Question

Given \(n+1\) data pairs \(\left(x_{0}, y_{0}\right),\left(x_{1}, y_{1}\right), \ldots,\left(x_{n}, y_{n}\right)\), define for \(j=0,1, \ldots, n\) the functions \(\rho_{j}=\prod_{i \neq j}\left(x_{j}-x_{i}\right)\), and let also \(\psi(x)=\prod_{i=0}^{n}\left(x-x_{i}\right)\) (a) Show that $$ \rho_{j}=\psi^{\prime}\left(x_{j}\right) . $$ (b) Show that the interpolating polynomial of degree at most \(n\) is given by $$ p_{n}(x)=\psi(x) \sum_{j=0}^{n} \frac{y_{j}}{\left(x-x_{j}\right) \psi^{\prime}\left(x_{j}\right)} $$

Short answer

Question: Prove (a) \(\rho_{j} = \psi'(x_{j})\) and (b) \(p_n(x) = \psi(x) \sum_{j=0}^n \frac{y_j}{(x-x_j)\psi'(x_j)}\) given the functions \(\rho_{j}\), \(\psi(x)\), and a set of \((x, y)\) data points. Solution: (a) We evaluated the derivative of \(\psi(x)\) and found that \(\psi'\left(x_j\right) = \sum_{i=0, i\neq j}^{n}\frac{\psi(x_j)}{x_j-x_i} = \rho_{j}\), proving the equality. (b) We showed that the interpolating polynomial \(p_n(x)\) can be represented as a linear combination of the given data points and proved that \(p_n(x) = \psi(x) \sum_{j=0}^n \frac{y_j}{(x-x_j)\psi'(x_j)}\), validating the given formula.

Step-by-step solution

(a) Proving \(\rho_{j} = \psi'(x_{j})\)

First, let's find the derivative of \(\psi(x)\): $$ \psi(x) = \prod_{i=0}^{n}(x-x_i) \\ \Rightarrow \psi'(x) = \sum_{j=0}^{n}\left(\frac{\psi(x)}{x-x_j}\right) $$ Now, let's evaluate it at \(x = x_j\): $$ \psi'\left(x_j\right) = \sum_{i=0}^{n}\frac{\psi(x_j)}{x_j-x_i} $$ Since we know that: $$ \rho_{j} = \prod_{i\neq j}(x_j-x_i) $$ So when we substitute \(x = x_j\), we get: $$ \psi'\left(x_j\right) = \sum_{i=0, i\neq j}^{n}\frac{\psi(x_j)}{x_j-x_i} = \rho_{j} \;\text{(Proved)} $$

(b) Proving the given formula for the interpolating polynomial

The interpolating polynomial \(p_n(x)\) can be represented as a linear combination of the given data points: $$ p_n(x) = \sum_{j=0}^n y_j L_j(x) $$ Where \(L_j(x)\) are the Lagrange basis polynomials defined as: $$ L_j(x) = \frac{\psi(x)}{(x-x_j)\psi'(x_j)} $$ So we need to prove that: $$ p_n(x) = \psi(x) \sum_{j=0}^n \frac{y_j}{(x-x_j)\psi'(x_j)} $$ Using the Lagrange basis polynomials: $$ p_n(x) = \sum_{j=0}^n y_j L_j(x) = \sum_{j=0}^n y_j \frac{\psi(x)}{(x-x_j)\psi'(x_j)} \\ \Rightarrow p_n(x) = \psi(x) \sum_{j=0}^n \frac{y_j}{(x-x_j)\psi'(x_j)} \;\text{(Proved)} $$ Thus, we have proved the given formula for the interpolating polynomial \(p_n(x)\).

Key concepts

Lagrange Basis Polynomials

When it comes to understanding the fundamentals of polynomial interpolation, the concept of Lagrange basis polynomials plays a pivotal role. These polynomials are used to form an interpolating polynomial that passes through a given set of points. Imagine you have a set of points on a graph and you want to draw a smooth curve that connects all the points without any breaks or sharp turns. This is essentially what the Lagrange basis polynomials allow you to do.

Lagrange basis polynomials are particularly nifty because they provide a way to construct such a curve in a piece-by-piece manner. For every data point \( x_j, y_j \), there's a corresponding Lagrange basis polynomial \( L_j(x) \). The magic formula for each of these polynomials is \[ L_j(x) = \frac{\prod_{i=0, ieq j}^{n}(x-x_i)}{\prod_{i=0, ieq j}^{n}(x_j-x_i)} \]. Intuitively, this formula ensures that \( L_j(x) \) is equal to 1 at \( x_j \) and 0 at all other data points \( x_i \) where \( i eq j\).

Using these individual pieces, one can assemble the full interpolating polynomial, which beautifully weaves through each data point, creating a curve that is as smooth as silk. The step-by-step solution earlier described how to derive the Lagrange basis polynomials, and their role is pivotal in constructing the overall solution to a polynomial interpolation problem.

Numerical Methods

Numerical methods are the tools that make it possible to solve problems like polynomial interpolation, where abstract mathematical functions may not suffice. In essence, they're a collection of techniques used to form approximate solutions to complex problems, often involving a large amount of data or calculations that would be impractical to compute analytically.

One of the golden stars in the arena of numerical methods is polynomial interpolation. The essence of this approach lies in taking a discrete set of data points and estimating the function that would best represent the underlying trend or pattern. It's like trying to find the perfect (mathematical) journey that visits all the 'cities' defined by your data points, without taking any detours.

The exercise above elegantly illustrates how to apply numerical methods, specifically the interpolation technique, to construct a function that honors the data points provided. The process of proving that \( \rho_{j} = \psi'(x_{j}) \) and establishing the formula for the interpolating polynomial draws deeply from numerical analysis concepts. The fact that these methods give us the power to construct complex models from relatively simple operations is a testament to their incredible utility in fields like engineering, physics, and computer science.

Polynomial Interpolation

Polynomial interpolation is a type of interpolation where a polynomial function is selected to approximate the relationship among a set of data points. It's like drawing a path that connects all your favorite spots in a park, making sure your journey is smooth and takes you through every spot without backtracking.

In polynomial interpolation, the goal is to create a polynomial that passes exactly through a set of given points \( (x_0, y_0), (x_1, y_1), \ldots, (x_n, y_n) \). The exercise we looked at provides a clear-cut example of how this is done by constructing the interpolating polynomial \( p_n(x) \). The process involves creating a sum of the products of data points \( y_j \) and the corresponding Lagrange basis polynomials.The elegance of this method is that it doesn't just give you a function that passes through the points but does so with the lowest possible degree, ensuring the smoothest and simplest curve. Additionally, the formula \[ p_n(x) = \psi(x) \sum_{j=0}^{n} \frac{y_j}{(x-x_j)\psi'(x_j)} \] is central to polynomial interpolation, tying together the concept of derivatives with the original function to ensure accuracy at the given data points. This method has a wide array of applications, from scientific computing to financial modeling, wherever the prediction of outcomes based on historical data is vital.