Question

Given the following data pairs, find the interpolating polynomial of degree at most 3 and estimate the value of \(y\) corresponding to \(x=1.5\). a. (0,1),(1,2),(2,5),(3,10) b. (0,1),(1,1.49),(2,-0.42),(3,-11.33) c. (0,2),(1,2.03),(2,-0.40),(-1,0.89)

Short answer

Use Lagrange interpolation for each dataset to estimate the polynomial, then evaluate at \(x=1.5\).

Step-by-step solution

Understand the Problem

We need to find an interpolating polynomial of degree at most 3 that fits the given data pairs, and then estimate the value of \(y\) for \(x=1.5\). This can be done using methods like Lagrange interpolation or Newton's divided differences.

Constructing Lagrange Interpolating Polynomial

The Lagrange polynomial for a set of points \((x_i, y_i)\) is given by: \[ P(x) = \sum_{i=0}^{n} y_i L_i(x) \] where \( L_i(x) = \prod_{\substack{j=0 \ j eq i}}^{n} \frac{x - x_j}{x_i - x_j} \). Use this formula to construct a degree 3 polynomial for each dataset.

Applying Lagrange Formula to Dataset a

For dataset a: points are (0,1), (1,2), (2,5), (3,10). Calculate the Lagrange basis polynomials \(L_i(x)\) and then use them to construct \(P(x)\).

Calculating L_A(x) for Dataset a

Compute \(P_A(x)\) as follows: \[ L_0(x) = \frac{(x-1)(x-2)(x-3)}{(0-1)(0-2)(0-3)} \] \[ L_1(x) = \frac{(x-0)(x-2)(x-3)}{(1-0)(1-2)(1-3)} \] \[ L_2(x) = \frac{(x-0)(x-1)(x-3)}{(2-0)(2-1)(2-3)} \] \[ L_3(x) = \frac{(x-0)(x-1)(x-2)}{(3-0)(3-1)(3-2)} \] Substitute back into \( P_A(x) = \sum y_i L_i(x) \) and simplify.

Estimate y for x=1.5 in Dataset a

Replace \(x\) with 1.5 in the polynomial \(P_A(x)\) calculated in Step 4 to estimate \(y\).

Repeat Steps for Datasets b and c

Repeat steps 3 to 5 for datasets b and c using their respective points (0,1),(1,1.49),(2,-0.42),(3,-11.33) and (0,2),(1,2.03),(2,-0.40),(-1,0.89).

Key concepts

Interpolating Polynomial

An interpolating polynomial is a special type of polynomial used to estimate or predict values within a range of known data points. It "fits" through a given set of data pairs, allowing us to estimate unknown values smoothly.
For instance, if you have data points like (0,1), (1,2), (2,5), and (3,10), you can create an interpolating polynomial of degree at most three that passes through all these points. The polynomial acts almost like a roadmap, guiding you to predict values like where the line might be at a point like 1.5. Here, we are requested to estimate the unknown value of \(y\) for a given \(x\) value using this polynomial.

Newton's Divided Differences

Newton's divided differences is a method for polynomial interpolation, often used to construct an interpolating polynomial in a form that easily allows for adding new points. This method isn't used explicitly here, but it's helpful to understand its basis.
In Newton's method, the polynomial is represented as a combination of terms that include successively higher powers of \(x\), each multiplied by coefficients calculated from the known data. These coefficients come from divided differences, offering a flexible approach to build the interpolating polynomial and easily modify it if new data becomes available.

Lagrange Basis Polynomials

Lagrange basis polynomials form the backbone of the Lagrange interpolation method. They act as building blocks, simplifying the construction of the interpolating polynomial.
The basic idea consists of using specific polynomials \(L_i(x)\) such that each \(L_i(x)\) is zero at all given \(x\) points except the corresponding \(x_i\). For a degree three polynomial, you'll have four of these basis polynomials if you're working with four data points:

• \(L_0(x)\) vanishes at all points except \(x_0\)
• \(L_1(x)\) vanishes at all points except \(x_1\)
• So on for \(L_2(x)\) and \(L_3(x)\)

These polynomials are then weighted by their corresponding \(y\) values. Finally, the weighted sum of these Lagrange basis polynomials gives the full interpolating polynomial.

Polynomial Estimation

Polynomial estimation is the final step, where the constructed interpolating polynomial is used to estimate an unknown value. In practice, this involves plugging a specific \(x\) value into the polynomial.
For example, after constructing the polynomial from your data points, you replace \(x\) with the desired value, like 1.5, to estimate the corresponding \(y\). This way, polynomial estimation helps predict unknown values based on existing data points, providing a powerful tool for analysis and forecasting.
Understanding how polynomial estimation works with interpolating polynomials is crucial, as it gives you a method to deduce or fill in gaps within your data.