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# psCompStat1Ans - Fall 2007 ARE211 Problem Set#12 Answer key...

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Unformatted text preview: Fall 2007 ARE211 Problem Set #12- Answer key First CompStat Problem Set (1) In problem 3 on your last problem set, you found the maximum and minimum distance from the origin to the ellipse x 2 1 + x 1 x 2 + x 2 2 = 3. Generalize this problem to “minimize/maximize the distance from the origin to the ellipse x 2 1 + x 1 x 2 + αx 2 2 = 3” and use the Envelope theorem (starting from α = 1) to estimate the maximum and minimum distance from the origin to the following ellipse, x 2 1 + x 1 x 2 + 0 . 9 x 2 2 = 3 . Ans: From problem 3 on the preceding problem set we know that the Lagrangian was: L = x 2 1 + x 2 2 + λ (3- x 2 1- x 1 x 2- x 2 2 ) Let’s parameterize the Lagrangian accordingly to estimate the change. L = x 2 1 + x 2 2 + λ (3- x 2 1- x 1 x 2- αx 2 2 ) Initially, α = 1 . From the envelope theorem we know that the first derivative of the squared distance M 2 ( α ) w.r.t the parameter α is dM 2 ( α ) dα = δL δα =- λx 2 2 Using a first order Taylor expansion we know that: M 2 ( α + dα ) ≈ M 2 ( α ) + δM 2 δα dα = M 2 ( α )- λx 2 2 dα In the following α = 1 ,dα =- . 1 (1) x (1) 1 = 1 , x (1) 2 = 1 , λ (1) = 2 3 , M ( α = 1) = √ 2 ≈ 1 . 4142 M 2 (0 . 9) ≈ 2- 2 3 * 1 2 * (- . 1) = 31 15 ⇒ M (0 . 9) = radicalbig M 2 (0 . 9) = radicalBig 31 15 ≈ 1 . 4376 (2) x (2) 1 =- 1 , x (2) 2 =- 1 , λ (2) = 2 3 , M ( α = 1) = √ 2 ≈ 1 . 4142 M 2 (0 . 9) ≈ 2- 2 3 * (- 1) 2 * (- . 1) = 31 15 ⇒ M (0 . 9) = radicalbig M 2 (0 . 9) = radicalBig 31 15 ≈ 1 . 4376 (3) x (3) 1 = √ 3 , x (3) 2 =- √ 3 , λ (3) = 2 , M ( α = 1) = √ 6 ≈ 2 . 4495 M 2 (0 . 9) ≈ 6- 2 * (- √ 3) 2 (- . 1) = 6 . 6 ⇒ M (0 . 9) = radicalbig M 2 (0 . 9) = √ 6 . 6 ≈ 2 . 5690 1 2 (4) x (4) 1 =- √ 3 , x (4) 2 = √ 3 , λ (4) = 2 , M ( α = 1) = √ 6 ≈ 2 . 4495 M 2 (0 . 9) ≈ 6- 2 * √ 3 2 (- . 1) = 6 . 6 ⇒ M (0 . 9) = radicalbig M 2 (0 . 9) = √ 6 . 6 ≈ 2 . 5690 3 (2) a) Prove that the expression α 2- αx 3 + x 5 = 17 defines x implicitly as a function of α . Ans: The partial derivative of the expression w.r.t. x is δf δx =- 3 αx 2 + 5 x 4 Evaluated at the point (¯ α, ¯ x ) = (5 , 2) : δf δx =- 3 * 5 * 4+5 * 16 = 20 negationslash = 0 Since δf (¯ α, ¯ x ) δx negationslash = 0 there exists a neighborhood around (¯ α, ¯ x ) where x can be written as an implicit function of α , i.e., x = g ( α ) . in a neighborhood of (¯ α, ¯ x ) = (5 , 2) b) Estimate the x-value which corresponds to α = 4 . 8 using a first order approximation. Ans: The partial derivative of f w.r.t. α is δf δα = 2 α- x 3 Evaluated at the point (¯ α, ¯ x ) = (5 , 2) : δf δα = 2 * 5- 8 = 2 From the implicit function theorem we know δg δα =- δf δα δf δx =- 2 20 =- . 1 Using a first order Taylor expansion: g ( α + dα ) ≈ g ( α ) + δg δα dα Hence for α = 5 ,dα =- . 2 , g (4 . 8) ≈ 2- . 1 * (- . 2) = 2 . 02 4 (3) Consider the equation α 3 1 + 3 α 2 2 + 4 α 1 x 2- 3 x 2 α 2 = 1. Does this equation define= 1....
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psCompStat1Ans - Fall 2007 ARE211 Problem Set#12 Answer key...

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