Problem 3 - A Book of Abstract Algebra

Question

Prove that there is one and only one polynomial \(p(x)\) of degree \(\leq n\) such that \(p\left(a_{0}\right)=b_{0}, \ldots, p\left(a_{n}\right)=b_{n}\)

Step-by-Step Solution

Answer

There is a unique polynomial \( p(x) \) of degree \( \leq n \) satisfying \( p(a_i) = b_i \) for given points.

Step 1: Understand the Polynomial Interpolation Problem

We are tasked to find a polynomial \( p(x) \) of degree \( \leq n \) such that it passes through given points \( (a_0, b_0), (a_1, b_1), \ldots, (a_n, b_n) \). This means \( p(a_i) = b_i \) for \( i = 0, 1, \ldots, n \). This is known as the Lagrange interpolation problem.

Step 2: Use the Method of Lagrange Interpolating Polynomial

Construct the Lagrange interpolating polynomial \( p(x) \) given by:\[ p(x) = \sum_{i=0}^{n} b_i L_i(x) \]where each \( L_i(x) \) is the Lagrange basis polynomial defined as:\[ L_i(x) = \prod_{\substack{0 \leq j \leq n \ j eq i}} \frac{x - a_j}{a_i - a_j} \].

Step 3: Show that the Polynomial Meets the Conditions

Substitute \( x = a_i \) into \( p(x) \). For \( p(a_i) \), besides \( L_i(a_i) = 1 \) (since every term in the product is zero except when \( j=i \)), every other \( L_j(a_i) = 0 \) for \( j eq i \). Therefore,\[ p(a_i) = b_i \times 1 + \sum_{j eq i} b_j \times 0 = b_i \].This confirms the polynomial satisfies \( p(a_i) = b_i \) for all \( i \).

Step 4: Establish Uniqueness of the Polynomial

Assume there is another polynomial \( q(x) \) of degree \( \leq n \) that also satisfies the conditions: \( q(a_i) = b_i \) for all \( i \). Consider the polynomial \( r(x) = p(x) - q(x) \). This new polynomial \( r(x) \) has degree \( \leq n \) and satisfies \( r(a_i) = 0 \) for all \( i \).

Step 5: Use the Root Properties of Polynomials

Since \( r(x) \) has more roots \( a_i \) \((0, 1, \ldots, n)\) than its degree \( \leq n \), by the fundamental theorem of algebra, it must be the zero polynomial \( r(x) = 0 \). Therefore, \( p(x) = q(x) \) for all \( x \), proving that the original polynomial \( p(x) \) is unique.

Key Concepts

Lagrange Interpolating Polynomial

The Lagrange Interpolating Polynomial is a special polynomial used to find a polynomial that exactly passes through a given set of points \((a_0, b_0), (a_1, b_1), \ldots, (a_n, b_n)\).
It is defined as a linear combination of Lagrange basis polynomials. This method simplifies polynomial approximation by ensuring that the polynomial not only fits these points but also has a degree of at most \(n\).
The formula for the Lagrange Interpolating Polynomial is:

• \( p(x) = \sum_{i=0}^{n} b_i L_i(x) \)

Here, the term \(b_i\) represents the known value of the polynomial \(p(x)\) at each point \(a_i\). This polynomial is crucial because it gives us a descriptive polynomial that meets strict conditions while having a defined structure.

Lagrange basis polynomials

Lagrange basis polynomials are the building blocks of the Lagrange Interpolating Polynomial.
Each Lagrange basis polynomial \(L_i(x)\) is constructed such that it is equal to 1 at \(x = a_i\), and 0 at all other given sample points.
The formula for the Lagrange basis polynomial \(L_i(x)\) is:

• \( L_i(x) = \prod_{\substack{0 \leq j \leq n \ j e i}} \frac{x - a_j}{a_i - a_j} \)

These equations create polynomials that focus on one specific data point while restricting influence elsewhere.
They help ensure the entirety of the Lagrange polynomial focuses on providing accuracy at specified points without unnecessary complexity. By multiplying each \(b_i\) with \(L_i(x)\), the interpolation focuses on the specific contributions to fit between each point \(a_i\). This ensures the conditions \(p(a_i) = b_i\) is satisfied for all specified points.

Fundamental Theorem of Algebra

The Fundamental Theorem of Algebra is a critical mathematical principle that relates to the roots of polynomials.
It states that every non-zero single-variable polynomial with complex coefficients has as many roots as its degree when counted with multiplicity.
This theorem ensures each polynomial of degree \(n\) has exactly \(n\) roots in the complex number set. In Lagrange interpolation, this theorem is used to prove uniqueness.
Consider a polynomial \(r(x) = p(x) - q(x)\) where the goal is to show uniqueness. This polynomial \(r(x)\) must have roots \(a_i\) for \(i = 0, 1, \ldots, n\) since \(p(a_i) = q(a_i)\).
Because \(r(x)\) has more roots than its degree, it must be zero per the Fundamental Theorem of Algebra, thus ensuring \(p(x) = q(x)\) and proving there's only one possible polynomial satisfying the conditions.

Degree of a Polynomial

The degree of a polynomial is an important concept that indicates the highest power of the variable present in the polynomial.
In polynomial interpolation, we are often looking for a polynomial of degree \( \leq n\) that fits a specific set of data points.
The degree influences the number of turning points and roots that a polynomial can have. Determining the degree is crucial, as it essentially sets the maximum number of roots the polynomial can possess based on the Fundamental Theorem of Algebra.
In the context of interpolation, understanding the degree ensures that the polynomial fits exactly through a given number of points without exceeding necessary complexity. This balance is vital in creating a polynomial that is efficient and accurate, corresponding exactly and only to those required points.