91Ó°ÊÓ

For Example \(4,\) the author wanted to find a polynomial \(p\) such that $$ p(1)=1, p(2)=4, p(3)=9, p(4)=16, p(5)=31 $$ Carry out the following steps to see how that polynomial was found. (a) Note that the polynomial $$ (x-2)(x-3)(x-4)(x-5) $$ is 0 for \(x=2,3,4,5\) but is not zero for \(x=1 .\) By dividing the polynomial above by a suitable number, find a polynomial \(p_{1}\) such that \(p_{1}(1)=1\) and $$ p_{1}(2)=p_{1}(3)=p_{1}(4)=p_{1}(5)=0 $$ (b) Similarly, find a polynomial \(p_{2}\) of degree 4 such that \(p_{2}(2)=1\) and $$ p_{2}(1)=p_{2}(3)=p_{2}(4)=p_{2}(5)=0 $$ (c) Similarly, find polynomials \(p_{j},\) for \(j=3,4,5,\) such that each \(p_{j}\) satisfies \(p_{j}(j)=1\) and \(p_{j}(k)=0\) for values of \(k\) in \\{1,2,3,4,5\\} other than \(j\) (d) Explain why the polynomial \(p\) defined by $$ p=p_{1}+4 p_{2}+9 p_{3}+16 p_{4}+31 p_{5} $$ satisfies $$ p(1)=1, p(2)=4, p(3)=9, p(4)=16, p(5)=31 $$

Step by step solution

01

Find polynomial \(p_1(x)\)

Divide the polynomial \((x-2)(x-3)(x-4)(x-5)\) by a suitable number to get \(p_1(x)\). To find the divisor, evaluate the polynomial at x = 1: \[ (1-2)(1-3)(1-4)(1-5) = -24 \] To make the value of \(p_1(1) = 1\), we need to divide by -24. So, we have: \[ p_1(x) = -\frac{1}{24}(x-2)(x-3)(x-4)(x-5) \]

02

Find polynomial \(p_2(x)\)

Use the same method to find a polynomial of degree 4 such that \(p_2(2)=1\) and \(p_2(1)=p_2(3)=p_2(4)=p_2(5)=0\). Evaluate the polynomial at x = 2: \[ (2-1)(2-3)(2-4)(2-5) = -6 \] So, we have: \[ p_2(x) = -\frac{1}{6}(x-1)(x-3)(x-4)(x-5) \]

03

Find the remaining polynomials \(p_j(x)\)

Similarly, find polynomials \(p_3(x), p_4(x),\) and \(p_5(x)\): \[ p_3(x) = \frac{1}{4}(x-1)(x-2)(x-4)(x-5) \] \[ p_4(x) = -\frac{1}{6}(x-1)(x-2)(x-3)(x-5) \] \[ p_5(x) = \frac{1}{120}(x-1)(x-2)(x-3)(x-4) \]

04

Combine all the polynomials \(p_j(x)\)

Create the polynomial p(x) by summing up the previous polynomials, each multiplied by their corresponding function values: \[ p(x) = p_1(x) + 4p_2(x) + 9p_3(x) + 16p_4(x) + 31p_5(x) \]

05

Verify that p(x) satisfies the conditions

Let's check whether p(x) meets the required conditions: \[ p(1) = p_1(1) + 4p_2(1) + 9p_3(1) + 16p_4(1) + 31p_5(1) = 1 + 0 + 0 + 0 + 0 = 1 \] \[ p(2) = p_1(2) + 4p_2(2) + 9p_3(2) + 16p_4(2) + 31p_5(2) = 0 + 4 + 0 + 0 + 0 = 4 \] \[ p(3) = p_1(3) + 4p_2(3) + 9p_3(3) + 16p_4(3) + 31p_5(3) = 0 + 0 + 9 + 0 + 0 = 9 \] \[ p(4) = p_1(4) + 4p_2(4) + 9p_3(4) + 16p_4(4) + 31p_5(4) = 0 + 0 + 0 + 16 + 0 = 16 \] \[ p(5) = p_1(5) + 4p_2(5) + 9p_3(5) + 16p_4(5) + 31p_5(5) = 0 + 0 + 0 + 0 + 31 = 31 \] Since all these conditions are satisfied, we have found the desired polynomial p(x).

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Key Concepts

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

Lagrange Polynomials

Lagrange polynomials are used to construct a polynomial that passes through a given set of points. These polynomials have a unique property: for each point, the polynomial equals one at that point and zero at all other specified points. In our example, individual polynomials \( p_1, p_2, ..., p_5 \) were constructed in such a way that each polynomial satisfies the required condition of being zero at certain x-values and one at another. For instance, \( p_1(x) \) was designed to be zero at \( x = 2, 3, 4, \) and \( 5 \) and equals one at \( x = 1 \).
Lagrange polynomials form the basis for Lagrange interpolation, which provides a straightforward method to find the polynomial of least degree that fits all the given data points. This is particularly helpful when trying to model a function that matches a set of specific values.

Polynomial Roots

Polynomial roots, also known as solutions or zeros, are the values for which the polynomial equals zero. Understanding the roots of a polynomial is fundamental because they tell us where the graph of the polynomial will cross the x-axis. In the example exercise, we used polynomial roots to determine the base of each \( p_j(x) \) polynomial through factors like \((x - 2), (x - 3)\), etc.
The significance of these roots lies in their symmetry or role in constructing polynomials that can serve specific conditions. For instance, the roots of \( (x-2)(x-3)(x-4)(x-5) \) were utilized to ensure that \( p_1(2), p_1(3), p_1(4), \) and \( p_1(5) \) become zero, aligning precisely with the desired behavior outlined by Lagrange interpolation for \( p_1(x) \).
Understanding this concept allows students to manipulate and construct polynomials effectively to match a given dataset.

Polynomial Construction

Polynomial construction involves creating a polynomial equation that passes through a set of given points or satisfies a specific condition. Combining all previously found Lagrange polynomials such as \( p_1(x), p_2(x), ..., p_5(x) \) formed the final polynomial \( p(x) \). Here, the combination is done by multiplying each \( p_j(x) \) by its respective function value and summing them up:

• \( p(x) = p_1(x) + 4p_2(x) + 9p_3(x) + 16p_4(x) + 31p_5(x) \)

This expression ensures that the constructed polynomial \( p \) satisfies the conditions \( p(1)=1, p(2)=4, p(3)=9, p(4)=16, \) and \( p(5)=31 \).
Constructing polynomials this way shows the power of blending simpler polynomials to achieve complex interpolation goals. This strategy of construction enables the designing of specific polynomial features, making polynomial interpolation a valuable tool in numerical analysis and various applications like computer graphics or data fitting.