(1 - N/N) + B(N) => 1 = A(0) + BN => BN = 1 => B = 1/NSo, the partial fractions decomposition is:1/P + 1/(N(1 - P/N)) = 1/P + 1/(N - P)Wait, actually, let me double-check that. If I have:1 / [P(1 - P/N)] = 1/P + 1/(N - P)Yes, that seems correct because:1/P + 1/(N - P) = (N - P + P) / [P(N - P)] = N / [P(N - P)] = 1 / [P(1 - P/N)]So, that works out. Therefore, the integral becomes:∫ [1/P + 1/(N - P)] dP = ∫ k dtIntegrating both sides:ln|P| - ln|N - P| = kt + CSimplify the left side:ln|P / (N - P)| = kt + CExponentiating both sides:P / (N - P) = e^{kt + C} = e^C e^{kt}Let me denote e^C as another constant, say, C'. So:P / (N - P) = C' e^{kt}Now, solve for P:P = C' e^{kt} (N - P)P = C' N e^{kt} - C' e^{kt} PBring the P terms to one side:P + C' e^{kt} P = C' N e^{kt}Factor out P:P (1 + C' e^{kt}) = C' N e^{kt}Therefore:P = [C' N e^{kt}] / [1 + C' e^{kt}]Now, apply the initial condition P(0) = P0. Let's plug in t = 0:P0 = [C' N e^{0}] / [1 + C' e^{0}] = [C' N] / [1 + C']Solving for C':P0 (1 + C') = C' NP0 + P0 C' = C' NP0 = C' N - P0 C'P0 = C' (N - P0)Thus, C' = P0 / (N - P0)So, substituting back into the expression for P(t):P(t) = [ (P0 / (N - P0)) N e^{kt} ] / [1 + (P0 / (N - P0)) e^{kt} ]Simplify numerator and denominator:Numerator: (P0 N / (N - P0)) e^{kt}Denominator: 1 + (P0 / (N - P0)) e^{kt} = [ (N - P0) + P0 e^{kt} ] / (N - P0)So, P(t) becomes:[ (P0 N / (N - P0)) e^{kt} ] / [ (N - P0 + P0 e^{kt}) / (N - P0) ) ] = [ P0 N e^{kt} ] / [ N - P0 + P0 e^{kt} ]Factor N in the denominator:= [ P0 N e^{kt} ] / [ N (1 - P0/N) + P0 e^{kt} ]But maybe it's cleaner to write it as:P(t) = [ P0 N e^{kt} ] / [ N - P0 + P0 e^{kt} ]Alternatively, factor N in the denominator:= [ P0 e^{kt} ] / [ (N - P0)/N + P0 e^{kt}/N ]But perhaps the standard form is better:P(t) = N / (1 + (N - P0)/P0 e^{-kt})Wait, let me check that. Let me rearrange the expression:Starting from P(t) = [ P0 N e^{kt} ] / [ N - P0 + P0 e^{kt} ]Divide numerator and denominator by e^{kt}:P(t) = [ P0 N ] / [ (N - P0) e^{-kt} + P0 ]Then, factor N in the denominator:= [ P0 N ] / [ N ( (N - P0)/N e^{-kt} + P0/N ) ]Wait, that might complicate things. Alternatively, factor out P0 in the denominator:Wait, perhaps it's better to leave it as:P(t) = [ P0 N e^{kt} ] / [ N - P0 + P0 e^{kt} ]Alternatively, factor N from numerator and denominator:= N [ P0 e^{kt} ] / [ N - P0 + P0 e^{kt} ]But I think the standard logistic growth solution is usually written as:P(t) = N / (1 + (N/P0 - 1) e^{-kt})Let me see if that's equivalent. Starting from my expression:P(t) = [ P0 N e^{kt} ] / [ N - P0 + P0 e^{kt} ]Let me factor N in the denominator:= [ P0 N e^{kt} ] / [ N (1 - P0/N) + P0 e^{kt} ]= [ P0 e^{kt} ] / [ (1 - P0/N) + (P0/N) e^{kt} ]Multiply numerator and denominator by e^{-kt}:= [ P0 ] / [ (1 - P0/N) e^{-kt} + P0/N ]= [ P0 ] / [ (N - P0)/N e^{-kt} + P0/N ]Factor 1/N:= [ P0 ] / [ (N - P0) e^{-kt} + P0 ) / N ]= [ P0 N ] / [ (N - P0) e^{-kt} + P0 ]Which is the same as:= N / [ (N - P0)/P0 e^{-kt} + 1 ]= N / [ 1 + (N - P0)/P0 e^{-kt} ]Yes, that's the standard form. So, P(t) = N / (1 + (N - P0)/P0 e^{-kt})So, that's the solution for part 1.Now, moving on to part 2. Dr. Smith has H(t) and T(t) varying periodically as H(t) = H0 + H1 sin(ωt) and T(t) = T0 + T1 cos(ωt). We need to find the average growth rate constant k_bar over one period, which is 2π/ω.Given that k(H, T) = a e^{(bH + cT)/d}So, k(t) = a e^{(bH(t) + cT(t))/d} = a e^{(b(H0 + H1 sin ωt) + c(T0 + T1 cos ωt))/d}Simplify the exponent:= a e^{ [bH0 + cT0 + bH1 sin ωt + cT1 cos ωt ] / d }= a e^{(bH0 + cT0)/d} e^{(bH1 sin ωt + cT1 cos ωt)/d }Let me denote (bH0 + cT0)/d as a constant, say, K0. So,k(t) = a e^{K0} e^{(bH1 sin ωt + cT1 cos ωt)/d }Let me denote the time-dependent exponent as (bH1 sin ωt + cT1 cos ωt)/d. Let's call this term E(t):E(t) = (bH1 sin ωt + cT1 cos ωt)/dSo, k(t) = a e^{K0} e^{E(t)} = a e^{K0} e^{(bH1 sin ωt + cT1 cos ωt)/d }We need to find the average of k(t) over one period. The average of a function over a period is (1/T) ∫_{0}^{T} k(t) dt, where T = 2π/ω.So,k_bar = (1/T) ∫_{0}^{T} a e^{K0} e^{(bH1 sin ωt + cT1 cos ωt)/d } dt= a e^{K0} (1/T) ∫_{0}^{T} e^{(bH1 sin ωt + cT1 cos ωt)/d } dtLet me make a substitution to simplify the integral. Let me set θ = ωt, so when t = 0, θ = 0, and when t = T = 2π/ω, θ = 2π. Also, dt = dθ / ω. So,k_bar = a e^{K0} (1/(2π/ω)) ∫_{0}^{2π} e^{(bH1 sin θ + cT1 cos θ)/d } (dθ / ω )Simplify the constants:= a e^{K0} (ω / (2π)) ∫_{0}^{2π} e^{(bH1 sin θ + cT1 cos θ)/d } (dθ / ω )= a e^{K0} (1/(2π)) ∫_{0}^{2π} e^{(bH1 sin θ + cT1 cos θ)/d } dθSo, the integral becomes (1/(2π)) ∫_{0}^{2π} e^{A sin θ + B cos θ} dθ, where A = bH1/d and B = cT1/d.I recall that the integral of e^{A sin θ + B cos θ} over 0 to 2π can be expressed using Bessel functions or perhaps using some identity. Let me think.Alternatively, we can write A sin θ + B cos θ as C sin(θ + φ), where C = sqrt(A² + B²) and φ = arctan(B/A) or something like that.Let me try that. Let me write A sin θ + B cos θ = C sin(θ + φ)Where C = sqrt(A² + B²), and φ is such that cos φ = A/C and sin φ = B/C.So, then:e^{A sin θ + B cos θ} = e^{C sin(θ + φ)}So, the integral becomes ∫_{0}^{2π} e^{C sin(θ + φ)} dθBut since the integral is over a full period, shifting θ by φ doesn't change the value. So, ∫_{0}^{2π} e^{C sin(θ + φ)} dθ = ∫_{0}^{2π} e^{C sin θ} dθAnd I remember that ∫_{0}^{2π} e^{C sin θ} dθ = 2π I_0(C), where I_0 is the modified Bessel function of the first kind of order 0.So, putting it all together:∫_{0}^{2π} e^{A sin θ + B cos θ} dθ = 2π I_0(C) where C = sqrt(A² + B²)Therefore, the average k_bar is:k_bar = a e^{K0} (1/(2π)) * 2π I_0(C) = a e^{K0} I_0(C)But let's substitute back A and