Question

Let \(F\) be a field, \(u_{0}, \ldots, u_{n-1} \in F \backslash\\{0\\}\) with \(u_{i} \neq \pm u_{j}\) for \(0 \leq i

Short answer

(i) Proved: all odd terms must cancel out. (ii) Even Lagrange basis. (iii) \(f(x) = g(x^2)\). (iv) Use \(f(x) = x \cdot g(x^2)\). (v) Evaluate trigonometric values at specified points.

Step-by-step solution

Understanding Even Polynomial

For part (i), the task is to show that if a polynomial \(f\) satisfies \(f(u_i) = f(-u_i)\) for distinct non-zero field elements \(u_i\) and its degree is less than \(2n\), then \(f(x) = f(-x)\) for all \(x\). Notice that the condition of equality at \(u_i\) and \(-u_i\) for a polynomial of degree less than 2n suggests that all odd terms must cancel out, leaving only even terms. This is because a polynomial can have at most \(2n - 1\) roots, and the given condition provides \(2n\) conditions (\(u_i\) and \(-u_i\) for each \(u_i\)) if it had any odd terms.

Using Lagrange Interpolation for Even Function

For part (ii), the Lagrange interpolation formula creates a polynomial that passes through given points. Since \(f(x) = f(-x)\), we construct Lagrange basis polynomials that are even by including terms \(L_i(x) = \prod_{j eq i} \frac{(x^2-u_j^2)}{(u_i^2-u_j^2)}\). This modification to squares ensures evenness. Since the function is even and polynomial is of degree less than \(2n\), it interpolates exactly at points \(u_i\) and \(-u_i\) with \(v_i\).

Relation of f and g through Degree Reduction

In part (iii), \(g\) is a polynomial of degree less than \(n\), satisfying \(g(u_i^2) = v_i\). The polynomial \(f\) mentioned in (ii) can be expressed as \(f(x) = g(x^2)\). This substitution ensures \(f\) is even, and it exactly correlates with \(g\)'s interpolation over squared terms.

Odd Interpolating Polynomials

For part (iv), to find an odd interpolating polynomial, the condition should be \(f(u_i) = -f(-u_i)\). Analogous to even interpolation, odd interpolation would include constructing a polynomial of the form \(f(x) = x \cdot g(x^2)\), where \(g(x)\) handles the interpolation through squared values of \(x\). The degree of \(f\) would be \(2n-1\).

Even and Odd Polynomial for Trigonometric Functions

In part (v), for even polynomial \(f_0\), we use known values of \(\cos(x)\) at \(x = \pi/6, \pi/3, \pi/2\), which results in points \((\pi/6, \sqrt{3}/2), (\pi/3, 1/2), (\pi/2, 0)\). We create an interpolating polynomial through these, ensuring even terms. Similarly, for odd \(f_1\), interpolating values from \(\sin(x)\) at the previous points.

Key concepts

Even Polynomials

Even polynomials are special types of polynomials where every term has an even degree. This means each term has exponents like 0, 2, 4, etc. For a polynomial function \( f(x) \), it is termed as "even" if it satisfies \( f(x) = f(-x) \) for all \( x \). This condition implies that the graph of an even polynomial is symmetric about the y-axis.

• This symmetry ensures that substituting \( x \) with \( -x \) doesn't change the value of the polynomial.
• Even polynomials include familiar functions like constant terms (degree zero) and quadratic polynomials (degree two).

Understanding even polynomials is crucial in many mathematical contexts, such as solving problems involving symmetry and even functions. They frequently appear in interpolation tasks, where symmetries simplify the calculation and understanding of the relationship between data points and the polynomial. Interpolating polynomials with even characteristics are constructed by ensuring that terms of odd degree cancel out, leaving only the even components.

Lagrange Interpolation

Lagrange interpolation is a popular technique for constructing a polynomial that passes precisely through a given set of data points. It is particularly useful in numerical analysis and approximation, as it builds a polynomial that perfectly fits specific input-output pairs.

• Given a set of points \( (x_0, y_0), (x_1, y_1), ..., (x_n, y_n) \), Lagrange interpolation forms the polynomial \( P(x) = \sum_{i=0}^{n} y_i L_i(x) \).
• The basis polynomial \( L_i(x) \) is calculated as \( \prod_{j eq i} \frac{(x-x_j)}{(x_i-x_j)} \), designed to equal one at \( x_i \) and zero at all other \( x_j \).
• In the context of even polynomials, the Lagrange basis can be adjusted to ensure evenness by constructing it with squares of the terms, ensuring symmetry in the resultant polynomial.

This method is powerful because it provides precise control over the polynomial's behavior at specific points. It is invaluable in numerical applications where interpolation between known data values is required, offering exact values at those points and a smooth estimate for values in between.

Odd Polynomials

Odd polynomials are characterized by their terms having odd degrees, such as 1, 3, 5, and so on. For a function \( f(x) \), it is termed "odd" if it satisfies the property \( f(-x) = -f(x) \) for all \( x \).

• This property indicates that the graph of an odd polynomial exhibits rotational symmetry around the origin.
• Typical examples include linear functions (degree one) and cubic functions (degree three).

In interpolation, odd polynomials are encountered when the data or function being approximated inherently shows anti-symmetric properties. This means the output at \( x \) is the opposite of that at \( -x \). To create an odd interpolating polynomial, methods similar to those for even polynomials are used, but ensuring that the polys remain anti-symmetric by incorporating odd degree terms such as \( x \cdot g(x^2) \). This construction maintains the desired characteristic of reversing sign at symmetric input points.