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

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SUMMARY

The discussion clarifies that the term ln(Y/X) is not included in the calculation of the degree of the numerator because it is a degree zero homogeneous function. According to the definition of homogeneous functions, for any real number t, the function g(tx, ty) simplifies to g(x, y) as the t's cancel out. This indicates that ln(Y/X) behaves like a constant, contributing no degree to the overall function.

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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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