Even and Odd Functions

Even Function

An even function is defined as a function \( f(x) \) that satisfies the condition:

\[ f(-x) = f(x), \quad \text{for all } x \text{ in the domain of } f. \]

For example, consider the function \( f(x) = x^3 \sin(x) \). To determine whether \( f(x) \) is even, replace \( x \) with \( -x \):

\[ f(-x) = (-x)^3 \sin(-x). \]

Using the properties of exponents and the sine function:

\[ (-x)^3 = -x^3 \quad \text{and} \quad \sin(-x) = -\sin(x), \]

we get:

\[ f(-x) = (-x^3)(-\sin(x)) = x^3 \sin(x). \]

Thus, \( f(-x) = f(x) \), confirming that \( f(x) = x^3 \sin(x) \) is an even function.

In general, to check whether a function \( f(x) \) is even, replace \( x \) with \( -x \) in \( f(x) \). If \( f(-x) = f(x) \), the function is even. If \( f(-x) \neq f(x) \), the function is not even.

The symmetry of even functions can be understood geometrically. For an even function, the output at a point \( a \) is \( f(a) \), and the output at \( -a \) is \( f(-a) = f(a) \). This implies that the points \( (a, f(a)) \) and \( (-a, f(a)) \) lie on the graph of \( f(x) \). Geometrically, \( (-a, f(a)) \) is the image of \( (a, f(a)) \) when reflected across the y-axis. Since this is true for any point \( a \) in the domain of \( f(x) \), the graph of \( f(x) \) must be symmetric about the y-axis.

To illustrate this, consider the function \( f(x) = x^3 \sin(x) \), which is symmetric about the y-axis. Below is its graph.

The graph above illustrates the function \( f(x) = x^3 \sin(x) \). It is symmetric about the y-axis, which confirms that \( f(x) \) is an even function. The symmetry occurs because for any point \( (a, f(a)) \), there exists a corresponding point \( (-a, f(a)) \), satisfying the condition \( f(-x) = f(x) \). This symmetry is a fundamental characteristic of all even functions.

Odd Function

An odd function is one where the rule \( f(-x) = -f(x) \) holds for all \( x \) in the domain. To see this in action, take \( f(x) = x^2 \sin(x) \). Replace \( x \) with \( -x \):

\[ f(-x) = (-x)^2 \sin(-x) = x^2(-\sin(x)) = -x^2 \sin(x). \]

Clearly, \( f(-x) = -f(x) \), so \( f(x) = x^2 \sin(x) \) is odd.

You can check for odd functions in general by replacing \( x \) with \( -x \) in \( f(x) \). If the result is \( -f(x) \), the function is odd. Simple.

Odd functions have a distinct symmetry. The output at \( a \) is \( f(a) \), and the output at \( -a \) is \( f(-a) = -f(a) \). This means \( (a, f(a)) \) and \( (-a, -f(a)) \) both lie on the graph. Geometrically, \( (-a, -f(a)) \) is the image of \( (a, f(a)) \) when reflected about the origin. So, the graph of an odd function always has origin symmetry.

For instance, look at \( f(x) = x^2 \sin(x) \). The graph below shows its symmetry about the origin, reinforcing that it’s odd.

The graph of \( f(x) = x^2 \sin(x) \) confirms its symmetry about the origin, a hallmark of odd functions. For every point \( (a, f(a)) \), there exists a corresponding point \( (-a, -f(a)) \), showing that \( f(-x) = -f(x) \). This origin symmetry makes \( f(x) \) an odd function.

For an odd function \( f(x) \), if it is defined at \( x = 0 \), then its value must be \( f(0) = 0 \). This can be explained as follows:

An odd function satisfies the property:

\[ f(-x) = -f(x) \quad \text{for all } x \text{ in its domain.} \]

Now, substitute \( x = 0 \) into this equation:

\[ f(-0) = -f(0). \]

Since \( -0 = 0 \), the equation becomes:

\[ f(0) = -f(0). \]

This implies:

\[ 2f(0) = 0 \quad \implies \quad f(0) = 0. \]

Thus, if an odd function is defined at \( x = 0 \), its value must be zero because it is the only value that satisfies the symmetry property of odd functions.

Geometrically, this makes sense because odd functions are symmetric about the origin. At \( x = 0 \), the point \( (0, f(0)) \) must align with itself under origin symmetry. The only way this can happen is if \( f(0) = 0 \), making the point \( (0, 0) \) lie on the graph of the function.

