Question

Given a sequence \(y_{0}, y_{1}, y_{2}, \ldots\), define the backward difference operator \(\nabla\) by $$ \nabla y_{i}=y_{i}-y_{i-1} $$ Powers of \(\nabla\) are defined recursively by $$ \begin{aligned} \nabla^{0} y_{i} &=y_{i} \\ \nabla^{j} y_{i} &=\nabla\left(\nabla^{j-1} y_{i}\right), \quad j=1,2, \ldots . \end{aligned} $$ Thus, \(\nabla^{2} y_{i}=\nabla\left(y_{i}-y_{i-1}\right)=y_{i}-2 y_{i-1}+y_{i-2}\), etc. Consider polynomial interpolation at equispaced points, \(x_{i}=x_{0}+i h, i=0,1, \ldots, n\) (a) Show that $$ f\left[x_{n}, x_{n-1}, \ldots, x_{n-j}\right]=\frac{1}{j ! h^{j}} \nabla^{j} f\left(x_{n}\right) $$ [Hint: Use mathematical induction.] (b) Show that the interpolating polynomial of degree at most \(n\) is given by the Newton backward difference formula $$ p_{n}(x)=\sum_{j=0}^{n}(-1)^{j}\left(\begin{array}{l} s \\ j \end{array}\right) \nabla^{j} f\left(x_{n}\right) $$ where \(s=\frac{x_{n}-x}{h}\) and \(\left(\begin{array}{l}s \\\ j\end{array}\right)=\frac{s(s-1) \cdots(s-j+1)}{j !}\) (with \(\left(\begin{array}{l}s \\ 0\end{array}\right)=1\) ).

Short answer

In this problem, we were given a sequence and the definition of a backward difference operator. We needed to find the polynomial interpolation for equispaced points using the Newton backward difference formula. We approached this in two parts - first, by finding a relationship between the divided difference and the backward difference operator, which we proved using mathematical induction. Then, we found the interpolating polynomial using the Newton backward difference formula and the relationship previously derived. The interpolating polynomial is given by: $$ p_{n}(x) = \sum_{j=0}^{n} \left(\begin{array}{l} s \\\ j \end{array}\right) (-1)^j \nabla^{j} f(x_n) $$ where \(s=\frac{x_{n}-x}{h}\) and $\left(\begin{array}{l}s \\\ j\end{array}\right)=\frac{s(s-1) \cdots(s-j+1)}{j!}$.

Step-by-step solution

Part (a) - Relationship between divided difference and backward difference operator

Using mathematical induction, we will show that $$ f[x_n, x_{n-1}, \ldots, x_{n-j}] = \frac{1}{j ! h^{j}} \nabla^{j} f(x_{n}) $$ Base case (j=0): $$ f[x_n] = \frac{1}{0 ! h^{0}} \nabla^{0} f(x_{n}) $$ Since \(\nabla^0y_i = y_i\), and \(0!h^0=1\), the base case is true. Inductive step: Assume that the statement is true for j=k: $$ f[x_n, x_{n-1}, \ldots, x_{n-k}] = \frac{1}{k ! h^{k}} \nabla^{k} f(x_{n}) $$ Now we need to show that it is true for j=k+1: $$ f[x_n, x_{n-1}, \ldots, x_{n-k-1}] = \frac{1}{(k+1) ! h^{k+1}} \nabla^{k+1} f(x_{n}) $$ By the definition of divided differences, we know: $$ f[x_n, x_{n-1}, \ldots, x_{n-k-1}] = \frac{f[x_{n}, x_{n-1}, \ldots, x_{n-k}] - f[x_{n-1}, x_{n-2}, \ldots, x_{n-k-1}]}{h} $$ Using the inductive hypothesis: $$ \frac{1}{h} \left[\frac{1}{k ! h^{k}} \nabla^{k} f(x_{n}) - \frac{1}{k ! h^{k}} \nabla^{k} f(x_{n-1}) \right] = \frac{1}{(k+1) ! h^{k+1}} \nabla^{k+1} f(x_{n}) $$ Since \(\nabla^{k+1} f(x_{n}) = \nabla^{k} f(x_{n}) - \nabla^{k} f(x_{n-1})\): $$ \frac{1}{k! h^{k+1}} \nabla^{k+1} f(x_{n}) = \frac{1}{(k+1)! h^{k+1}} \nabla^{k+1} f(x_{n}) $$ This shows that the statement holds for j=k+1. By mathematical induction, the statement is true for all integer values of j.

