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Chapter 5: Problem 4

Show that every polynomial of odd degree with real coefficients has at least one real root.

Short Answer

Expert verified
The problem is proven by the fact that an odd degree polynomial with real coefficients will approach \(-\infty\) and \(+\infty\) as \(x\) approaches \(-\infty\) and \(+\infty\) respectively, and due to the Intermediate Value Theorem, there will always be some \(x\) which makes \(P(x) = 0\), indicating a real root.

Step by step solution

01

Understand the properties of Odd degree polynomials

A polynomial of odd degree has the form \(P(x) = a_nx^n + a_{n-1}x^{n-1} + \ldots + a_1x + a_0\), where \(a_n \neq 0\), \(a_i\) are real numbers and \(n\) is an odd integer. With a non-zero leading coefficient, the polynomial is well-defined and has degree \(n\). Since \(n\) is odd, the degree of the polynomial is also odd.
02

Analyze End Behavior

Assuming \(a_n > 0\), the end behavior of an odd degree polynomial can be analyzed by taking the limit as \(x\) approaches \(+\infty\) and \(-\infty\), we get: \(\lim_{x \to +\infty} P(x) = +\infty\) and \(\lim_{x \to -\infty} P(x) = -\infty\). This is due to the fact that the highest power term dominates for large \(x\), and the fact that an odd power of a negative number is negative and of a positive number is positive.
03

Apply Intermediate Value Theorem

The Intermediate Value Theorem states that if a real-valued function is continuous in the interval \([a, b]\), and \(d\) is any number between \(f(a)\) and \(f(b)\), then there exists some \(c \in [a, b]\) such that \(f(c) = d\). In this case, the polynomial function is continuous everywhere, and since \(P(x)\) approaches \(-\infty\) as \(x \to -\infty\) and \(+\infty\) as \(x \to +\infty\), for every value of \(y\), there exists some \(x\) such that \(P(x) = y\). Particularly, this is true for \(y = 0\), therefore, there exists at least one real root.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Odd Degree Polynomial
When discussing the topic of polynomials, we come across different types of degrees. The degree of a polynomial is simply the highest power of the variable present in the polynomial. An odd degree polynomial is one where this degree is an odd number, such as 1, 3, 5, etc.
This means that if we denote a polynomial by \(P(x)\), it will have a general form:
  • \(P(x) = a_nx^n + a_{n-1}x^{n-1} + \ldots + a_1x + a_0\)
where \(n\) is an odd integer, and all \(a_i\) are coefficients that are real numbers. The coefficient \(a_n\), associated with the highest power \(n\), must not be zero because it defines the polynomial's degree.
A special characteristic of odd degree polynomials is their end behavior. As \(x\) approaches positive infinity, \(P(x)\) also reaches positive or negative infinity depending on the sign of \(a_n\). This end behavior helps explain why odd degree polynomials always have at least one real root.
Intermediate Value Theorem
The Intermediate Value Theorem is a fundamental concept in calculus and is particularly useful when we discuss continuous functions. This theorem states that if \(f\) is a continuous function on a closed interval \([a, b]\), and if \(d\) is a value between \(f(a)\) and \(f(b)\), there exists an \(c\) within the interval \([a, b]\) such that \(f(c) = d\).
When applied to polynomials, which are continuous everywhere on the real number line, the theorem implies a crucial realization. Given that the end behavior of an odd degree polynomial moves from negative infinity to positive infinity (or vice versa), it inherently crosses the x-axis at least once. Thus, there must be at least one real number \(c\) where \(P(c) = 0\), hence proving there is at least one real root. This makes the Intermediate Value Theorem an essential tool in establishing the existence of real roots for odd degree polynomials.
Polynomials with Real Coefficients
Polynomials with real coefficients are equations where all coefficients \(a_i\) are real numbers. This characteristic is important because it ensures that the polynomial's behavior and roots follow certain predictable patterns on the real number line.
Real coefficients maintain the polynomial's continuity, making it possible to apply the Intermediate Value Theorem, among other mathematical principles. Additionally, polynomials with real coefficients will not exhibit complex behavior that occurs with imaginary coefficients.
For instance, an odd degree polynomial with real coefficients is not only defined robustly in real numbers but also guarantees that the function will have real roots. The real coefficients align with the polynomial's end behavior so that when one evaluates the limits as \(x\) approaches positive or negative infinity, the results align with real-life expectations of growth and decline, thus reinforcing the existence of real roots.

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Most popular questions from this chapter

Let \(A \subseteq \mathbb{R}\) and suppose that \(f: A \rightarrow \mathbb{R}\) has the following property: for each \(\varepsilon>0\) there exists a function \(g_{\varepsilon}: A \rightarrow \mathbb{R}\) such that \(g_{\varepsilon}\) is uniformly continuous on \(A\) and \(\left|f(x)-g_{\varepsilon}(x)\right|<\varepsilon\) for all \(x \in A .\) Prove that \(f\) is uniformly continuous on \(A\).

Let \(\delta\) be a gauge on \(I:=[a, b]\) and suppose that \(I\) does not have a \(\delta\) -fine partition. (a) Let \(c:=\frac{1}{2}(a+b)\). Show that at least one of the intervals \([a, c]\) and \([c, b]\) does not have a \(\delta\) - fine partition. (b) Construct a nested sequence \(\left(I_{n}\right)\) of subintervals with the length of \(I_{n}\) equal to \((b-a) / 2^{n}\) such that \(I_{n}\) does not have a \(\delta\) -fine partition. (c) Let \(\xi \in \cap_{n=1}^{\infty} I_{n}\) and let \(p \in \mathbb{N}\) be such that \((b-a) / 2^{p}<\delta(\xi)\). Show that \(I_{p} \subseteq[\xi-\delta(\xi)\), \(\xi+\delta(\xi)]\), so the pair \(\left(I_{p}, \xi\right)\) is a \(\delta\) -fine partition of \(I_{p}\).

Let \(g\) be defined on \(\mathbb{R}\) by \(g(1):=0\), and \(g(x):=2\) if \(x \neq 1\), and let \(f(x):=x+1\) for all \(x \in \mathbb{R}\). Show that \(\lim g \circ f \neq(g \circ f)(0)\). Why doesn't this contradict Theorem 5.2.6?

Prove that if \(f\) and \(g\) are each uniformly continuous on \(\mathbb{R}\), then the composite function \(f \circ g\) is uniformly continuous on \(\mathbb{R}\).

Let \(I:=[a, b]\) and let \(f: I \rightarrow \mathbb{R}\) be a (not necessarily continuous) function with the property that for every \(x \in I\), the function \(f\) is bounded on a neighborhood \(V_{\delta_{\lambda}}(x)\) of \(x\) (in the sense of Definition 4.2.1). Prove that \(f\) is bounded on \(I\).
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