Properties of Even and Odd Functions with Respect to Algebra of Functions

Let \( f(x) \) and \( g(x) \) be two functions defined on a common domain. A function \( f(x) \) is even if \( f(-x) = f(x) \), and odd if \( f(-x) = -f(x) \). Using these definitions, we rigorously establish the behavior of even and odd functions under various operations.

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Addition and Subtraction

Let \( h(x) = f(x) + g(x) \).

1. If \( f(x) \) and \( g(x) \) are both even, then:

   \[ h(-x) = f(-x) + g(-x) = f(x) + g(x) = h(x). \]

   Thus, the sum (or difference) of two even functions is even.

2. If \( f(x) \) and \( g(x) \) are both odd, then:

   \[ h(-x) = f(-x) + g(-x) = -f(x) - g(x) = -(f(x) + g(x)) = -h(x). \]

   Thus, the sum (or difference) of two odd functions is odd.

3. If \( f(x) \) is even and \( g(x) \) is odd, then:

   \[ h(-x) = f(-x) + g(-x) = f(x) - g(x). \]

   Since \( h(-x) \neq h(x) \) and \( h(-x) \neq -h(x) \) in general, the result is neither even nor odd unless one of \(f\) or \(g\) is a zero fucntion.

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Multiplication

Let \( h(x) = f(x)g(x) \).

1. If \( f(x) \) and \( g(x) \) are both even, then:

   \[ h(-x) = f(-x)g(-x) = f(x)g(x) = h(x). \]

   Thus, the product of two even functions is even.

2. If \( f(x) \) and \( g(x) \) are both odd, then:

   \[ h(-x) = f(-x)g(-x) = (-f(x))(-g(x)) = f(x)g(x) = h(x). \]

   Thus, the product of two odd functions is even.

3. If \( f(x) \) is even and \( g(x) \) is odd, then:

   \[ h(-x) = f(-x)g(-x) = f(x)(-g(x)) = -f(x)g(x) = -h(x). \]

   Thus, the product of an even function and an odd function is odd.

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Division

Let \( h(x) = \frac{f(x)}{g(x)} \), assuming \( g(x) \neq 0 \).

1. If \( f(x) \) and \( g(x) \) are both even, then:

   \[ h(-x) = \frac{f(-x)}{g(-x)} = \frac{f(x)}{g(x)} = h(x). \]

   Thus, the quotient of two even functions is even.

2. If \( f(x) \) and \( g(x) \) are both odd, then:

   \[ h(-x) = \frac{f(-x)}{g(-x)} = \frac{-f(x)}{-g(x)} = \frac{f(x)}{g(x)} = h(x). \]

   Thus, the quotient of two odd functions is even.

3. If \( f(x) \) is even and \( g(x) \) is odd, then:

   \[ h(-x) = \frac{f(-x)}{g(-x)} = \frac{f(x)}{-g(x)} = -\frac{f(x)}{g(x)} = -h(x). \]

   Thus, the quotient of an even function and an odd function is odd.

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Composition

Let \( h(x) = f(g(x)) \).

1. If \( f(x) \) and \( g(x) \) are both even, then:

   \[ h(-x) = f(g(-x)) = f(g(x)) = h(x). \]

   Thus, the composition of two even functions is even.

2. If \( f(x) \) and \( g(x) \) are both odd, then:

   \[ h(-x) = f(g(-x)) = f(-g(x)) = -f(g(x)) = -h(x). \]

   Thus, the composition of two odd functions is odd.

3. If \( f(x) \) is even and \( g(x) \) is odd, then:

   \[ h(-x) = f(g(-x)) = f(-g(x)) = f(g(x)) = h(x). \]

   Thus, the composition of an even function with an odd function is even.

4. If \( f(x) \) is odd and \( g(x) \) is even, then:

   \[ h(-x) = f(g(-x)) = f(g(x)) = h(x). \]

   Thus, the composition of an odd function with an even function is even.

5. For the composition \( f(g(x)) \), if \( g(x) \) is an even function (i.e., \( g(-x) = g(x) \)), then \( f(g(x)) \) is always even, regardless of whether \( f(x) \) is odd, even, or neither.