B:A = bH1/d, B = cT1/dSo, C = sqrt( (bH1/d)^2 + (cT1/d)^2 ) = (1/d) sqrt( (bH1)^2 + (cT1)^2 )Let me denote sqrt( (bH1)^2 + (cT1)^2 ) as D, so C = D/dTherefore, k_bar = a e^{(bH0 + cT0)/d} I_0(D/d)But let me write it out explicitly:k_bar = a e^{(bH0 + cT0)/d} I_0( sqrt( (bH1)^2 + (cT1)^2 ) / d )Alternatively, since D = sqrt( (bH1)^2 + (cT1)^2 ), we can write:k_bar = a e^{(bH0 + cT0)/d} I_0( sqrt( (bH1)^2 + (cT1)^2 ) / d )So, that's the expression for the average growth rate constant over one period.Let me recap:1. Solved the logistic equation to get P(t) = N / (1 + (N - P0)/P0 e^{-kt})2. For the average k, expressed k(t) in terms of H(t) and T(t), recognized the integral over a period involves the exponential of a sinusoidal function, used the identity to express it in terms of Bessel functions, and arrived at the average k_bar as a product of the exponential term and the modified Bessel function I_0 evaluated at the amplitude of the sinusoidal terms divided by d.I think that's it. I should double-check the integral part because sometimes I might mix up the exact form, but I recall that the integral of e^{a sin θ} over 0 to 2π is 2π I_0(a), so that part seems correct.Also, the substitution θ = ωt was correct, and the scaling with dt = dθ/ω was handled properly, leading to the factor of 1/(2π) times the integral over 0 to 2π.So, I think the solution is solid.

answer：Okay, so I have two problems here about fetal heart rate monitoring. Let me try to tackle them one by one.Starting with problem 1. The nurse is monitoring the FHR of a fetus, specifically fetus number 3, and the data is modeled by the function:[ F_3(t) = 12 sinleft(frac{pi}{30} t + frac{pi}{6}right) + 140 ]She wants to find the average FHR over one hour. Hmm, okay. So, average value of a function over an interval. I remember that the average value of a function ( f(t) ) over an interval ([a, b]) is given by:[ text{Average} = frac{1}{b - a} int_{a}^{b} f(t) , dt ]In this case, the interval is from 0 to 60 minutes. So, the average FHR would be:[ text{Average FHR} = frac{1}{60 - 0} int_{0}^{60} F_3(t) , dt ]Substituting the given function:[ text{Average FHR} = frac{1}{60} int_{0}^{60} left[12 sinleft(frac{pi}{30} t + frac{pi}{6}right) + 140right] dt ]I can split this integral into two parts:[ frac{1}{60} left[ int_{0}^{60} 12 sinleft(frac{pi}{30} t + frac{pi}{6}right) dt + int_{0}^{60} 140 , dt right] ]Let me compute each integral separately.First, the integral of the sine function. The integral of ( sin(k t + c) ) with respect to t is ( -frac{1}{k} cos(k t + c) ). So, applying that here:Let ( k = frac{pi}{30} ) and ( c = frac{pi}{6} ).So,[ int 12 sinleft(frac{pi}{30} t + frac{pi}{6}right) dt = 12 left( -frac{30}{pi} cosleft(frac{pi}{30} t + frac{pi}{6}right) right) + C ][ = -frac{360}{pi} cosleft(frac{pi}{30} t + frac{pi}{6}right) + C ]Now, evaluating from 0 to 60:At t = 60:[ -frac{360}{pi} cosleft(frac{pi}{30} times 60 + frac{pi}{6}right) ][ = -frac{360}{pi} cosleft(2pi + frac{pi}{6}right) ]But ( cos(2pi + theta) = costheta ), so:[ = -frac{360}{pi} cosleft(frac{pi}{6}right) ][ = -frac{360}{pi} times frac{sqrt{3}}{2} ][ = -frac{360 sqrt{3}}{2pi} ][ = -frac{180 sqrt{3}}{pi} ]At t = 0:[ -frac{360}{pi} cosleft(frac{pi}{30} times 0 + frac{pi}{6}right) ][ = -frac{360}{pi} cosleft(frac{pi}{6}right) ][ = -frac{360}{pi} times frac{sqrt{3}}{2} ][ = -frac{180 sqrt{3}}{pi} ]So, the definite integral from 0 to 60 is:[ left[ -frac{180 sqrt{3}}{pi} right] - left[ -frac{180 sqrt{3}}{pi} right] = 0 ]Wait, that's interesting. The integral of the sine function over one period is zero. That makes sense because the positive and negative areas cancel out over a full period. So, the integral of the sine term is zero.Now, the second integral is straightforward:[ int_{0}^{60} 140 , dt = 140t bigg|_{0}^{60} = 140 times 60 - 140 times 0 = 8400 ]So, putting it all together:[ text{Average FHR} = frac{1}{60} (0 + 8400) = frac{8400}{60} = 140 ]So, the average FHR is 140 beats per minute. That makes sense because the sine function oscillates around the midline, which is 140 in this case. So, the average should just be the midline value.Moving on to problem 2. The nurse wants to compare the variability of FHR among the 10 fetuses using variance. She mentions that the variance of a sinusoidal function ( A sin(omega t + phi) + B ) over one period is ( frac{A^2}{2} ). So, for each fetus, the variance is ( frac{A_i^2}{2} ).We have the amplitudes ( A_1 ) through ( A_{10} ) given as:10, 15, 12, 8, 20, 5, 18, 14, 9, 11.We need to compute the total variance across all fetuses. I think this means we need to sum up the variances for each fetus.So, total variance ( V ) is:[ V = sum_{i=1}^{10} frac{A_i^2}{2} ]Let me compute each ( A_i^2 ) first:- ( A_1 = 10 ): ( 10^2 = 100 )- ( A_2 = 15 ): ( 15^2 = 225 )- ( A_3 = 12 ): ( 12^2 = 144 )- ( A_4 = 8 ): ( 8^2 = 64 )- ( A_5 = 20 ): ( 20^2 = 400 )- ( A_6 = 5 ): ( 5^2 = 25 )- ( A_7 = 18 ): ( 18^2 = 324 )- ( A_8 = 14 ): ( 14^2 = 196 )- ( A_9 = 9 ): ( 9^2 = 81 )- ( A_{10} = 11 ): ( 11^2 = 121 )Now, let's sum these up:100 + 225 = 325325 + 144 = 469469 + 64 = 533533 + 400 = 933933 + 25 = 958958 + 324 = 12821282 + 196 = 14781478 + 81 = 15591559 + 121 = 1680So, the sum of ( A_i^2 ) is 1680.Now, total variance ( V = frac{1680}{2} = 840 ).So, the total variance across all 10 fetuses is 840.Interpreting this result in the context of prenatal care: Variance measures how much the FHR deviates from the mean. A higher variance indicates more variability or fluctuations in the FHR, which could be a sign of potential issues. Conversely, lower variance suggests more stable FHR. In this case, the total variance is 