/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 20 Let \(g \in R[X],\) and let \(x ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 16: Problem 20

Let \(g \in R[X],\) and let \(x \in R\) be a root of \(g .\) Show that \(x\) is a multiple root of \(g\) if and only if \(x\) is also a root of \(\mathbf{D}(g)\) (see Exercise 7.18 ).

Short Answer

Expert verified
Question: Prove that \(x\) is a multiple root of \(g\) if and only if \(x\) is also a root of \(\mathbf{D}(g)\), where \(\mathbf{D}(g)\) is the first derivative of the polynomial \(g\). Solution: We can prove this by showing that 1. If \(x\) is a root of \(g\) and \(\mathbf{D}(g)\), then it's a multiple root, and 2. If \(x\) is a multiple root of \(g\), then it's a root of \(\mathbf{D}(g)\). If \(x\) is a root of \(g\) and \(\mathbf{D}(g)\), it implies that the highest power term of \(g(X)\) has a factor of \((X-x)^n\), meaning \(x\) is a multiple root of \(g\). On the other hand, if \(x\) is a multiple root of \(g\), then \(g(X) = (X-x)^k h(X)\), where \(k \geq 2\) and \(h\) is a polynomial. When calculating \(\mathbf{D}(g)\), we find that \(\mathbf{D}(g)(x) = 0\), meaning \(x\) is a root of \(\mathbf{D}(g)\). Thus, \(x\) is a multiple root of \(g\) if and only if \(x\) is a root of \(\mathbf{D}(g)\).

Step by step solution

01

Part 1: If \(x\) is a root of \(g\) and \(\mathbf{D}(g)\), then it's a multiple root

Assume \(x\) is a root of \(g\) and \(\mathbf{D}(g)\). Now, let \(g(X) = a_n(X-x)^n + a_{n-1}(X-x)^{n-1} + \cdots + a_1(X-x) + a_0\) for a polynomial \(g \in R[X]\). Since \(x\) is a root of \(g\), \(g(x)=0\). Now, we find the derivative of \(g\), using the power rule for derivatives: \(\mathbf{D}(g) = na_n(X-x)^{n-1} + (n-1)a_{n-1}(X-x)^{n-2} + \cdots + a_1\) \(x\) being a root of \(\mathbf{D}(g)\) implies that \(\mathbf{D}(g)(x) = 0\). Thus, \(na_n(x-x)^{n-1} + (n-1)a_{n-1}(x-x)^{n-2} + \cdots + a_1 = 0\) Notice that the first term of the derivative has a factor of \((X-x)^{n-1}\) since the highest power term of \(g(X)\) was raised to \(n\). In other words, \(g(X)\) has a factor of \((X-x)^n\) if and only if \(\mathbf{D}(g)(x)=0\). Therefore, \(x\) is a multiple root of \(g\).
02

Part 2: If \(x\) is a multiple root of \(g\), then it's a root of \(\mathbf{D}(g)\)

Assume \(x\) is a multiple root of \(g\). This means there exists an integer \(k \geq 2\) such that \(g(X) = (X-x)^k h(X)\) for some polynomial \(h \in R[X]\). Now, let's find the derivative of \(g\) using the product rule for derivatives: \(\mathbf{D}(g) = \mathbf{D}((X-x)^k h(X)) = (X-x)^k \mathbf{D}(h) + k(X-x)^{k-1}h(X)\) Now, evaluate \(\mathbf{D}(g)\) at \(x\): \(\mathbf{D}(g)(x) = (x-x)^k \mathbf{D}(h(x)) + k(x-x)^{k-1}h(x) = 0\) Thus, \(x\) is a root of \(\mathbf{D}(g)\). Therefore, we conclude that \(x\) is a multiple root of \(g\) if and only if \(x\) is a root of \(\mathbf{D}(g)\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Derivative
A derivative is a concept that helps us understand how a function changes at any given point. Think of it as a way to measure how fast something is growing or shrinking. When dealing with polynomials, derivatives are particularly useful because they allow us to explore the nature of the roots of the polynomial. For a polynomial function, its derivative is another polynomial of a lower degree.
  • For example, if you have a polynomial like \( f(x) = x^3 - 3x + 2 \), taking the derivative would give you \( f'(x) = 3x^2 - 3 \).
  • This new function \( f'(x) \) provides information about where \( f(x) \) increases or decreases and can identify potential multiple roots.
When we say that a number \( x \) is a root of a polynomial, it means that the polynomial equals zero when \( x \) is substituted into it. However, if \( x \) is also a root of the polynomial's derivative, it indicates that the root might appear more than once, making it a multiple root.
Polynomial Roots
Understanding the roots of a polynomial is essential in fields like engineering, physics, and computer science. A root of a polynomial is simply a value for which the polynomial equals zero. But what about multiple roots? Multiple roots, also known as repeated roots, occur when a root appears more than once in the polynomial.
  • Consider the polynomial \( g(x) = (x - 2)^3 \). Here, \( x = 2 \) is a root that appears three times, making it a multiple root.
  • This means that if you set \( g(x) = 0 \), the solution \( x = 2 \) satisfies the equation three times.
Multiple roots often impact the behavior of the graph of a polynomial. For instance, in the above example, the graph would just touch the x-axis at \( x = 2 \) but never cross it. Detecting such roots is critical in various mathematical applications and can be done using derivatives.
Power Rule for Derivatives
The power rule for derivatives is a straightforward yet powerful concept when dealing with polynomials. It states that if you have a term like \( x^n \), its derivative is \( nx^{n-1} \). This rule helps simplify the process of finding derivatives, especially for functions that are sums of power terms.
  • For instance, for the term \( 4x^5 \), the derivative is \( 20x^4 \).
  • Applying the power rule repeatedly allows you to find the derivative of even complex polynomials easily.
The power rule is especially crucial when determining the nature of polynomial roots. By easily finding the derivative of a polynomial, one can verify whether a given root is a multiple root. Since multiple roots will also be roots of the derivative, using the power rule effectively helps identify these intricate relationships within polynomials.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

