Functional Equations
Functional Equation
A functional equation is an equation where the unknown is a function, and it describes a relationship between the values of the function at different points. For example, \( f(x + y) = f(x) + f(y) \) is a functional equation.
In this section, we will learn some basic functional equations and explore a few key ideas. It is a vast topic, and we will delve deeper into it in the chapters on derivatives and differential equations.
A functional equation contains information about the outputs of a function, where the function itself is unknown. These equations allow us to extract significant information about the behavior of the function.
For example, let a function \( f: \mathbb{R} \to \mathbb{R} \) satisfy the equation:
for all \( x, y \in \mathbb{R} \).
Since this is true for all \( x \) and \( y \), and \( x \) and \( y \) are independent, we can substitute any real numbers in place of \( x \) and \( y \).
Suppose we put \( x = 0 \) and \( y = 0 \). Then:
Simplifying:
This tells us that \( f(0) = 0 \). This is just the beginning; by substituting other values, we can gather more information about \( f \).
If we substitute \( y = x \) into the equation \( f(x + y) = f(x) + f(y) \), we get:
Next, if we substitute \( y = -x \), we get:
Since we already know that \( f(0) = 0 \), this simplifies to:
Thus, \( f \) is an odd function, meaning \( f(-x) = -f(x) \) for all \( x \in \mathbb{R} \).
By substituting different values into the functional equation, we continue to uncover key properties of the unknown function \( f \).
The functional equation \( f(x + y) = f(x) + f(y) \) is satisfied by a family of functions. If we assume that \( f \) is a differentiable function (don’t worry about this assumption for now), we will prove later in the chapter on differential equations that the general solution to this functional equation is:
where \( a \) is a constant.
This means it is not just one specific function, but a family of linear functions of the form \( f(x) = ax \). For example, functions like \( f(x) = 2x \), \( f(x) = -100x \), and \( f(x) = 0x \) all satisfy this equation.
Let us verify this result. Assume \( f(x) = ax \), where \( a \) is a constant. Substituting this into the functional equation:
Expanding:
Thus, the functional equation is satisfied, confirming that \( f(x) = ax \) is a solution. This verification demonstrates that all linear functions with constant slope satisfy the given functional equation.
Solving Functional Equations via Substitution
Some functional equations can be solved by applying simple substitutions to generate new equations, which, when combined with the original, help determine the function.
Example
Consider a function \( f: \mathbb{R} \to \mathbb{R} \) that satisfies:
We want to find \(f\) that satisfies this.
Substitute \( x \mapsto 1-x \) into the equation
Replacing \( x \) with \( 1-x \), we obtain:
Simplify \( 1-(1-x) \) to \( x \), giving:
We now have the original equation:
and the new equation:
Multiply equation (1) by 2 to align coefficients of \( f(x) \):
Divide by 3:
Let us also verify the solution. Substitute \( f(x) = \frac{x^2 - 4x + 2}{3} \) into the original equation:
Compute \( f(1-x) = \frac{(1-x)^2 - 4(1-x) + 2}{3} \):
Now check:
Simplify:
Thus, the solution satisfies the given functional equation.
Final Answer:
Example
Solving Functional Equations via Substitution
Consider the function \( f: \mathbb{R} \setminus \{0\} \to \mathbb{R} \) that satisfies the equation:
This equation relates \( f(x) \) to \( f\left(\frac{1}{x}\right) \). To uncover more information about the structure of \( f \), let us replace \( x \) by \( \frac{1}{x} \). Doing so, the equation becomes:
Now we have two equations involving \( f(x) \) and \( f\left(\frac{1}{x}\right) \):
From the first equation, solve explicitly for \( f\left(\frac{1}{x}\right) \):
Substitute this expression for \( f\left(\frac{1}{x}\right) \) into the second equation:
Thus, the function \( f(x) \) is determined to be: