Odd functions and Even functions are distinguished by their symmetries: even functions are symmetric with respect to the y-axis, while odd functions are symmetric with respect to the origin. We will also explore the behavior of function sums: the sum of two even functions is still an even function, just as the sum of two odd functions is still odd. Finally, we will see how to decompose a function into its even part and its odd part.
- Even Functions
- Sum of two even functions
- Odd Functions
- Sum of two odd functions
- Functions that are neither even nor odd
- Integration over symmetric intervals
- Decomposition of a function into even and odd parts
- Proof of uniqueness of the decomposition
Even Functions
Definition. A function \( f : X \to Y \) is called even if:
\[ \forall x \in X, \quad -x \in X \quad \text{and} \quad f(-x) = f(x) \]
In other words, a function is even if its domain is symmetric with respect to the origin and the function takes the same value at \( x \) and \( -x \).
Example. The function \( f(x) = x^2 \) is even, since for every \( x \in \mathbb{R} \), we have \( (-x)^2 = x^2 \).
Example. The function \( f(x) = \cos(x) \) is even, since for every \( x \in \mathbb{R} \), we have \( \cos(-x) = \cos(x) \).
Example. The function \( f(x) = \cosh(x) \) is even, since for every \( x \in \mathbb{R} \), we have \( \cosh(-x) = \cosh(x) \).
Example. The function \( f(x) = e^{-x^2} \) is even, since for every \( x \in \mathbb{R} \), we have \( e^{-(-x)^2} = e^{-x^2} \).
Sum of two even functions
Let \( f \) and \( g \) be two even functions, defined on \( D_f \) and \( D_g \) respectively. Consider the sum function \( f + g \), defined on the domain \( D = D_f \cap D_g \). Since \( D_f \) and \( D_g \) are symmetric, their intersection \( D \) is also symmetric. For every \( x \in D \), we have:
\[ (f + g)(-x) = f(-x) + g(-x) = f(x) + g(x) = (f + g)(x) \]
therefore \( f + g \) is an even function.
An even function exhibits symmetry with respect to the y-axis. Therefore, to analyze its graph, it suffices to study it for \( x \geq 0 \); the portion for \( x < 0 \) is obtained by symmetry.
Odd Functions
Definition. A function \( f : X \to Y \) is called odd if:
\[ \forall x \in X, \quad -x \in X \quad \text{and} \quad f(-x) = -f(x) \]
In other words, a function is odd if its domain is symmetric with respect to the origin and the function takes opposite values at \( x \) and \( -x \).
Example. The function \( f(x) = \sin(x) \) is odd, since for every \( x \in \mathbb{R} \), we have \( \sin(-x) = -\sin(x) \).
Example. The function \( f(x) = x^3 \) is odd, since for every \( x \in \mathbb{R} \), we have \( (-x)^3 = -x^3 \).
An odd function is symmetric with respect to the origin: rotating its graph 180° around the origin yields the same graph.
Sum of two odd functions
Let \( f \) and \( g \) be two odd functions, defined on domains \( D_f \) and \( D_g \). Then the sum \( f + g \), defined on the common domain \( D = D_f \cap D_g \), is also odd.
Indeed, for every \( x \in D \), we have:
\[ (f + g)(-x) = f(-x) + g(-x) = -f(x) - g(x) = -(f(x) + g(x)) = -(f + g)(x) \]
Therefore \( f + g \) is an odd function.
Functions that are neither even nor odd
If a function satisfies neither the relation \( f(-x) = f(x) \) nor \( f(-x) = -f(x) \), then it is neither even nor odd.
Example. The functions \( f(x) = e^x \) and \( f(x) = x + 1 \) are neither even nor odd.
Integration over symmetric intervals
Even and odd functions are particularly useful in calculating definite integrals over intervals that are symmetric with respect to the origin.
If \( f \) is even, then:
\[ \int_{-a}^{a} f(x) \, dx = 2 \int_{0}^{a} f(x) \, dx \]
If \( f \) is odd, then:
\[ \int_{-a}^{a} f(x) \, dx = 0 \]
Decomposition of a function into even and odd parts
Let \( f : X \to \mathbb{R} \) be a function defined on a set that is symmetric with respect to the origin, that is, such that \( x \in X \implies -x \in X \). In this case, the function \( f \) can be written uniquely as the sum of an even function and an odd function.
Specifically, we define:
\[ f_e(x) = \frac{f(x) + f(-x)}{2} \quad \text{(even part of \( f \))} \]
\[ f_o(x) = \frac{f(x) - f(-x)}{2} \quad \text{(odd part of \( f \))} \]
It is straightforward to verify that \( f_e(-x) = f_e(x) \) and \( f_o(-x) = -f_o(x) \), so \( f_e \) is even and \( f_o \) is odd. Moreover, adding the two expressions gives:
\[ f(x) = f_e(x) + f_o(x) \]
that is, every function \( f \) (defined on a symmetric set) can be uniquely decomposed into the sum of an even function and an odd function.
Example. Consider the function \( f(x) = e^x \), defined on all of \( \mathbb{R} \). The even part is:
\[ f_e(x) = \frac{e^x + e^{-x}}{2} = \cosh x \]
while the odd part is:
\[ f_o(x) = \frac{e^x - e^{-x}}{2} = \sinh x \]
We thus have the decomposition:
\[ e^x = \cosh x + \sinh x \]
Remark: The decomposition is possible only if the domain of the function is symmetric with respect to the origin. Furthermore, the representation \( f = f_e + f_o \) is unique: there exists exactly one even function and one odd function whose sum gives \( f \).
Proof of uniqueness of the decomposition
Suppose that a function \( f : X \to \mathbb{R} \), defined on a set symmetric with respect to the origin, can be written in two distinct ways as the sum of an even function and an odd function:
\[ f(x) = f_e(x) + f_o(x) = g_e(x) + g_o(x), \]
where \( f_e \), \( g_e \) are even and \( f_o \), \( g_o \) are odd.
Then:
\[ f_e(x) - g_e(x) = g_o(x) - f_o(x) \]
Let \( h(x) = f_e(x) - g_e(x) = g_o(x) - f_o(x) \). The function \( h \) is therefore even (as the difference of two even functions) and odd (as the difference of two odd functions).
But if a function is simultaneously even and odd, then:
\[ h(x) = h(-x) \quad \text{and} \quad h(x) = -h(-x) \implies h(x) = -h(x) \]
that is, \( h(x) = 0 \) for every \( x \in X \).
Therefore \( f_e(x) = g_e(x) \) and \( f_o(x) = g_o(x) \), which proves the uniqueness of the decomposition.