3 a prove that b e x e x e x e x forms an orthogonal

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a) Prove that B = { e x + e x , e x e x } forms an orthogonal basis of V . Solution Write E 1 ( x ) := e x + e x and E 2 ( x ) := e x e x . Note that E 1 is an even function and E 2 is an odd function. In this inner product, an even function and an odd function are always orthogonal. To show they are a basis, we need to show they are both linearly independent and span. One way to see that they are linearly independent is to use that they are orthogonal (see Problem A-3d above). There are many other approaches. They span V since e x = E 1 + E 2 2 and e x = E 1 E 2 2 . Thus any linear combination of e x and e x can also be written as a linear combination of E 1 and E 2 . b) Find the best approximation in V (with respect to the L 2 inner product) of the func- tion g ( x ) = x . [ Hint: think orthogonal projection. Leave your answer in terms of integrals that could be evaluated easily using a computer program. ] Solution: To find the best approximation of g ( x ) = x in V , we use E 1 and E 2 as an orthogonal basis for V and want to write x = aE 1 + bE 2 + h ( x ) , (1) where a and b are constants and h ( x ) is orthogonal to both E 1 and E 2 . Then aE 1 + bE 2 is the orthogonal projection of x into V and is the desired best approximation of x in V : Proj V ( x ) = aE 1 ( x ) + bE 2 ( x ) . Our task is to find a and b . To compute a take the inner product of both sides of equation (1) with E 1 and use that E 1 is orthogonal to both E 2 and h . Thus ( x, E 1 ) = a bardbl E 1 bardbl 2 so a = ( x, E 1 ) bardbl E 1 bardbl 2 . Similarly ( x, E 2 ) = b bardbl E 2 bardbl 2 so b = ( x, E 2 ) bardbl E 2 bardbl 2 . These formulas for a and b just involve the inner product and hence straightforward but tedious integrals. [One can save time by observing that a = 0 because the numerator is the inner product of an odd and an even function. ] B–3. In a large city, a car rental company has three locations: the Airport, the City, and the Suburbs. One has data on which location the cars are returned daily: Rented at Airport: 2% are returned to the City and 25% to the Suburbs. The rest are returned to the Airport. 4
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Rented in City : 10% returned to Airport, 10% returned to Suburbs. The rest are returned to the City. Rented in Suburbs: 25% are returned to the Airport and 2% to the city. The rest are returned to the Suburbs. If initially there are 35 cars at the Airport, 150 in the city, and 35 in the suburbs, what is the long-term distribution of the cars? Solution: Let A k , C k , and S k denote the number of cars at the Airport, City, and Suburbs, respectively, on day k . The above data tells us that on day k + 1 A k +1 = . 73 A k + . 10 C k + . 25 S k C k +1 = . 02 A k + . 80 C k + . 02 S k S k +1 = . 25 A k + . 10 C k + . 73 S k and write D k = A k C k S k so D 0 = 35 150 35 . The vector D k gives the distribution of the cars on day k . Note for every day the sum of the components of D k is always 220 since the same cars just get moved around. The transition matrix for this Markov chain is T := . 73 . 10 . 25 . 02 . 80 . 02 . 25 . 10 . 73 The long-term distribution is D := lim k →∞ D k . From the theory, D = TD so D is an eigenvector of T associated with the eigenvalue 1. Thus we solve the equations (
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