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Chapter 48: Problem 8

Die Nullstelle \(x_{0}\) des Polynoms \(p\) vom Grade \(\geqslant 1\) besitzt genau dann die Vielfachheit wenn \(p\left(x_{0}\right)=p^{\prime}\left(x_{0}\right)=\cdots=p^{(e-1)}\left(x_{0}\right)=0\), aber \(p^{(v)}\left(x_{0}\right) \neq 0\) ist.

Short Answer

Expert verified
The root's multiplicity \( e \) is where \( p(x_0), p'(x_0), \ldots, p^{(e-1)}(x_0) = 0 \) and \( p^{(e)}(x_0) \neq 0 \).

Step by step solution

01

Understanding the Problem Statement

The problem is asking about the multiplicity of a root \( x_0 \) of a polynomial \( p \) of degree \( \geq 1 \). The multiplicity is the number \( e \) such that the polynomial and its derivatives up to order \( e-1 \) are zero at \( x_0 \), but the derivative of order \( e \) is not zero at \( x_0 \). This essentially means that \( x_0 \) is a root of multiplicity \( e \).
02

Concept of Multiplicity

For a root of multiplicity \( e \), the polynomial can be expressed in the form \( p(x) = (x - x_0)^e q(x) \), where \( q(x_0) eq 0 \). This implies that the root \( x_0 \) appears \( e \) times in the factorization of the polynomial.
03

Check the Condition

According to the condition given, the root \( x_0 \) satisfies: \( p(x_0) = p'(x_0) = \cdots = p^{(e-1)}(x_0) = 0 \), but \( p^{(v)}(x_0) eq 0 \). This indicates that each derivative up to \( e-1 \) is zero, matching the root's multiplicity. The condition \( p^{(e)}(x_0) eq 0 \) confirms that \( e \) is the root's exact multiplicity.
04

Verify the Final Derivative

The derivative \( p^{(e)}(x_0) eq 0 \) confirms that the first non-zero derivative is of order \( e \), which doesn't happen for derivatives less than \( e \). This final derivative condition is essential to distinguish the root's multiplicity and ensure it doesn't have higher multiplicity.

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Key Concepts

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

Polynomial Roots
In mathematics, a polynomial is an algebraic expression made up of variables and coefficients. The roots of a polynomial are the values for which the polynomial evaluates to zero. For a polynomial function \( p(x) \), a root \( x_0 \) satisfies \( p(x_0) = 0 \). The number of roots a polynomial can have is equal to its degree, though some roots may repeat multiple times.
  • A root of a polynomial is where the graph of the polynomial crosses the X-axis.
  • Real roots are points where the function intersects with the real number line.
  • Complex roots occur in conjugate pairs when the coefficients of the polynomial are real numbers.
Understanding polynomial roots is crucial for solving equations, graphing functions, and analyzing the behavior of mathematical models. Recognizing root multiplicities helps reveal deeper insights into polynomial functions.
Derivatives
Derivatives play a significant role in understanding the behavior of functions, including polynomials. The derivative of a function \( p(x) \), denoted as \( p'(x) \), measures the rate at which the function's value changes as its input changes.
  • The first derivative \( p'(x) \) provides information about the slope of the function at any given point.
  • Higher-order derivatives (like \( p''(x) \), \( p'''(x) \), etc.) provide further insights into the curvature and the rate of change of the slope.
  • In the context of roots, derivatives help us determine the multiplicity, as a root of multiplicity \( e \) will make the first \( e-1 \) derivatives zero.
Analyzing the derivatives of a polynomial at a given root allows us to specify the root's behavior, such as confirming whether the root repeats in the factorization.
Polynomial Factorization
Polynomial factorization refers to expressing a polynomial as a product of its factors. For a polynomial with roots \( x_0, x_1, ..., x_n \), factorization helps in breaking it down into simpler multiplied expressions. These factors are usually binomials of the form \( (x-x_i) \), where each root contributes to deciding the multiplicity.
  • The factorization process simplifies polynomial equations and helps in solving them analytically.
  • Understanding the multiplicity of a root affects how many times a factor \( (x-x_0) \) appears in the factorized form of the polynomial.
  • Factorization is essential for integration, finding polynomial limits, and simplifying polynomial expressions in calculus.
With factorization, polynomials become easier to analyze and solve, as it reveals all root multiplicities and simplifies the computation for various applications.
Root of a Polynomial
The concept of a "root" in polynomials is central to understanding equations. A root \( x_0 \) of a polynomial \( p(x) \) is a value that makes the polynomial equal zero, i.e., \( p(x_0) = 0 \). Each root can potentially affect the overall behavior and graph of the polynomial.
  • Roots can be real or complex, influencing the polynomial鈥檚 appearance on a graph and its factorization.
  • The multiplicity of a root plays a critical part, determining how the polynomial factors and behaves near the root.
  • To find a root, various mathematical techniques like factoring, synthetic division, or using the quadratic formula may be employed.
Recognizing and calculating the roots of a polynomial allow mathematicians to solve polynomial equations and understand the polynomial鈥檚 structure and dynamics.