   Proof:

   Let \( h(x) = f(g(x)) \). To check whether \( h(x) \) is even, we compute \( h(-x) \):

   \[ h(-x) = f(g(-x)). \]

   Since \( g(x) \) is even, \( g(-x) = g(x) \). Substituting this:

   \[ h(-x) = f(g(x)) = h(x). \]

   Thus, \( h(x) \) satisfies the condition \( h(-x) = h(x) \), and \( f(g(x)) \) is even.

   Example:

   Let \( f(x) = \sqrt{x} \) (neither even nor odd) and \( g(x) = \cos(x) \) (even). Then:

   \[ f(g(x)) = \sqrt{\cos(x)}. \]

   For \( h(x) = \sqrt{\cos(x)} \), we compute \( h(-x) \):

   \[ h(-x) = \sqrt{\cos(-x)}. \]

   Since \( \cos(-x) = \cos(x) \), we have:

   \[ h(-x) = \sqrt{\cos(x)} = h(x). \]

   Thus, \( h(x) = \sqrt{\cos(x)} \) is even.

Decomposition of a Function into Even and Odd Parts

Any function \( f(x) \) can be expressed as the sum of an even function and an odd function. This decomposition is given by:

\[ f(x) = f_E(x) + f_O(x), \]

where:

\[ f_E(x) = \frac{f(x) + f(-x)}{2} \quad \text{and} \quad f_O(x) = \frac{f(x) - f(-x)}{2}. \]

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Why Does This Work?

To understand why any function \( f(x) \) can be expressed as the sum of an even function \( f_E(x) \) and an odd function \( f_O(x) \), let us verify the decomposition:

\[ f(x) = f_E(x) + f_O(x), \]

where \( f_E(x) = \frac{f(x) + f(-x)}{2} \) and \( f_O(x) = \frac{f(x) - f(-x)}{2} \).

First, consider the even part \( f_E(x) \). To verify that it is indeed even, compute \( f_E(-x) \):

\[ f_E(-x) = \frac{f(-x) + f(-(-x))}{2} = \frac{f(-x) + f(x)}{2}. \]

Since the right-hand side simplifies back to \( f_E(x) \), we conclude that \( f_E(x) \) satisfies the condition \( f_E(-x) = f_E(x) \), making it even.

Next, examine the odd part \( f_O(x) \). To verify that it is odd, compute \( f_O(-x) \):

\[ f_O(-x) = \frac{f(-x) - f(-(-x))}{2} = \frac{f(-x) - f(x)}{2}. \]

Factoring out a negative sign, we get:

\[ f_O(-x) = -\frac{f(x) - f(-x)}{2} = -f_O(x). \]

Thus, \( f_O(x) \) satisfies the condition \( f_O(-x) = -f_O(x) \), confirming that it is odd.

Finally, consider the sum of the parts \( f_E(x) + f_O(x) \). Substituting their definitions:

\[ f_E(x) + f_O(x) = \frac{f(x) + f(-x)}{2} + \frac{f(x) - f(-x)}{2}. \]

Combining terms, we find:

\[ f_E(x) + f_O(x) = \frac{2f(x)}{2} = f(x). \]

This shows that \( f(x) \) can indeed be written as the sum of its even part \( f_E(x) \) and its odd part \( f_O(x) \). The decomposition works because the even and odd parts collectively account for both the symmetric and antisymmetric properties of \( f(x) \).

Extension and Restriction of a Function

Sometimes, we define a function \( f(x) \) on a specific domain, say \( X \). For practical reasons, we may want to extend this function to a larger domain \( A \), ensuring that the extended function \( g(x) \) behaves consistently with \( f(x) \) on \( X \). On the other hand, if a function \( g(x) \) is defined on a larger domain \( A \), we might want to restrict it to a smaller subset \( X \), focusing only on part of its behavior. These ideas—extension and restriction—allow us to carefully expand or limit the scope of a function while maintaining mathematical rigor.

Definition

Let \( f : X \to Y \) be a function, where \( X \subseteq A \) and \( Y \subseteq B \). We define extension and restriction as follows:

1. Extension:

   A function \( g : A \to B \) is said to be an extension of \( f \) if:

   \[ g(x) = f(x) \quad \text{for all } x \in X. \]

   This means \( g(x) \) agrees with \( f(x) \) on the original domain \( X \) and may define \( g(x) \) for \( x \in A \setminus X \).

