Question

$$ \begin{aligned} &\text { Bestimmen Sie das LAGRANGE-Polynom zur Interpolation der Messwerte }\\\ &\begin{array}{ccccc} \hline x_{i} & 1 & 2 & 3 & 5 \\ \hline y_{i} & 0 & 3 & 2 & 1 \\ \hline \end{array} \end{aligned} $$

Short answer

The Lagrange interpolating polynomial is \(-\frac{x^3}{4} + \frac{13x^2}{12} - \frac{71x}{24} + \frac{25}{12}\).

Step-by-step solution

Identify Lagrange Polynomial Form

A Lagrange polynomial is given by \( L(x) = \sum_{i=0}^{n} y_i \cdot l_i(x) \), where each \( l_i(x) \) is a Lagrange basis polynomial calculated for the given data points.

Define Lagrange Basis Polynomials

The Lagrange basis polynomial for a particular index \( i \) is \( l_i(x) = \prod_{\substack{0 \leq j \leq n \ j eq i}} \frac{x-x_j}{x_i-x_j} \). For our data points (1, 0), (2, 3), (3, 2), and (5, 1), calculate \( l_0(x), l_1(x), l_2(x), \) and \( l_3(x) \).

Calculate \( l_0(x) \)

Using the formula \( l_0(x) = \frac{(x-2)(x-3)(x-5)}{(1-2)(1-3)(1-5)} = \frac{(x-2)(x-3)(x-5)}{-24} \), expand to obtain the polynomial form: \( l_0(x) = -\frac{1}{24}(x-2)(x-3)(x-5) \).

Calculate \( l_1(x) \)

Using \( l_1(x) = \frac{(x-1)(x-3)(x-5)}{(2-1)(2-3)(2-5)} = \frac{(x-1)(x-3)(x-5)}{6} \), expand to obtain the polynomial form: \( l_1(x) = \frac{1}{6}(x-1)(x-3)(x-5) \).

Calculate \( l_2(x) \)

For \( l_2(x) = \frac{(x-1)(x-2)(x-5)}{(3-1)(3-2)(3-5)} = -\frac{1}{4}(x-1)(x-2)(x-5) \), expand to obtain the polynomial form: \( l_2(x) = -\frac{1}{4}(x-1)(x-2)(x-5) \).

Calculate \( l_3(x) \)

Calculate \( l_3(x) = \frac{(x-1)(x-2)(x-3)}{(5-1)(5-2)(5-3)} = \frac{1}{24}(x-1)(x-2)(x-3) \), expand to obtain the polynomial form: \( l_3(x) = \frac{1}{24}(x-1)(x-2)(x-3) \).

Assemble the Lagrange Polynomial

Combine the basis polynomials to form \( L(x) = y_0 \cdot l_0(x) + y_1 \cdot l_1(x) + y_2 \cdot l_2(x) + y_3 \cdot l_3(x) \). Substitute \( y_0 = 0, y_1 = 3, y_2 = 2, y_3 = 1 \) and expressions found into the formula.

Simplify the Expression

Simplify the expression step-by-step to obtain the polynomial. It results in \( L(x) = -\frac{x^3}{4} + \frac{13x^2}{12} - \frac{71x}{24} + \frac{25}{12} \).

Key concepts

Polynomial Interpolation

Polynomial interpolation is a method of estimating values between known data points. It's like drawing a smooth curve to connect the dots in a scatter plot. In the context of numerical analysis, polynomial interpolation is used to construct a polynomial that passes through a given set of points. This is particularly helpful when you have a handful of data points and wish to infer or predict values that are not explicitly given.

When you engage in polynomial interpolation, you aim to find a polynomial of the lowest possible degree that fits the known data perfectly. For instance, given points \(x_1, y_1\), \(x_2, y_2\), ..., etc., the problem is to find a polynomial \(P(x)\) such that \(P(x_i) = y_i\) for each \(i\). This requires solving a set of equations, often resulting in a unique polynomial if the number of data points is respected.

Basis Polynomials

Basis polynomials are the building blocks of Lagrange polynomial interpolation. In Lagrange interpolation, every data point is associated with its own basis polynomial. The Lagrange basis polynomial is designed such that it equals one at its own data point and zero at all other data points.

The mathematical definition for a Lagrange basis polynomial \(l_i(x)\) is given by:

• \( l_i(x) = \prod_{\substack{j=0 \ j eq i}}^{n} \frac{x-x_j}{x_i-x_j} \)

This formula will yield a polynomial that will be 1 at \(x = x_i\) and 0 at all other inputs. This property makes it extremely useful when assembling the final Lagrange polynomial since it ensures each term only contributes to its corresponding y-value. Basis polynomials serve the crucial role of isolating the influence of each data point in the interpolation process.

Numerical Analysis

Numerical analysis is a field of mathematics that focuses on finding approximate solutions to complex problems. Polynomial interpolation fits right into this field, as it allows for approximations of functions based on discrete data samples.

This technique is fundamental in numerical analysis because it provides a way to estimate values and evaluate functions that are otherwise difficult to compute directly. For example, if you have data that represents the changing temperature on a day, numerical analysis can help you interpolate and understand the likely temperature at any given time. Polynomial interpolation in particular is favored for its computational simplicity and effectiveness in various applications such as engineering and physical sciences. It serves as a key tool for extracting meaningful information from experimental data.

Interpolation Formula

The interpolation formula in Lagrange polynomial interpolation combines all the Lagrange basis polynomials multiplied by their respective y-values. This approach ensures that the polynomial accurately goes through each specified data point.

Defined more formally, for data points \(x_0, y_0\), \(x_1, y_1\), \(x_2, y_2\), ..., \(x_n, y_n\), the Lagrange interpolation polynomial is given by:

• \( L(x) = \sum_{i=0}^{n} y_i \cdot l_i(x) \)

Each term \(y_i \cdot l_i(x)\) makes sure that the polynomial achieves the right value \(y_i\) when \(x = x_i\). As demonstrated, calculating each \(l_i(x)\) using the corresponding differences and combining them provides a comprehensive solution to precisely interpolate the data. By using the final interpolation formula, the polynomial predictions beyond the provided data can be made reliably.