Problem 5 - A Book of Abstract Algebra

Question

If \(t(x)\) is any polynomial in \(F[x]\), and \(a_{0}, \ldots, a_{n} \in F\), the unique polynomial \(p(x)\) of degree \(\leq n\) such that \(p\left(a_{0}\right)=t\left(a_{0}\right), \ldots, p\left(a_{n}\right)=t\left(a_{n}\right)\) is called the Lagrange interpolator for \(t(x)\) and \(a_{0}, \ldots, a_{n} .\) Prove that the remainder, when \(t(x)\) is divided by \(\left(x-a_{0}\right)\left(x-a_{1}\right) \cdots\left(x-a_{n}\right)\), is the Lagrange interpolator.

Step-by-Step Solution

Answer

The remainder is the Lagrange interpolating polynomial for \( t(x) \) and points \( a_0, ..., a_n \).

Step 1: Understanding the Problem

We need to show that the remainder when a polynomial \( t(x) \) is divided by another polynomial \( (x - a_0)(x - a_1)...(x - a_n) \) is the Lagrange interpolating polynomial \( p(x) \). This polynomial \( p(x) \) satisfies \( p(a_0) = t(a_0), ..., p(a_n) = t(a_n) \).

Step 2: Division of Polynomials

When dividing the polynomial \( t(x) \) by \((x-a_0)(x-a_1)...(x-a_n)\), there exists a quotient polynomial \( q(x) \) and a remainder polynomial \( r(x) \) such that:\[ t(x) = q(x)(x-a_0)(x-a_1)...(x-a_n) + r(x) \]The degree of \( r(x) \) must be less than or equal to \( n \) because \((x-a_0)(x-a_1)...(x-a_n)\) has degree \( n+1 \).

Step 3: Lagrange Interpolator Properties

The remainder \( r(x) \) is of degree \( \leq n \), and we need to show that \( r(x) \) satisfies \( r(a_i) = t(a_i) \) for each \( i = 0, 1, ..., n \). This is precisely the condition for the Lagrange interpolating polynomial of \( t(x) \) and the points \( a_0, a_1, ..., a_n \).

Step 4: Satisfying Conditions at Interpolation Points

Evaluating the division expression at any interpolation point \( a_i \) gives:\[ t(a_i) = q(a_i)(a_i-a_0)(a_i-a_1)...(a_i-a_n) + r(a_i) \]Since each \((x-a_i)\) evaluates to zero at \( x = a_i \) and \( q(x)\) is finite, \( q(a_i)(a_i-a_0)(a_i-a_1)...(a_i-a_n) = 0 \, \). Hence, \( r(a_i) = t(a_i) \) for each \( i \).

Step 5: Conclusion

Since \( r(x) \) meets the conditions \( r(a_i) = t(a_i) \) for each \( i \) and is of degree \( \leq n \), it must be the unique Lagrange interpolating polynomial \( p(x) \). This proves that the remainder, when \( t(x) \) is divided by \( (x-a_0)(x-a_1)...(x-a_n) \), is indeed the Lagrange interpolator.

Key Concepts

Polynomial Division

Polynomial division is very similar to the long division we learn in arithmetic. When we perform polynomial division, we divide one polynomial, say \( t(x) \), by another polynomial. This process splits \( t(x) \) into two other polynomials: a quotient \( q(x) \) and a remainder \( r(x) \). The result is written in the form of:

• \( t(x) = q(x) \, \cdot \, d(x) + r(x) \)

where \( d(x) \) is the divisor. In the context of the exercise, our divisor is \((x-a_0)(x-a_1)...(x-a_n)\). Understanding polynomial division helps us break complex expressions into simpler parts, paving the way for polynomial interpolation and remainder theorem discussions.
Remember, the degree of the remainder \( r(x) \) should be less than the degree of the divisor, ensuring the remainder is indeed a smaller part of the original polynomial.

Remainder Theorem

The Remainder Theorem is a powerful tool in polynomial arithmetic. It states that if you divide a polynomial \( t(x) \) by a linear term \( x - a \), the remainder of that division is simply \( t(a) \). This theorem simplifies checking whether a number is a root of a polynomial, as if \( t(a) = 0 \), it implies \( x = a \) is a root of \( t(x) \). In the exercise we discussed, the polynomial division outcome is used to relate the remainder, when \( t(x) \) is divided by a polynomial of multiple linear factors, to the Lagrange interpolator. Each component, \( (x-a_i) \), ensures that when evaluated at this point, the remainder becomes equivalent to evaluating \( t(x) \) itself, as the multiplied components nullify the quotient term.
This understanding of the Remainder Theorem enriches the knowledge about polynomial division and shows why the remainder meets specific value conditions at interpolation points.

Polynomial Interpolation

Polynomial interpolation refers to the method of estimating values between unknown data points. The goal of polynomial interpolation is to find a polynomial that exactly goes through a set of known points. In our exercise, Lagrange Interpolation is used to achieve this. It seeks a polynomial, \( p(x) \), of a certain degree that matches the values of another polynomial at specific points, \( a_0, a_1, ..., a_n \). The interpolator \( p(x) \) has the characteristics that \( p(a_i) = t(a_i) \) for all given points, ensuring it provides an accurate pathway through the data.
The interpolation, when correctly applied, uses the nature of divisions and remainders to find the appropriate polynomial that fulfills the roles established by the given conditions. Thus, polynomial interpolation effectively bridges the gap between separately known values, presenting a coherent picture of the data as a whole.

Degree of a Polynomial

The degree of a polynomial indicates the highest power of the variable in the polynomial. This concept is crucial as it determines many properties of the polynomial, including the number of roots and the shape of its graph. In the context of this exercise, understanding the degree is essential when discussing the division and interpolation of polynomials. The divisor's degree, comprising factors \((x-a_0)(x-a_1)...(x-a_n)\), is \( n+1 \), enforcing that the remainder \( r(x) \) has a degree constraint of less than \( n+1 \).
This degree condition ensures that any polynomial created or found, such as the Lagrange interpolating polynomial, respects this boundary and possesses the suitability to model the problem accurately. Fixing the degree ensures that the interpolating polynomial uniquely fits the data, harmonizing with the guiding rules of interpolation in a structured manner.