Problem 55 - Calculus Single Variable

Question

Suppose \(p(x)\) is a polynomial of degree \(n>1\). Let \(I(x)=x,\) and define \(f=p \circ(I+p) .\) What is deg \((f) ?\) How are the roots of \(p\) related to the roots of \(f\) ? Show that there is a polynomial \(q\) such that \(f=p \circ q .\) Illustrate with \(p(x)=x^{2}-3 x-4\)

Step-by-Step Solution

Answer

\(\deg(f) = 2n\). Roots of \(p\) correspond to transformations in roots of \(f\). Polynomial \(q(x) = x + p(x)\) satisfies \(f = p \circ q\). For \(p(x) = x^2 - 3x - 4\), \(f(x) = (x^2 - 2x - 4)^2 - 3(x^2 - 2x - 4) - 4\).

Step 1: Understand the Composition

The function \(f = p \circ (I + p)\) means that for \(f(x)\), we substitute \(I(x) + p(x)\) into \(p(x)\). Thus, \(f(x) = p(x + p(x))\). Evaluate how this impacts the degree of \(f\).

Step 2: Determining the Degree of \(f\)

\(p(x)\) is a polynomial of degree \(n\). Substituting \(x + p(x)\) into \(p\), the degree of the new polynomial potentially doubles. Essentially, the degree of \(f\) is twice the degree of \(p\), so \(\deg(f) = 2n\).

Step 3: Relate Roots of \(p\) to Roots of \(f\)

The roots of \(p(x)\) are those values where \(p(c) = 0\). For \(f(x) = p(x + p(x))\), the roots will shift according to the values where \(x + p(x)\) roots correspond to the transformations in \(f(x)\). Specifically, \(p(x) = 0\) implies \(f(x) = p(p(x))\) which is non-necessarily zero unless composed with \(I(x)\) or similar transformations.

Step 4: Show there exists a polynomial \(q\) such that \(f = p \circ q\)

Notice that \(f(x) = p(x + p(x))\) implies there could be \(q(x)\) such that \(q(x) = x + p(x)\). That is, \(f(x) = p(q(x))\). As \(q(x) = x + p(x)\) is polynomial of degree \(n\), \(q(x)\) works here.

Step 5: Illustrate with \(p(x) = x^2 - 3x - 4\)

Given \(p(x) = x^2 - 3x - 4\), find \(f(x) = p(x + p(x))\). First find \(q(x) = x + (x^2 - 3x - 4) = x^2 - 2x - 4\). Therefore, \(f(x) = p(q(x)) = (x^2 - 2x - 4)^2 - 3(x^2 - 2x - 4) - 4\). Expand to confirm the degree remains consistent.

Key Concepts

Degree of a Polynomial

The degree of a polynomial is an essential concept that tells us the highest power of the variable in the polynomial expression. For example, in a polynomial like \( p(x) = x^2 - 3x - 4 \), the degree is 2 because the highest power of \( x \) is 2.

Understanding the degree is important as it helps anticipate the behavior and key characteristics of the polynomial function, such as its number of roots or intersections with the x-axis. When we perform operations like polynomial composition, the degree can change, and predicting this change helps in understanding the overall structure of the function.

In the original exercise, we start with a polynomial \( p(x) \) of degree \( n \) and find that after composing \( p \) with \( I + p \), the degree doubles, resulting in \( \deg(f) = 2n \). This doubling occurs because we're essentially nesting the polynomials inside each other, which means you multiply their degrees.

Polynomial Composition

Polynomial composition involves creating a new polynomial by plugging one polynomial into another. Think of it as substituting one polynomial expression in place of the variable in another polynomial.

For instance, if we have polynomials \( p(x) = x^2 - 3x - 4 \) and \( q(x) = x + 2 \), composing \( p \) and \( q \), denoted by \( p \circ q \), means replacing every \( x \) in \( p(x) \) with \( q(x) \). This results in \( p(q(x)) = (x+2)^2 - 3(x+2) - 4 \). After simplifying, you end up with a new polynomial, potentially of different degree and characteristics than \( p(x) \) or \( q(x) \).

In the exercise, the composition \( f = p \circ (I + p) \) involved substituting \( x + p(x) \) back into \( p(x) \). This combination results in a polynomial with twice the degree of the original \( p(x) \), illustrating how polynomial composition can significantly change the form and properties of functions.

Roots of Equations

The roots of a polynomial equation are the solutions for which the polynomial equals zero. They are essentially the x-intercepts of the polynomial graph.

For example, the polynomial \( p(x) = x^2 - 3x - 4 \) has roots when \( x^2 - 3x - 4 = 0 \). Solving this equation gives the roots or values of \( x \) where the polynomial touches or crosses the x-axis.

In relation to the given exercise, understanding how the roots of the original polynomial \( p(x) \) relate to the composed polynomial \( f(x) \) is key. For the composed polynomial \( f(x) = p(x + p(x)) \), roots are found based on how the transformations inside \( f \) affect the x-values.

Moreover, recognizing these roots can inform about the behavior of the polynomial, such as where it becomes zero and insights into its graphical representation. Yet, it’s crucial to note that transformations in composition may lead those roots to shift or form differently compared to the original polynomial.