Problem 77 - Precalculus Mathematics for Calculus

Question

Recall that a function \(f\) is odd if \(f(-x)=-f(x)\) or even if \(f(-x)=f(x)\) for all real \(x\). (a) Show that a polynomial \(P(x)\) that contains only odd powers of \(x\) is an odd function. (b) Show that a polynomial \(P(x)\) that contains only even powers of \(x\) is an even function. (c) Show that if a polynomial \(P(x)\) contains both odd and even powers of \(x,\) then it is neither an odd nor an even function. (d) Express the function $$P(x)=x^{5}+6 x^{3}-x^{2}-2 x+5$$ as the sum of an odd function and an even function.

Step-by-Step Solution

Answer

(a) Odd; (b) Even; (c) Neither; (d) \(P(x) = (x^5 + 6x^3 - 2x) + (-x^2 + 5)\).

Step 1: Analyze Odd Powers Polynomial

Let the polynomial be \(P(x) = a_n x^n + a_{n-2} x^{n-2} + \ldots + a_1 x\). Each power of \(x\) is an odd number. By definition, a function is odd if \(f(-x) = -f(x)\). Substitute \(-x\) into the polynomial to get \(P(-x) = a_n (-x)^n + a_{n-2} (-x)^{n-2} + \ldots + a_1 (-x)\). Each power remains odd, so \((-x)^n = -(x^n)\). Thus, \(P(-x) = -[a_n x^n + a_{n-2} x^{n-2} + \ldots + a_1 x] = -P(x)\). Hence, \(P(x)\) is an odd function.

Step 2: Analyze Even Powers Polynomial

Let the polynomial be \(P(x) = a_n x^n + a_{n-2} x^{n-2} + \ldots + a_0\), where each power is even. By definition, \(f(-x) = f(x)\) for an even function. Substitute \(-x\) to get \(P(-x) = a_n (-x)^n + a_{n-2} (-x)^{n-2} + \ldots + a_0\). For even powers, \((-x)^n = x^n\). Thus, \(P(-x) = a_n x^n + a_{n-2} x^{n-2} + \ldots + a_0 = P(x)\). Hence, \(P(x)\) is an even function.

Step 3: Mixed Powers Analysis

Consider polynomial \(P(x)\) that includes both odd and even powers: \(a_n x^n + a_{n-1} x^{n-1} + \ldots + a_0\). Test \(P(-x)\). The odd power terms will satisfy \((-)\) sign change, while even power terms will not change. Therefore, \(P(-x)\) will result neither solely negative nor positive of \(P(x)\), confirming it is neither odd nor even.

Step 4: Express \(P(x)\) as Sum of Odd and Even Functions

Given \(P(x) = x^5 + 6x^3 - x^2 - 2x + 5\). The odd part \(P_{odd}(x)\) takes odd powers: \(x^5 + 6x^3 - 2x\). The even part \(P_{even}(x)\) takes even powers: \(-x^2 + 5\). Verify: \(P_{odd}(-x) = -x^5 - 6x^3 + 2x = -P_{odd}(x)\) and \(P_{even}(-x) = -(-x)^2 + 5 = -x^2 + 5 = P_{even}(x)\). Thus, \(P(x) = P_{odd}(x) + P_{even}(x)\).

Key Concepts

Polynomials

Polynomials are mathematical expressions consisting of variables and coefficients, combined through operations of addition, subtraction, and multiplication. A polynomial is usually expressed in the form:

• \(P(x) = a_n x^n + a_{n-1} x^{n-1} + \ldots + a_1 x + a_0\)

where \(a_n, a_{n-1}, \ldots, a_0\) are constants (coefficients) and \(n\) represents the degree of the polynomial, which is determined by the highest power of \(x\).
Polynomials can contain either odd or even powers of \(x\), or both. Each polynomial behaves differently depending on these powers and their combination.

Odd and Even Powers

A polynomial with only odd powers of \(x\) might look like \(x^3 + 2x\). In contrast, one with only even powers might look like \(x^4 + 3x^2 + 1\). When both odd and even powers are present, such as in \(x^5 + x^4 + x\), different function properties emerge, influencing whether the polynomial is classified as odd, even, or neither.

Function Properties

The symmetry of functions determines whether they are classified as odd or even. These properties are crucial in understanding their behavior and how they simplify many mathematical processes.

Odd Functions

A function \(f(x)\) is said to be odd if it satisfies the condition:

• \(f(-x) = -f(x)\)

for all real values of \(x\). Graphically, this means the function is symmetric about the origin.
For polynomials, if every term's power of \(x\) is odd, as in \(x^3 + x\), it typically results in an odd function.

Even Functions

A function is even if:

• \(f(-x) = f(x)\)

This indicates that the function is symmetric about the y-axis. If all terms in a polynomial have even powers, such as \(x^2 + 4\), the polynomial is generally even.

Neither Odd Nor Even

When a polynomial contains a mix of odd and even powers, as in \(x^3 + x^2\), it usually does not exhibit symmetry about the origin or y-axis, thus it is neither odd nor even.

Mathematical Proofs

Proving the properties of odd and even polynomials involves verifying their symmetry by substitution and transformation. These proofs are foundational in understanding these function types and their real-world applications.

Analyzing Power-Based Symmetry

To prove a polynomial is odd, replace \(x\) with \(-x\) in the polynomial and show that

• \(P(-x) = -P(x)\).

For instance, for \(P(x) = x^3 + x\), substituting \(-x\) gives \(-x^3 - x\), which equals \(-P(x)\). Thus, it is odd.
For an even polynomial, the verification is

• \(P(-x) = P(x)\).

As with \(P(x) = x^2 + 5\), substituting gives \((-x)^2 + 5 = x^2 + 5\). This proves it is even.

Expressing Polynomials as Sum

A polynomial like \(P(x) = x^5 + 6x^3 - x^2 - 2x + 5\) can be decomposed into odd and even functions. Separate the odd power terms into one group and even power terms into another:

• Odd Part: \(x^5 + 6x^3 - 2x\)
• Even Part: \(-x^2 + 5\)

This division highlights the distinct symmetry properties while retaining the polynomial's overall shape, aiding in analysis and other mathematical operations.