840, which is a measure of overall variability across all fetuses. This could help the nurse identify which fetuses have higher variability, potentially requiring closer monitoring. However, since this is a total variance across all, it might not directly indicate individual variability but rather the combined variability. The nurse might need to look at individual variances to assess each fetus's condition.Wait, actually, hold on. The problem says "total variance of the FHR across all fetuses." So, is this the sum of variances for each fetus, or is it the variance of all FHR data points combined?Hmm, the problem says, "the variance of a sinusoidal function... over one period is ( frac{A^2}{2} )", so for each fetus, their FHR has a variance of ( frac{A_i^2}{2} ). So, if we want the total variance across all fetuses, it's the sum of individual variances.So, yes, 840 is the total variance. But in statistics, variance isn't typically additive unless we're considering independent variables. But in this context, since each fetus is independent, adding their variances makes sense if we're considering the total variability across all.Alternatively, if we were to compute the variance of all FHR data points combined, that would be different. But given the problem states each fetus has its own variance, and we need the total variance, I think summing them is correct.So, the total variance is 840. This tells the nurse that, collectively, the fetuses have a significant amount of variability in their heart rates. Fetuses with higher A_i contribute more to this total variance. For example, fetus 5 with A_5=20 contributes ( frac{400}{2}=200 ) to the variance, which is the highest. So, this fetus has the most variable FHR, which might be a concern and require closer attention.On the other hand, fetus 6 with A_6=5 contributes only 12.5 to the variance, the least. So, its FHR is the most stable.Therefore, this total variance gives the nurse a way to assess which fetuses might be at higher risk based on their FHR variability.**Final Answer**1. The average FHR over the one-hour period for fetus 3 is boxed{140} beats per minute.2. The total variance of the FHR across all 10 fetuses is boxed{840}.

answer：Alright, so I have these two problems to solve. Let me take them one at a time. Starting with the first problem: Alex and Jamie have a special clock that measures time using a function related to the Fibonacci sequence. The function is given as ( f(t) = F_{n+2} ), where ( F_n ) is the nth Fibonacci number. They mention that the current time ( t ) is such that ( f(t) = 144 ). I need to find the value of ( n ).Okay, so I remember that the Fibonacci sequence starts with ( F_0 = 0 ), ( F_1 = 1 ), and each subsequent term is the sum of the two preceding ones. So, ( F_2 = 1 ), ( F_3 = 2 ), ( F_4 = 3 ), ( F_5 = 5 ), and so on. Given that ( f(t) = F_{n+2} = 144 ), I need to find the index ( n ) such that the (n+2)th Fibonacci number is 144. So, first, I should figure out which Fibonacci number is 144.Let me list out the Fibonacci numbers until I reach 144.Starting from ( F_0 ):- ( F_0 = 0 )- ( F_1 = 1 )- ( F_2 = 1 ) (since ( F_2 = F_1 + F_0 = 1 + 0 = 1 ))- ( F_3 = F_2 + F_1 = 1 + 1 = 2 )- ( F_4 = F_3 + F_2 = 2 + 1 = 3 )- ( F_5 = F_4 + F_3 = 3 + 2 = 5 )- ( F_6 = F_5 + F_4 = 5 + 3 = 8 )- ( F_7 = F_6 + F_5 = 8 + 5 = 13 )- ( F_8 = F_7 + F_6 = 13 + 8 = 21 )- ( F_9 = F_8 + F_7 = 21 + 13 = 34 )- ( F_{10} = F_9 + F_8 = 34 + 21 = 55 )- ( F_{11} = F_{10} + F_9 = 55 + 34 = 89 )- ( F_{12} = F_{11} + F_{10} = 89 + 55 = 144 )Ah, there we go. So, ( F_{12} = 144 ). Given that ( f(t) = F_{n+2} = 144 ), we have ( F_{n+2} = F_{12} ). Therefore, ( n + 2 = 12 ), which implies ( n = 10 ).Wait, let me double-check that. If ( n + 2 = 12 ), then subtracting 2 from both sides gives ( n = 10 ). Yeah, that seems right.So, the value of ( n ) is 10.Moving on to the second problem: Alex and Jamie have defined a sequence ( C_n ) for their companionship moments. The recurrence relation is given as ( C_n = C_{n-1} + C_{n-2} + 2C_{n-3} ) with initial conditions ( C_0 = 2 ), ( C_1 = 3 ), and ( C_2 = 5 ). I need to determine ( C_5 ).Alright, so this is a linear recurrence relation. Let me write down the given information:- ( C_0 = 2 )- ( C_1 = 3 )- ( C_2 = 5 )- For ( n geq 3 ), ( C_n = C_{n-1} + C_{n-2} + 2C_{n-3} )I need to find ( C_5 ). So, I can compute ( C_3 ), ( C_4 ), and then ( C_5 ) step by step.Let's compute ( C_3 ):( C_3 = C_2 + C_1 + 2C_0 = 5 + 3 + 2*2 = 5 + 3 + 4 = 12 )So, ( C_3 = 12 ).Next, ( C_4 ):( C_4 = C_3 + C_2 + 2C_1 = 12 + 5 + 2*3 = 12 + 5 + 6 = 23 )So, ( C_4 = 23 ).Now, ( C_5 ):( C_5 = C_4 + C_3 + 2C_2 = 23 + 12 + 2*5 = 23 + 12 + 10 = 45 )Therefore, ( C_5 = 45 ).Wait, let me verify each step again to make sure I didn't make a calculation error.For ( C_3 ):- ( C_2 = 5 )- ( C_1 = 3 )- ( 2C_0 = 2*2 = 4 )- Sum: 5 + 3 + 4 = 12. Correct.For ( C_4 ):- ( C_3 = 12 )- ( C_2 = 5 )- ( 2C_1 = 2*3 = 6 )- Sum: 12 + 5 + 6 = 23. Correct.For ( C_5 ):- ( C_4 = 23 )- ( C_3 = 12 )- ( 2C_2 = 2*5 = 10 )- Sum: 23 + 12 + 10 = 45. Correct.So, yeah, ( C_5 = 45 ).Just to be thorough, maybe I can write out the sequence up to ( C_5 ):- ( C_0 = 2 )- ( C_1 = 3 )- ( C_2 = 5 )- ( C_3 = 12 )- ( C_4 = 23 )- ( C_5 = 45 )Yep, that seems consistent.Alternatively, I can think about solving the recurrence relation more formally, maybe using characteristic equations or generating functions, but since the problem only asks for ( C_5 ), computing each term step by step is straightforward and less error-prone for me.So, I think I've got both problems solved.**Final Answer**1. The value of ( n ) is boxed{10}.2. The value of ( C_5 ) is boxed{45}.

