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# 1 Definition of homogeneous functions

A function $$f$$ of variable $$x$$ is said to be homogeneous of degree $$m$$ if it satisfies the following equality:

$\begin{equation} f\left(\lambda x\right) = \lambda^m \times f\left(x\right) \tag{1.1} \end{equation}$

As an example, the function $$f(x)=x$$ is homogeneous of degree $$1$$ because $$f(\lambda x) = \lambda x$$.

This definition can be extended to multi-variable functions. Let us consider a function $$f$$ of variables $$x$$ and $$y$$. Then $$f$$ is said to be homogeneous of degree $$m$$ with respect to variable $$x$$ if it satisfies the following equality:

$\begin{equation} f\left(\lambda x, y\right) = \lambda^m \times f\left(x, y\right) \tag{1.2} \end{equation}$

For example, the function $$f(x, y)=xy$$ is homogeneous of degree $$1$$ with respect to variable $$x$$ because $$f(\lambda x, y) = \lambda xy$$. One could also say that the function $$f(x, y) = xy$$ is homogeneous of degree $$1$$ with respect to variable $$y$$ because $$f(x, \lambda y) = \lambda xy$$. Finally, one could also say that the function $$f(x, y)=xy$$ is homogeneous of degree $$2$$ with respect to variables $$x$$ and $$y$$ because $$f(\lambda x, \lambda y) = \lambda^2 xy$$. Thus, it is important to specify both (i) the homogeneous degree and (ii) the considered variables.

The variables which can be multiplied by a factor $$\lambda$$ and that leads to a homogeneous function are called scalable parameters (because multiplying them by $$\lambda$$ leads to increase the function value). Other variables are said to be non-scalable variables. As we will see soon, Euler’s theorem application depends on the considered scalable variables.

Finally, for functions of many variables, it is convenient to define vectors as variables: first variable/vector is called $$A$$ and is filled with all scalable variables while the second vector/variable is called $$a$$ and is filled with all non-scalable variables.

Let us consider the function $$f(x,y,z) = xy \times \exp\left(z\right)$$. Because $$f(\lambda x,\lambda y,z) = \lambda^2 f(x,y,z)$$, the scalable variables are $$x$$ and $$y$$ while the non-scalable parameter is $$z$$. Thus, in a simpler way, the $$f$$ function could write $$f\left(A,a\right)$$ with $$A=(x,y)$$ and $$a=z$$. The homogeneity condition writes:

$\begin{equation} f\left(a,\lambda A\right) = \lambda^m \times f\left(a, A\right) \tag{1.3} \end{equation}$

# 2 Derivation of Euler’s theorem

Let us consider a homogeneous function of degree $$m$$ of variables $$A$$. Therefore, this function satisfies equation (1.3). Euler’s theorem states that function $$f$$ satisfies the following equation:

$\begin{equation} f\left(a, \lambda A\right) = \lambda^m \times f\left(a, A\right) \Leftrightarrow \sum_i \left(\frac{\partial f}{\partial A_i}\right)_{a, A_{j \neq i}} \times A_i = m \times f\left(a, A\right) \tag{2.1} \end{equation}$

## 2.1 Left to right implication

To obtain this equality, let us write the differential of equation (1.3).

• First, let define the function $$g\left(\lambda, A\right) = \lambda \times A$$. Differential of LHS then writes:

$\begin{equation} df\left(a,g\right) = \sum_i \left(\frac{\partial f}{\partial a_i}\right)_{a_{j \neq i}, g} da_i + \sum_i \left(\frac{\partial f}{\partial g_i}\right)_{a, g_{j \neq i}} dg_i \tag{2.2} \end{equation}$

The differential element $$dg_i$$ can be written with respect to variables $$\lambda$$ and $$A_i$$:

$\begin{equation} dg_i = \lambda dA_i + A_i d\lambda \tag{2.3} \end{equation}$

Thus, equation (2.2) writes:

$\begin{equation} df\left(a,A, \lambda\right) = \sum_i \left(\frac{\partial f}{\partial a_i}\right)_{a_{j \neq i}, g} da_i + \lambda \times \sum_i \left(\frac{\partial f}{\partial g_i}\right)_{a, g_{j \neq i}} dA_i + \sum_i \left(\frac{\partial f}{\partial g_i}\right)_{a, g_{j \neq i}} A_i d\lambda \tag{2.4} \end{equation}$

• Differential of the RHS writes:

$\begin{equation} d\left[\lambda^m \times f\left(a, A\right)\right] = m \times \lambda^{m-1} f\left(a, A\right) \times d\lambda + \lambda^m \left[\sum_i \left(\frac{\partial f}{\partial a_i}\right)_{a_{j \neq i}, A} da_i + \sum_i \left(\frac{\partial f}{\partial A_i}\right)_{a, A_{j \neq i}} dA_i\right] \tag{2.5} \end{equation}$

By identifying coefficients before $$da_i$$, $$dA_i$$ and $$d\lambda$$ elements, one can write:

