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

In summary, the degree of the numerator in this case is considered to be 0 because $\ln(y/x)$ is a degree zero homogeneous function, meaning that it behaves like a constant when multiplied by a scalar.
  • #1
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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  • #2
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.
 
  • #3
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.
 

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

What is a homogeneous function?

A homogeneous function is a mathematical function where each term has the same degree. In other words, when you multiply all the variables in the function by a constant, the result is the same as multiplying the function by that constant raised to the degree of the function. For example, x^2 + 3xy + 2y^2 is a homogeneous function of degree 2, while x^2 + 3xy + y^3 is not homogeneous.

What is the purpose of solving homogeneous function confusion?

The purpose of solving homogeneous function confusion is to simplify the function and make it easier to work with. By rearranging the terms and factoring out common factors, we can reduce the degree of the function and potentially find a more elegant solution.

How do I solve a homogeneous function with ln(Y/X) in the numerator?

To solve a homogeneous function with ln(Y/X) in the numerator, we can use the property of logarithms that states ln(a/b) = ln(a) - ln(b). We can then use this property to rewrite the function in terms of ln(Y) and ln(X), which will allow us to factor out common terms and simplify the function.

Can I use any value for X and Y when solving a homogeneous function?

No, the values for X and Y must be non-zero in order for the function to be homogeneous. If either X or Y is equal to 0, the function will not have the same degree when multiplied by a constant.

Are there any other methods for solving homogeneous functions?

Yes, there are other methods such as using substitution or using the Euler's homogeneous function theorem. However, using the property of logarithms is often the simplest and most efficient method for solving homogeneous functions that involve ln(Y/X) in the numerator.

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