answer：Alright, so I have two questions to solve here. Let me take them one by one.Starting with the first question: It's about calculating the probability that Team A scores more than 110 points in a game against Team B. Hmm, okay. Let me parse the information given.Team A's average points per game, P, follows a normal distribution with a mean of 102.5 and a standard deviation of 8.3. Team B's average points allowed per game, Q, is normally distributed with a mean of 98.7 and a standard deviation of 7.4. The variables P and Q are independent.Wait, so we need to find the probability that Team A scores more than 110 points in a game against Team B. But hold on, is Team B's points allowed affecting Team A's points scored? Or are they independent? The problem says they are independent random variables, so P and Q don't influence each other. So, actually, the performance of Team B doesn't affect how many points Team A scores. So, does that mean that the probability that Team A scores more than 110 is just based on their own distribution?Let me think. If they are independent, then Team A's points are solely dependent on their own distribution, regardless of Team B's performance. So, to find the probability that Team A scores more than 110, I can just use Team A's normal distribution parameters.So, Team A's points, P ~ N(102.5, 8.3²). We need P(P > 110). To calculate this, I can standardize the value 110.The formula for the z-score is z = (X - μ) / σ. Plugging in the numbers: z = (110 - 102.5) / 8.3.Calculating that: 110 - 102.5 is 7.5. Then, 7.5 divided by 8.3. Let me compute that. 7.5 / 8.3 is approximately 0.9036.So, z ≈ 0.9036. Now, I need to find the probability that Z > 0.9036, where Z is the standard normal variable.Looking at the standard normal distribution table, the area to the left of z = 0.90 is approximately 0.8159, and for z = 0.91, it's about 0.8186. Since 0.9036 is between 0.90 and 0.91, I can interpolate.Difference between 0.90 and 0.91 is 0.01 in z, which corresponds to a difference of 0.8186 - 0.8159 = 0.0027 in probability.Our z is 0.9036, which is 0.0036 above 0.90. So, the fraction is 0.0036 / 0.01 = 0.36.Therefore, the area to the left of z = 0.9036 is approximately 0.8159 + 0.36 * 0.0027 ≈ 0.8159 + 0.000972 ≈ 0.8169.Hence, the area to the right (which is what we need) is 1 - 0.8169 = 0.1831.So, the probability that Team A scores more than 110 points is approximately 18.31%.Wait, let me double-check my calculations. Maybe I should use a calculator for the z-score.Alternatively, using a calculator: z = (110 - 102.5) / 8.3 = 7.5 / 8.3 ≈ 0.9036.Looking up 0.9036 in the z-table, or using a calculator for the cumulative distribution function (CDF). If I recall, the CDF for 0.90 is about 0.8159, as I mentioned. For 0.9036, it's slightly higher. Maybe using linear approximation is okay.Alternatively, I can use the formula for the standard normal distribution:P(Z > z) = 1 - Φ(z), where Φ(z) is the CDF.Using a calculator, if I compute Φ(0.9036), it's approximately 0.8164. So, P(Z > 0.9036) ≈ 1 - 0.8164 = 0.1836, which is about 18.36%. So, my initial approximation was pretty close.Therefore, the probability is approximately 18.4%.Wait, but the question says "in a randomly selected game where these two teams play against each other." Hmm, does that change anything? Because if they are playing against each other, is there any dependency? But the problem states that P and Q are independent, so regardless of who they are playing against, Team A's points are independent of Team B's points allowed.So, yeah, the probability is just based on Team A's distribution.So, I think 18.4% is the correct answer.Moving on to the second question: The executive wants to determine the correlation between the number of assists (A) and the number of turnovers (T) for Team A. We have the covariance of A and T, which is 15.2, the standard deviation of assists is 4.1, and the standard deviation of turnovers is 5.3.Correlation coefficient, ρ, is calculated as the covariance divided by the product of the standard deviations of the two variables.So, ρ = Cov(A, T) / (σ_A * σ_T).Plugging in the numbers: Cov(A, T) = 15.2, σ_A = 4.1, σ_T = 5.3.So, ρ = 15.2 / (4.1 * 5.3).First, compute the denominator: 4.1 * 5.3.Calculating that: 4 * 5 = 20, 4 * 0.3 = 1.2, 0.1 * 5 = 0.5, 0.1 * 0.3 = 0.03. Adding up: 20 + 1.2 + 0.5 + 0.03 = 21.73.Wait, that's not correct. Wait, 4.1 * 5.3 is actually calculated as (4 + 0.1) * (5 + 0.3) = 4*5 + 4*0.3 + 0.1*5 + 0.1*0.3 = 20 + 1.2 + 0.5 + 0.03 = 21.73. So, that's correct.So, denominator is 21.73.So, ρ = 15.2 / 21.73 ≈ ?Calculating that: 15.2 divided by 21.73.Well, 21.73 goes into 15.2 approximately 0.7 times because 21.73 * 0.7 = 15.211. So, that's almost exactly 0.7.Wait, 21.73 * 0.7 = 15.211, which is just a bit more than 15.2. So, 0.7 is slightly higher than the actual value. So, maybe 0.699 or something.But since 21.73 * 0.7 = 15.211, which is very close to 15.2, the difference is 15.211 - 15.2 = 0.011.So, 0.011 / 21.73 ≈ 0.0005. So, subtracting that from 0.7 gives approximately 0.6995.So, approximately 0.7.Therefore, the correlation coefficient is approximately 0.7.But let me verify with a calculator.15.2 / 21.73.Let me compute 15.2 divided by 21.73.21.73 goes into 15.2 how many times?21.73 * 0.7 = 15.211, as above. So, 0.7 is just a bit more than 15.2.So, 15.2 / 21.73 ≈ 0.6995, which is approximately 0.7.So, the correlation coefficient is approximately 0.7.But since the question didn't specify rounding, maybe we should present it as 0.7 or perhaps more accurately.Wait, 15.2 / 21.73 is exactly equal to 1520 / 2173. Let me compute that fraction.Divide numerator and denominator by GCD(1520,2173). Let's see, 2173 divided by 1520 is 1 with remainder 653. Then, 1520 divided by 653 is 2 with remainder 214. 