Interpolating polynomial for sin([itex]\pi{x}[/itex])

[alphabeta89]

Homework Statement

Consider the function sin([itex]\pi[/itex][itex]x[/itex]) on [-1,1] and its approximations by interpolating polynomials. For integer [itex]n[/itex][itex]\geq[/itex]1, let [itex]x_{n,j}=-1+\frac{2j}{n}[/itex] for [itex]j=0,1,...,n[/itex], and let [itex]p_{n}(x)[/itex] be the [itex]n[/itex]th-degree polynomial interpolating sin([itex]\pi[/itex][itex]x[/itex]) at the nodes [itex]x_{n,0},...,x_{n,n}[/itex]. Prove that

[itex]\max_{x\in[-1,1]}\left | {sin}{(\pi{x})-p_{n}{(x)}} \right | \to 0[/itex] as [itex]n \to \infty[/itex]

Homework Equations

The Attempt at a Solution

I have no idea how to start!

 

[mighty2000]

Hello there! This is a very interesting problem that involves understanding the concept of polynomial interpolation and its relationship to the sine function. Let's break down the problem step by step.

First, we need to understand what polynomial interpolation is. Polynomial interpolation is a method of approximating a function by using a polynomial that passes through a given set of points. In this case, we are looking at interpolating the sine function at the points x_{n,0},...,x_{n,n}.

Next, we need to understand the notation used in the problem. The notation \max_{x\in[-1,1]} represents the maximum value of the function over the interval [-1,1]. So, we are essentially trying to find the maximum difference between the sine function and the nth-degree polynomial approximation over the interval [-1,1].

Now, let's take a closer look at the nodes x_{n,0},...,x_{n,n}. These nodes are equally spaced points on the interval [-1,1]. In other words, the distance between any two consecutive nodes is \frac{2}{n}. This will be important in our proof.

Next, we need to understand what it means for a sequence of polynomials to converge to a function. In this case, as n \to \infty, we want to show that the interpolating polynomials p_{n}(x) converge to the sine function over the interval [-1,1]. In other words, as n \to \infty, the maximum difference between the sine function and the nth-degree polynomial approximation over the interval [-1,1] approaches 0.

To prove this, we can use the Weierstrass approximation theorem, which states that any continuous function on a closed interval can be approximated by a polynomial with arbitrary precision. Since the sine function is continuous on the interval [-1,1], we can use this theorem to show that as n \to \infty, the nth-degree polynomial approximation p_{n}(x) will converge to the sine function over the interval [-1,1].

Now, let's look at the maximum difference between the sine function and the nth-degree polynomial approximation over the interval [-1,1]. As mentioned earlier, the distance between any two consecutive nodes is \frac{2}{n}. As n \to \infty, this distance approaches 0, which means that the nodes x_{n,0},...,x_{n,n}