2. Restriction:

   Conversely, \( f(x) \) is a restriction of \( g(x) \) to \( X \) if:

   \[ f(x) = g(x) \quad \text{for all } x \in X. \]

   In this case, \( f(x) \) inherits its values and behavior from \( g(x) \), but only on the smaller domain \( X \).

If \( g(x) \) is an extension of \( f(x) \), then \( f(x) \) is a restriction of \( g(x) \), and vice versa.

Example

Let \( f : [-\pi, \pi] \to \mathbb{R} \) be a function defined as \( f(x) = |x| \). Consider the function \( g : \mathbb{R} \to \mathbb{R} \), defined by:

\[ g(x) = \cos^{-1}(\cos x). \]

1. On \( x \in [-\pi, \pi] \), we compute:

   \[ g(x) = \cos^{-1}(\cos x) = |x|, \]

   since \( \cos^{-1}(\cos x) \) reduces to the principal value, which equals \( |x| \) in this interval. Thus, \( g(x) = f(x) \) for \( x \in [-\pi, \pi] \).

2. For \( x \notin [-\pi, \pi] \), the function \( g(x) = \cos^{-1}(\cos x) \) extends \( f(x) \) to all \( x \in \mathbb{R} \), leveraging the periodicity of \( \cos x \).

Since \( g(x) \) agrees with \( f(x) \) on \( [-\pi, \pi] \) and defines values beyond this domain, \( g(x) \) is an extension of \( f(x) \). Conversely, \( f(x) \) is the restriction of \( g(x) \) to \( [-\pi, \pi] \).

The concepts of extension and restriction are fundamental when we deal with:

1. Expanding a function's domain while maintaining consistency in its behavior.
2. Focusing on a subset of a function's domain to study localized behavior.

Even Extension of a Function

Suppose there is a function \( f(x) \) defined on the interval \( [0, a] \). The even extension of \( f(x) \) to the interval \( [-a, a] \) is a new function \( f_E(x) \), defined as:

\[ f_E(x) = \begin{cases} f(x), & x \in [0, a], \\ f(-x), & x \in [-a, 0]. \end{cases} \]

This definition ensures that \( f_E(x) \) satisfies the property of even functions:

\[ f_E(-x) = f_E(x), \quad \text{for all } x \in [-a, a]. \]

• For \( x \in [0, a] \), \( f_E(x) \) is the same as \( f(x) \). This retains the original values of the function on the interval where it was initially defined.
• For \( x \in [-a, 0] \), \( f_E(x) = f(-x) \), which mirrors the values of \( f(x) \) from \( [0, a] \) into the negative part of the domain. This creates symmetry about the y-axis.

Since \( f_E(x) = f_E(-x) \), the function \( f_E(x) \) is an even function.

Geometric Interpretation

The even extension of \( f(x) \) reflects the graph of \( f(x) \) from \( [0, a] \) into \( [-a, 0] \) by flipping it across the y-axis. For \( x \in [-a, 0] \), the function value is \( f(-x) \), which corresponds to the reflection of \( f(x) \) in the y-axis.

Example

Given the function \( f(x) = x - x^3 \) defined on \( [0, 2] \), find its even extension \( f_E(x) \) on the interval \( [-2, 2] \).

Solution:

The even extension of \( f(x) \) is defined as:

\[ f_E(x) = \begin{cases} f(x), & x \in [0, 2], \\ f(-x), & x \in [-2, 0]. \end{cases} \]

1. For \( x \in [0, 2] \), the function remains the same:

   \[ f_E(x) = f(x) = x - x^3. \]

2. For \( x \in [-2, 0] \), the function is reflected about the y-axis:

   \[ f_E(x) = f(-x) = (-x) - (-x)^3 = -x + x^3. \]

Thus, the even extension \( f_E(x) \) is:

\[ f_E(x) = \begin{cases} x - x^3, & x \in [0, 2], \\ -x + x^3, & x \in [-2, 0]. \end{cases} \]

Odd Extension of a Function

Suppose there is a function \( f(x) \) defined on the interval \( [0, a] \). The odd extension of \( f(x) \) to the interval \( [-a, a] \) is a new function \( f_O(x) \), defined as:

\[ f_O(x) = \begin{cases} f(x), & x \in [0, a], \\ -f(-x), & x \in [-a, 0]. \end{cases} \]

Geometric Interpretation

The odd extension of \( f(x) \) reflects the graph of \( f(x) \) from \( [0, a] \) into \( [-a, 0] \) by flipping it across the origin. For \( x \in [-a, 0] \), the function value is \( -f(-x) \), which corresponds to the reflection of \( f(x) \) in the origin.

