Problem 55 - Calculus with Concepts in Calculus

Question

Let \(f\) be a function, and let \(g(x)=\frac{1}{2}[f(x)+f(-x)]\) and \(h(x)=\frac{1}{2}[f(x)-f(-x)]\) a. Show that \(g\) is an even function. b. Show that \(h\) is an odd function. c. Show that \(f=g+h .\) (Thus every function can be written as the sum of an even and an odd function.)

Step-by-Step Solution

Answer

a. \(g(x)\) is even. b. \(h(x)\) is odd. c. \(f(x) = g(x) + h(x)\).

Step 1: Define an Even Function

A function \( g(x) \) is considered even if for all \( x \), \( g(-x) = g(x) \). This symmetry means the function behaves identically whether you input \( x \) or \( -x \).

Step 2: Prove g(x) is Even

Substitute \(-x\) into \(g(x)\): \[ g(-x) = \frac{1}{2}[f(-x) + f(-(-x))] = \frac{1}{2}[f(-x) + f(x)]. \] This equals \(g(x)\). Thus, \(g(x)\) is even.

Step 3: Define an Odd Function

A function \( h(x) \) is odd if for all \( x \), \( h(-x) = -h(x) \). This means the function is symmetric about the origin.

Step 4: Prove h(x) is Odd

Substitute \(-x\) into \(h(x)\): \[ h(-x) = \frac{1}{2}[f(-x) - f(-(-x))] = \frac{1}{2}[f(-x) - f(x)]. \] This is \(-\frac{1}{2}[f(x) - f(-x)] = -h(x)\). Thus, \(h(x)\) is odd.

Step 5: Express f as the Sum of g and h

Now consider \(g(x) + h(x)\): \[ g(x) + h(x) = \frac{1}{2}[f(x) + f(-x)] + \frac{1}{2}[f(x) - f(-x)] = f(x). \] So \(f(x) = g(x) + h(x)\).

Key Concepts

Function Symmetry

Understanding the concept of function symmetry can greatly help in identifying even and odd functions. In mathematics, symmetry refers to a precise and balanced correspondence in size and layout. For functions, two types of symmetry are particularly important: even symmetry and odd symmetry.

• Even functions have symmetrical graphs with respect to the y-axis. This means if you flip the graph over the y-axis, it looks the same.
• Odd functions have graphs that are symmetrical about the origin. If you rotate the graph 180 degrees about the origin, it remains unchanged.

To check whether a function is even, you look for equivalency between the function values of positive and negative inputs: \[ g(x) = g(-x) \]For example, the function \( g(x) = \frac{1}{2}[f(x) + f(-x)] \) proved to be even because substituting \(-x\) results in the original function form. Meanwhile, an odd function satisfies:\[ h(-x) = -h(x) \]This was shown in the function \( h(x) = \frac{1}{2}[f(x) - f(-x)] \), where the substitution results in the negation of the function itself, confirming its odd symmetry.

Function Decomposition

Function decomposition involves breaking down a complex function into simpler, more manageable parts. This method is not only useful for theoretical mathematics but also offers practical applications in fields like engineering and computer science.

• Decomposing a function helps in analyzing its properties more easily.
• It allows you to understand how different components of a function contribute to its overall behavior.

In the exercise, the function \( f(x) \) was decomposed into two parts: \( g(x) \) and \( h(x) \), which are even and odd functions respectively. This decomposition simplifies the exploration of the function's properties:\[ f(x) = g(x) + h(x) \]This idea shows that any function can be expressed as the sum of an even and an odd function. This fundamental property is pivotal in various mathematical analyses, including Fourier series and other areas where tracing symmetry behaviors is essential.

Algebraic Function Properties

Algebraic function properties are rules and concepts that define how functions behave under various operations. Grasping these properties allows deeper insights into the functionality and flexibility of equations. Understanding the behavior of even and odd functions is crucial:

• An even function \( g(x) \) will satisfy \( g(x) = g(-x) \), ensuring identical outputs for opposite inputs.
• An odd function \( h(x) \) fulfills \( h(-x) = -h(x) \), reflecting a change in sign when the input is flipped.

These properties naturally lead to further revelations, such as function decomposition, where a more complex function can be split into these two symmetrical components. Using these principles, you can identify and utilize the characteristics of functions more effectively. This understanding enhances your ability to manipulate and transform functions in algebraic expressions, solving complex mathematical problems.