Problem 75 - Thomas Calculus

Question

a. Show that if a function \(f\) is defined on an interval symmetric about the origin (so that \(f\) is defined at \(-x\) whenever it is defined at \(x )\) , then $$ f(x)=\frac{f(x)+f(-x)}{2}+\frac{f(x)-f(-x)}{2} $$ Then show that \((f(x)+f(-x)) / 2\) is even and that \((f(x)-f(-x)) / 2\) is odd. b. Equation (1) simplifies considerably if \(f\) itself is (i) even or (ii) odd. What are the new equations? Give reasons for your answers.

Step-by-Step Solution

Answer

The expression splits a function into even and odd components; if \( f \) is even the odd component is 0, and if \( f \) is odd the even component is 0.

Step 1: Express Function as Sum

The exercise requires us to express a function \( f(x) \) as a sum of two components. Let's begin by writing:\[f(x)=\frac{f(x)+f(-x)}{2}+\frac{f(x)-f(-x)}{2}.\]This formula expresses \( f(x) \) in terms of its symmetrically defined values at \( x \) and \( -x \).

Step 2: Prove Even Component

The expression \( \frac{f(x)+f(-x)}{2} \) is an even function. To prove this, consider evaluating at \( -x \):\[\frac{f(-x) + f(x)}{2} = \frac{f(x) + f(-x)}{2}.\]Thus, the function is identical when \( x \) is replaced by \( -x \), satisfying the condition of an even function.

Step 3: Prove Odd Component

Now, consider \( \frac{f(x)-f(-x)}{2} \), which is an odd function. To prove this, evaluate it at \( -x \):\[\frac{f(-x) - f(x)}{2} = -\frac{f(x) - f(-x)}{2}.\]The change of sign confirms that this component is an odd function.

Step 4: Simplification if \(f\) is Even

If \( f \) is an even function, then by definition \( f(x) = f(-x) \). Substitute this into the original equation:\[f(x) = \frac{f(x) + f(x)}{2} + \frac{f(x) - f(x)}{2} = f(x).\]Thus, for even \( f \), the formula simplifies to \( f(x) = f(x) \), confirming its evenness.

Step 5: Simplification if \(f\) is Odd

If \( f \) is an odd function, then by definition \( f(x) = -f(-x) \). Substitute this into the original equation:\[f(x) = \frac{f(x) - f(x)}{2} + \frac{f(x) + f(x)}{2} = f(x).\]Thus, for odd \( f \), the formula also simplifies to \( f(x) = f(x) \), confirming its oddness.

Key Concepts

Symmetry of Functions

Functions can exhibit different types of symmetry, which helps us understand their behavior better. Specifically, functions can be classified as even or odd depending on their symmetry characteristics.

• Even Functions: These functions have symmetry with respect to the y-axis. A function \( f(x) \) is even if \( f(-x) = f(x) \) for every \( x \) in the domain. A geometric reflection across the y-axis leaves their graph unchanged.
• Odd Functions: These functions have symmetry with respect to the origin. A function \( f(x) \) is odd if \( f(-x) = -f(x) \). Their graphs remain unchanged when rotated 180 degrees around the origin.

To utilize this knowledge in calculus, solving problems involving symmetric intervals becomes easier. In such problems, understanding how these symmetries affect the function's expression is crucial.

Function Decomposition

Decomposing a function involves breaking it down into simpler parts, making analysis easier. In the given exercise, the function \( f(x) \) is split into two parts: an even component and an odd component. This decomposition is expressed as:\[f(x) = \frac{f(x) + f(-x)}{2} + \frac{f(x) - f(-x)}{2}\]

• The first part, \( \frac{f(x) + f(-x)}{2} \), represents the even component.
• The second part, \( \frac{f(x) - f(-x)}{2} \), represents the odd component.

This decomposition helps in identifying inherent symmetries in the function, facilitating both theoretical analysis and practical calculus operations. By understanding the role of each component, students can gain insights into the behavior and properties of the function as a whole.

Mathematical Proof

Proof is a fundamental aspect of mathematics, providing a convincing and logical conclusion to support assertions. In the problem, the proof involves showing the even and odd nature of certain components.

• For the even component \( \frac{f(x) + f(-x)}{2} \), we show that substituting \(-x\) results in the original expression, confirming it remains unchanged, and thus is even.
• For the odd component \( \frac{f(x) - f(-x)}{2} \), substituting \(-x\) produces the negation of the original expression, confirming it changes signs and thus is odd.

Learning to perform these proofs sharpens analytical thinking and deepens understanding of function properties, both key skills in advanced mathematics.

Calculus Concepts

Calculus often deals with complex functions, and understanding the properties of even and odd functions is crucial. In particular, the behavior of these functions has implications on their derivatives and integrals, which are core calculus operations.

• Derivatives: The derivative of an even function is odd, and the derivative of an odd function is even. Understanding these patterns helps anticipate the behavior of derivatives.
• Integrals: When integrating over symmetric intervals, even functions simplify the computation by doubling the integral from 0 to a positive endpoint, while odd functions can yield zero due to symmetrical cancellation over a symmetric interval.

The exercise you observed highlights how function decomposition and symmetry can simplify calculus operations. By mastering these concepts, students can tackle more challenging calculus problems with ease.