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Euler’s Theorem for Homogeneous Functions

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Last updated date: 17th Apr 2024
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An Introduction to Euler’s Theorem

Euler’s theorem for Homogeneous Functions is used to derive a relationship between the product of the function with its degree and partial derivatives of it.


Euler’s Theorem


Euler’s Theorem


In this article, we will first discuss the statement of the theorem followed by the mathematical expression of Euler’s theorem and prove the theorem. We will also discuss the things for which Euler’s Theorem is used and is applicable. A brief history of mathematician Leonhard Euler will also be discussed after whom the theorem is named so.


History of Leonhard Euler


Image:


Leonhard Euler,


  • Name: Leonhard Euler

  • Born: 15 April 1707

  • Died: 18 September 1783


Statement of Euler’s Theorem

Euler's theorem states that if $(f$) is a homogeneous function of the degree $n$ of $k$ variables $x_{1}, x_{2}, x_{3}, \ldots \ldots, x_{k}$, then

$x_{1} \dfrac{\partial f}{\partial x_{1}}+x_{2} \dfrac{\partial f}{\partial x_{2}}+x_{3} \dfrac{\partial f}{\partial x_{3}}+\ldots \ldots+x_{k} \dfrac{\partial f}{\partial x_{k}}=n f$

Here, we will be discussing 2 variables only. So, if $f$ is a homogeneous function of degree $n$ of variables $x$ and $y$, then from Euler's Theorem, we get

$x \dfrac{\partial f}{\partial x}+y \dfrac{\partial f}{\partial y}=n f$


Proof of Euler’s Theorem

Proof:

Let $f=u[x, y]$ be a homogenous function of degree $n$ of the variables $x, y$.

$f=u[x, y] \ldots \ldots \ldots[1]$

Now, we know that

$u[X, Y]=t^{n} u[x, y] \ldots \ldots \ldots[2]$

This is because when $u$ is a function of $X, Y$, then it becomes a function of $x, y, t$ because $X, Y$ are a function of $t$.

This means that

$X=x t^{n} Y=y t^{n}$

Now, we will consider the above equations for degree 1.

$X=x t \ldots \ldots \ldots[3] \\$

$Y=y t \ldots \ldots \ldots[4]$

where $t$ is an arbitrary parameter.

Now, we will differentiate the equation [3] partially w.r.t. $t$.

$X=x t \\$

$\Rightarrow \dfrac{\partial}{\partial t} X=\dfrac{\partial}{\partial t} x t \\$

$\Rightarrow \dfrac{\partial X}{\partial t}=x \dfrac{\partial t}{\partial t} \\$

$\Rightarrow \dfrac{\partial X}{\partial t}=x \ldots \ldots \ldots[5]$

Again, we will differentiate the equation [4] partially w.r.t. $t$.

$Y=y t \\$

$\Rightarrow \dfrac{\partial}{\partial t} Y=\dfrac{\partial}{\partial t} y t \\$

$\Rightarrow \dfrac{\partial Y}{\partial t}=y \ldots \ldots \ldots[6]$

Now, we will substitute $t=1$ in equation [3].

$X=x t \\$

$\Rightarrow X=x[1]$

Multiplying the terms, we get

$X=x \ldots \ldots \ldots \ldots \text { [7] }$

Now, we will substitute $t=1$ in equation [4].

$Y=y t \\$

$\Rightarrow Y=y[1]$

Multiplying the terms, we get

$Y=y \ldots \ldots \ldots \ldots[8]$

We will now differentiate the equation [1] partially w.r.t. $x$.

$f=u[x, y] \\$

$\Rightarrow \dfrac{\partial}{\partial x} f=\dfrac{\partial}{\partial x} u[x, y] \\$

$\Rightarrow \dfrac{\partial f}{\partial x}=\dfrac{\partial u}{\partial x} \ldots \ldots \ldots \ldots .[9]$

Partial differentiation of $y$ w.r.t. $x$ is $\underline{\underline{0} \text {. }}$

Now, we will differentiate the equation [1] partially w.r.t. $y$.