653 divided by 214 is 3 with remainder 11. 214 divided by 11 is 19 with remainder 5. 11 divided by 5 is 2 with remainder 1. 5 divided by 1 is 5 with remainder 0. So, GCD is 1. Therefore, the fraction is 1520/2173.Calculating 1520 ÷ 2173:2173 goes into 1520 zero times. Add decimal: 15200 ÷ 2173.2173 * 7 = 15211, which is just above 15200. So, 6 times: 2173 * 6 = 13038.15200 - 13038 = 2162.Bring down a zero: 21620.2173 goes into 21620 approximately 9 times (2173*9=19557). 21620 - 19557=2063.Bring down a zero: 20630.2173 goes into 20630 about 9 times (2173*9=19557). 20630 - 19557=1073.Bring down a zero: 10730.2173 goes into 10730 about 4 times (2173*4=8692). 10730 - 8692=2038.Bring down a zero: 20380.2173 goes into 20380 about 9 times (2173*9=19557). 20380 - 19557=823.Bring down a zero: 8230.2173 goes into 8230 about 3 times (2173*3=6519). 8230 - 6519=1711.Bring down a zero: 17110.2173 goes into 17110 about 7 times (2173*7=15211). 17110 - 15211=1899.Bring down a zero: 18990.2173 goes into 18990 about 8 times (2173*8=17384). 18990 - 17384=1606.Bring down a zero: 16060.2173 goes into 16060 about 7 times (2173*7=15211). 16060 - 15211=849.At this point, I can see the decimal is approximately 0.6995, which is roughly 0.70.So, the correlation coefficient is approximately 0.70.Therefore, the answers are approximately 18.4% for the first question and 0.70 for the second.**Final Answer**1. The probability that Team A scores more than 110 points is boxed{0.184}.2. The correlation coefficient between assists and turnovers for Team A is boxed{0.70}.

answer：Okay, so I have this problem about eX-Girl's music, which is known for being eclectic and experimental. The problem involves a piecewise function representing the rhythmic intensity of a song. The function f(t) is defined differently on two intervals: from 0 to 5 seconds, it's 2e^(-t/2) sin(2πt), and from 5 to 10 seconds, it's 1/(t-4) + cos(3πt). The first part asks me to determine the point of discontinuity in the interval [0,10] and classify the type of discontinuity. The second part is about computing the Fourier series representation of f(t) over [0,10]. Starting with the first part. I need to check if there's a discontinuity at t=5 because that's where the function changes its definition. So, I should check the left-hand limit as t approaches 5 from below and the right-hand limit as t approaches 5 from above, and see if they are equal to each other and to f(5). If they aren't, then there's a discontinuity.Let me compute the left-hand limit first. As t approaches 5 from the left, f(t) is 2e^(-t/2) sin(2πt). So, plugging t=5 into that expression:Left-hand limit = 2e^(-5/2) sin(2π*5)Calculating that: 2π*5 is 10π, and sin(10π) is 0 because sine of any integer multiple of π is 0. So, the left-hand limit is 2e^(-5/2)*0 = 0.Now, the right-hand limit as t approaches 5 from the right. Here, f(t) is 1/(t-4) + cos(3πt). Plugging t=5 into this:Right-hand limit = 1/(5-4) + cos(3π*5) = 1/1 + cos(15π)Cos(15π) is equal to cos(π) because cosine has a period of 2π, so 15π is 7*2π + π, so cos(15π) = cos(π) = -1. Therefore, right-hand limit is 1 + (-1) = 0.So both the left-hand limit and the right-hand limit as t approaches 5 are 0. Now, what is f(5)? Since t=5 is included in the second piece, f(5) is 1/(5-4) + cos(15π) = 1 + (-1) = 0. Wait, so both the left and right limits are 0, and f(5) is also 0. So, is there a discontinuity? It seems like f(t) is continuous at t=5 because the left limit, right limit, and the function value all equal 0. Hmm, that's interesting. So maybe there's no discontinuity at t=5? But let me double-check my calculations.For the left-hand limit: 2e^(-5/2) sin(10π). Sin(10π) is indeed 0, so that term is 0. For the right-hand limit: 1/(5-4)=1, cos(15π)=cos(π)= -1, so 1 + (-1)=0. And f(5)=0. So, all three are equal. Therefore, f(t) is continuous at t=5.Wait, but is that the only point where a discontinuity could occur? Let me think. The function is defined piecewise, so the only potential point of discontinuity is at t=5. Since both pieces are continuous on their own intervals, the only possible issue is at t=5. Since we've established that the left and right limits match and equal f(5), the function is continuous there. Therefore, there is no point of discontinuity in [0,10]. So, the answer to part 1 is that there is no discontinuity in the interval [0,10].Wait, but hold on. Let me check the function definitions again. For the first piece, t is from 0 to 5, not including 5. The second piece is from 5 to 10, including 5. So, at t=5, the function is defined by the second piece. But since the left limit and the right limit both equal f(5), it's continuous. So, no discontinuity.So, part 1: There is no discontinuity in the interval [0,10].Moving on to part 2: Compute the Fourier series representation of f(t) over [0,10]. Fourier series are typically computed over an interval, usually of length 2L, so that the period is 2L. Here, the interval is [0,10], which is 10 units long. So, if we consider this as a periodic function with period 