<p>You are probably talking about Euler's homogeneous function theorem <a href="http://mathworld.wolfram.com/EulersHomogeneousFunctionTheorem.html" rel="nofollow">http://mathworld.wolfram.com/EulersHomogeneousFunctionTheorem.html</a> , which shows that you can decompose a function into its partial derivatives based on its homogeneity. </p>

<p>So a typical Cobb Douglass $U(x,y)=x^ay^{1-a}$</p>

<p>This function is homogeneous of degree 1 because when we multply each argument by a scalar:</p>

<p>$U(cx,cy)=(cx)^a(cy)^{1-a}=cx^ay^{1-a}=cU(x,y)$</p>

<p>So if we apply the theorem to this function we get:</p>

<p>$U(x,y)=xU_x+yU_y=xax^{a-1}y^{1-a}+x^ay(1-a)y^{-a}=x^ay^{1-a}$</p>

<p>So the result that we get back the original utility is expected. </p>

CC BY-SA 3.0
2016-09-04T13:13:50.347Z
2016-09-04T13:13:50.347Z
VCG
<p>When I apply Euler's Theorem onto the Cobb-Douglas equation I receive the Cobb-Douglas equation back. That is, there is no change after applying the theorem. Can someone please explain what the intent and point of this is, seeing no difference? </p>

CC BY-SA 3.0
2016-09-04T04:59:01.457Z
Mathematical operations do not exist in a vacuum. Why are you using Euler's Theorem? Is it maybe that something is equal to the full derivative of the function? One might also say that there is no point in differentiating $e^x$ as the derivative is the same. True, unless one actually needs the derivative for something. Are you doing a PhD? Rather than trying to cram through the math try to think about what you would do and why. Mathematics is merely a tool.
Giskard
<microeconomics>
Euler's theorem and Cobb-Douglas
Irina
# Euler's theorem and Cobb-Douglas

(Asked by: username_0 on 2016-09-04)

When I apply Euler's Theorem onto the Cobb-Douglas equation I receive the Cobb-Douglas equation back. That is, there is no change after applying the theorem. Can someone please explain what the intent and point of this is, seeing no difference?

**Tags:** microeconomics
> **Comments:**
> **username_1**: Mathematical operations do not exist in a vacuum. Why are you using Euler's Theorem? Is it maybe that something is equal to the full derivative of the function? One might also say that there is no point in differentiating $e^x$ as the derivative is the same. True, unless one actually needs the derivative for something. Are you doing a PhD? Rather than trying to cram through the math try to think about what you would do and why. Mathematics is merely a tool.  

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## 1 answers

### Answer by username_2 on 2016-09-04. Score: 2 | Accepted answer

You are probably talking about Euler's homogeneous function theorem <http://mathworld.wolfram.com/EulersHomogeneousFunctionTheorem.html> , which shows that you can decompose a function into its partial derivatives based on its homogeneity.

So a typical Cobb Douglass $U(x,y)=x^ay^{1-a}$

This function is homogeneous of degree 1 because when we multply each argument by a scalar:

$U(cx,cy)=(cx)^a(cy)^{1-a}=cx^ay^{1-a}=cU(x,y)$

So if we apply the theorem to this function we get:

$U(x,y)=xU\_x+yU\_y=xax^{a-1}y^{1-a}+x^ay(1-a)y^{-a}=x^ay^{1-a}$

So the result that we get back the original utility is expected.


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