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The pre-order being complete has no topological meaning, but purely set-theoretic. It means that for any two points $x,y$ in the domain of the pre-order, we must have $x \succsim y$ or $y \succsim x$ (or possibly both here, because we have a pre-order, so we can have both at the same time (invariant goods (?), or some such thing, economics is not my field, ...
Your textbook isn't wrong. $N$ is a fixed number (6 in the example), so the sum is over a finite number of terms, which is perfectly ok. Besides, the sum isn't the harmonic series... It is $$\frac{1}{N}+ \frac{1}{N}+ ... + \frac{1}{N}$$ where there are $N$ summands. So in the example $$\frac{1}{6}+ \frac{1}{6}+ \frac{1}{6}+ \frac{1}{6}+ \frac{1}{6}+ \... 2 Use Lagrange Multipliers. \mathcal{L}=50L^{0.2}K^{0.8}-\lambda (1000-L-K) Take first order condition with respect to K and set it to zero: 50\times0.8\times L^{0.2}K^{-0.2}=-\lambda Take first order condition with respect to L and set it to zero: 50\times0.2\times L^{-0.8}K^{0.8}=-\lambda Take first order condition with respect to \lambda ... 2 Instead of payout, think in terms of the slope of the payout (i.e the delta). If you long a call at A and cover it by shorting a call at B > A, the slope will be 1 between A and B and 0 otherwise. If you long a put at D and cover it by shorting a put at C < D, the slope will be -1 between C and D and 0 otherwise. In the ... 2 The inequality is false. Take K=2, \gamma=1/2 and$$ C_1=100,\quad C_2=1,\quad N_1=1,\quad N_2=100. $$Then$$ \frac{\sum_{i=1}C_{i}}{\sum_{i=1} {N_i}^\gamma {C_{i}}^{1-\gamma}}=\frac{101}{20}>5 $$and$$ \frac{K\max_iC_{i}}{\sum_{i=1} {N_i}^\gamma \max_k\{{C_{k}}\}^{1-\gamma}}=\frac{200}{10\cdot(10+1)}<2 $$PS. A commentator correctly remarked ... 2 You have p+7=\sqrt{\frac{4000}{Q}}. Increasing Q by 1 decreases the price that can be charged (on all sales) by \Delta p, where p-\Delta p+7=\sqrt{\frac{4000}{Q+1}}. So we have \Delta p=\sqrt\frac{4000}{Q}-\sqrt{\frac{4000}{Q+1}}=\sqrt{\frac{4000}{Q}}\left(1-\sqrt{\frac{Q}{Q+1}}\right) =(p+7)\left(1-\sqrt{\frac{1}{1+\frac{1}{Q}}}\right). We now ... 2 The production function states the quantity that a firm can produce. So if it produces 10 units:$$10=f(L,M)=L^{1/2}M^{1/2}$$Hence:$$100=LM\frac{100}{L}=M$$We know the cost will be:$$C=9L+81M=9L+\frac{8100}{L}$$Minimizing this using standard procedures C'(L)=0.. gives L=30 C=540 so I think you typed something wrong. I have ... 1 (1) does not imply (2), and (1) and (2) together do not imply (3). There are many non-dictatorial methods that fail to satisfy (2); see this Wikipedia article. For a simple method that satisfies (1) and (2) but not three, let u and v be two designated voters. If u and v have the same preferences, their preferences are adopted by society; if not, ... 1 First we translate:$$\frac{dP}{dt}=-\frac{1}{2}(S(P)-D(p))$$Then we substitute:$$\frac{dP}{dt}=-\frac{1}{2}(5P-60) $$But now you have a first order differential equation. hint: You can seperate them. you're going to have something involving e, and something constant, involving your initial condition. Here are some notes 1 First step is to solve for the rate of p, which is p'.$$S-D=80+3p-(140-2p)=5p-60\frac{S-D}{2}=\frac{5p-60}{2}$$Since p is decreasing at the rate, p' needs to be negative.So$$p'=\frac{60-5p}{2}$$Now it's easy to get$$2p'+5p=60$$It's a non-homogeneous first-order differential equation.$$2D+5=0,D=-\frac52$$Therefore you get the general solution... 1 Let's phrase the problem mathematically. You are being asked to solve:: \frac{dp}{dt} = -\frac{1}{2} (S(p)-D(t)) So substituting in the given supply and demand equations: \frac{dp}{dt} = -\frac{1}{2} (5p-60) Separating the variables: \frac{dp}{p-12} = -\frac{5}{2} dt Integrating both sides, we get the general result: \ln|p-12| = -\frac{5}{2}t+C... 1 You appear in effect to be assuming that \succsim is the same as ordinary \ge. This need not be the case. In fact, the whole point of the argument is that the lower contour sets for any continuous preorder on X are closed, so that in this sense all continuous preorders on X behave like the familiar natural order \ge. Of course the steps of the ... 1 It seems that posers of these problems sometimes delight in expressing the marginal functions in terms of  \ p \  , rather than in terms of  \ q \  which is used in the definitions. On the other hand, it is a reasonable way to do things frequently, since the business has direct control (generally) over the price set, but not over the demand. One way to ... 1 You can look up the mathematical definitions in any game theory text book. The interpretation is basically that the players' actions reinforce/offset (complement/substitute) each other. For example, if a buyer is more likely to buy a good at some price, when it is more likely that all other buyers are taking that price, then buyers are strategic complements.... 1 As x,y goes to infinity, profit goes to negative infinity. So, the maximum exists. We need to check the critical values. As economics only cares about positive numbers, critical values are x=0,y=0 and the values that make the partial derivatives 0. For x=0, the profit is$$p(y)=1200-10y^2+96y $p'(y)=96-20y$, thus, $p$ will be maximum at $y_0=24/5$....