🧠:<think>First, I need to determine the dimensions of the mural. The area of the mural is given as 120 square feet, and the length is 1.5 times the width. Let's denote the width as ( w ) and the length as ( l = 1.5w ).Using the area formula for a rectangle, ( text{Area} = text{length} times text{width} ), I can set up the equation:[1.5w times w = 120]Simplifying this, I get:[1.5w^2 = 120]To solve for ( w ), I divide both sides by 1.5:[w^2 = frac{120}{1.5} = 80]Taking the square root of both sides gives:[w = sqrt{80} approx 8.94 text{ feet}]Then, the length ( l ) is:[l = 1.5 times 8.94 approx 13.41 text{ feet}]Next, I need to calculate the total amount of decorative material required for the border. The border is 2 feet wide on all sides, so the overall dimensions including the border are:[text{Total width} = w + 2 times 2 = w + 4][text{Total length} = l + 2 times 2 = l + 4]Substituting the values of ( w ) and ( l ):[text{Total width} = 8.94 + 4 = 12.94 text{ feet}][text{Total length} = 13.41 + 4 = 17.41 text{ feet}]The total amount of decorative material needed is the perimeter of the larger rectangle:[text{Perimeter} = 2 times (text{Total width} + text{Total length}) = 2 times (12.94 + 17.41) = 2 times 30.35 = 60.7 text{ feet}]</think>

🧠:<think>Alright, so I have this problem about genetic impact on brain health, specifically looking at a rare disorder affecting neuron connectivity. The problem is divided into two parts: first, deriving the expected value of the connectivity strength ( S ), and second, using the Central Limit Theorem to estimate the probability that ( S ) exceeds a threshold ( T ).Let me start with the first part. The connectivity strength ( S ) is given by the sum:[ S = sum_{i=1}^{n} frac{1}{1+e^{-k_i}} ]Each ( k_i ) is normally distributed with mean ( mu ) and variance ( sigma^2 ). So, I need to find the expected value ( E(S) ).Since expectation is linear, I can write:[ E(S) = Eleft( sum_{i=1}^{n} frac{1}{1+e^{-k_i}} right) = sum_{i=1}^{n} Eleft( frac{1}{1+e^{-k_i}} right) ]So, I just need to compute ( Eleft( frac{1}{1+e^{-k_i}} right) ) for each ( i ). Since each ( k_i ) is identically distributed, this expectation will be the same for all ( i ), so I can compute it once and multiply by ( n ).Let me denote ( X = k_i ), so ( X sim N(mu, sigma^2) ). Then, I need to compute:[ Eleft( frac{1}{1 + e^{-X}} right) ]Hmm, this looks familiar. The function ( frac{1}{1 + e^{-x}} ) is the logistic function, often denoted as ( sigma(x) ). So, we're looking for the expectation of the logistic function of a normal random variable.I remember that the expectation of the logistic function of a normal variable doesn't have a closed-form solution, but perhaps there's an approximation or a known result.Wait, let me think. If ( X ) is normally distributed, then ( sigma(X) ) is a probit-like function. But I don't recall a simple expression for its expectation.Alternatively, maybe I can express it in terms of the cumulative distribution function (CDF) of another distribution. Let me recall that:[ frac{1}{1 + e^{-x}} = sigma(x) = text{CDF of logistic distribution} ]But ( X ) is normal, not logistic. Hmm, maybe I can relate this expectation to the CDF of a different distribution.Alternatively, perhaps using a Taylor series expansion or a moment-generating function approach? Let me consider the moment-generating function of ( X ).Wait, but the expectation of ( sigma(X) ) might not have a simple closed-form. Maybe I can express it in terms of the error function or something similar?Alternatively, perhaps I can use a probabilistic interpretation. Let me consider that ( sigma(X) ) is the probability that a logistic random variable is less than or equal to ( X ). But I'm not sure if that helps.Wait, another thought: if ( X ) is normal, then ( sigma(X) ) is the probability that a standard logistic variable is less than ( X ). But I don't think that directly helps.Alternatively, maybe I can use a transformation. Let me let ( Y = sigma(X) ), so ( Y = frac{1}{1 + e^{-X}} ). Then, ( Y ) is a function of ( X ), which is normal. So, perhaps I can find the expectation of ( Y ) by integrating over the distribution of ( X ).So, formally:[ E(Y) = Eleft( frac{1}{1 + e^{-X}} right) = int_{-infty}^{infty} frac{1}{1 + e^{-x}} cdot frac{1}{sqrt{2pisigma^2}} e^{-frac{(x - mu)^2}{2sigma^2}} dx ]Hmm, that integral looks complicated. Maybe I can make a substitution to simplify it.Let me set ( z = x - mu ), so ( x = z + mu ). Then, the integral becomes:[ E(Y) = int_{-infty}^{infty} frac{1}{1 + e^{-(z + mu)}} cdot frac{1}{sqrt{2pisigma^2}} e^{-frac{z^2}{2sigma^2}} dz ]Simplify the exponent in the logistic term:[ frac{1}{1 + e^{-(z + mu)}} = frac{e^{mu + z}}{1 + e^{mu + z}} = frac{1}{1 + e^{-(mu + z)}} ]Wait, that doesn't seem to help much. Maybe another substitution. Let me set ( t = z + mu ), but that just brings me back to the original variable.Alternatively, perhaps I can write the logistic function as ( sigma(x) = frac{1}{2} left(1 + tanhleft(frac{x}{2}right)right) ). Is that correct?Wait, let me check:[ tanhleft(frac{x}{2}right) = frac{e^{x/2} - e^{-x/2}}{e^{x/2} + e^{-x/2}} = frac{e^{x} - 1}{e^{x} + 1} ]So, ( frac{1}{2} left(1 + tanhleft(frac{x}{2}right)right) = frac{1}{2} left(1 + frac{e^{x} - 1}{e^{x} + 1}right) = frac{1}{2} left( frac{(e^{x} + 1) + (e^{x} - 1)}{e^{x} + 1} right) = frac{1}{2} left( frac{2e^{x}}{e^{x} + 1} right) = frac{e^{x}}{e^{x} + 1} = frac{1}{1 + e^{-x}} ).Yes, that works. So, ( sigma(x) = frac{1}{2} left(1 + tanhleft(frac{x}{2}right)right) ).Therefore, ( E(Y) = Eleft( frac{1}{2} left(1 + tanhleft(frac{X}{2}right)right) right) = frac{1}{2} left(1 + Eleft( tanhleft( frac{X}{2} right) right) right) ).So, now I need to compute ( Eleft( tanhleft( frac{X}{2} right) right) ) where ( X sim N(mu, sigma^2) ).Hmm, the expectation of the hyperbolic tangent of a normal variable. I think there's a known result for this. Let me recall.I remember that for a normal variable ( X sim N(mu, sigma^2) ), the expectation ( E[tanh(aX)] ) can be expressed in terms of the derivative of the log partition function or something related to the cumulants.Alternatively, perhaps using the moment-generating function.Wait, the moment-generating function of ( X ) is ( M(t) = e^{mu t + frac{1}{2} sigma^2 t^2} ).But ( tanh(aX) ) is a nonlinear function, so it's not straightforward.Alternatively, perhaps using a series expansion for ( tanh ). Let me recall that:[ tanh(z) = sum_{n=1}^{infty} frac{B_{2n} (-4)^n (1 - 4^n)}{(2n)!} z^{2n - 1} ]But that might not be helpful here.Wait, another approach: use the fact that ( tanh(z) = frac{d}{dz} ln cosh(z) ). Hmm, but I don't see how that helps.Alternatively, perhaps express ( tanh(z) ) in terms of exponentials:[ tanh(z) = frac{e^{z} - e^{-z}}{e^{z} + e^{-z}} ]So, ( tanhleft( frac{X}{2} right) = frac{e^{X/2} - e^{-X/2}}{e^{X/2} + e^{-X/2}} ).Hmm, maybe I can write this as ( frac{e^{X} - 1}{e^{X} + 1} ), but that might not help.Alternatively, perhaps express ( tanhleft( frac{X}{2} right) ) as ( frac{sinh(X)}{cosh(X)} ), but that's the same as ( tanh(X) ), which isn't helpful.Wait, perhaps I can use the identity:[ tanhleft( frac{X}{2} right) = frac{sinh(X)}{1 + cosh(X)} ]But I'm not sure if that helps.Alternatively, perhaps consider that ( tanhleft( frac{X}{2} right) ) is the derivative of ( ln coshleft( frac{X}{2} right) ), but again, not sure.Wait, maybe I can use the fact that for a normal variable ( X ), ( E[e^{tX}] ) is known, and perhaps relate ( E[tanh(aX)] ) to derivatives of the moment-generating function.Let me think. Let me denote ( a = 1/2 ), so ( E[tanh(aX)] ).We know that ( tanh(aX) = frac{e^{2aX} - 1}{e^{2aX} + 1} ). Hmm, maybe not helpful.Alternatively, perhaps write ( tanh(aX) = frac{d}{da} ln cosh(aX) ). Let me check:[ frac{d}{da} ln cosh(aX) = frac{X sinh(aX)}{cosh(aX)} = X tanh(aX) ]So, ( tanh(aX) = frac{1}{X} frac{d}{da} ln cosh(aX) ).But then, taking expectation:[ E[tanh(aX)] = Eleft[ frac{1}{X} frac{d}{da} ln cosh(aX) right] ]But this seems complicated because of the ( 1/X ) term, which complicates things, especially since ( X ) can be zero.Alternatively, perhaps I can use the fact that ( tanh(aX) ) is an odd function if ( X ) is symmetric around zero, but in our case, ( X ) has a mean ( mu ), so it's not symmetric unless ( mu = 0 ).Wait, maybe I can use a Taylor expansion of ( tanh(aX) ) around ( X = mu ). Let me consider expanding ( tanh(aX) ) as a series and then taking the expectation term by term.So, let me write:[ tanh(aX) = tanh(amu + a(X - mu)) ]Let me set ( Y = X - mu ), so ( Y sim N(0, sigma^2) ). Then,[ tanh(amu + aY) = tanh(amu + aY) ]Using the expansion for ( tanh(b + cY) ) where ( b = amu ) and ( c = a ).The Taylor expansion of ( tanh(b + cY) ) around ( Y = 0 ) is:[ tanh(b) + c text{sech}^2(b) Y - frac{c^2}{3} tanh(b) text{sech}^2(b) Y^2 + cdots ]So, up to the second order, we have:[ tanh(amu + aY) approx tanh(amu) + a text{sech}^2(amu) Y - frac{a^2}{3} tanh(amu) text{sech}^2(amu) Y^2 ]Now, taking expectation:[ E[tanh(amu + aY)] approx tanh(amu) + a text{sech}^2(amu) E[Y] - frac{a^2}{3} tanh(amu) text{sech}^2(amu) E[Y^2] ]But since ( Y ) is zero-mean, ( E[Y] = 0 ), and ( E[Y^2] = sigma^2 ). So,[ E[tanh(amu + aY)] approx tanh(amu) - frac{a^2}{3} tanh(amu) text{sech}^2(amu) sigma^2 ]Therefore, plugging back ( a = 1/2 ):[ E[tanhleft( frac{X}{2} right)] approx tanhleft( frac{mu}{2} right) - frac{1}{12} tanhleft( frac{mu}{2} right) text{sech}^2left( frac{mu}{2} right) sigma^2 ]So, this is an approximation of the expectation of ( tanh(X/2) ) for a normal variable ( X ) with mean ( mu ) and variance ( sigma^2 ).Therefore, going back to ( E(Y) ):[ E(Y) = frac{1}{2} left(1 + Eleft( tanhleft( frac{X}{2} right) right) right) approx frac{1}{2} left(1 + tanhleft( frac{mu}{2} right) - frac{1}{12} tanhleft( frac{mu}{2} right) text{sech}^2left( frac{mu}{2} right) sigma^2 right) ]Simplifying:[ E(Y) approx frac{1}{2} + frac{1}{2} tanhleft( frac{mu}{2} right) - frac{1}{24} tanhleft( frac{mu}{2} right) text{sech}^2left( frac{mu}{2} right) sigma^2 ]But this is getting quite involved, and I'm not sure if this is the best approach. Maybe there's a better way.Wait, perhaps I can use the fact that for a normal variable ( X sim N(mu, sigma^2) ), the expectation ( E[sigma(X)] ) can be expressed in terms of the error function.Let me recall that:[ E[sigma(X)] = int_{-infty}^{infty} frac{1}{1 + e^{-x}} cdot frac{1}{sqrt{2pisigma^2}} e^{-frac{(x - mu)^2}{2sigma^2}} dx ]Let me make a substitution: let ( z = frac{x - mu}{sigma} ), so ( x = mu + sigma z ), and ( dx = sigma dz ). Then, the integral becomes:[ E[sigma(X)] = int_{-infty}^{infty} frac{1}{1 + e^{-(mu + sigma z)}} cdot frac{1}{sqrt{2pi}} e^{-frac{z^2}{2}} dz ]Simplify the exponent in the logistic term:[ frac{1}{1 + e^{-(mu + sigma z)}} = frac{e^{mu + sigma z}}{1 + e^{mu + sigma z}} = frac{1}{1 + e^{-(mu + sigma z)}} ]Wait, that's the same as before. Maybe I can express this as:[ frac{1}{1 + e^{-(mu + sigma z)}} = frac{1}{2} left(1 + tanhleft( frac{mu + sigma z}{2} right) right) ]So, substituting back:[ E[sigma(X)] = frac{1}{2} int_{-infty}^{infty} left(1 + tanhleft( frac{mu + sigma z}{2} right) right) cdot frac{1}{sqrt{2pi}} e^{-frac{z^2}{2}} dz ]Which simplifies to:[ E[sigma(X)] = frac{1}{2} left(1 + int_{-infty}^{infty} tanhleft( frac{mu + sigma z}{2} right) cdot frac{1}{sqrt{2pi}} e^{-frac{z^2}{2}} dz right) ]So, now I need to compute:[ I = int_{-infty}^{infty} tanhleft( frac{mu + sigma z}{2} right) cdot frac{1}{sqrt{2pi}} e^{-frac{z^2}{2}} dz ]This integral might be related to the expectation of ( tanh ) of a normal variable, which I tried earlier.Alternatively, perhaps using a substitution. Let me set ( w = frac{mu + sigma z}{2} ), so ( z = frac{2w - mu}{sigma} ), and ( dz = frac{2}{sigma} dw ).Then, the integral becomes:[ I = int_{-infty}^{infty} tanh(w) cdot frac{1}{sqrt{2pi}} e^{-frac{(frac{2w - mu}{sigma})^2}{2}} cdot frac{2}{sigma} dw ]Simplify the exponent:[ -frac{(frac{2w - mu}{sigma})^2}{2} = -frac{(2w - mu)^2}{2sigma^2} ]So,[ I = frac{2}{sigma sqrt{2pi}} int_{-infty}^{infty} tanh(w) e^{-frac{(2w - mu)^2}{2sigma^2}} dw ]This still looks complicated. Maybe I can express ( tanh(w) ) in terms of exponentials:[ tanh(w) = frac{e^{w} - e^{-w}}{e^{w} + e^{-w}} = frac{e^{2w} - 1}{e^{2w} + 1} ]But I don't see a straightforward way to integrate this.Alternatively, perhaps consider that ( tanh(w) ) can be written as ( 1 - frac{2}{e^{2w} + 1} ), but again, not sure.Wait, another thought: perhaps express ( tanh(w) ) as a derivative. Since ( frac{d}{dw} ln cosh(w) = tanh(w) ), maybe integrating by parts.Let me try integrating by parts. Let me set ( u = tanh(w) ) and ( dv = e^{-frac{(2w - mu)^2}{2sigma^2}} dw ). Then, ( du = text{sech}^2(w) dw ), and ( v ) would be the integral of ( dv ), which is related to the error function.But integrating ( dv ) would result in:[ v = int e^{-frac{(2w - mu)^2}{2sigma^2}} dw ]Let me make a substitution: let ( t = frac{2w - mu}{sigma sqrt{2}} ), so ( w = frac{sigma sqrt{2} t + mu}{2} ), and ( dw = frac{sigma sqrt{2}}{2} dt ).Then,[ v = int e^{-t^2} cdot frac{sigma sqrt{2}}{2} dt = frac{sigma sqrt{2}}{2} int e^{-t^2} dt = frac{sigma sqrt{2}}{2} sqrt{pi} text{erf}(t) + C ]But this is getting too involved, and I'm not sure if integrating by parts will lead to a solution.Maybe I need to accept that this integral doesn't have a closed-form solution and instead look for an approximation or a known result.Wait, perhaps I can use the fact that for a normal variable ( Z sim N(0,1) ), the expectation ( E[tanh(aZ + b)] ) can be expressed in terms of the error function or the cumulative distribution function.I found a reference that says:For ( Z sim N(0,1) ), ( E[tanh(aZ + b)] = tanh(b) cdot text{erf}left( frac{a}{sqrt{pi}} cdot text{sech}^2(b) right) )But I'm not sure if that's correct. Let me check.Alternatively, perhaps using a series expansion for ( tanh(aZ + b) ) around ( Z = 0 ), but that might not be helpful.Wait, another approach: use the fact that ( tanh(w) ) is an odd function, but in our case, the integral is over all ( w ), so maybe symmetry can help. But since ( mu ) is not necessarily zero, the integral isn't symmetric.Alternatively, perhaps use a saddle-point approximation or some other asymptotic method, but that might be beyond the scope here.Wait, perhaps I can use the fact that ( tanh(w) ) can be expressed as a sum of exponentials and then use the moment-generating function.But I'm not making progress here. Maybe I should look for a different approach.Wait, going back to the original problem: the function ( S ) is the sum of ( n ) logistic functions of normal variables. Since each term is ( frac{1}{1 + e^{-k_i}} ), and each ( k_i ) is normal, perhaps the sum ( S ) can be approximated as a normal distribution when ( n ) is large, using the Central Limit Theorem.But that's part 2 of the problem. For part 1, I just need the expectation.Wait, maybe I can use the fact that for a normal variable ( X sim N(mu, sigma^2) ), the expectation ( E[sigma(X)] ) can be expressed as:[ E[sigma(X)] = Phileft( frac{mu}{sqrt{1 + sigma^2}} right) ]Where ( Phi ) is the standard normal CDF. Is that correct?Wait, let me think. If ( X sim N(mu, sigma^2) ), then ( sigma(X) ) is the probability that a standard logistic variable is less than ( X ). But I don't think that directly relates to the standard normal CDF.Alternatively, perhaps using a probit approximation. The expectation ( E[sigma(X)] ) can be approximated by ( Phileft( frac{mu}{sqrt{1 + sigma^2}} right) ). I think this is a known approximation in some contexts, like in dichotomous choice models.Let me check the logic. If ( X sim N(mu, sigma^2) ), then ( sigma(X) ) is the probability that a logistic variable is less than ( X ). But if we approximate the logistic distribution with a normal distribution, then the expectation ( E[sigma(X)] ) can be approximated by ( Phileft( frac{mu}{sqrt{1 + sigma^2}} right) ).Wait, actually, the variance of the logistic distribution is ( frac{pi^2}{3} ), so if we approximate it with a normal distribution with the same variance, the standard deviation would be ( sqrt{frac{pi^2}{3}} ). But I'm not sure if that's directly applicable here.Alternatively, perhaps using a different scaling. Let me consider that the logistic function ( sigma(x) ) is similar to the standard normal CDF ( Phi(x) ), but scaled differently.In fact, ( sigma(x) ) can be approximated by ( Phileft( frac{x}{sqrt{3/pi^2}} right) ), since the logistic distribution has a scale parameter related to the normal distribution.But I'm not sure if that helps in computing the expectation.Wait, perhaps using a delta method approximation. The delta method is used to approximate the expectation of a function of a random variable.So, if ( X sim N(mu, sigma^2) ), then ( E[g(X)] approx g(mu) + frac{1}{2} g''(mu) sigma^2 ).Let me apply this to ( g(X) = sigma(X) = frac{1}{1 + e^{-X}} ).First, compute ( g(mu) = frac{1}{1 + e^{-mu}} ).Next, compute the second derivative ( g''(X) ).First derivative:[ g'(X) = frac{e^{-X}}{(1 + e^{-X})^2} = sigma(X) (1 - sigma(X)) ]Second derivative:[ g''(X) = frac{d}{dX} [ sigma(X) (1 - sigma(X)) ] = sigma'(X) (1 - sigma(X)) - sigma(X) sigma'(X) = sigma'(X) (1 - 2sigma(X)) ]But ( sigma'(X) = sigma(X) (1 - sigma(X)) ), so:[ g''(X) = sigma(X) (1 - sigma(X)) (1 - 2sigma(X)) ]Therefore, evaluating at ( X = mu ):[ g''(mu) = sigma(mu) (1 - sigma(mu)) (1 - 2sigma(mu)) ]So, the delta method approximation is:[ E[g(X)] approx g(mu) + frac{1}{2} g''(mu) sigma^2 ]Plugging in:[ E[sigma(X)] approx sigma(mu) + frac{1}{2} sigma(mu) (1 - sigma(mu)) (1 - 2sigma(mu)) sigma^2 ]Simplify:[ E[sigma(X)] approx sigma(mu) + frac{1}{2} sigma(mu) (1 - sigma(mu)) (1 - 2sigma(mu)) sigma^2 ]This gives an approximation of the expectation in terms of ( mu ), ( sigma^2 ), and ( sigma(mu) ).So, putting it all together, the expected value of ( S ) is:[ E(S) = n cdot Eleft( frac{1}{1 + e^{-k_i}} right) approx n left[ sigma(mu) + frac{1}{2} sigma(mu) (1 - sigma(mu)) (1 - 2sigma(mu)) sigma^2 right] ]Where ( sigma(mu) = frac{1}{1 + e^{-mu}} ).Alternatively, if we use the earlier approximation with the hyperbolic tangent, we had:[ E(Y) approx frac{1}{2} + frac{1}{2} tanhleft( frac{mu}{2} right) - frac{1}{24} tanhleft( frac{mu}{2} right) text{sech}^2left( frac{mu}{2} right) sigma^2 ]But since ( sigma(mu) = frac{1}{2} left(1 + tanhleft( frac{mu}{2} right) right) ), we can write:[ E(Y) approx sigma(mu) - frac{1}{24} tanhleft( frac{mu}{2} right) text{sech}^2left( frac{mu}{2} right) sigma^2 ]Which is another approximation.Given that both methods give similar forms, perhaps the delta method is more straightforward.Therefore, summarizing, the expected value of ( S ) is approximately:[ E(S) approx n left[ sigma(mu) + frac{1}{2} sigma(mu) (1 - sigma(mu)) (1 - 2sigma(mu)) sigma^2 right] ]But I'm not entirely confident about this result. Maybe I should check with a simpler case.Suppose ( sigma^2 = 0 ), so ( k_i = mu ) almost surely. Then, ( E(S) = n cdot frac{1}{1 + e^{-mu}} ), which is correct.If ( mu = 0 ), then ( sigma(0) = 0.5 ), and the second term becomes ( 0.5 cdot 0.5 cdot (1 - 2 cdot 0.5) cdot sigma^2 = 0.5 cdot 0.5 cdot 0 cdot sigma^2 = 0 ). So, ( E(S) approx n cdot 0.5 ), which makes sense because if ( mu = 0 ), each term is symmetric around 0.5.If ( mu ) is large positive, ( sigma(mu) approx 1 ), and the second term becomes ( 1 cdot (1 - 1) cdot (1 - 2 cdot 1) cdot sigma^2 = 0 ), so ( E(S) approx n cdot 1 ), which is correct because if ( mu ) is large, each term is almost 1.Similarly, if ( mu ) is large negative, ( sigma(mu) approx 0 ), and the second term is 0, so ( E(S) approx 0 ), which is correct.Therefore, the approximation seems reasonable.So, for part 1, the expected value of ( S ) is approximately:[ E(S) approx n left[ frac{1}{1 + e^{-mu}} + frac{1}{2} cdot frac{1}{1 + e^{-mu}} cdot left(1 - frac{1}{1 + e^{-mu}}right) cdot left(1 - 2 cdot frac{1}{1 + e^{-mu}}right) cdot sigma^2 right] ]Simplifying the terms:Let me denote ( p = frac{1}{1 + e^{-mu}} ), so ( 1 - p = frac{e^{-mu}}{1 + e^{-mu}} ), and ( 1 - 2p = 1 - frac{2}{1 + e^{-mu}} = frac{1 + e^{-mu} - 2}{1 + e^{-mu}} = frac{e^{-mu} - 1}{1 + e^{-mu}} = -frac{1 - e^{-mu}}{1 + e^{-mu}} = - (1 - p) ).Therefore, the expression becomes:[ E(S) approx n left[ p + frac{1}{2} p (1 - p) (- (1 - p)) sigma^2 right] = n left[ p - frac{1}{2} p (1 - p)^2 sigma^2 right] ]So,[ E(S) approx n p left(1 - frac{1}{2} (1 - p)^2 sigma^2 right) ]This seems like a reasonable approximation.Alternatively, if we use the hyperbolic tangent approximation, we had:[ E(Y) approx frac{1}{2} + frac{1}{2} tanhleft( frac{mu}{2} right) - frac{1}{24} tanhleft( frac{mu}{2} right) text{sech}^2left( frac{mu}{2} right) sigma^2 ]But since ( p = frac{1}{2} left(1 + tanhleft( frac{mu}{2} right) right) ), we can write:[ tanhleft( frac{mu}{2} right) = 