Example

Let \( f(x) = x^2 \), defined on \( [0, 2] \). The odd extension \( f_O(x) \) is:

\[ f_O(x) = \begin{cases} x^2, & x \in [0, 2], \\ -(-x)^2, & x \in [-2, 0]. \end{cases} \]

Simplifying:

\[ f_O(x) = \begin{cases} x^2, & x \in [0, 2], \\ -x^2, & x \in [-2, 0]. \end{cases} \]

The resulting function \( f_O(x) \) is symmetric about the origin.

Even Periodic Extension of a Function

The even periodic extension of a function \( f(x) \), originally defined on the interval \( [0, a] \), creates a function \( f_{EP}(x) \) that is both even and periodic with a period of \( 2a \). It is constructed by reflecting \( f(x) \) symmetrically about the y-axis in the interval \( [-a, a] \) and repeating this pattern periodically across the entire real line.

Formally, the even periodic extension \( f_{EP}(x) \) is defined as:

\[ f_{EP}(x) = \begin{cases} f(x), & x \in [0, a], \\ f(-x), & x \in [-a, 0], \end{cases} \]

with the periodic property:

\[ f_{EP}(x + 2a) = f_{EP}(x), \quad \text{for all } x \in \mathbb{R}. \]

For \( x \in [0, a] \), \( f_{EP}(x) = f(x) \), preserving the original function on its domain. For \( x \in [-a, 0] \), \( f_{EP}(x) = f(-x) \), ensuring even symmetry about the y-axis. This construction makes \( f_{EP}(x) \) repeat symmetrically and periodically with period \( 2a \).

For example, consider \( f(x) = x \), defined on \( [0, 1] \). The even periodic extension of \( f(x) \) is:

\[ f_{EP}(x) = \begin{cases} x, & x \in [0, 1], \\ -x, & x \in [-1, 0], \end{cases} \]

which satisfies \( f_{EP}(x + 2) = f_{EP}(x) \) for all \( x \). This results in a trnagular wave pattern that is symmetric about the y-axis and repeats with a period of 2.

Odd Periodic Extension of a Function

The odd periodic extension of a function \( f(x) \), originally defined on the interval \( [0, a] \), constructs a new function \( f_{OP}(x) \) that is both odd and periodic, with a period of \( 2a \). It is defined as:

\[ f_{OP}(x) = \begin{cases} f(x), & x \in [0, a], \\ -f(-x), & x \in [-a, 0], \end{cases} \]

and satisfies the periodicity condition:

\[ f_{OP}(x + 2a) = f_{OP}(x), \quad \text{for all } x \in \mathbb{R}. \]

For \( x \in [0, a] \), the function \( f_{OP}(x) \) retains the values of the original function \( f(x) \). For \( x \in [-a, 0] \), the function is defined as \( f_{OP}(x) = -f(-x) \), ensuring odd symmetry about the origin. This symmetry means that for any \( x \), \( f_{OP}(-x) = -f_{OP}(x) \). The periodicity ensures that the pattern repeats every \( 2a \), creating a fully odd and periodic extension of the original function.

Example: Odd Periodic Extension of \( f(x) = \sqrt{x - x^2} \) on \( [0, 1] \)

Let \( f(x) = x - x^2 \) be defined on \( [0, 1] \). The odd periodic extension \( f_{OP}(x) \) is constructed as:

\[ f_{OP}(x) = \begin{cases} f(x) = x - x^2, & x \in [0, 1], \\ -f(-x) = -(-x - (-x)^2) = x + x^2, & x \in [-1, 0]. \end{cases} \]

The odd periodicity ensures that for all \( x \in \mathbb{R} \), \( f_{OP}(x + 2) = f_{OP}(x) \).

Thus, \( f_{OP}(x) \) is defined for one period \( [-1, 1] \) as:

\[ f_{OP}(x) = \begin{cases} x - x^2, & x \in [0, 1], \\ x + x^2, & x \in [-1, 0]. \end{cases} \]

This pattern repeats with a period of 2, ensuring both odd symmetry (\( f_{OP}(-x) = -f_{OP}(x) \)) and periodicity.