MHB Solving Homogeneous Function Confusion: ln(Y/X) in Numerator

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The discussion clarifies that the term ln(Y/X) is not considered in the calculation of the degree of the numerator because it behaves like a constant. When applying the definition of a homogeneous function, the transformation shows that ln(Y/X) remains unchanged when scaling by t, leading to the conclusion that it has a degree of zero. This means it does not contribute to the overall degree of homogeneity in the function. Participants confirm that ln(Y/X) is effectively a degree zero homogeneous function. Understanding this concept is essential for grasping the properties of homogeneous functions.
zkee
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Hey people!

I'm confused as to why the ln(Y/X) part of the numerator is not considered in the calculation of the degree of numerator.

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Any help or websites to browse through for the answer would be appreciated!
 

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The definition of an homogeneous function means that if $t \in \mathbb{R}$ then $f(tx,ty) = t^n f(x,y)$, where $n$ is the degree of homogeneity. When you do this to $g(x,y) = \ln (y/x)$ what you get is

$$g(tx,ty) = \ln \left( \frac{ty}{tx} \right) = \ln \left( \frac{y}{x} \right) = g(x,y).$$

The $t$'s cancel, therefore it makes no contribution. This is a degree zero homogeneous function. :)

Best wishes,

Fantini.
 
zkee said:
Hey people!

I'm confused as to why the ln(Y/X) part of the numerator is not considered in the calculation of the degree of numerator.

Any help or websites to browse through for the answer would be appreciated!

Welcome to MHB, zkee! :)

The degree of y/x is 0.
Or in other words, $\ln(y/x)$ behaves like a constant.
 

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