2p - 1 ]And,[ text{sech}^2left( frac{mu}{2} right) = 1 - tanh^2left( frac{mu}{2} right) = 1 - (2p - 1)^2 = 1 - (4p^2 - 4p + 1) = 4p(1 - p) ]Therefore, substituting back:[ E(Y) approx frac{1}{2} + frac{1}{2}(2p - 1) - frac{1}{24}(2p - 1)(4p(1 - p)) sigma^2 ]Simplify:[ E(Y) approx frac{1}{2} + p - frac{1}{2} - frac{1}{24} cdot 4p(1 - p)(2p - 1) sigma^2 ]Which simplifies to:[ E(Y) approx p - frac{1}{6} p(1 - p)(2p - 1) sigma^2 ]Comparing this with the delta method result:From delta method:[ E(Y) approx p - frac{1}{2} p(1 - p)^2 sigma^2 ]From hyperbolic tangent approximation:[ E(Y) approx p - frac{1}{6} p(1 - p)(2p - 1) sigma^2 ]These are different, but both are approximations. It's unclear which one is more accurate without further analysis.Given that the delta method is a standard approach for such approximations, I think it's reasonable to use that result.Therefore, the expected value of ( S ) is approximately:[ E(S) approx n left[ frac{1}{1 + e^{-mu}} - frac{1}{2} cdot frac{1}{1 + e^{-mu}} cdot left(1 - frac{1}{1 + e^{-mu}}right)^2 cdot sigma^2 right] ]Simplifying:Let ( p = frac{1}{1 + e^{-mu}} ), then:[ E(S) approx n left[ p - frac{1}{2} p (1 - p)^2 sigma^2 right] ]This is the expected value of ( S ).Now, moving on to part 2: estimating the probability that ( S ) exceeds a threshold ( T ) using the Central Limit Theorem, assuming ( n ) is large.Given that ( S ) is the sum of ( n ) independent random variables ( Y_i = frac{1}{1 + e^{-k_i}} ), each with mean ( mu_Y = E(Y_i) ) and variance ( sigma_Y^2 = text{Var}(Y_i) ).By the Central Limit Theorem, for large ( n ), ( S ) is approximately normally distributed with mean ( n mu_Y ) and variance ( n sigma_Y^2 ).Therefore, the probability that ( S > T ) can be approximated as:[ P(S > T) approx Pleft( Z > frac{T - n mu_Y}{sqrt{n sigma_Y^2}} right) = 1 - Phileft( frac{T - n mu_Y}{sqrt{n sigma_Y^2}} right) ]Where ( Z ) is a standard normal variable and ( Phi ) is its CDF.But to write this, I need expressions for ( mu_Y ) and ( sigma_Y^2 ).From part 1, we have an approximation for ( mu_Y = E(Y_i) approx p - frac{1}{2} p (1 - p)^2 sigma^2 ), but actually, for the CLT, we need the exact mean and variance, but since we don't have a closed-form for ( E(Y_i) ), perhaps we can use the delta method to approximate the variance as well.Wait, actually, the variance of ( Y_i ) can be approximated using the delta method as well.Given ( Y_i = sigma(k_i) ), and ( k_i sim N(mu, sigma^2) ), then:[ text{Var}(Y_i) approx [g'(mu)]^2 sigma^2 ]Where ( g(k) = sigma(k) ).We already computed ( g'(k) = sigma(k)(1 - sigma(k)) ).Therefore,[ text{Var}(Y_i) approx [ sigma(mu)(1 - sigma(mu)) ]^2 sigma^2 ]So,[ sigma_Y^2 = text{Var}(Y_i) approx p^2 (1 - p)^2 sigma^2 ]Therefore, the variance of ( S ) is:[ text{Var}(S) = n sigma_Y^2 approx n p^2 (1 - p)^2 sigma^2 ]Thus, the standard deviation of ( S ) is:[ sqrt{text{Var}(S)} approx sqrt{n} p (1 - p) sigma ]Therefore, the probability that ( S > T ) is approximately:[ P(S > T) approx 1 - Phileft( frac{T - n p}{sqrt{n} p (1 - p) sigma} right) ]Where ( p = frac{1}{1 + e^{-mu}} ).Alternatively, using the more precise mean from part 1, which includes the correction term, but for simplicity, perhaps it's better to use the leading term ( n p ) as the mean, especially since the correction term is of order ( sigma^2 ), which might be small if ( sigma ) is not too large.Therefore, the probability is approximately:[ P(S > T) approx 1 - Phileft( frac{T - n p}{sqrt{n} p (1 - p) sigma} right) ]Where ( p = frac{1}{1 + e^{-mu}} ).So, summarizing:1. The expected value of ( S ) is approximately:[ E(S) approx n left[ frac{1}{1 + e^{-mu}} - frac{1}{2} cdot frac{1}{1 + e^{-mu}} cdot left(1 - frac{1}{1 + e^{-mu}}right)^2 cdot sigma^2 right] ]2. The probability that ( S > T ) is approximately:[ P(S > T) approx 1 - Phileft( frac{T - n cdot frac{1}{1 + e^{-mu}}}{sqrt{n} cdot frac{1}{1 + e^{-mu}} cdot left(1 - frac{1}{1 + e^{-mu}}right) cdot sigma} right) ]But to make it more concise, let me denote ( p = frac{1}{1 + e^{-mu}} ), then:1. ( E(S) approx n left[ p - frac{1}{2} p (1 - p)^2 sigma^2 right] )2. ( P(S > T) approx 1 - Phileft( frac{T - n p}{sqrt{n} p (1 - p) sigma} right) )I think this is a reasonable answer given the constraints.</think>

🧠:<think>Okay, so I have this problem where I need to solve a system of differential equations modeling the influence of Dostoevsky and Tolstoy over time. The equations are:[ frac{dD}{dt} = aD + bT + c ][ frac{dT}{dt} = dD + eT + f ]And I need to find the general solution first, and then plug in specific constants and initial conditions to get the specific solutions. Hmm, okay. Let me start by recalling how to solve systems of linear differential equations.First, I remember that systems like this can be written in matrix form. Let me write it as:[ begin{pmatrix} frac{dD}{dt} frac{dT}{dt} end{pmatrix} = begin{pmatrix} a & b d & e end{pmatrix} begin{pmatrix} D T end{pmatrix} + begin{pmatrix} c f end{pmatrix} ]So, this is a nonhomogeneous system because of the constant terms c and f. To solve this, I think I need to find the homogeneous solution first and then find a particular solution.The homogeneous system would be:[ frac{dD}{dt} = aD + bT ][ frac{dT}{dt} = dD + eT ]Which can be written as:[ mathbf{X}' = mathbf{A}mathbf{X} ]Where (mathbf{X} = begin{pmatrix} D T end{pmatrix}) and (mathbf{A} = begin{pmatrix} a & b d & e end{pmatrix}).To solve this, I need to find the eigenvalues and eigenvectors of matrix A. The eigenvalues λ satisfy the characteristic equation:[ det(mathbf{A} - lambda mathbf{I}) = 0 ]So, computing the determinant:[ (a - lambda)(e - lambda) - bd = 0 ]Expanding that:[ lambda^2 - (a + e)lambda + (ae - bd) = 0 ]So, the eigenvalues are:[ lambda = frac{(a + e) pm sqrt{(a + e)^2 - 4(ae - bd)}}{2} ]Simplify the discriminant:[ (a + e)^2 - 4(ae - bd) = a^2 + 2ae + e^2 - 4ae + 4bd = a^2 - 2ae + e^2 + 4bd = (a - e)^2 + 4bd ]So, the eigenvalues are:[ lambda = frac{a + e pm sqrt{(a - e)^2 + 4bd}}{2} ]Depending on whether the discriminant is positive, zero, or negative, we'll have real distinct, repeated, or complex eigenvalues.Once we have the eigenvalues, we can find the corresponding eigenvectors and write the homogeneous solution.But since the system is nonhomogeneous, we also need a particular solution. For that, I can use the method of undetermined coefficients. Since the nonhomogeneous terms are constants (c and f), I can assume a particular solution is a constant vector, say (mathbf{X_p} = begin{pmatrix} D_p T_p end{pmatrix}).Plugging this into the system:[ 0 = aD_p + bT_p + c ][ 0 = dD_p + eT_p + f ]So, we have a system of equations:1. ( aD_p + bT_p = -c )2. ( dD_p + eT_p = -f )We can solve this system for ( D_p ) and ( T_p ). Let's write it in matrix form:[ begin{pmatrix} a & b d & e end{pmatrix} begin{pmatrix} D_p T_p end{pmatrix} = begin{pmatrix} -c -f end{pmatrix} ]Assuming the matrix is invertible (i.e., determinant is not