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.
The statements that describe Euler’s theorem on homogeneous functions are as follows:
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:
x.du/dx + y.du/dy = n.u
x.du/dx + y.du/dy + z.du/dz = n.u
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:
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.
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.