$f=u[x, y] \\$

$\Rightarrow \dfrac{\partial}{\partial y} f=\dfrac{\partial}{\partial y} u[x, y] \\$

$\Rightarrow \dfrac{\partial f}{\partial y}=\dfrac{\partial u}{\partial y} \ldots \cdots \cdots[10]$

Partial differentiation of $x$ w.r.t. $y$ is 0 .

Now, we will differentiate the equation [2] partially w.r.t. $t$.

Since $u$ is a function of $X, Y$ and $X, Y$ are a function of $x, y$, and $t$, thus apply the chain rule of partial differentiation to differentiate.

$u[X, Y]=t^{n} u[x, y] \\$

$\Rightarrow \dfrac{\partial u}{\partial x} \cdot \dfrac{\partial X}{\partial t}+\dfrac{\partial u}{\partial y} \cdot \dfrac{\partial Y}{\partial t}=n t^{n-1} u[x, y] \ldots \ldots \ldots[\text { [11] }$

We will substitute $x$ for $\dfrac{\partial x}{\partial t}, y$ for $\dfrac{\partial Y}{\partial t}, \dfrac{\partial f}{\partial x}$ for $\dfrac{\partial u}{\partial x}, \dfrac{\partial f}{\partial y}$ for $\dfrac{\partial u}{\partial y}$ in equation [11]. Thus, the equation [11] becomes:

$\Rightarrow \dfrac{\partial u}{\partial x} \cdot \dfrac{\partial X}{\partial t}+\dfrac{\partial u}{\partial y} \cdot \dfrac{\partial Y}{\partial t}=n t^{n-1} u[x, y]$

Using the result of equations [7] and [8], we can write the above equation as

$\Rightarrow \dfrac{\partial f}{\partial x} \cdot x+\dfrac{\partial f}{\partial y} \cdot y=n t^{n-1} u[x, y] \ldots \ldots \ldots \text { [12] }$

We will again substitute $t=1$ and $f$ for $u[x, y]$ in the equation [12], we get

$\Rightarrow \dfrac{\partial f}{\partial x} \cdot x+\dfrac{\partial f}{\partial y} \cdot y=n[1]^{n-1} f \\$

$\Rightarrow \dfrac{\partial f}{\partial x} \cdot x+\dfrac{\partial f}{\partial y} \cdot y=n f$

Hence proved.


Limitations of Euler's Theorem


Application of Euler’s Theorem

  • Euler’s theorem has wide application in electronic devices which work on the AC principle.

  • Euler’s formula is used by scientists to perform various calculations and research.


Solved Examples

1. If $u(x, y)=\dfrac{x^{2}+y^{2}}{\sqrt{x+y}}$, prove that $x \dfrac{\partial u}{\partial x}+y \dfrac{\partial u}{\partial y}=\dfrac{3}{2} u$.

Ans:

Given $u(x, y)=\dfrac{x^{2}+y^{2}}{\sqrt{x+y}}$

We can say that

$\Rightarrow u(\lambda x, \lambda y)=\dfrac{\lambda^{2} x^{2}+\lambda^{2} y^{2}}{\sqrt{\lambda x+\lambda y}} \\$

$\Rightarrow u(\lambda x, \lambda y)=\dfrac{\lambda^{2}\left(x^{2}+y^{2}\right)}{\lambda^{1 / 2} \sqrt{x+y}} \\$

$\Rightarrow u(\lambda x, \lambda y)=\dfrac{\lambda^{3 / 2}\left(x^{2}+y^{2}\right)}{\sqrt{x+y}} u$

is a homogeneous function of degree $\dfrac{3}{2}$

By Euler's Theorem,

$x \dfrac{\partial u}{\partial x}+y \dfrac{\partial u}{\partial y}=\dfrac{3}{2} u$


2. If $v(x, y)=\log \left(\dfrac{x^{2}+y^{2}}{x+y}\right)$, prove that $x \dfrac{\partial v}{\partial x}+y \dfrac{\partial u}{\partial y}=1$

Ans:

Given that

$v(x, y)=\log \left(\dfrac{x^{2}+y^{2}}{x+y}\right)$

Change into an exponential function.

