{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

M25008SelSol3

# M25008SelSol3 - Mathematics 250 Fall 2008 Selected...

This preview shows pages 1–2. Sign up to view the full content.

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 ( tx ) = 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 ( tx ) as a function of t . Then it is the composition of the mapping g : t tx and of f . Since the derivative of g is just x , we compute D ( f g )( t ) = Df ( g ( t )) Dg ( t ) = Df ( tx )( 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 ( tx )( 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 ( tx ) = 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 ( tx ) + t - k ( Df ( tx ) x ) = - kt - k - 1 f ( tx ) + t - k - 1 ( Df ( tx )( tx )) = - kt - k - 1 f ( tx ) + t - k - 1 ( kf ( tx )) = 0 . The second equation in the first line uses the Chain Rule. The equality of the expressions in the second line follows because Df ( tx ) is a linear function in the direction: t ( Df ( tx ) x ) = Df ( tx )( tx ). The equation from the second line to the third follows by applying Euler’s equation at the point tx . This computation shows that t - k f ( tx ), as a function of t , has derivative identically zero. Therefore, by the Mean Value Theorem, it is constant. Therefore, t - k f ( tx ) = 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 : a) Compute the derivative of the identity mapping. b) Given Φ : D E , an invertible mapping between two domains in R k , compute the derivative of Φ - 1 in terms of the derivative of Φ.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern