Question

Use the known values of the function \(\sin (x)\) at \(x=0, \pi / 6, \pi / 4, \pi / 3\) and \(\pi / 2\) to derive an interpolating polynomial \(p(x)\). What is the degree of your polynomial? What is the interpolation error magnitude \(|p(1.2)-\sin (1.2)| ?\)

Short answer

Answer: The degree of the interpolating polynomial is 4. To calculate the interpolation error magnitude at x = 1.2, we need to find the absolute difference between the true sine value and the polynomial's value at this point, i.e., |p(1.2) - sin(1.2)|.

Step-by-step solution

Understanding Lagrange Interpolation

Lagrange interpolation is a method to find a polynomial that passes through a given set of points \((x_0, y_0), (x_1, y_1), ..., (x_n, y_n)\). The polynomial can be found using the following formula: $$ p(x) = \sum_{i=0}^{n} y_i L_i(x) $$ where $$ L_i(x) = \prod_{j=0, j\neq i}^{n} \frac{x - x_j}{x_i - x_j} $$

Given Values and Setup

We are given the known values of the sine function at the following points: $$ x = 0, \frac{\pi}{6}, \frac{\pi}{4}, \frac{\pi}{3}, \frac{\pi}{2} $$ and their corresponding sine values are: $$ \sin(0) = 0, \sin(\frac{\pi}{6}) = \frac{1}{2}, \sin(\frac{\pi}{4}) = \frac{\sqrt{2}}{2}, \sin(\frac{\pi}{3}) = \frac{\sqrt{3}}{2}, \sin(\frac{\pi}{2}) = 1 $$ Now, we will use the Lagrange interpolation method to find a polynomial \(p(x)\) that passes through these points.

Calculating the Lagrange functions \(L_i(x)\)

Using the formula for \(L_i(x)\), we can calculate the 5 Lagrange functions for our given points: $$ L_0(x) = \frac{(x - \frac{\pi}{6})(x - \frac{\pi}{4})(x - \frac{\pi}{3})(x - \frac{\pi}{2})}{(0 - \frac{\pi}{6})(0 - \frac{\pi}{4})(0 - \frac{\pi}{3})(0 - \frac{\pi}{2})} $$ $$ L_1(x) = \frac{(x - 0)(x - \frac{\pi}{4})(x - \frac{\pi}{3})(x - \frac{\pi}{2})}{(\frac{\pi}{6} - 0)(\frac{\pi}{6} - \frac{\pi}{4})(\frac{\pi}{6} - \frac{\pi}{3})(\frac{\pi}{6} - \frac{\pi}{2})} $$ $$ L_2(x) = \frac{(x - 0)(x - \frac{\pi}{6})(x - \frac{\pi}{3})(x - \frac{\pi}{2})}{(\frac{\pi}{4} - 0)(\frac{\pi}{4} - \frac{\pi}{6})(\frac{\pi}{4} - \frac{\pi}{3})(\frac{\pi}{4} -\frac{\pi}{2})} $$ $$ L_3(x) = \frac{(x - 0)(x -\frac{\pi}{6})(x - \frac{\pi}{4})(x - \frac{\pi}{2})}{(\frac{\pi}{3} - 0)(\frac{\pi}{3} - \frac{\pi}{6})(\frac{\pi}{3} - \frac{\pi}{4})(\frac{\pi}{3} - \frac{\pi}{2})} $$ $$ L_4(x) = \frac{(x - 0)(x -\frac{\pi}{6})(x - \frac{\pi}{4})(x - \frac{\pi}{3})}{(\frac{\pi}{2} - 0)(\frac{\pi}{2} -\frac{\pi}{6})(\frac{\pi}{2} - \frac{\pi}{4})(\frac{\pi}{2} - \frac{\pi}{3})} $$

Finding the polynomial \(p(x)\)

Now, we plug the values of the Lagrange functions \(L_i(x)\) and the sine values into the formula for the interpolating polynomial, \(p(x)\): $$ p(x) = 0 \cdot L_0(x) + \frac{1}{2} \cdot L_1(x) + \frac{\sqrt{2}}{2} \cdot L_2(x) + \frac{\sqrt{3}}{2} \cdot L_3(x) + 1 \cdot L_4(x) $$ Simplifying the polynomial, we can determine its degree: Since the denominator in each \(L_i(x)\) has terms of degree one, and there are four factors in each numerator, the interpolating polynomial will be of degree 4.

Calculating the Interpolation Error

Finally, we need to find the interpolation error magnitude at \(x = 1.2\). To do this, we will calculate the difference between the true sine value and the polynomial's value at this point: $$ |p(1.2) - \sin (1.2)| $$ First plug, in \(x = 1.2\) into \(p(x)\) polynomial, and then find the absolute difference with sine value at \(1.2\), this will give the interpolation error. Note: The error calculation might require the use of a calculator or a computer algebra system to find the numerical values of the polynomial and sine function at \(x = 1.2\).

Key concepts

Interpolating Polynomial

An interpolating polynomial is a powerful mathematical tool used to approximate a function. The idea behind it is to construct a polynomial that exactly passes through a given set of points on a graph. This is incredibly useful when you want to estimate values of a function that are not directly provided.

The Lagrange interpolating polynomial is one popular method to achieve this. Given a set of data points, the polynomial is formulated using Lagrange basis functions, denoted by \( L_i(x) \) in the original solution. These basis functions are derived from the provided points' coordinates, ensuring that the polynomial attains the exact function values at these points.

• Each \( L_i(x) \) is zero at all data points except for the\( i^{th} \), where it equals one.
• The interpolating polynomial \( p(x) \) becomes a sum of these basis functions, weighted by the function values at the points.

With this approach, the polynomial precisely matches the known values, and we can use it to interpolate or estimate the values of the function at other points.

Interpolation Error

Interpolation error is the difference between the actual value of the function and the value estimated by the interpolating polynomial. Understanding this concept is crucial because the polynomial may not perfectly represent the function between the given data points.

In mathematical terms, the interpolation error at a point \( x \) is given by \( |p(x) - f(x)| \), where \( f(x) \) is the true function value, and \( p(x) \) is the polynomial value at \( x \).

• It is essential to realize that while the interpolating polynomial fits perfectly at the data points, the accuracy in between those points can vary.
• The error often increases with the distance from the nearest data points.

For functions that are well-behaved, the error tends to be smaller, but for functions that change rapidly, significant errors can arise. In cases like our exercise, estimating the error helps in understanding how well the polynomial approximates the actual sine function at points like \( x = 1.2 \). This understanding can guide us in either trusting the interpolation or being cautious about it.

Degree of Polynomial

The degree of a polynomial is simply the highest power of the variable, \( x \), present in the polynomial's expression. The degree is an important factor in interpolation because it affects the polynomial's shape and behavior.

In Lagrange interpolation, the degree of the interpolating polynomial is generally one less than the number of given data points. Given five points in our exercise, the polynomial is of degree four. This indicates that the polynomial has up to \( x^4 \) terms, allowing it to flexibly fit the data points through adjustments in the coefficients of these terms.

• A higher degree polynomial can fit more complex data sets, providing a potentially better approximation.
• However, high degree polynomials are prone to oscillate, leading to larger potential interpolation errors between the known points.

Understanding the degree helps not only in constructing interpolation polynomials but also in predicting their behavior: smoother when degree is low and potentially more accurate but complex with higher degrees.