zero), we can solve for ( D_p ) and ( T_p ):[ begin{pmatrix} D_p T_p end{pmatrix} = frac{1}{ae - bd} begin{pmatrix} e & -b -d & a end{pmatrix} begin{pmatrix} -c -f end{pmatrix} ]Calculating this:First, compute the determinant ( Delta = ae - bd ).Then,[ D_p = frac{e(-c) + (-b)(-f)}{Delta} = frac{-ec + bf}{Delta} ][ T_p = frac{-d(-c) + a(-f)}{Delta} = frac{dc - af}{Delta} ]So, the particular solution is:[ mathbf{X_p} = begin{pmatrix} frac{bf - ec}{Delta} frac{dc - af}{Delta} end{pmatrix} ]Therefore, the general solution is the homogeneous solution plus the particular solution:[ mathbf{X}(t) = mathbf{X_h}(t) + mathbf{X_p} ]Now, the homogeneous solution depends on the eigenvalues. Let me outline the cases:1. Distinct Real Eigenvalues (λ₁ ≠ λ₂): The solution is a combination of terms involving ( e^{lambda_1 t} ) and ( e^{lambda_2 t} ), each multiplied by their respective eigenvectors.2. Repeated Real Eigenvalue (λ₁ = λ₂): The solution involves terms like ( e^{lambda t} ) and ( t e^{lambda t} ), multiplied by eigenvectors and generalized eigenvectors.3. Complex Eigenvalues (α ± βi): The solution can be written using Euler's formula, involving sine and cosine terms with exponential decay/growth.Since the problem doesn't specify the nature of the eigenvalues, the general solution will depend on these cases. However, for the specific constants given in part 2, I can compute the eigenvalues and proceed accordingly.But before moving on to part 2, let me summarize the steps for the general solution:1. Find the eigenvalues of matrix A.2. Find the corresponding eigenvectors.3. Write the homogeneous solution based on the eigenvalues.4. Find the particular solution using the method above.5. Combine them for the general solution.So, for part 1, the general solution is:[ D(t) = text{Homogeneous part for D} + frac{bf - ec}{Delta} ][ T(t) = text{Homogeneous part for T} + frac{dc - af}{Delta} ]But to write it explicitly, I need to consider the eigenvalues and eigenvectors.Alternatively, another method is to use integrating factors or matrix exponentials, but that might be more complicated.Wait, maybe using Laplace transforms? Hmm, but that might also be involved.Alternatively, I can write the system as:[ frac{dD}{dt} - aD - bT = c ][ frac{dT}{dt} - dD - eT = f ]This is a linear system, so perhaps I can solve one equation for one variable and substitute into the other.Let me try solving for T from the first equation:From the first equation:[ frac{dD}{dt} - aD - bT = c ][ Rightarrow -bT = frac{dD}{dt} - aD - c ][ Rightarrow T = frac{aD + c - frac{dD}{dt}}{b} ]Assuming b ≠ 0.Then, substitute this into the second equation:[ frac{dT}{dt} - dD - eT = f ]First, compute dT/dt:[ T = frac{aD + c - frac{dD}{dt}}{b} ][ Rightarrow frac{dT}{dt} = frac{a frac{dD}{dt} - frac{d^2 D}{dt^2}}{b} ]So, plugging into the second equation:[ frac{a frac{dD}{dt} - frac{d^2 D}{dt^2}}{b} - dD - e left( frac{aD + c - frac{dD}{dt}}{b} right) = f ]Multiply through by b to eliminate denominators:[ a frac{dD}{dt} - frac{d^2 D}{dt^2} - b d D - e(aD + c - frac{dD}{dt}) = b f ]Expand the terms:[ a frac{dD}{dt} - frac{d^2 D}{dt^2} - b d D - e a D - e c + e frac{dD}{dt} = b f ]Combine like terms:- The second derivative term: ( - frac{d^2 D}{dt^2} )- The first derivative terms: ( a frac{dD}{dt} + e frac{dD}{dt} = (a + e) frac{dD}{dt} )- The D terms: ( -b d D - e a D = (-b d - e a) D )- Constants: ( -e c )- Right-hand side: ( b f )So, putting it all together:[ - frac{d^2 D}{dt^2} + (a + e) frac{dD}{dt} + (-b d - e a) D = b f + e c ]Multiply both sides by -1:[ frac{d^2 D}{dt^2} - (a + e) frac{dD}{dt} + (b d + e a) D = -b f - e c ]So, now we have a second-order linear differential equation for D(t):[ frac{d^2 D}{dt^2} - (a + e) frac{dD}{dt} + (b d + a e) D = -b f - e c ]This is a linear nonhomogeneous ODE. The general solution will be the homogeneous solution plus a particular solution.First, solve the homogeneous equation:[ frac{d^2 D}{dt^2} - (a + e) frac{dD}{dt} + (b d + a e) D = 0 ]The characteristic equation is:[ r^2 - (a + e) r + (b d + a e) = 0 ]Solving for r:[ r = frac{(a + e) pm sqrt{(a + e)^2 - 4(b d + a e)}}{2} ]Simplify the discriminant:[ (a + e)^2 - 4(b d + a e) = a^2 + 2 a e + e^2 - 4 b d - 4 a e = a^2 - 2 a e + e^2 - 4 b d = (a - e)^2 - 4 b d ]So, the roots are:[ r = frac{a + e pm sqrt{(a - e)^2 - 4 b d}}{2} ]Depending on the discriminant, we have different cases.Now, for the particular solution, since the nonhomogeneous term is a constant (-b f - e c), we can assume a constant particular solution D_p.Let me denote D_p as a constant. Then, plug into the ODE:[ 0 - (a + e) cdot 0 + (b d + a e) D_p = -b f - e c ][ (b d + a e) D_p = -b f - e c ][ D_p = frac{-b f - e c}{b d + a e} ]Assuming that ( b d + a e neq 0 ).So, the general solution for D(t) is:[ D(t) = C_1 e^{r_1 t} + C_2 e^{r_2 t} + D_p ]Where ( r_1 ) and ( r_2 ) are the roots from above, and ( C_1 ), ( C_2 ) are constants determined by initial conditions.Once we have D(t), we can substitute back into the expression for T(t):[ T(t) = frac{a D(t) + c - frac{d D(t)}{dt}}{b} ]So, that's the general approach.But this seems a bit involved. Maybe another approach is better, like diagonalizing the matrix or using eigenvalues directly.Alternatively, since the system is linear, we can write it in terms of matrix exponentials.The general solution is:[ mathbf{X}(t) = e^{mathbf{A} t} mathbf{X}(0) + int_0^t e^{mathbf{A}(t - tau)} mathbf{B} dtau ]Where (mathbf{B} = begin{pmatrix} c f end{pmatrix}).But computing the matrix exponential might be complicated without knowing the specific eigenvalues.Alternatively, since I already have the second-order equation for D(t), maybe it's easier to proceed with that.So, to summarize, for the general solution:1. Solve the second-order ODE for D(t), which involves finding the roots of the characteristic equation.2. Find D(t) as a combination of exponential terms based on the roots.3. Use D(t) to find T(t) via the expression derived earlier.This seems manageable.Now, moving on to part 2, where specific constants are given:a = 1, b = -1, c = 0.5, d = 0.5, e = -1.5, f = 1.So, let's plug these into the equations.First, let's write down the system:[ frac{dD}{dt} = D - T + 0.5 ][ frac{dT}{dt} = 0.5 D - 1.5 T + 1 ]With initial conditions D(0) = 2, T(0) = 3.I need to solve this system.Let me try the approach of converting it into a second-order ODE for D(t).From the first equation:[ frac{dD}{dt} = D - T + 0.5 ][ Rightarrow -T = frac{dD}{dt} - D - 0.5 ][ Rightarrow T = D + 0.5 - frac{dD}{dt} ]Now, substitute this into the second equation:[ frac{dT}{dt} = 0.5 D - 1.5 T + 1 ]First, compute dT/dt:[ T = D + 0.5 - frac{dD}{dt} ][ Rightarrow frac{dT}{dt} = frac{dD}{dt} - frac{d^2 D}{dt^2} ]So, plug into the second equation:[ frac{dD}{dt} - frac{d^2 D}{dt^2} = 0.5 D - 1.5 left( D + 0.5 - frac{dD}{dt} right) + 1 ]Let me expand the right-hand side:First, compute the term inside the brackets:[ -1.5 left( D + 0.5 - frac{dD}{dt} right) = -1.5 D - 0.75 + 1.5 frac{dD}{dt} ]So, the entire right-hand side becomes:[ 0.5 D - 1.5 D - 0.75 + 1.5 frac{dD}{dt} + 1 ][ = (0.5 - 1.5) D + (-0.75 + 1) + 1.5 frac{dD}{dt} ][ = (-1) D + (0.25) + 1.5 frac{dD}{dt} ]So, putting it all together:Left-hand side: ( frac{dD}{dt} - frac{d^2 D}{dt^2} )Right-hand side: ( -D + 0.25 + 1.5 frac{dD}{dt} )Bring all terms to the left-hand side:[ frac{dD}{dt} - frac{d^2 D}{dt^2} + D - 0.25 - 1.5 frac{dD}{dt} = 0 ]Combine like terms:- ( - frac{d^2 D}{dt^2} )- ( frac{dD}{dt} - 1.5 frac{dD}{dt} = -0.5 frac{dD}{dt} )- ( D )- ( -0.25 )So, the equation becomes:[ - frac{d^2 D}{dt^2} - 0.5 frac{dD}{dt} + D - 0.25 = 0 ]Multiply through by -1 to make it more standard:[ frac{d^2 D}{dt^2} + 0.5 frac{dD}{dt} - D + 0.25 = 0 ]So, the ODE is:[ frac{d^2 D}{dt^2} + 0.5 frac{dD}{dt} - D = -0.25 ]This is a linear nonhomogeneous second-order ODE.First, solve the homogeneous equation:[ frac{d^2 D}{dt^2} + 0.5 frac{dD}{dt} - D = 0 ]The characteristic equation is:[ r^2 + 0.5 r - 1 = 0 ]Solving for r:[ r = frac{-0.5 pm sqrt{(0.5)^2 + 4}}{2} ][ = frac{-0.5 pm sqrt{0.25 + 4}}{2} ][ = frac{-0.5 pm sqrt{4.25}}{2} ][ = frac{-0.5 pm frac{sqrt{17}}{2}}{2} ][ = frac{-1 pm sqrt{17}}{4} ]So, the roots are:[ r_1 = frac{-1 + sqrt{17}}{4} ][ r_2 = frac{-1 - sqrt{17}}{4} ]These are real and distinct roots.Therefore, the homogeneous solution is:[ D_h(t) = C_1 e^{r_1 t} + C_2 e^{r_2 t} ]Now, find a particular solution for the nonhomogeneous equation. The nonhomogeneous term is a constant (-0.25). So, we can assume a constant particular solution ( D_p = K ).Plug into the ODE:[ 0 + 0 - K = -0.25 ][ -K = -0.25 ][ K = 0.25 ]So, the particular solution is ( D_p(t) = 0.25 ).Therefore, the general solution for D(t) is:[ D(t) = C_1 e^{r_1 t} + C_2 e^{r_2 t} + 0.25 ]Now, we can find T(t) using the earlier expression:[ T(t) = D(t) + 0.5 - frac{dD}{dt} ]Compute ( frac{dD}{dt} ):[ frac{dD}{dt} = C_1 r_1 e^{r_1 t} + C_2 r_2 e^{r_2 t} ]So,[ T(t) = C_1 e^{r_1 t} + C_2 e^{r_2 t} + 0.25 + 0.5 - (C_1 r_1 e^{r_1 t} + C_2 r_2 e^{r_2 t}) ][ = (C_1 e^{r_1 t} - C_1 r_1 e^{r_1 t}) + (C_2 e^{r_2 t} - C_2 r_2 e^{r_2 t}) + 0.75 ][ = C_1 e^{r_1 t}(1 - r_1) + C_2 e^{r_2 t}(1 - r_2) + 0.75 ]Now, apply the initial conditions to find C₁ and C₂.Given D(0) = 2:[ D(0) = C_1 + C_2 + 0.25 = 2 ][ C_1 + C_2 = 2 - 0.25 = 1.75 ]And T(0) = 3:From T(t):[ T(0) = C_1 (1 - r_1) + C_2 (1 - r_2) + 0.75 = 3 ][ C_1 (1 - r_1) + C_2 (1 - r_2) = 3 - 0.75 = 2.25 ]So, we have the system:1. ( C_1 + C_2 = 1.75 )2. ( C_1 (1 - r_1) + C_2 (1 - r_2) = 2.25 )Let me compute ( 1 - r_1 ) and ( 1 - r_2 ):Given:[ r_1 = frac{-1 + sqrt{17}}{4} ][ 1 - r_1 = 1 - frac{-1 + sqrt{17}}{4} = frac{4 + 1 - sqrt{17}}{4} = frac{5 - sqrt{17}}{4} ]Similarly,[ r_2 = frac{-1 - sqrt{17}}{4} ][ 1 - r_2 = 1 - frac{-1 - sqrt{17}}{4} = frac{4 + 1 + sqrt{17}}{4} = frac{5 + sqrt{17}}{4} ]So, the second equation becomes:[ C_1 left( frac{5 - sqrt{17}}{4} right) + C_2 left( frac{5 + sqrt{17}}{4} right) = 2.25 ]Multiply both sides by 4 to eliminate denominators:[ C_1 (5 - sqrt{17}) + C_2 (5 + sqrt{17}) = 9 ]Now, we have:1. ( C_1 + C_2 = 1.75 ) (Equation 1)2. ( C_1 (5 - sqrt{17}) + C_2 (5 + sqrt{17}) = 9 ) (Equation 2)Let me denote ( C_1 = x ) and ( C_2 = y ) for simplicity.So,1. ( x + y = 1.75 )2. ( x (5 - sqrt{17}) + y (5 + sqrt{17}) = 9 )From Equation 1, ( y = 1.75 - x ). Substitute into Equation 2:[ x (5 - sqrt{17}) + (1.75 - x)(5 + sqrt{17}) = 9 ]Expand:[ 5x - x sqrt{17} + 1.75(5 + sqrt{17}) - x(5 + sqrt{17}) = 9 ]Simplify term by term:First term: ( 5x - x sqrt{17} )Second term: ( 1.75 times 5 + 1.75 sqrt{17} = 8.75 + 1.75 sqrt{17} )Third term: ( -5x - x sqrt{17} )Combine all terms:( 5x - x sqrt{17} + 8.75 + 1.75 sqrt{17} - 5x - x sqrt{17} = 9 )Simplify:- ( 5x - 5x = 0 )- ( -x sqrt{17} - x sqrt{17} = -2x sqrt{17} )- Constants: ( 8.75 + 1.75 sqrt{17} )So, the equation becomes:[ -2x sqrt{17} + 8.75 + 1.75 sqrt{17} = 9 ]Bring constants to the right:[ -2x sqrt{17} = 9 - 8.75 - 1.75 sqrt{17} ][ -2x sqrt{17} = 0.25 - 1.75 sqrt{17} ]Solve for x:[ x = frac{0.25 - 1.75 sqrt{17}}{-2 sqrt{17}} ][ = frac{0.25}{-2 sqrt{17}} + frac{1.75 sqrt{17}}{2 sqrt{17}} ][ = -frac{0.25}{2 sqrt{17}} + frac{1.75}{2} ][ = -frac{0.125}{sqrt{17}} + 0.875 ]Rationalizing the denominator for the first term:[ -frac{0.125}{sqrt{17}} = -frac{0.125 sqrt{17}}{17} = -frac{sqrt{17}}{136} ]So,[ x = -frac{sqrt{17}}{136} + 0.875 ]Similarly, since ( y = 1.75 - x ):[ y = 1.75 - left( -frac{sqrt{17}}{136} + 0.875 right) ][ = 1.75 + frac{sqrt{17}}{136} - 0.875 ][ = 0.875 + frac{sqrt{17}}{136} ]So, ( C_1 = x = 0.875 - frac{sqrt{17}}{136} ) and ( C_2 = y = 0.875 + frac{sqrt{17}}{136} ).But let me express these in fractions to be more precise.0.875 is 7/8, and 0.125 is 1/8.So,[ x = frac{7}{8} - frac{sqrt{17}}{136} ][ y = frac{7}{8} + frac{sqrt{17}}{136} ]Simplify ( frac{sqrt{17}}{136} ):Note that 136 = 8 * 17, so:[ frac{sqrt{17}}{136} = frac{sqrt{17}}{8 times 17} = frac{1}{8 sqrt{17}} ]But perhaps it's better to leave it as is.So, putting it all together, the solutions are:[ D(t) = left( frac{7}{8} - frac{sqrt{17}}{136} right) e^{r_1 t} + left( frac{7}{8} + frac{sqrt{17}}{136} right) e^{r_2 t} + 0.25 ]And,[ T(t) = left( frac{7}{8} - frac{sqrt{17}}{136} right) e^{r_1 t} left( frac{5 - sqrt{17}}{4} right) + left( frac{7}{8} + frac{sqrt{17}}{136} right) e^{r_2 t} left( frac{5 + sqrt{17}}{4} right) + 0.75 ]This seems quite messy, but let me see if I can simplify it.Alternatively, perhaps I made a miscalculation earlier. Let me double-check the steps.Wait, when I substituted ( y = 1.75 - x ) into Equation 2, let me verify that step.Equation 2 after substitution:[ x (5 - sqrt{17}) + (1.75 - x)(5 + sqrt{17}) = 9 ]Expanding:[ 5x - x sqrt{17} + 1.75 times 5 + 1.75 sqrt{17} - 5x - x sqrt{17} = 9 ]Simplify:- ( 5x - 5x = 0 )- ( -x sqrt{17} - x sqrt{17} = -2x sqrt{17} )- Constants: ( 8.75 + 1.75 sqrt{17} )So,[ -2x sqrt{17} + 8.75 + 1.75 sqrt{17} = 9 ]Then,[ -2x sqrt{17} = 9 - 8.75 - 1.75 sqrt{17} ][ -2x sqrt{17} = 0.25 - 1.75 sqrt{17} ]So,[ x = frac{0.25 - 1.75 sqrt{17}}{-2 sqrt{17}} ][ = frac{0.25}{-2 sqrt{17}} + frac{1.75 sqrt{17}}{2 sqrt{17}} ][ = -frac{0.25}{2 sqrt{17}} + frac{1.75}{2} ][ = -frac{0.125}{sqrt{17}} + 0.875 ]Yes, that seems correct.Alternatively, perhaps I can express 0.25 as 1/4 and 1.75 as 7/4 to keep fractions consistent.So,[ x = frac{1/4 - (7/4) sqrt{17}}{-2 sqrt{17}} ][ = frac{1}{4} cdot frac{1 - 7 sqrt{17}}{-2 sqrt{17}} ][ = frac{1 - 7 sqrt{17}}{-8 sqrt{17}} ][ = frac{7 sqrt{17} - 1}{8 sqrt{17}} ]Multiply numerator and denominator by ( sqrt{17} ):[ = frac{(7 sqrt{17} - 1) sqrt{17}}{8 times 17} ][ = frac{7 times 17 - sqrt{17}}{136} ][ = frac{119 - sqrt{17}}{136} ][ = frac{119}{136} - frac{sqrt{17}}{136} ][ = frac{7}{8} - frac{sqrt{17}}{136} ]Yes, same as before.Similarly, ( y = 1.75 - x = frac{7}{4} - left( frac{7}{8} - frac{sqrt{17}}{136} right) )[ = frac{14}{8} - frac{7}{8} + frac{sqrt{17}}{136} ][ = frac{7}{8} + frac{sqrt{17}}{136} ]So, that's consistent.Therefore, the solutions are as above.But perhaps I can write them in a more compact form.Let me denote ( C_1 = frac{7}{8} - frac{sqrt{17}}{136} ) and ( C_2 = frac{7}{8} + frac{sqrt{17}}{136} ).So,[ D(t) = C_1 e^{r_1 t} + C_2 e^{r_2 t} + 0.25 ][ T(t) = C_1 e^{r_1 t} left( frac{5 - sqrt{17}}{4} right) + C_2 e^{r_2 t} left( frac{5 + sqrt{17}}{4} right) + 0.75 ]Alternatively, factor out the constants:For D(t):[ D(t) = left( frac{7}{8} - frac{sqrt{17}}{136} right) e^{frac{-1 + sqrt{17}}{4} t} + left( frac{7}{8} + frac{sqrt{17}}{136} right) e^{frac{-1 - sqrt{17}}{4} t} + 0.25 ]For T(t):[ T(t) = left( frac{7}{8} - frac{sqrt{17}}{136} right) e^{frac{-1 + sqrt{17}}{4} t} cdot frac{5 - sqrt{17}}{4} + left( frac{7}{8} + frac{sqrt{17}}{136} right) e^{frac{-1 - sqrt{17}}{4} t} cdot frac{5 + sqrt{17}}{4} + 0.75 ]This is quite involved, but it's the specific solution given the constants.Alternatively, perhaps I can write the solutions in terms of the matrix exponential, but that might not necessarily simplify things.Alternatively, maybe I can use another method, like decoupling the equations, but I think the approach I took is correct.To recap, I converted the system into a second-order ODE for D(t), solved it, found the constants using initial conditions, and then expressed T(t) in terms of D(t).Therefore, the specific solutions are as above.Final Answer1. The general solution is given by the homogeneous solution plus the particular solution, which involves exponential terms based on the eigenvalues of the system matrix.2. The specific solutions are:[ D(t) = left( frac{7}{8} - frac{sqrt{17}}{136} right) e^{frac{-1 + sqrt{17}}{4} t} + left( frac{7}{8} + frac{sqrt{17}}{136} right) e^{frac{-1 - sqrt{17}}{4} t} + frac{1}{4} ][ T(t) = left( frac{7}{8} - frac{sqrt{17}}{136} right) e^{frac{-1 + sqrt{17}}{4} t} cdot frac{5 - sqrt{17}}{4} + left( frac{7}{8} + frac{sqrt{17}}{136} right) e^{frac{-1 - sqrt{17}}{4} t} cdot frac{5 + sqrt{17}}{4} + frac{3}{4} ]So, boxed:For ( D(t) ):[ boxed{D(t) = left( frac{7}{8} - frac{sqrt{17}}{136} right) e^{frac{-1 + sqrt{17}}{4} t} + left( frac{7}{8} + frac{sqrt{17}}{136} right) e^{frac{-1 - sqrt{17}}{4} t} + frac{1}{4}} ]For ( T(t) ):[ boxed{T(t) = left( frac{7}{8} - frac{sqrt{17}}{136} right) cdot frac{5 - sqrt{17}}{4} e^{frac{-1 + sqrt{17}}{4} t} + left( frac{7}{8} + frac{sqrt{17}}{136} right) cdot frac{5 + sqrt{17}}{4} e^{frac{-1 - sqrt{17}}{4} t} + frac{3}{4}} ]</think>

🧠:<think>Okay, so I have this problem about calculating energy expenditure during a workout session. It's split into two sub-problems. Let me start with Sub-problem 1.First, the intensity ( I(t) ) is given as a sinusoidal function: ( I(t) = 3 + 2sin(pi t) ). The energy expenditure ( E(t) ) is proportional to the square of the intensity, so ( E(t) = kI(t)^2 ). I need to find the total energy expenditure over a 2-hour period. That means I need to integrate ( E(t) ) from 0 to 2 hours.Let me write down the integral:[text{Total Energy} = int_{0}^{2} E(t) , dt = int_{0}^{2} kI(t)^2 , dt = k int_{0}^{2} (3 + 2sin(pi t))^2 , dt]Alright, so I can factor out the constant ( k ) and focus on expanding the square inside the integral. Let me expand ( (3 + 2sin(pi t))^2 ):[(3 + 2sin(pi t))^2 = 9 + 12sin(pi t) + 4sin^2(pi t)]So now the integral becomes:[k int_{0}^{2} left(9 + 12sin(pi t) + 4sin^2(pi t)right) dt]I can split this into three separate integrals:1. ( int_{0}^{2} 9 , dt )2. ( int_{0}^{2} 12sin(pi t) , dt )3. ( int_{0}^{2} 4sin^2(pi t) , dt )Let me compute each one step by step.First integral: ( int_{0}^{2} 9 , dt ). That's straightforward. The integral of a constant is the constant times the interval length.[9 times (2 - 0) = 18]Second integral: ( int_{0}^{2} 12sin(pi t) , dt ). The integral of ( sin(pi t) ) with respect to ( t ) is ( -frac{1}{pi}cos(pi t) ). So:[12 left[ -frac{1}{pi}cos(pi t) right]_0^{2} = 12 left( -frac{1}{pi}cos(2pi) + frac{1}{pi}cos(0) right)]We know that ( cos(2pi) = 1 ) and ( cos(0) = 1 ), so:[12 left( -frac{1}{pi}(1) + frac{1}{pi}(1) right) = 12 times 0 = 0]So the second integral is zero. That makes sense because the sine function is symmetric over the interval, so the positive and negative areas cancel out.Third integral: ( int_{0}^{2} 4sin^2(pi t) , dt ). Hmm, integrating ( sin^2 ) can be tricky. I remember that there's a power-reduction identity for ( sin^2 theta ):[sin^2(theta) = frac{1 - cos(2theta)}{2}]So applying that here:[4 int_{0}^{2} sin^2(pi t) , dt = 4 int_{0}^{2} frac{1 - cos(2pi t)}{2} , dt = 2 int_{0}^{2} (1 - cos(2pi t)) , dt]Now split this into two integrals:1. ( 2 int_{0}^{2} 1 , dt )2. ( -2 int_{0}^{2} cos(2pi t) , dt )First part: ( 2 times 2 = 4 ).Second part: The integral of ( cos(2pi t) ) is ( frac{1}{2pi}sin(2pi t) ). So:[-2 left[ frac{1}{2pi}sin(2pi t) right]_0^{2} = -2 left( frac{1}{2pi}sin(4pi) - frac{1}{2pi}sin(0) right)]But ( sin(4pi) = 0 ) and ( sin(0) = 0 ), so this whole part is zero.Therefore, the third integral is 4.Putting it all together:Total integral = 18 (from first integral) + 0 (second integral) + 4 (third integral) = 22.So the total energy expenditure is ( 22k ).Wait, let me double-check my calculations. The first integral is 9 over 2 hours, so 18. The second integral is zero, that seems right. The third integral: I used the power-reduction formula, which is correct. Then I split it into two parts, the first part was 2*2=4, the second part was zero. So 4. So 18 + 4 is 22. Yeah, that seems correct.So for Sub-problem 1, the total energy expenditure is ( 22k ).Moving on to Sub-problem 2. Now, we have an adjustment factor ( A(t) = 1 + 0.5t^2 ). So the new energy expenditure model is ( E(t) = kI(t)^2A(t) ). So we need to compute:[text{Total Energy} = int_{0}^{2} kI(t)^2A(t) , dt = k int_{0}^{2} (3 + 2sin(pi t))^2 (1 + 0.5t^2) , dt]This seems more complicated because now we have a product of two functions: the squared intensity and the quadratic adjustment factor. I need to expand this product before integrating.First, let me expand ( (3 + 2sin(pi t))^2 ) as before, which is 9 + 12 sin(πt) + 4 sin²(πt). Then multiply this by ( 1 + 0.5t^2 ):[(9 + 12sin(pi t) + 4sin^2(pi t))(1 + 0.5t^2) = 9(1 + 0.5t^2) + 12sin(pi t)(1 + 0.5t^2) + 4sin^2(pi t)(1 + 0.5t^2)]So expanding each term:1. ( 9(1 + 0.5t^2) = 9 + 4.5t^2 )2. ( 12sin(pi t)(1 + 0.5t^2) = 12sin(pi t) + 6t^2sin(pi t) )3. ( 4sin^2(pi t)(1 + 0.5t^2) = 4sin^2(pi t) + 2t^2sin^2(pi t) )So putting it all together, the integrand becomes:[9 + 4.5t^2 + 12sin(pi t) + 6t^2sin(pi t) + 4sin^2(pi t) + 2t^2sin^2(pi t)]Therefore, the integral is:[k int_{0}^{2} left(9 + 4.5t^2 + 12sin(pi t) + 6t^2sin(pi t) + 4sin^2(pi t) + 2t^2sin^2(pi t)right) dt]This integral has six terms. Let me handle each term separately.1. ( int_{0}^{2} 9 , dt )2. ( int_{0}^{2} 4.5t^2 , dt )3. ( int_{0}^{2} 12sin(pi t) , dt )4. ( int_{0}^{2} 6t^2sin(pi t) , dt )5. ( int_{0}^{2} 4sin^2(pi t) , dt )6. ( int_{0}^{2} 2t^2sin^2(pi t) , dt )Let me compute each integral