91Ó°ÊÓ

Es ist \(\left.p(x):=a_{4} x^{4}+a_{3} x^{3}+a_{2} x^{2}+a_{1} x+a_{0}=\left\\{\left(a_{4} x+a_{3}\right) x+a_{2}\right] x+a_{1}\right\\} x+a_{0}\) Infolgedessen l?ßt sich \(p\left(x_{0}\right)\) bequem nach dem leicht verständichen Hornerschen Schema\\} (William G. Horner, 1756-1837; 81) berechnen: \begin{tabular}{lllll} \hline\(x_{0}\) & \(a_{3}\) & \(a_{2}\) & \(a_{1}\) & \(a_{0}\) \\ & \(\alpha_{4} x_{0}\) & \(\alpha_{3} x_{0}\) & \(\alpha_{2} x_{0}\) & \(\alpha_{1} x_{0}\) \\ \hline \end{tabular} \(\alpha_{4}:=a_{4} \quad \alpha_{3}:=\alpha_{4} x_{0}+a_{3} \quad \alpha_{2}:=\alpha_{3} x_{0}+a_{2} \quad \alpha_{1}:=\alpha_{2} x_{0}+a_{1} \quad \alpha_{0}:=\alpha_{1} x_{0}+a_{0}=p\left(x_{0}\right)\) Entwirf ein Hornersches Schema für Polynome beliebigen Grades und berechne die Werte der folgenden Polynome an den Stellen \(x_{0} ;=1,2,3\) : a) \(p(x):=x^{3}-2 x^{2}+x-1, \quad\) b) \(q(x):=x^{5}+x^{2}-x+2\) (beachte, daß verschwindende Koeffizienten in dem Hornerschen Schema aufgeführt werden müssen).

Step by step solution

01

Introduction to Horner's Scheme

In this exercise, we will use Horner's Scheme, a method for evaluating polynomials efficiently. Horner's method reorganizes the polynomial into nested factors, allowing the computation of the polynomial's value at a specific point by performing a series of multiplications and additions.

02

Organize Polynomial Coefficients for p(x)

For the polynomial \( p(x) = x^3 - 2x^2 + x - 1 \), identify the coefficients: \( a_3 = 1 \), \( a_2 = -2 \), \( a_1 = 1 \), and \( a_0 = -1 \). Prepare to calculate \( p(x_0) \) at \( x_0 = 1, 2, 3 \).

03

Calculate p(x_0) using Horner's Scheme

Use the recursive formulas: \( \alpha_3 = a_3 \), \( \alpha_2 = \alpha_3 x_0 + a_2 \), \( \alpha_1 = \alpha_2 x_0 + a_1 \), and \( \alpha_0 = \alpha_1 x_0 + a_0 \). Perform the calculations:- For \( x_0 = 1 \): - \( \alpha_3 = 1 \) - \( \alpha_2 = 1 \cdot 1 + (-2) = -1 \) - \( \alpha_1 = -1 \cdot 1 + 1 = 0 \) - \( \alpha_0 = 0 \cdot 1 - 1 = -1 \)- For \( x_0 = 2 \): - \( \alpha_3 = 1 \) - \( \alpha_2 = 1 \cdot 2 + (-2) = 0 \) - \( \alpha_1 = 0 \cdot 2 + 1 = 1 \) - \( \alpha_0 = 1 \cdot 2 - 1 = 1 \)- For \( x_0 = 3 \): - \( \alpha_3 = 1 \) - \( \alpha_2 = 1 \cdot 3 + (-2) = 1 \) - \( \alpha_1 = 1 \cdot 3 + 1 = 4 \) - \( \alpha_0 = 4 \cdot 3 - 1 = 11 \)

04

Organize Polynomial Coefficients for q(x)

For the polynomial \( q(x) = x^5 + x^2 - x + 2 \), note all coefficients including zeros: \( a_5 = 1 \), \( a_4 = 0 \), \( a_3 = 0 \), \( a_2 = 1 \), \( a_1 = -1 \), \( a_0 = 2 \). Calculate \( q(x_0) \) at \( x_0 = 1, 2, 3 \).

05

Calculate q(x_0) using Horner's Scheme

Use the recursive formulas for calculation:- For \( x_0 = 1 \): - \( \alpha_5 = 1 \) - \( \alpha_4 = 1 \cdot 1 + 0 = 1 \) - \( \alpha_3 = 1 \cdot 1 + 0 = 1 \) - \( \alpha_2 = 1 \cdot 1 + 1 = 2 \) - \( \alpha_1 = 2 \cdot 1 - 1 = 1 \) - \( \alpha_0 = 1 \cdot 1 + 2 = 3 \)- For \( x_0 = 2 \): - \( \alpha_5 = 1 \) - \( \alpha_4 = 1 \cdot 2 + 0 = 2 \) - \( \alpha_3 = 2 \cdot 2 + 0 = 4 \) - \( \alpha_2 = 4 \cdot 2 + 1 = 9 \) - \( \alpha_1 = 9 \cdot 2 - 1 = 17 \) - \( \alpha_0 = 17 \cdot 2 + 2 = 36 \)- For \( x_0 = 3 \): - \( \alpha_5 = 1 \) - \( \alpha_4 = 1 \cdot 3 + 0 = 3 \) - \( \alpha_3 = 3 \cdot 3 + 0 = 9 \) - \( \alpha_2 = 9 \cdot 3 + 1 = 28 \) - \( \alpha_1 = 28 \cdot 3 - 1 = 83 \) - \( \alpha_0 = 83 \cdot 3 + 2 = 251 \)

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Key Concepts

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

Polynomial Evaluation

Polynomial evaluation is a crucial technique in mathematics. It involves determining the value of a polynomial at a particular point. A polynomial, generally expressed as \( p(x) = a_nx^n + a_{n-1}x^{n-1} + \ldots + a_0 \), needs to be computed by substituting a specific value into \( x \).
However, evaluating a polynomial using the straightforward method of calculating each power of \( x \) and then combining them might not be efficient, especially for polynomials of high degree. That's where methods like Horner's Scheme come into play.
This method reorganizes the polynomial into a nested form, minimizing the number of multiplications and additions required. For example, a polynomial \( p(x) = x^3 - 2x^2 + x - 1 \) can be rewritten as \(((1x - 2)x + 1)x - 1\), simplifying the calculation process significantly.

Polynomial Coefficients

Polynomial coefficients are the numerical factors of each term in a polynomial. For the polynomial \( p(x) = a_nx^n + a_{n-1}x^{n-1} + \ldots + a_0 \), each \( a_i \) represents a coefficient corresponding to the term \( x^i \).
These coefficients play a vital role in polynomial evaluation and help define the polynomial's shape and behavior. In the polynomial \( p(x) = x^3 - 2x^2 + x - 1 \), the coefficients are \( 1 \), \(-2 \), \( 1 \), and \(-1 \), respectively.
In exercises employing Horner's Method, it is important to note all coefficients, including zeros. For example, if a term like \( x^4 \) is missing, it implicitly means \( a_4 = 0 \). Including zero coefficients in Horner's Scheme ensures accurate evaluation.

Efficient Computation

Efficient computation is a goal in polynomial evaluation that is achieved through methods like Horner's Scheme. This method reduces the computational complexity by rewriting polynomials into nested forms.
Consider the polynomial \( q(x) = x^5 + x^2 - x + 2 \). Evaluating this directly at multiple points would require numerous calculations. Using Horner's Scheme, it's rewritten and calculated as:

• \( \alpha_5 = a_5 \)
• \( \alpha_4 = \alpha_5 \cdot x_0 + a_4 \)
• \( \alpha_3 = \alpha_4 \cdot x_0 + a_3 \)
• \( \alpha_2 = \alpha_3 \cdot x_0 + a_2 \)
• \( \alpha_1 = \alpha_2 \cdot x_0 + a_1 \)
• \( \alpha_0 = \alpha_1 \cdot x_0 + a_0 \)

This process incorporates fewer multiplications and additions, making the computation faster and less error-prone.
Especially for computers and calculators, Horner's Scheme significantly reduces the time and resources needed to evaluate polynomials.

Recursive Calculation

Recursive calculation is a technique where a repetitive process is applied to break a complex problem into simpler parts. Horner's Scheme employs recursion implicitly by applying a sequence of operations repeatedly.
In this method, each coefficient calculation depends on the result of the previous one. For example, when evaluating the polynomial \( p(x) \) at \( x_0 = 1 \):

• Start with \( \alpha_3 = a_3 \)
• Compute \( \alpha_2 = \alpha_3 \cdot x_0 + a_2 \)
• Proceed with \( \alpha_1 = \alpha_2 \cdot x_0 + a_1 \)
• Finally, find \( \alpha_0 = \alpha_1 \cdot x_0 + a_0 \)

Each step builds on the previous, showcasing the recursive nature where the results of earlier steps are inputs for subsequent calculations.
This allows complex polynomial evaluations to become a systematic sequence of simpler calculations. This recursive approach enhances understanding and efficiency, especially for large-scale problems.