What is Lagrange’s interpolation polynomial for the following data?x 2 4 f(x) 3 5
x + 1
Finding Lagrange’s Interpolation Polynomial
Lagrange’s interpolation polynomial is a way to find a polynomial that passes through a given set of data points. For a set of \(n+1\) distinct points \((x_0, y_0), (x_1, y_1), \ldots, (x_n, y_n)\), the Lagrange polynomial \(P(x)\) is given by:
\[P(x) = \sum_{j=0}^{n} y_j L_j(x)\]
where \(L_j(x)\) is the Lagrange basis polynomial, defined as:
\[L_j(x) = \prod_{i=0, i \neq j}^{n} \frac{x - x_i}{x_j - x_i}\]
Calculating the Polynomial for Given Data
We are given two data points:
| x | f(x) |
|---|---|
| 2 | 3 |
| 4 | 5 |
Let these points be \((x_0, y_0) = (2, 3)\) and \((x_1, y_1) = (4, 5)\). Since we have two points, \(n=1\). The Lagrange polynomial for two points is:
\[P(x) = y_0 \frac{x - x_1}{x_0 - x_1} + y_1 \frac{x - x_0}{x_1 - x_0}\]
Step-by-Step Calculation
Now, we substitute the values of \(x_0, y_0, x_1, y_1\) into the formula:
Substitute \(x_0 = 2\), \(y_0 = 3\), \(x_1 = 4\), \(y_1 = 5\):
\[P(x) = 3 \left(\frac{x - 4}{2 - 4}\right) + 5 \left(\frac{x - 2}{4 - 2}\right)\]
Simplify the denominators:
\[P(x) = 3 \left(\frac{x - 4}{-2}\right) + 5 \left(\frac{x - 2}{2}\right)\]
Rewrite the terms:
\[P(x) = -\frac{3}{2}(x - 4) + \frac{5}{2}(x - 2)\]
Distribute the constants:
\[P(x) = -\frac{3}{2}x + \frac{12}{2} + \frac{5}{2}x - \frac{10}{2}\]
Simplify the fractions:
\[P(x) = -\frac{3}{2}x + 6 + \frac{5}{2}x - 5\]
Combine like terms (terms with \(x\) and constant terms):
\[P(x) = \left(\frac{5}{2}x - \frac{3}{2}x\right) + (6 - 5)\]
\[P(x) = \left(\frac{5 - 3}{2}\right)x + 1\]
\[P(x) = \left(\frac{2}{2}\right)x + 1\]
\[P(x) = 1x + 1\]
\[P(x) = x + 1\]
Thus, the Lagrange’s interpolation polynomial for the given data is \(x + 1\).
Revision Table: Key Concepts in Lagrange Interpolation
| Concept | Description |
|---|---|
| Lagrange Polynomial | A unique polynomial of degree at most \(n\) that passes through \(n+1\) distinct data points. |
| Basis Polynomials \(L_j(x)\) | Polynomials that are 1 at \(x_j\) and 0 at all other \(x_i\) (\(i \neq j\)). |
| Interpolation | The process of estimating values between known data points. |
Additional Information on Interpolation Polynomials
Lagrange’s interpolation is a fundamental method in numerical analysis for constructing a polynomial that takes specified values at specified points. It is useful for approximating functions when only a finite number of data points are known.
- One of the advantages of Lagrange interpolation is that the formula is symmetric and does not require solving a system of linear equations, unlike other methods like finding the coefficients of a standard polynomial form directly.
- However, adding a new data point requires recalculating the entire polynomial, which can be computationally expensive. Methods like Newton's form of the interpolating polynomial are more efficient for adding points.
- The degree of the resulting polynomial is at most \(n\), where \(n+1\) is the number of data points. In this case, with two points, the polynomial is of degree at most 1, which is a linear function (\(x+1\)).
Important Questions from Numerical Methods
If f(x) is a polynomial of degree n in x, then nth difference of this polynomial is
If f(1) = 4 and f(5) = 6, then what is the value of f(3) using Lagrange’s interpolation?
Which theorem states that "An integral function attains every finite value with atmost one possible exception"?
Let h be defined in finite-difference fraction notation as follows.
The order of convergence of Newton Raphson method is:
