Problem 16 - The Calculus with Analytic Geometry

Question

Prove that any function can be expressed as the sum of an even function and an odd function by writing $$ f(x)=\frac{1}{2}[f(x)+f(-x)]+\frac{1}{2}[f(x)-f(-x)] $$ and showing that the function having function values \(\frac{1}{2}[f(x)+f(-x)]\) is an even function and the function having function values \(\frac{1}{2}[f(x)-f(-x)]\) is an odd function.

Step-by-Step Solution

Answer

Any function can be expressed as \(f(x) = \frac{1}{2}[f(x) + f(-x)] + \frac{1}{2}[f(x) - f(-x)]\) where \(\frac{1}{2}[f(x) + f(-x)]\) is even and \(\frac{1}{2}[f(x) - f(-x)]\) is odd.

Step 1 - Decompose the function

Express the function as the sum of two parts: \(f(x) = \frac{1}{2}[f(x) + f(-x)] + \frac{1}{2}[f(x) - f(-x)]\)

Step 2 - Identify the even function part

Consider \(g(x) = \frac{1}{2}[f(x)+f(-x)]\). Let's check if \(g(x)\) is an even function by evaluating \(g(-x)\). Substitute \(-x\) into \(g(x)\): \(g(-x) = \frac{1}{2}[f(-x) + f(x)] = \frac{1}{2}[f(x) + f(-x)] = g(x)\). Hence, \(g(x)\) is an even function.

Step 3 - Identify the odd function part

Consider \(h(x) = \frac{1}{2}[f(x)-f(-x)]\). Let's check if \(h(x)\) is an odd function by evaluating \(h(-x)\). Substitute \(-x\) into \(h(x)\): \(h(-x) = \frac{1}{2}[f(-x) - f(x)] = -\frac{1}{2}[f(x) - f(-x)] = -h(x)\). Hence, \(h(x)\) is an odd function.

Step 4 - Combine the parts

Since \(g(x)\) is even and \(h(x)\) is odd, \(f(x)\) can be expressed as the sum of \(g(x)\) and \(h(x)\). Thus, \(f(x) = g(x) + h(x)\) or \(f(x) = \frac{1}{2}[f(x) + f(-x)] + \frac{1}{2}[f(x) - f(-x)]\) where \(g(x)\) is an even function and \(h(x)\) is an odd function.

Key Concepts

Even Functions

In mathematics, an even function is a function that satisfies the property \(f(x) = f(-x)\). This means the function's graph is symmetric with respect to the y-axis. To check if a function is even, you can substitute \(-x\) for \(x\) in the function and see if the function remains unchanged. For example, let's consider \(g(x) = \frac{1}{2}[f(x) + f(-x)]\). Substituting \(-x\) into \(g(x)\): \(g(-x) = \frac{1}{2}[f(-x) + f(x)] = \frac{1}{2}[f(x) + f(-x)] = g(x)\). Hence, \(g(x)\) is an even function because \(g(-x) = g(x)\). This type of symmetry can be found in many everyday applications such as in physics and engineering.

Odd Functions

An odd function is a function that satisfies the property \(f(-x) = -f(x)\). This implies the function's graph is symmetric with respect to the origin. To verify if a function is odd, you can replace \(x\) with \(-x\) and check if you get the negative of the original function. For instance, consider \(h(x) = \frac{1}{2}[f(x) - f(-x)]\). If we substitute \(-x\) into \(h(x)\): \(h(-x) = \frac{1}{2}[f(-x) - f(x)] = -\frac{1}{2}[f(x) - f(-x)] = -h(x)\). Therefore, \(h(x)\) is an odd function because \(h(-x) = -h(x)\). Odd functions often appear in scenarios related to alternating currents and waveforms.

Function Properties

Functions can be described by several properties. Two crucial types of properties are those related to even and odd functions. Here are some key points to remember:

• Even Functions: Symmetric about the y-axis. \(f(x) = f(-x)\).
• Odd Functions: Symmetrical about the origin. \(f(-x) = -f(x)\).
• Any function can be decomposed into the sum of an even function and an odd function. For any function \(f(x)\), we can write \(f(x) = \frac{1}{2} [f(x) + f(-x)] + \frac{1}{2} [f(x) - f(-x)]\).

Understanding these properties allows for the analysis, simplification, and synthesis of more complex functions. They can be particularly useful in calculus and signal processing.

Calculus Proof

The decomposition of any function into an even and an odd function can be proven using calculus. Consider the function \(f(x)\). It can be split into two functions as follows:

• \(g(x) = \frac{1}{2}[f(x)+f(-x)]\), proving that \(g(x)\) is even: \(g(-x) = \frac{1}{2}[f(-x)+f(x)] = g(x)\).
• \(h(x) = \frac{1}{2}[f(x)-f(-x)]\), proving that \(h(x)\) is odd: \(h(-x) = \frac{1}{2}[f(-x)-f(x)] = -h(x)\).

The function \(f(x)\) can thus be expressed as the sum of these two parts: \(f(x) = g(x) + h(x)\) or more explicitly: \(f(x) = \frac{1}{2}[f(x) + f(-x)] + \frac{1}{2}[f(x) - f(-x)]\). This separation helps in analyzing complex functions and solving differential equations more effectively.