In the field \(E\) in Example 16.16, find all the elements of degree 2 over \(\mathbb{Z}_{2}\)

Let \(F\) be a field, and consider the ring of bivariate polynomials \(F[X, Y]\). Show that in this ring, the polynomial \(X^{2}+Y^{2}-1\) is irreducible, provided \(F\) does not have characteristic 2 . What happens if \(F\) has characteristic \(2 ?\)

Consider the polynomial $$ X^{3}-1=(X-1)\left(X^{2}+X+1\right) $$ Over \(\mathbb{C},\) the roots of \(X^{3}-1\) are \(1,(-1 \pm \sqrt{-3}) / 2 .\) Let \(\omega:=(-1+\sqrt{-3}) / 2,\) and note that \(\omega^{2}=-1-\omega=(-1-\sqrt{-3}) / 2,\) and \(\omega^{3}=1\). (a) Show that the ring \(\mathbb{Z}[\omega]\) consists of all elements of the form \(a+b \omega,\) where \(a, b \in \mathbb{Z},\) and is an integral domain. This ring is called the ring of Eisenstein integers. (b) Show that the only units in \(\mathbb{Z}[\omega]\) are \(\pm 1, \pm \omega,\) and \(\pm \omega^{2}\). (c) Show that \(\mathbb{Z}[\omega]\) is a Euclidean domain.

Suppose \(p\) is a prime, \(g \in \mathbb{Z}[X],\) and \(x \in \mathbb{Z},\) such that \(g(x) \equiv 0(\bmod p)\) and \(\mathbf{D}(g)(x) \not \equiv 0(\bmod p)\). Show that for every positive integer \(e,\) there exists an integer \(\hat{x}\) such that \(g(\hat{x}) \equiv 0\left(\bmod p^{e}\right),\) and give an efficient procedure to compute such an \(\hat{x}\), given \(p, g, x,\) and \(e\). Hint: mimic the "lifting" procedure discussed in \(\S 12.5 .2\)

(a) Show that the "is associate to" relation is an equivalence relation. (b) Consider an equivalence class \(C\) induced by the "is associate to" relation. Show that if \(C\) contains an irreducible element, then all elements of \(C\) are irreducible. (c) Suppose that for every equivalence class \(C\) that contains irreducibles, we choose one element of \(C,\) and call it a distinguished irreducible. Show that \(D\) is a UFD if and only if every non-zero element of \(D\) can be expressed as \(u p_{1}^{e_{1}} \cdots p_{r}^{e_{r}},\) where \(u\) is a unit, \(p_{1}, \ldots, p_{r}\) are distinguished irreducibles, and this expression is unique up to a reordering of the \(p_{i}\) 's.
See all solutions

Recommended explanations on Math Textbooks

Discrete Mathematics

Read Explanation

Statistics

Read Explanation

Applied Mathematics

Read Explanation

Calculus

Read Explanation

Decision Maths

Read Explanation

Logic and Functions

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.