one by one.1. First term: ( int_{0}^{2} 9 , dt = 9 times 2 = 18 ). Same as before.2. Second term: ( int_{0}^{2} 4.5t^2 , dt ). The integral of ( t^2 ) is ( frac{t^3}{3} ). So:[4.5 left[ frac{t^3}{3} right]_0^{2} = 4.5 times left( frac{8}{3} - 0 right) = 4.5 times frac{8}{3} = 1.5 times 8 = 12]3. Third term: ( int_{0}^{2} 12sin(pi t) , dt ). As before, the integral of ( sin(pi t) ) is ( -frac{1}{pi}cos(pi t) ). So:[12 left[ -frac{1}{pi}cos(pi t) right]_0^{2} = 12 left( -frac{1}{pi}cos(2pi) + frac{1}{pi}cos(0) right) = 12 times 0 = 0]Same result as before because the sine function over a full period integrates to zero.4. Fourth term: ( int_{0}^{2} 6t^2sin(pi t) , dt ). This looks more complicated. It's a product of a polynomial and a sine function. I think I need to use integration by parts here.Integration by parts formula is ( int u , dv = uv - int v , du ). Let me set:Let ( u = t^2 ), so ( du = 2t , dt ).Let ( dv = sin(pi t) , dt ), so ( v = -frac{1}{pi}cos(pi t) ).So applying integration by parts:[int t^2 sin(pi t) , dt = -frac{t^2}{pi}cos(pi t) + frac{2}{pi} int t cos(pi t) , dt]Now, the remaining integral ( int t cos(pi t) , dt ) also requires integration by parts.Let me set:Let ( u = t ), so ( du = dt ).Let ( dv = cos(pi t) , dt ), so ( v = frac{1}{pi}sin(pi t) ).So,[int t cos(pi t) , dt = frac{t}{pi}sin(pi t) - frac{1}{pi} int sin(pi t) , dt = frac{t}{pi}sin(pi t) + frac{1}{pi^2}cos(pi t) + C]Putting it back into the previous expression:[int t^2 sin(pi t) , dt = -frac{t^2}{pi}cos(pi t) + frac{2}{pi} left( frac{t}{pi}sin(pi t) + frac{1}{pi^2}cos(pi t) right ) + C]Simplify:[= -frac{t^2}{pi}cos(pi t) + frac{2t}{pi^2}sin(pi t) + frac{2}{pi^3}cos(pi t) + C]So, the definite integral from 0 to 2 is:[left[ -frac{t^2}{pi}cos(pi t) + frac{2t}{pi^2}sin(pi t) + frac{2}{pi^3}cos(pi t) right ]_0^{2}]Let me compute this at t=2 and t=0.At t=2:- ( cos(2pi) = 1 )- ( sin(2pi) = 0 )So,[-frac{4}{pi}(1) + frac{4}{pi^2}(0) + frac{2}{pi^3}(1) = -frac{4}{pi} + 0 + frac{2}{pi^3}]At t=0:- ( cos(0) = 1 )- ( sin(0) = 0 )So,[-0 + 0 + frac{2}{pi^3}(1) = frac{2}{pi^3}]Subtracting the lower limit from the upper limit:[left( -frac{4}{pi} + frac{2}{pi^3} right ) - left( frac{2}{pi^3} right ) = -frac{4}{pi} + frac{2}{pi^3} - frac{2}{pi^3} = -frac{4}{pi}]Therefore, the definite integral ( int_{0}^{2} t^2sin(pi t) , dt = -frac{4}{pi} ).But remember, the fourth term is ( 6 times ) this integral:[6 times left( -frac{4}{pi} right ) = -frac{24}{pi}]5. Fifth term: ( int_{0}^{2} 4sin^2(pi t) , dt ). I remember from Sub-problem 1 that ( int sin^2(pi t) , dt ) can be simplified using the power-reduction formula.Using the same identity:[sin^2(pi t) = frac{1 - cos(2pi t)}{2}]So,[4 int_{0}^{2} sin^2(pi t) , dt = 4 times frac{1}{2} int_{0}^{2} (1 - cos(2pi t)) , dt = 2 int_{0}^{2} (1 - cos(2pi t)) , dt]Which is similar to before. Splitting into two integrals:1. ( 2 int_{0}^{2} 1 , dt = 2 times 2 = 4 )2. ( -2 int_{0}^{2} cos(2pi t) , dt )The integral of ( cos(2pi t) ) is ( frac{1}{2pi}sin(2pi t) ). Evaluating from 0 to 2:[-2 left[ frac{1}{2pi}sin(2pi t) right ]_0^{2} = -2 left( frac{1}{2pi}sin(4pi) - frac{1}{2pi}sin(0) right ) = -2 times 0 = 0]So the fifth term is 4.6. Sixth term: ( int_{0}^{2} 2t^2sin^2(pi t) , dt ). This looks complicated. Let me see. I can use the power-reduction formula again for ( sin^2(pi t) ):[sin^2(pi t) = frac{1 - cos(2pi t)}{2}]So,[2 int_{0}^{2} t^2 sin^2(pi t) , dt = 2 int_{0}^{2} t^2 left( frac{1 - cos(2pi t)}{2} right ) dt = int_{0}^{2} t^2 (1 - cos(2pi t)) , dt]So, split into two integrals:1. ( int_{0}^{2} t^2 , dt )2. ( -int_{0}^{2} t^2 cos(2pi t) , dt )Compute the first integral:( int_{0}^{2} t^2 , dt = left[ frac{t^3}{3} right ]_0^{2} = frac{8}{3} - 0 = frac{8}{3} )Second integral: ( -int_{0}^{2} t^2 cos(2pi t) , dt ). This requires integration by parts as well.Let me set:Let ( u = t^2 ), so ( du = 2t , dt ).Let ( dv = cos(2pi t) , dt ), so ( v = frac{1}{2pi}sin(2pi t) ).Applying integration by parts:[int t^2 cos(2pi t) , dt = frac{t^2}{2pi}sin(2pi t) - frac{1}{pi} int t sin(2pi t) , dt]Now, the remaining integral ( int t sin(2pi t) , dt ) also needs integration by parts.Let me set:Let ( u = t ), so ( du = dt ).Let ( dv = sin(2pi t) , dt ), so ( v = -frac{1}{2pi}cos(2pi t) ).So,[int t sin(2pi t) , dt = -frac{t}{2pi}cos(2pi t) + frac{1}{2pi} int cos(2pi t) , dt = -frac{t}{2pi}cos(2pi t) + frac{1}{(2pi)^2}sin(2pi t) + C]Putting it back into the previous expression:[int t^2 cos(2pi t) , dt = frac{t^2}{2pi}sin(2pi t) - frac{1}{pi} left( -frac{t}{2pi}cos(2pi t) + frac{1}{(2pi)^2}sin(2pi t) right ) + C]Simplify:[= frac{t^2}{2pi}sin(2pi t) + frac{t}{2pi^2}cos(2pi t) - frac{1}{4pi^3}sin(2pi t) + C]Now, evaluate the definite integral from 0 to 2.At t=2:- ( sin(4pi) = 0 )- ( cos(4pi) = 1 )So,[frac{4}{2pi}(0) + frac{2}{2pi^2}(1) - frac{1}{4pi^3}(0) = 0 + frac{1}{pi^2} - 0 = frac{1}{pi^2}]At t=0:- ( sin(0) = 0 )- ( cos(0) = 1 )So,[0 + 0 - frac{1}{4pi^3}(0) = 0]Subtracting the lower limit from the upper limit:[frac{1}{pi^2} - 0 = frac{1}{pi^2}]Therefore, the definite integral ( int_{0}^{2} t^2 cos(2pi t) , dt = frac{1}{pi^2} ).But remember, the second integral is negative:[-int_{0}^{2} t^2 cos(2pi t) , dt = -frac{1}{pi^2}]So, putting it all together, the sixth term is:[frac{8}{3} - frac{1}{pi^2}]Therefore, the sixth term is ( frac{8}{3} - frac{1}{pi^2} ).Now, let me sum up all six terms:1. 182. 123. 04. ( -frac{24}{pi} )5. 46. ( frac{8}{3} - frac{1}{pi^2} )Adding them together:First, add the constants:18 + 12 + 4 + ( frac{8}{3} ) = 34 + ( frac{8}{3} ) = ( frac{102}{3} + frac{8}{3} = frac{110}{3} )Now, the terms with ( pi ):( -frac{24}{pi} - frac{1}{pi^2} )So, the total integral is:( frac{110}{3} - frac{24}{pi} - frac{1}{pi^2} )Therefore, the total energy expenditure is:( k times left( frac{110}{3} - frac{24}{pi} - frac{1}{pi^2} right ) )Let me compute the numerical values for better understanding.First, ( frac{110}{3} approx 36.6667 )( frac{24}{pi} approx frac{24}{3.1416} approx 7.6394 )( frac{1}{pi^2} approx frac{1}{9.8696} approx 0.1013 )So,Total ≈ 36.6667 - 7.6394 - 0.1013 ≈ 36.6667 - 7.7407 ≈ 28.926So approximately 28.926k.But since the question asks for an exact expression, I should keep it symbolic.So, the exact total energy expenditure is:( k left( frac{110}{3} - frac{24}{pi} - frac{1}{pi^2} right ) )Let me double-check my calculations for each term to make sure I didn't make a mistake.First term: 18, correct.Second term: 12, correct.Third term: 0, correct.Fourth term: -24/π, correct.Fifth term: 4, correct.Sixth term: 8/3 - 1/π², correct.Adding them: 18 + 12 + 4 = 34; 34 + 8/3 = 34 + 2.6667 ≈ 36.6667. Then subtract 24/π ≈ 7.6394 and 1/π² ≈ 0.1013, so total ≈ 36.6667 - 7.7407 ≈ 28.926. That seems consistent.So, I think my calculations are correct.Final AnswerSub-problem 1: The total energy expenditure is boxed{22k}.Sub-problem 2: The total energy expenditure is boxed{k left( dfrac{110}{3} - dfrac{24}{pi} - dfrac{1}{pi^2} right)}.</think>