Part (b) - Interpolating polynomial using the Newton backward difference formula

To find the interpolating polynomial of degree at most n, we need to find \(p_n(x)\) using the Newton backward difference formula: $$ p_{n}(x)=\sum_{j=0}^{n}(-1)^{j}\left(\begin{array}{l} s \\\ j \end{array}\right) \nabla^{j} f\left(x_{n}\right) $$ where \(s=\frac{x_{n}-x}{h}\) and $\left(\begin{array}{l}s \\\ j\end{array}\right)=\frac{s(s-1) \cdots(s-j+1)}{j!}$ From part (a), we know that $$ f[x_n, x_{n-1}, \ldots, x_{n-j}] = \frac{1}{j! h^{j}} \nabla^j f(x_n) $$ By replacing the divided difference in the Newton backward difference formula: $$ p_n(x) = f(x_n) + (x - x_n)\frac{1}{h}\nabla f(x_n) + \frac{(x - x_n)(x - x_n + h)}{2!h^2}\nabla^2 f(x_n) + \cdots + \frac{(x - x_n)\cdots(x - x_n + (n-1)h)}{n!h^n} \nabla^n f(x_n) $$ As the problem asks for the interpolating polynomial in terms of s, we substitute s back: $$ p_{n}(x) = \sum_{j=0}^{n} \left(\begin{array}{l} s \\\ j \end{array}\right) (-1)^j \nabla^{j} f(x_n) $$ This is the interpolating polynomial using the Newton backward difference formula.

Key concepts

Polynomial Interpolation

Polynomial interpolation is a numerical method used to estimate values between known data points. When we have a set of points, and we want to find a polynomial that passes through all these points, we are engaging in what's called interpolation. Often, the points are given in the form of discrete data, like a table of values or results from an experiment or simulation. The goal is to construct a single polynomial that approximates the function which generated the data.

Interpolation is particularly useful because it allows us to predict or estimate the value of the function at points where we do not have data. We assume that the polynomial captures the underlying trend of the dataset. Polynomial interpolation can be done using various methods, including Lagrange and Newton forms, which are particularly handy for theoretical work and computational applications, respectively. Each of these methods has its unique advantages depending on the context in which it's being used.

For equispaced points, which are points spaced at regular intervals along the x-axis, polynomial interpolation becomes a bit more straightforward because the distance between each point, known as 'h', is constant. This consistency can simplify computations and is particularly relevant to the Newton backward difference formula, which we'll explore further in the context of the Newton backward difference formula.

Numerical Methods

Numerical methods are techniques used to obtain approximate solutions to mathematical problems that may not be solvable by exact, analytical means. They play a critically important role in engineering, physics, finance, and many more fields where precise solutions are often out of reach.

Numerical methods cover a vast range of algorithms and techniques, such as root-finding for non-linear equations, numerical integration and differentiation, and solving systems of linear equations. A key feature of these methods is iteration — many numerical methods rely on an iterative process to improve on a guess of the solution. This is because iterative methods often start with an initial approximation and refine this guess to get closer and closer to the desired solution.

The backward difference operator, a concept from numerical methods, is instrumental when dealing with discrete data. It is especially useful in numerical differentiation and integration, which predicts the rate of change and the accumulation of quantities, respectively. The simplicity and versatility of this operator illustrate the power of numerical methods in solving practical problems. Students often encounter numerical methods because they allow for the practical application of mathematical concepts to real-world problems, enabling them to grasp these concepts more effectively.

Newton Backward Difference Formula

The Newton backward difference formula is a powerful tool in polynomial interpolation for estimating the shape of a function based on known data points. This formula is particularly beneficial when interpolating at the end of a set of data points, which is where the 'backward' part comes in — it works from the last known value and estimates the function going backward.

The formula is a finite difference version of a Taylor series expansion around a particular point and leverages the backward difference operator, (abla). Due to the formula's structure, it's ideally suited for instances where we have equispaced points as it simplifies the representation and calculation of the polynomial coefficients.

When applying the Newton backward difference formula, the term 's' represents the distance, in units of 'h', between the x-value for which we seek an estimated y-value and the last known x-value, (x_n). The binomial coefficients that appear in the formula are a function of 's' and have a practical interpretation: they determine the weight each successive term in the polynomial has when estimating the value of the interpolating polynomial at a given point.

An Example in Action

Using the given exercise, students can witness how the values of the backward differences directly impact the resulting polynomial, giving concrete insight into how the formula constructs the estimate. The Newton backward difference formula illustrates the underlying concept of polynomial interpolation, using backward differences to simplify the calculation and presentation of the interpolating polynomial.