10, then L=5. Alternatively, if we are to compute the Fourier series over [0,10], it can be considered as a function with period 10.But wait, the function f(t) is defined on [0,10], so if we are to compute its Fourier series, we can represent it as a Fourier series over [0,10], either as a sine-cosine series or as a complex exponential series. Since the function is piecewise defined, it might be more straightforward to compute the Fourier series over [0,10] as a function with period 10.But first, let's recall that the Fourier series of a function f(t) over an interval [0, T] is given by:f(t) = a0/2 + Σ [an cos(nωt) + bn sin(nωt)], where ω = 2π/T.For T=10, ω=π/5.The coefficients are given by:a0 = (2/T) ∫₀^T f(t) dtan = (2/T) ∫₀^T f(t) cos(nωt) dtbn = (2/T) ∫₀^T f(t) sin(nωt) dtSo, in this case, T=10, ω=π/5.Therefore, we need to compute a0, an, and bn.But since f(t) is defined piecewise, we'll have to split the integrals into two parts: from 0 to 5 and from 5 to 10.So, let's write down the expressions:a0 = (2/10)[∫₀^5 2e^(-t/2) sin(2πt) dt + ∫₅^10 [1/(t-4) + cos(3πt)] dt]Similarly,an = (2/10)[∫₀^5 2e^(-t/2) sin(2πt) cos(nπt/5) dt + ∫₅^10 [1/(t-4) + cos(3πt)] cos(nπt/5) dt]bn = (2/10)[∫₀^5 2e^(-t/2) sin(2πt) sin(nπt/5) dt + ∫₅^10 [1/(t-4) + cos(3πt)] sin(nπt/5) dt]So, we have to compute these integrals for a0, an, bn.This seems quite involved, especially because some of these integrals might not have elementary antiderivatives. Let's tackle them one by one.Starting with a0:a0 = (1/5)[∫₀^5 2e^(-t/2) sin(2πt) dt + ∫₅^10 [1/(t-4) + cos(3πt)] dt]Let me compute each integral separately.First integral: I1 = ∫₀^5 2e^(-t/2) sin(2πt) dtSecond integral: I2 = ∫₅^10 [1/(t-4) + cos(3πt)] dtCompute I1:I1 = 2 ∫₀^5 e^(-t/2) sin(2πt) dtThis integral can be solved using integration by parts or by using the formula for ∫ e^{at} sin(bt) dt.Recall that ∫ e^{at} sin(bt) dt = e^{at} [a sin(bt) - b cos(bt)] / (a² + b²) + CHere, a = -1/2, b = 2πSo, applying the formula:I1 = 2 [ e^(-t/2) [ (-1/2) sin(2πt) - 2π cos(2πt) ] / ( (-1/2)^2 + (2π)^2 ) ] evaluated from 0 to 5.Simplify the denominator:(-1/2)^2 = 1/4(2π)^2 = 4π²So denominator = 1/4 + 4π² = (1 + 16π²)/4Therefore, I1 = 2 [ e^(-t/2) [ (-1/2) sin(2πt) - 2π cos(2πt) ] * 4 / (1 + 16π²) ] from 0 to 5.Simplify:I1 = 2 * 4 / (1 + 16π²) [ e^(-t/2) [ (-1/2) sin(2πt) - 2π cos(2πt) ] ] from 0 to 5I1 = 8 / (1 + 16π²) [ e^(-5/2) [ (-1/2) sin(10π) - 2π cos(10π) ] - e^(0) [ (-1/2) sin(0) - 2π cos(0) ] ]Compute each term:At t=5:sin(10π) = 0cos(10π) = 1So, first part: e^(-5/2) [ 0 - 2π*1 ] = -2π e^(-5/2)At t=0:sin(0)=0cos(0)=1So, second part: e^0 [ 0 - 2π*1 ] = -2πTherefore, I1 = 8 / (1 + 16π²) [ (-2π e^(-5/2)) - (-2π) ] = 8 / (1 + 16π²) [ -2π e^(-5/2) + 2π ] = 8 / (1 + 16π²) * 2π (1 - e^(-5/2)) = 16π (1 - e^(-5/2)) / (1 + 16π²)So, I1 = 16π (1 - e^(-5/2)) / (1 + 16π²)Now, compute I2:I2 = ∫₅^10 [1/(t-4) + cos(3πt)] dtSplit into two integrals:I2 = ∫₅^10 1/(t-4) dt + ∫₅^10 cos(3πt) dtCompute first integral: ∫ 1/(t-4) dt from 5 to 10Let u = t - 4, then du = dt. When t=5, u=1; when t=10, u=6.So, ∫₁^6 1/u du = ln|u| from 1 to 6 = ln(6) - ln(1) = ln(6)Second integral: ∫₅^10 cos(3πt) dtIntegrate cos(3πt):∫ cos(3πt) dt = (1/(3π)) sin(3πt) + CEvaluate from 5 to 10:(1/(3π)) [ sin(30π) - sin(15π) ]But sin(30π)=0 and sin(15π)=0, since sine of any integer multiple of π is 0.Therefore, the second integral is 0.Thus, I2 = ln(6) + 0 = ln(6)Therefore, a0 = (1/5)(I1 + I2) = (1/5)[16π (1 - e^(-5/2)) / (1 + 16π²) + ln(6)]So, that's a0.Now, moving on to an:an = (2/10)[∫₀^5 2e^(-t/2) sin(2πt) cos(nπt/5) dt + ∫₅^10 [1/(t-4) + cos(3πt)] cos(nπt/5) dt]Simplify:an = (1/5)[2 ∫₀^5 e^(-t/2) sin(2πt) cos(nπt/5) dt + ∫₅^10 [1/(t-4) + cos(3πt)] cos(nπt/5) dt]Let me denote the first integral as J1 and the second as J2.So, an = (1/5)[2 J1 + J2]Compute J1: ∫₀^5 e^(-t/2) sin(2πt) cos(nπt/5) dtThis integral looks complicated. Maybe we can use a trigonometric identity to simplify the product of sines and cosines.Recall that sin A cos B = [sin(A+B) + sin(A-B)] / 2So, sin(2πt) cos(nπt/5) = [sin(2πt + nπt/5) + sin(2πt - nπt/5)] / 2Therefore, J1 = ∫₀^5 e^(-t/2) [sin((2π + nπ/5)t) + sin((2π - nπ/5)t)] / 2 dtSo, J1 = (1/2)[∫₀^5 e^(-t/2) sin((2π + nπ/5)t) dt + ∫₀^5 e^(-t/2) sin((2π - nπ/5)t) dt]Let me denote the two integrals as J1a and J1b.J1a = ∫₀^5 e^(-t/2) sin((2π + nπ/5)t) dtJ1b = ∫₀^5 e^(-t/2) sin((2π - nπ/5)t) dtEach of these integrals can be solved using the formula for ∫ e^{at} sin(bt) dt, which we used earlier.Recall that ∫ e^{at} sin(bt) dt = e^{at} [a sin(bt) - b cos(bt)] / (a² + b²) + CSo, for J1a:a = -1/2, b = 2π + nπ/5Similarly, for J1b:a = -1/2, b = 2π - nπ/5Therefore, let's compute J1a:J1a = [ e^(-t/2) [ (-1/2) sin((2π + nπ/5)t) - (2π + nπ/5) cos((2π + nπ/5)t) ] / ( (-1/2)^2 + (2π + nπ/5)^2 ) ] from 0 to 5Similarly for J1b:J1b = [ e^(-t/2) [ (-1/2) sin((2π - nπ/5)t) - (2π - nπ/5) cos((2π - nπ/5)t) ] / ( (-1/2)^2 + (2π - nπ/5)^2 ) ] from 0 to 5This is getting quite involved, and the expressions will be quite lengthy. Maybe we can denote some terms to simplify.Let me denote for J1a:b1 = 2π + nπ/5Denominator1 = (1/4) + b1² = (1 + 4b1²)/4Similarly, for J1b:b2 = 2π - nπ/5Denominator2 = (1/4) + b2² = (1 + 4b2²)/4Therefore, J1a = [ e^(-t/2) [ (-1/2) sin(b1 t) - b1 cos(b1 t) ] * 4 / (1 + 4b1²) ] from 0 to 5Similarly, J1b = [ e^(-t/2) [ (-1/2) sin(b2 t) - b2 cos(b2 t) ] * 4 / (1 + 4b2²) ] from 0 to 5So, J1a = (4 / (1 + 4b1²)) [ e^(-5/2) [ (-1/2) sin(5b1) - b1 cos(5b1) ] - e^0 [ (-1/2) sin(0) - b1 cos(0) ] ]Similarly for J1b.Compute J1a:First, evaluate at t=5:sin(5b1) = sin(5*(2π + nπ/5)) = sin(10π + nπ) = sin(nπ) = 0 because sin(kπ)=0 for integer k.cos(5b1) = cos(10π + nπ) = cos(nπ) = (-1)^nSimilarly, at t=0:sin(0)=0cos(0)=1Therefore, J1a becomes:(4 / (1 + 4b1²)) [ e^(-5/2) [ 0 - b1*(-1)^n ] - [ 0 - b1*1 ] ]Simplify:= (4 / (1 + 4b1²)) [ e^(-5/2) (b1 (-1)^{n+1}) + b1 ]Similarly, J1b:Evaluate at t=5:sin(5b2) = sin(5*(2π - nπ/5)) = sin(10π - nπ) = sin(-nπ) = 0cos(5b2) = cos(10π - nπ) = cos(nπ) = (-1)^nAt t=0:sin(0)=0cos(0)=1Therefore, J1b becomes:(4 / (1 + 4b2²)) [ e^(-5/2) [ 0 - b2*(-1)^n ] - [ 0 - b2*1 ] ]= (4 / (1 + 4b2²)) [ e^(-5/2) (b2 (-1)^{n+1}) + b2 ]Therefore, J1 = (1/2)[ J1a + J1b ]= (1/2) [ (4 / (1 + 4b1²)) (b1 (-1)^{n+1} e^(-5/2) + b1 ) + (4 / (1 + 4b2²)) (b2 (-1)^{n+1} e^(-5/2) + b2 ) ]Factor out the 4 and b terms:= (1/2) [ 4 b1 / (1 + 4b1²) ( (-1)^{n+1} e^(-5/2) + 1 ) + 4 b2 / (1 + 4b2²) ( (-1)^{n+1} e^(-5/2) + 1 ) ]= 2 [ b1 / (1 + 4b1²) ( (-1)^{n+1} e^(-5/2) + 1 ) + b2 / (1 + 4b2²) ( (-1)^{n+1} e^(-5/2) + 1 ) ]Now, substituting back b1 and b2:b1 = 2π + nπ/5b2 = 2π - nπ/5So,J1 = 2 [ (2π + nπ/5) / (1 + 4(2π + nπ/5)^2 ) ( (-1)^{n+1} e^(-5/2) + 1 ) + (2π - nπ/5) / (1 + 4(2π - nπ/5)^2 ) ( (-1)^{n+1} e^(-5/2) + 1 ) ]This is a very complicated expression, but it's manageable.Now, moving on to J2:J2 = ∫₅^10 [1/(t-4) + cos(3πt)] cos(nπt/5) dtAgain, split into two integrals:J2 = ∫₅^10 1/(t-4) cos(nπt/5) dt + ∫₅^10 cos(3πt) cos(nπt/5) dtLet me denote these as J2a and J2b.Compute J2a:J2a = ∫₅^10 1/(t-4) cos(nπt/5) dtLet u = t - 4, then du = dt, and when t=5, u=1; t=10, u=6.So, J2a = ∫₁^6 1/u cos(nπ(u + 4)/5) du= ∫₁^6 (1/u) cos(nπ u /5 + 4nπ/5) duThis integral doesn't have an elementary antiderivative, so we might need to express it in terms of special functions or leave it as is. Alternatively, if n is an integer, perhaps we can find a pattern or use orthogonality, but I don't think that's the case here.Similarly, J2b = ∫₅^10 cos(3πt) cos(nπt/5) dtAgain, we can use a trigonometric identity:cos A cos B = [cos(A+B) + cos(A-B)] / 2So, cos(3πt) cos(nπt/5) = [cos(3πt + nπt/5) + cos(3πt - nπt/5)] / 2Therefore, J2b = (1/2)[∫₅^10 cos((3π + nπ/5)t) dt + ∫₅^10 cos((3π - nπ/5)t) dt]Let me denote these as J2b1 and J2b2.Compute J2b1:J2b1 = ∫₅^10 cos((3π + nπ/5)t) dt= [ sin((3π + nπ/5)t) / (3π + nπ/5) ] from 5 to 10Similarly, J2b2 = ∫₅^10 cos((3π - nπ/5)t) dt= [ sin((3π - nπ/5)t) / (3π - nπ/5) ] from 5 to 10So, evaluating J2b1:sin((3π + nπ/5)*10) - sin((3π + nπ/5)*5) divided by (3π + nπ/5)Similarly for J2b2.Compute sin((3π + nπ/5)*10):= sin(30π + 2nπ) = sin(30π) cos(2nπ) + cos(30π) sin(2nπ) = 0 + 1*0 = 0Similarly, sin((3π + nπ/5)*5) = sin(15π + nπ) = sin(15π)cos(nπ) + cos(15π)sin(nπ) = 0 + (-1)^n *0 = 0Therefore, J2b1 = [0 - 0] / (3π + nπ/5) = 0Similarly, J2b2:sin((3π - nπ/5)*10) - sin((3π - nπ/5)*5) divided by (3π - nπ/5)Compute sin((3π - nπ/5)*10) = sin(30π - 2nπ) = sin(30π)cos(2nπ) - cos(30π)sin(2nπ) = 0 - 1*0 = 0sin((3π - nπ/5)*5) = sin(15π - nπ) = sin(15π)cos(nπ) - cos(15π)sin(nπ) = 0 - (-1)^n *0 = 0Therefore, J2b2 = [0 - 0] / (3π - nπ/5) = 0Thus, J2b = (1/2)(0 + 0) = 0Therefore, J2 = J2a + J2b = J2a + 0 = J2aSo, J2 = ∫₁^6 (1/u) cos(nπ u /5 + 4nπ/5) duThis integral is still complicated. Maybe we can make a substitution or express it in terms of cosine integrals, but I don't think it simplifies nicely. So, perhaps we can leave it as is or express it in terms of the cosine integral function, but since this is a Fourier series, we might need to keep it in integral form.Alternatively, if we consider that n is an integer, perhaps we can find some orthogonality or periodicity, but I don't see an immediate simplification.Therefore, putting it all together:an = (1/5)[2 J1 + J2] = (1/5)[2 * [complicated expression] + J2a]Given the complexity, it's clear that computing an explicitly for general n is going to be very involved, and likely not expressible in a simple closed-form. Similarly, bn will involve similar integrals, which might also be complicated.Given the time constraints and the complexity, perhaps the problem expects us to recognize that the Fourier series can be expressed as a sum of integrals, but without computing them explicitly. Alternatively, maybe there's a simplification I'm missing.Wait, perhaps I made a mistake in assuming the period is 10. The function is defined on [0,10], but if we're computing the Fourier series over [0,10], it's typically considered as a function with period 10, so the Fourier series will have terms with frequencies that are multiples of ω=π/5.But given the complexity of the integrals, especially J2a, which involves 1/(t-4) multiplied by a cosine term, which doesn't have an elementary antiderivative, I think the Fourier series can't be expressed in a simple closed-form. Therefore, the answer would be expressed in terms of these integrals.Similarly, for bn, we would have to compute:bn = (2/10)[∫₀^5 2e^(-t/2) sin(2πt) sin(nπt/5) dt + ∫₅^10 [1/(t-4) + cos(3πt)] sin(nπt/5) dt]Again, this would involve similar steps as an, leading to complicated expressions.Given that, perhaps the answer is to express the Fourier series in terms of these integrals, acknowledging that they can't be simplified further without more advanced techniques or numerical methods.Alternatively, maybe the function is even or odd, but looking at f(t), it doesn't seem to be symmetric in any obvious way, so we can't exploit that to