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Most popular questions from this chapter

Bestimme die Gleichung der Tangente an das Schaubild der folgenden Funktionen inden angegebenen Punkten \(P\) : a) \(f(x):=1 / x, P:=(1,1)\). b) \(f(x):=\mathrm{e}^{x}, P:=(0,1)\). c) \(f(x):=\sin x, P:=(0,0)\)

Beweise die folgenden Formeln f眉r nat眉rliches \(n \geq 2\) : a) \(\sum_{k=1}^{n} k\left(\begin{array}{l}n \\ k\end{array}\right)=n 2^{n-1}\) b) \(\sum_{k=1}^{n}(-1)^{k-1} k\left(\begin{array}{l}n \\\ k\end{array}\right)=0\), c) \(\sum_{k-2}^{n} k(k-1)\left(\begin{array}{l}n \\\ k\end{array}\right)=n(n-1) 2^{n-2}\) Hinweis: \(\sum_{k=0}^{n}\left(\begin{array}{l}n \\ k\end{array}\right) x^{k}=(1+x)^{n}\)

a) \(\left(x^{2} \mathrm{e}^{x}\right)^{\prime}=\left(x^{2}+2 x\right) \mathrm{e}^{x}\) b) \(\left(\cos x \cdot \mathrm{e}^{-x}\right)^{(4)}=-4 \cos x \cdot \mathrm{e}^{-x}\) c) \((\ln (\ln x))^{\prime}=1 /(x \ln x)\) d) \(\frac{d}{d x}\left(1+x^{2}\right)^{\sin x}=\left(1+x^{2}\right)^{\sin x}\left(\frac{2 x \sin x}{1+x^{2}}+\ln \left(1+x^{2}\right) \cdot \cos x\right)\), e) \(\frac{\mathrm{d}}{\mathrm{d} x}\left(\frac{1+x}{1-x}\right)^{x^{2}}=\left(\frac{1+x}{1-x}\right)^{x^{2}} 2 x\left(\ln \frac{1+x}{1-x}+\frac{x}{1-x^{2}}\right)\) f) \(\frac{\mathrm{d}}{\mathrm{d} x} \sqrt{\mathrm{e}^{x^{2+x+1}}}=\left(x+\frac{1}{2}\right) \sqrt{\mathrm{e}^{x_{2} x+1}}\) g) \(\frac{d}{d x} \frac{\sqrt{x} \sin x}{\ln x}=\frac{(2 x \cos x+\sin x) \ln x-2 \sin x}{2 \sqrt{x}(\ln x)^{2}}\) h) \(\frac{\mathrm{d}}{\mathrm{d} x} \sqrt{\mathrm{e}^{\sin \sqrt{\mathrm{x}}}}=\frac{\cos \sqrt{x}}{4 \sqrt{x}} \sqrt{\mathrm{e}^{\sin \sqrt{x}}}\).

F眉r alle \(x\) ist \(\sin 2 x=2 \sin x \cos x\) und \(\cos 2 x=\cos ^{2} x-\sin ^{2} x\).

Sei \(p(x):=\sum_{k=0}^{n} a_{k} x^{k}\). Zeige: a) \(a_{k}=p^{(k)}(0) / k !\), also \(p(x)=p(0)+\frac{p^{\prime}(0)}{1 !} x+\frac{p^{\prime \prime}(0)}{2 !} x^{2}+\cdots+\frac{p^{(n)}(0)}{n !} x^{n}\) b) \(p\left(x_{0}+h\right)=p\left(x_{0}\right)+\frac{p^{\prime}\left(x_{0}\right)}{1 !} h+\frac{p^{\prime \prime}\left(x_{0}\right)}{2 !} h^{2}+\cdots+\frac{p^{(n)}\left(x_{0}\right)}{n !} h^{n}\) f眉r jedes \(h\) Ein Polynom \(n\)-ten Grades wird also eindeutig durch die \(n+1\) Werte \(p^{(k)}\left(x_{0}\right), k=0\) \(1, \ldots, n\), bestimmt
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