Euler’s Theorem on Homogeneous Functions

Euler’s theorem on homogeneous functions is a very convoluted, Complex, and very elaborated theorem. The identification of homogeneous and non-homogeneous functions was made easier with this theorem.

Euler’s Theorem

The statements that describe Euler’s theorem on homogeneous functions are as follows:

• Euler’s theorem is used to establish a relationship between the partial derivatives and the function product with its degree. 
• A homogeneous function of degree n, with x,y & z variables is a function in which all terms are of degree n. 

Euler’s Theorem Formula:

A function f(x,y) will be a homogeneous function in x and y of degree n if:

f(tx,ty) = t^n.f(x,y)

Following are the Euler’s theorem formula for two and three variables:

• For two variables, ‘u’ is a homogeneous function in x and y of degree n, then the Euler’s theorem is as follows:

x.du/dx + y.du/dy = n.u

• For three variables, ‘u’ is a homogeneous function of three variables x,y, and z of degree n, then the Euler’s theorem is following:

x.du/dx + y.du/dy + z.du/dz = n.u

Euler’s Theorem Calculator

To understand the theorem, you must first learn the fundamentals of solving the homogeneous function theorem. The most important part of solving the theorem is differentiation.

Following are things to remember before you start differentiation:

• Multiply the power of the number with that of its variable number. 
• If there is no number, we are multiplying the power with the number 1. 
• After multiplying, we decrease 1 from power (-1). 
• If there is no power on the variable, then (-1) would make it zero, and hence only the variable number would be left. 
• If there is no variable to a number, the number would simply vanish. 
• If the variable is in a division state, multiply it with the numerator and convert the denominator’s power to minus. 
• If the variable is under root, you remove the root and decrease the power by ½. 
• If there is no power to the variable while converting from division to multiplication, you put the power with the minus sign (-). 

Let us learn how to differentiate through the following example:

Example: 2√x – 5/x^2

Solution: Differentiating “√”, hence decreasing the power by ½. 

2x^½ – 5/x^2

Now we see that x^2 is in a division state hence converting it to multiplication and converting the power to negative. 

2x^½ – 5(x^-2)

x^-½ – 10x^-3

Now let us understand the equation of two variables and understand Euler’s theorem with an example:

Example: x^3 + y^3 + 3xy^2 

Solution: Euler’s formula for two variables is-

x.du/dx + y.du/dy = n.u —eq no. 1

First, we have to differentiate for du/dx. 

du/dx = 3x^2 + 3y^2  —eq no. 2

Now, differentiating for du/dy. 

du/dy = 3y^2 + 3X2xy —eq no. 3

Putting equation numbers 2 & 3 in equation number 1 (L.H.S.). 

We get,

x (3x^2 + 3y^2) + y (3y^2 + 3X2xy)

We’ll take 3 as a common,

3 (x^3 + xy^2 + y^3 + 2xy^2)

3 (x^3 + y^3 + 3xy^2) —eq no. 4

Equation number 4 is equal to equation number 1 (R.H.S.). 

3 (x^3 + y^3 + 3xy^2) = n.u

Hence, 3 is the degree of the function. The function is also proven to be homogeneous. 

Now let us understand the equation of three variables and understand Euler’s theorem with an example:

Example: 3x^2 + 6y^2 + 3z^2

Solution: Euler’s formula for three variables is-

x.du/dx + y.du/dy + z.du/dz = n.u —eq no. 1

First, we have to differentiate for du/dx. 

du/dx = 6x —eq no. 2

Now, differentiating for du/dy. 

du/dy = 12y —eq no. 3

Lastly, differentiating for du/dz. 

du/dz = 6z —eq no. 4

Putting equation numbers 2,3 & 4 in equation number 1 (L.H.S.). 

We get, 

x (6x) + y (12y) + z (6z)

We’ll take 2 as a common,

2 (3x^2 + 6y^2 + 3z^2) —eq no. 5

Equation number 5 is equal to equation number 1 (R.H.S.). 

2 (3x^2 + 6y^2 + 3z^2) = n.u

Hence, 2 is the degree of the function. The function is also proven to be homogeneous. 

Non-homogeneous Example: x^4 + y^4 +2

Solution: Euler’s formula for two variables is-

x.du/dx + y.du/dy = n.u —eq no. 1

First, we have to differentiate for du/dx. 

du/dx = 4x^3 —eq no. 2

Now, differentiating for du/dy. 

du/dy = 4y^3 —eq no. 3

Putting equation numbers 2 & 3 in equation number 1 (L.H.S.). 

We get,

x (4x^3) + y (4y^3)

We’ll take 4 as a common. 

4 (x^4 + y^4) —eq no. 4

Equation number 4 is not equal to equation number 1 (R.H.S.). 

4 (x^4 + y^4) ≠ n.u

Hence, there is no degree of function. The function is also proven to be non-homogeneous. 

Conclusion:

Euler‘s theorem on homogeneous functions is quite popular though its objectives are always always left unnoticed. When inputs are multiplied by any number, the output increases by some factor to—making the theorem useful in fields like economics. It helps to determine aggregate scales.