M25008SelSol3

M25008SelSol3 - Mathematics 250 Fall 2008 Selected...

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Mathematics 250 Fall 2008 Selected Solutions, Assignment 3 MVM, Exercise 7.3.11 : Derive Euler’s formula for homogeneous functions. A function f : R m R is called homogeneous of degree k if , for any vector ~x and any scalar t , we have the equality f ( t~x ) = t k f ( ~x ). Show that a function is homogeneous of degree k if and only if Df ( ~x )( ~x ) = x 1 ∂f ∂x 1 + x 2 ∂f ∂x 2 + · · · + x k ∂f ∂x k = kf ( ~x ) . Response: This is an application of the Chain Rule. We think of f ( t~x ) as a function of t . Then it is the composition of the mapping g : t t~x and of f . Since the derivative of g is just ~x , we compute D ( f g )( t ) = Df ( g ( t )) Dg ( t ) = Df ( t~x )( ~x ). On the other hand, from one variable calculus, we know that d dt ( t k f ( ~x ) = kt k - 1 f ( x ), since f ( ~x ) is just a constant in this calculation. Equating the two formulas for D ( f g ) gives the formula Df ( t~x )( ~x ) = t k - 1 f ( ~x ). Setting t = 1 then gives Euler’s formula. We are also asked to show the converse, that if f satisfies Euler’s equation, then f is homogeneous of degree k . For this, it is convenient to consider the function t - k f ( t~x ) = t - k ( f g )( t ). If we differentiate this using the product rule, and use Euler’s formula, we find that d dt ( t - k ( f g )( t )) = - kt - k - 1 ( f g )( t )) + t - k d dt ( f g )( t ) = - kt - k - 1 f ( ~x ) + t - k ( Df ( g ( t )) Dg ( t )) = - kt - k - 1 f ( t~x ) + t - k ( Df ( t~x ) ~x ) = - kt - k - 1 f ( t~x ) + t - k - 1 ( Df ( t~x )( t~x )) = - kt - k - 1 f ( t~x ) + t - k - 1 ( kf ( t~x )) = 0 . The second equation in the first line uses the Chain Rule. The equality of the expressions in the second line follows because Df ( t~x ) is a linear function in the direction: t ( Df ( t~x ) ~x ) = Df ( t~x )( t~x ). The equation from the second line to the third follows by applying Euler’s equation at the point t~x . This computation shows that t - k f ( t~x ), as a function of t , has derivative identically zero. Therefore, by the Mean Value Theorem, it is constant. Therefore, t - k f ( t~x ) = 1 - k f (1 · ~x ) = f ( ~x ) . Multiplying by t k shows that f satisfies the equation defining a homogeneous function of degree k . AS3.1
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This note was uploaded on 09/24/2009 for the course MATH 250 taught by Professor Rogerhowe during the Fall '06 term at Yale.

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M25008SelSol3 - Mathematics 250 Fall 2008 Selected...

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