Problem 86 - Calculus Single Variable

Question

Let \(p\) be any polynomial. a. Show that i. \(q(x)=(p(x)+p(-x)) / 2\) is an even polynomial, and ii. \(r(x)=(p(x)-p(-x)) / 2\) is an odd polynomial. Because \(p=q+r,\) this shows that every polynomial can be written as the sum of an even polynomial and an odd polynomial. b. Show that \(q\) contains only even powers of \(x,\) and \(r\) contains only odd powers. c. If \(p\) is even, deduce that the coefficient of each odd power of \(x\) in \(p(x)\) is zero. If \(p\) is odd, deduce that the coefficient of each even power of \(x\) in \(p(x)\) is zero. d. If \(p\) is even, show that there is a polynomial \(s(x)\) such $$ \text { that } p(x)=s\left(x^{2}\right) $$ e. If \(p\) is odd, show that \(p(0)=0 .\) Deduce that there is an even polynomial \(t(x)\) such that \(p(x)=x \cdot t(x)\)

Step-by-Step Solution

Answer

Every polynomial can be expressed as the sum of an even and an odd polynomial. Even polynomials contain only even degree terms and odd polynomials only odd degree terms. For an odd polynomial, \(p(0)=0\).

Step 1: Understanding Even and Odd Polynomials

Before we dive into solving these parts, let's recall that an even polynomial has the property that for every term in the polynomial, the degree of the term is even. For odd polynomials, each term has an odd degree.

Step 1: Proving q(x) is an Even Polynomial

For a polynomial to be even, it must satisfy the condition \(q(x) = q(-x)\). Let's begin by calculating \(q(-x)\):\[q(-x) = \frac{p(-x) + p(x)}{2}\]By rearranging terms, you'll notice:\[q(-x) = \frac{p(x) + p(-x)}{2} = q(x)\]Therefore, \(q(x)\) is indeed an even polynomial.

Step 2: Proving r(x) is an Odd Polynomial

For a polynomial to be odd, it must satisfy the condition \(r(x) = -r(-x)\). Calculate \(r(-x)\):\[r(-x) = \frac{p(-x) - p(x)}{2}\]You can write this as:\[r(-x) = -\left(\frac{p(x) - p(-x)}{2}\right) = -r(x)\]Hence, \(r(x)\) is an odd polynomial.

Step 3: Showing q and r only have Even and Odd Powers

Observe that \(q(x)\) only involves terms like \(a_n x^n + a_n (-x)^n\) where \(n\) is even, yielding terms like \(2a_n x^n\). Thus, \(q(x)\) consists of even powers. Similarly, \(r(x)\) involves terms like \(a_n x^n - a_n (-x)^n\) where \(n\) is odd, yielding terms \(2a_n x^n\). Thus, \(r(x)\) has only odd powers.

Step 4: Deductions from Even and Odd Properties

If \(p\) is even, it implies \(p(x) = p(-x)\), so the odd power terms cancel out, making their coefficients zero. If \(p\) is odd, \(p(x) = -p(-x)\), canceling out even power terms, hence their coefficients are zero.

Step 5: Expressing an Even Polynomial as s(x²)

If \(p(x)\) is even, it can be expressed in terms of even powers only, say \(p(x) = a_n x^n + ... + a_0\) where all powers are even. Thus, it can be written as \(p(x) = s(x^2)\) where \(s(x)\) is another polynomial.

Step 6: Proving p(x)=x⋅t(x) for an Odd Polynomial

For an odd polynomial \(p(x)\), we have \(p(0) = 0\), since all terms are odd-powered. Therefore, we can factor out \(x\) and the remaining expression \(t(x)\) is an even polynomial.

Key Concepts

Even Polynomial

An even polynomial is characterized by all its terms having even degrees. This means each power of the variable, usually denoted as \( x \), is an even number like 0, 2, 4, etc.
Understanding even polynomials begins with recognizing that they are symmetric around the y-axis. For any even polynomial \( p(x) \), it satisfies the condition: \( p(x) = p(-x) \) for all values of \( x \). This symmetry is key because, no matter which direction you approach it from on the x-axis, it retains the same value.
For example, \( f(x) = x^4 + 2x^2 + 1 \) is an even polynomial because substituting \(-x\) results in the same expression, indicating no directional change in value along the x-axis.

Odd Polynomial

Odd polynomials have the unique property that all terms have odd powers, like 1, 3, 5, etc. The defining characteristic of an odd polynomial is that it satisfies the condition \( p(x) = -p(-x) \).
This means that these polynomials are symmetric about the origin. For any odd polynomial, if you replace \( x \) with \( -x \), you will get the negative of the original polynomial. This reflects their symmetry, creating a mirror image across the origin.
A simple example is \( g(x) = x^3 + 3x \). Substituting \( -x \) gives \(-x^3 - 3x\), which is the negative of the original polynomial, showing it is truly odd.

Polynomial Factorization

Factorization is a method used to express a polynomial as a product of its factors, which are usually lower degree polynomials. This is closely related to dividing numbers into their prime components.
For polynomials, factorization can simplify solving polynomial equations and analyzing their behavior. One way to factor polynomials is by identifying common terms or using special products and identities.
For instance, if you know a polynomial is even, you can express it as a function of \( x^2 \), such as \( p(x) = s(x^2) \), providing an approach to simplify factors. Similarly, if a polynomial is odd, you can express it as \( x \cdot t(x) \), factoring out \( x \) because of the property \( p(0) = 0 \).
Factorization reveals hidden relationships between variables which can be pivotal in equations and derivations.

Polynomial Coefficients

The coefficients of a polynomial are the numerical factors that multiply each term, and they play a critical role in determining the polynomial's value and behavior. They tell us how steep or flat the curve of the polynomial is at any given point.
The coefficient of a term is often denoted by letters such as \( a_n \), where \( n \) signifies the degree of the corresponding term. In even polynomials like \( x^4 + 2x^2 + 1 \), the coefficients are \( 1 \), \( 2 \), and \( 1 \). Meanwhile, for odd polynomials like \( x^3 + 3x \), the coefficients are \( 1 \) and \( 3 \).
Understanding coefficients helps determine whether a polynomial is likely to be even or odd, based on the presence or absence of terms with even or odd powers. Managing these coefficients properly aides in simplifying and solving polynomial expressions effectively.