$\begin{equation} \begin{cases} \left(\frac{\partial f}{\partial a_i}\right)_{a_{j \neq i}, g} = \lambda^m \times \left(\frac{\partial f}{\partial a_i}\right)_{a_{j \neq i}, A} \\ \lambda \times \left(\frac{\partial f}{\partial g_i}\right)_{a, g_{j \neq i}} = \lambda^m \times \left(\frac{\partial f}{\partial A_i}\right)_{a, A_{j \neq i}} \\ \sum_i \left(\frac{\partial f}{\partial g_i}\right)_{a, g_{j \neq i}} A_i = m \times \lambda^{m-1} f\left(a, A\right) \end{cases} \tag{2.6} \end{equation}$

Second equality of (2.6) rewrites:

$\begin{equation} \left(\frac{\partial f}{\partial g_i}\right)_{a, g_{j \neq i}} = \lambda^{m-1} \times \left(\frac{\partial f}{\partial A_i}\right)_{a, A_{j \neq i}} \tag{2.7} \end{equation}$

Injecting this expression of $$\left(\frac{\partial f}{\partial g_i}\right)_{a,g_{j \neq i}}$$ in the third equality of (2.6), one obtains:

$\begin{equation} \lambda^{m-1} \sum_i \left(\frac{\partial f}{\partial A_i}\right)_{a, A_{j \neq i}} A_i = m \times \lambda^{m-1} f\left(a, A\right) \tag{2.8} \end{equation}$

Dividing equation (2.8) by $$\lambda^{m-1}$$, the Euler theorem is obtained:

$\begin{equation} \sum_i \left(\frac{\partial f}{\partial A_i}\right)_{a, A_{j \neq i}} A_i = m \times f\left(a, A\right) \tag{2.9} \end{equation}$

## 2.2 Right to left implication

So far, one has only demonstrated that a homogeneous function leads to Euler’s theorem. However, one might be interested to derive the opposite, i.e. that a function is homogeneous if it satisfies Euler’s theorem. First, we note that equation (2.1) allows us to write:

$\begin{equation} \sum_i \left(\frac{\partial f}{\partial \lambda A_i}\right)_{a, \lambda A_{j \neq i}}\left(a, \lambda A\right) \times \lambda A_i = m \times f\left(a, \lambda A\right) \tag{2.10} \end{equation}$

Setting $$g = \lambda \times A$$, one gets:

$\begin{equation} \sum_i \left(\frac{\partial f}{\partial g_i}\right)_{a, g_{j \neq i}}\left(a, g\right) \times g_i = m \times f\left(a, g\right) \tag{2.11} \end{equation}$

From (2.4), one gets:

$\begin{equation} \left(\frac{\partial f}{\partial \lambda}\right)_{a, A} = \sum_i \left(\frac{\partial f}{\partial g_i}\right)_{a, g_{j \neq i}}\left(a, g\right) \times A_i \tag{2.12} \end{equation}$

Combining equations (2.11) and (2.12), one finds that:

$\begin{equation} \lambda \times \left(\frac{\partial f}{\partial \lambda}\right)_{a, A} = m \times f\left(a, \lambda A\right) \tag{2.13} \end{equation}$

If one separates the variables and performs integration:

$\begin{equation} \int_{f\left(a, A\right)}^{f\left(a, \lambda A\right)} \frac{df}{f} = m \times \int_{1}^{\lambda} \frac{d \lambda}{\lambda} \tag{2.14} \end{equation}$

After integration:

$\begin{equation} \ln\left[\frac{f\left(a, \lambda A\right)}{f\left(a, A\right)}\right] = m \times \ln\left(\frac{\lambda}{1}\right) \tag{2.15} \end{equation}$

Finally, one obtains that the $$f$$ function satisfies the homogeneity condition:

$\begin{equation} f\left(a, \lambda A\right) = \lambda^m \times f\left(a, A\right) \tag{2.16} \end{equation}$

# 3 Higher degree Euler theorem

As seen in equation (2.1), Euler’s theorem is primarily intended to give a mathematical relation between the first derivatives of the $$f$$ function with respect to its scalable variables and the $$f$$ function. As we will see, it is possible to derive a mathematical constraint involving the $$f$$ function higher derivatives.

Let us start by deriving equation (2.1) with respect to variable $$A_j$$:

$\begin{equation} \sum_{i \neq j} \left(\frac{\partial^2 f}{\partial A_j \partial A_i}\right)_{a, A_{j \neq i}} A_i + \left(\frac{\partial^2 f}{\partial A_i^2}\right)_{a, A_{j \neq i}} A_i + \left(\frac{\partial f}{\partial A_i}\right)_{a, A_{j \neq i}} = m \times \left(\frac{\partial f}{\partial A_i}\right) _{a, A_{j \neq i}} \tag{3.1} \end{equation}$

After some rearrangements, one obtains:

$\begin{equation} \sum_{i} \left(\frac{\partial^2 f}{\partial A_j \partial A_i}\right)_{a, A_{j \neq i}} A_i = \left(m-1\right) \times \left(\frac{\partial f}{\partial A_j}\right) _{a, A_{i \neq j}} \tag{3.2} \end{equation}$

Because of Euler’s theorem, the functions $$\left(\frac{\partial f}{\partial A_j}\right)_{a, A_{i \neq j}}$$ are homogeneous of degree $$m-1$$.