simplify the Fourier series.Therefore, in conclusion, the Fourier series representation of f(t) over [0,10] is given by:f(t) = a0/2 + Σ [an cos(nπt/5) + bn sin(nπt/5)]where the coefficients a0, an, bn are given by the integrals computed above, which involve complex expressions that may not have elementary closed-form solutions.So, summarizing:1. The function f(t) is continuous at t=5, hence there is no discontinuity in [0,10].2. The Fourier series exists but its coefficients involve integrals that are complicated and may not have simple closed-form expressions, so the series is represented in terms of these integrals.But wait, the problem says "compute the Fourier series representation". If it's expecting a specific form, perhaps I need to proceed differently. Maybe instead of computing over [0,10], we can consider extending the function periodically, but since the function is defined on [0,10], it's already a period of 10.Alternatively, perhaps the function can be expressed as a sum of exponentials, but given the piecewise nature, it's not straightforward.Alternatively, maybe using complex Fourier series, but that might not simplify things.Alternatively, perhaps the function is already a combination of sinusoids and exponentials, so the Fourier series would involve those terms, but I'm not sure.Alternatively, perhaps the function is smooth except at t=5, but since it's continuous there, the Fourier series should converge to the function.But given the complexity, I think the answer is that the Fourier series is given by the sum with coefficients a0, an, bn as computed above, involving those integrals.Therefore, the final answer for part 2 is that the Fourier series is expressed as:f(t) = a0/2 + Σ [an cos(nπt/5) + bn sin(nπt/5)]wherea0 = (1/5)[16π (1 - e^(-5/2)) / (1 + 16π²) + ln(6)]an = (1/5)[2 J1 + J2]bn = (1/5)[2 K1 + K2]with J1, J2, K1, K2 being the respective integrals computed above.But since the problem asks to compute the Fourier series, perhaps it's acceptable to leave it in terms of these integrals, acknowledging that they can't be simplified further.Alternatively, maybe the problem expects a different approach, such as using the fact that the function is piecewise defined and using the Fourier series for each piece separately, but I don't think that's standard.Alternatively, perhaps the function is even or odd, but as I checked earlier, it's neither.Alternatively, maybe the function is periodic with period 10, so we can write the Fourier series as a sum over n of c_n e^{i n π t /5}, but again, the integrals would be similar.Given that, I think the answer is as above, expressing the Fourier series in terms of the integrals for a0, an, bn.So, to summarize:1. There is no discontinuity in [0,10].2. The Fourier series is given by the expression above, with coefficients involving the integrals computed.But since the problem is in Chinese and the user is asking for the answer in a box, perhaps for part 1, the answer is no discontinuity, and for part 2, the Fourier series is expressed as the sum with coefficients a0, an, bn as computed.But given the complexity, maybe the problem expects a different approach or perhaps the answer is that the Fourier series cannot be expressed in a simple closed form due to the complexity of the integrals involved.Alternatively, perhaps the function is already a Fourier series, but that doesn't seem to be the case.Alternatively, maybe the function is a combination of sinusoids and exponentials, so its Fourier series would include those terms, but I don't think that's the case.Alternatively, perhaps the function is a sum of sinusoids with different frequencies, so the Fourier series would pick up those frequencies.But given the time, I think I have to conclude that the Fourier series is as above, with the coefficients expressed in terms of the integrals.Therefore, the final answers are:1. No discontinuity in [0,10].2. The Fourier series is given by the expression with coefficients a0, an, bn as computed, involving the integrals.But since the problem is in Chinese, perhaps the user expects the answer in Chinese, but the user's instruction is in English, so I think it's okay to proceed as above.But wait, the user wrote the problem in English, so the answer should be in English.Therefore, the final answers are:1. There is no point of discontinuity in the interval [0,10].2. The Fourier series representation of f(t) over [0,10] is given by:f(t) = a0/2 + Σ [an cos(nπt/5) + bn sin(nπt/5)]where the coefficients a0, an, bn are computed as above, involving the integrals which may not have simple closed-form expressions.But since the problem asks to compute the Fourier series, perhaps it's acceptable to leave it in terms of the integrals.Alternatively, maybe the problem expects a different approach, such as recognizing that the function is a combination of known functions whose Fourier series are known, but given the piecewise nature, it's not straightforward.Alternatively, perhaps the function is already a Fourier series, but that's not the case.Alternatively, perhaps the function is a sum of sinusoids and exponentials, so the Fourier series would include those terms, but I don't think that's the case.Given that, I think the answer is as above.Therefore, the final answers are:1. There is no discontinuity in the interval [0,10].2. The Fourier series is expressed as above, with coefficients involving the integrals computed.