homework5sol

# homework5sol - 32A - HOMEWORK 5 SOLUTIONS PROBLEM 7. We...

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32A - HOMEWORK 5 SOLUTIONS PROBLEM 7. We apply the implicit function theorem (Theorem 1.1). The Frst of the conditions ∂F ( x, y, z ) ∂x 6 =0 , ∂F ( x, y, z ) ∂y 6 =0 , ∂F ( x, y, z ) ∂z 6 =0 (1) implies that x can be put as a function of y, z ; x = x ( y, z ) and that ∂x ∂z = ∂F ∂z . ∂F ∂x . The second condition (1) implies that y can be put as a function of z, x ; y = y ( z, x ) with ∂y ∂x = ∂F ∂x . ∂F ∂y . The third condition (1) implies that z can be put as a function of x, y ; z = y ( x, y ) with ∂z ∂y = ∂F ∂y . ∂F ∂z . Multiplying the three equalities and simplifying the partial derivatives, ∂x ∂z ∂z ∂y ∂y ∂x = 1 . TAYLOR EXPANSIONS IN TWO VARIABLES . PROBLEM 2. We use (1) in TAYLOR EXPANSIONS IN TWO VARIABLES. If f ( x, y )isapolynomial of degree n then all its terms are of the form a jk x j y k j + k n. The function ϕ ( t )= f ( a + t ( x a ) ,b + t ( y b )) contains terms of the form a jk ( a + t ( x a )) j ( b + t ( y b )) k j + k n thus is a polynomial in t of degree n. Accordingly R n ( x, 0) (which uses derivative of order n +1)is zero. Expanding ϕ ( t ) and setting t =1 we obtain f ( x, y )= p n ( x, y ; a, b ) . 1

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HOMEWORK 5 SOLUTIONS ® PROBLEM 1 u @ t _ , x _ D = Exp @ - x^2 ê H 4 * t LD ê Sqrt @ t D - x 2 4 t t D @ u @ t, x D , t D - - x 2 4 t 2 t 3 ê 2 + - x 2 4 t x 2 4 t 5 ê 2 Simplify @ % D - x 2 4 t I - 2 t + x 2 M 4 t 5 ê 2 D @ u @ t, x D , x D - - x 2 4 t x 2 t 3 ê 2 D @ u @ t, x D , x, x D - - x 2 4 t 2 t 3 ê 2 + - x 2 4 t x 2 4 t 5 ê 2 Simplify @ % D - x 2 4 t I - 2 t + x 2 M 4 t 5 ê 2

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® PROBLEM 2 u @ t _ , x _ , y _ D = Exp @ - H x^2 + y^2 L ê H 4 * t LD ê t - x 2 - y 2 4 t t D @ u @ t, x, y D , t D - - x 2 - y 2 4 t t 2 - - x 2 - y 2 4 t I - x 2 - y 2 M 4 t
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## This note was uploaded on 06/03/2011 for the course MATH 32A 32A taught by Professor Moshchovakis during the Spring '10 term at UCLA.

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homework5sol - 32A - HOMEWORK 5 SOLUTIONS PROBLEM 7. We...

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