Let $e^{v}=\dfrac{x^{2}+y^{2}}{x+y}=f(x, y)$

$f(x, y)=\dfrac{\lambda^{2} x^{2}+\lambda^{2} y^{2}}{\lambda x+\lambda y}$

$\Rightarrow f(x, y)=\dfrac{\lambda^{2}\left(x^{2}+y^{2}\right)}{\lambda(x+y)} f$ is a homogeneous function of degree 1.

By Euler's Theorem,

$x \dfrac{\partial f}{\partial x}+y \dfrac{\partial f}{\partial y}=1 \times f$

$\Rightarrow f x \dfrac{\partial}{\partial x} e^{v}+y \dfrac{\partial}{\partial y} e^{v}=e^{v}$ exists.

$\Rightarrow x^{v} \dfrac{\partial f}{\partial x}+y e^{v} \dfrac{\partial f}{\partial y}=e^{v} \\$

$\Rightarrow x \dfrac{\partial f}{\partial x}+y \dfrac{\partial f}{\partial y}=\dfrac{e^{v}}{e^{v}}=1$

Hence Proved.


3. If $u=\tan ^{-1}(x+y)$, then $x \dfrac{\partial u}{\partial x}+y \dfrac{\partial u}{\partial y}=$

Ans:

Given that

$\tan u=x+y \tan u$ is a homogeneous of degree $=1$

$\sum_{l} \dfrac{x d \tan u}{d x}=n \tan u \sec ^{2} u \\$

$\Rightarrow \dfrac{x d u}{d x}=\tan u \\$

$\Rightarrow \sum_{a} \dfrac{x d u}{d x}=\dfrac{1}{2} \sin 2 u \\$

$\Rightarrow x \dfrac{\partial u}{\partial x}+y \dfrac{\partial u}{\partial y}=\dfrac{1}{2} \sin 2 u$


Important Formulas to Remember

The function $f(x, y)$ is said to be a homogeneous function if

$f(\lambda x, \lambda y)=\lambda^{n} f(x, y), \text { for any non zero constant } \lambda .$

Here, $n$ is the degree of homogeneous function $f$.

If $f$ is a homogeneous function of degree $n$ of variables $x$ and $y$, then

$x \dfrac{\partial f}{\partial x}+y \dfrac{\partial f}{\partial y}=n f$


Conclusion

Euler’s Theorem has a wide range of applications in daily life and it is a fundamental tool of algebra. In this article, we have discussed Euler's theorem and its proof along with its applications of it in daily life.

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FAQs on Euler’s Theorem for Homogeneous Functions

1. What is the difference between Euler Formula and Euler’s Formula for Homogeneous Equations? Discuss their similarity if any.

Euler's Formula explains the relationship between trigonometric functions and complex exponentials, i.e., $e^{i x}=\cos x+i \sin x$, where $\cos \mathrm{x}$ and $\sin \mathrm{x}$ represent the complex exponentials, whereas Euler's theorem for homogeneous equations explains that the homogeneous equation is the solution of the partial differential equation. Despite differences, both share a lot of similarities. Both form a fundamental part of calculus and have applications in Artificial Intelligence and research.

2. Is there any Euler Formula for cubes? If yes, is it related to Euler’s Formula for Homogeneous Equations?

Yes, there is an Euler formula for cube. No, it is not related to Euler’s formula for homogeneous equations. We have different Euler’s theorems and formulas and the Euler formula for cube is used in the case of complex polyhedron 3-D figures. Euler’s formula for cube is used to derive relationships between edges, vertices, and faces of 3-D figures. The relation is given as F+V=E+2, where F, E, and V represent the Faces, Edges, and Vertices, respectively, of the cube.

3. Is Euler’s Formula only applicable for 2 variable homogeneous equations?

No, Euler’s Formula can be applied for homogeneous equations in any number of independent variables. The only condition to satisfy Euler’s formula is that the equation should be homogenous and the variables should be independent of each other. Proof of the theorem for any number of variables can be obtained using the generalisation of proof for two independent variables which we have discussed in the above article. Proof using generalisation can be obtained using the principle of Principal Method Induction.