What is Lagrange Interpolation Theorem?
A polynomial is an algebraic expression that can have one or more terms. The meaning of “Poly” is “many” and “nominal” means “terms,” so, in other words, a polynomial is “many terms.” It can have constants, variables, and exponents. They all can be combined using mathematical operations like additions, subtraction, multiplication, and division except that a division by a variable is not allowed in a polynomial expression.
Example of a polynomial – 2xy2 + 4x – 6 –> This polynomial has 3 terms which are 2xy2, 4x, and 6
A polynomial can have one or more terms, but infinite numbers of terms are not allowed as well as the exponent can only be positive integers (0, 1, 2, …); hence 3xy-2 is not a polynomial.
Interpolation
Within a range of a discrete set of data points, interpolation is the method of finding new data points. It is a technique in which an estimate of a mathematical expression is found, taking any intermediate value for the independent variable. The main use of interpolation is to figure out what other data can exist outside of their collected data. Many professionals like photographers, scientists, mathematicians, or engineers use this method for their experiments. A common use is in the scaling of images when one interpolates the next position of a pixel based on the given positions of pixels in an image.
Lagrange Interpolation Theorem
This theorem is a means to construct a polynomial that goes through a desired set of points and takes certain values at arbitrary points. If a function f(x) is known at discrete points xi, i = 0, 1, 2,… then this theorem gives the approximation formula for nth degree polynomials to the function f(x). More so, it gives a constructive proof of the theorem below:
For a point p(2,4), how do we represent it as a polynomial?
P(x) = 3
P(1) = 3
Similarly for a sequence of points, (2,3), (4,5) how can we find a polynomial to represent it?
P(x) = (x-4)/(2-4) * 3 + (x-2)/(4-2) * 5
P(2) = 3 and P(4) = 5
Going by the above examples, the general form of Lagrange Interpolation theorem can be gives as:
P(x) = (x – x2) (x-x3)/(x1 – x2) (x1 – x3) * y1 + (x – x1) (x-x3)/(x2 – x1) (x2 – x3) * y2 + (x – x1) (x-x2)/(x3 – x1) (x3 – x2) * y3
Or P (x)= \[\sum_{i=1}^{3}\] \[P_{i}\](x) \[Y_{i}\]
The Theorem – For n real values which are distinct: x1, x2, x3, x4,..xn, and n real values (might not be distinct) y1, y2, y3, y4… yn, a unique polynomial exists with real coefficients which satisfies the formula:
P(xi) = yi, i ϵ (1, 2,3, …, n) so that deg(P) < n
Proof of Lagrange Theorem
Let us consider an nth degree polynomial which is given by the below expression:
F(x) = A0 (x-x1) (x-x2) (x-x3)….(x-xn) + A1 (x-x0) (x-x2) (x-x3)….(x-xn) + ……..+ An (x-x1) (x-x2) (x-x3)….(x-xn-1)
Now we substitute values of our observations i.e. xi and we obtain the values of Ai:
So we put x = x0 and we get A0 as below:
f(x0) = y0 = A0 (x0-x1) (x0-x2) (x0-x2)….(x0-xn), the other terms become 0
Hence A0 = y0/(x0-x1) (x0-x2) (x0-x3)….(x0-xn)
Similarly for x1 we would get
f(x1) = y1 = = A1 (x1-x0) (x1-x2) (x1-x3)….(x1-xn), the other terms become 0
Hence A1 = y1/(x1-x0) (x1-x2) (x1-x3)….(x1-xn)
In this way we can obtain all the values of As from A2, A3,… An
An = yn/(xn-x0) (xn-x2) (xn-x3)….(xn-xn-1)
Now if we substitute all the values of As in the main function, we get Lagrange’s interpolation theorem:
F(x) = y0 * (x-x1) (x-x2) (x-x3)….(x-xn)/ (x0-x1) (x0-x2) (x0-x3)….(x0-xn) + y1 * (x-x0) (x-x2) (x-x3)….(x-xn)/ (x1-x0) (x1-x2) (x1-x3)….(x1-xn) + ……..+ yn (x-x1) (x-x2) (x-x3)….(x-xn-1)/ (xn-x0) (xn-x2) (xn-x3)….(xn-xn-1)
Note: Lagrange’s theorem applies to both equally and non-equally spaced points. This means that all the values of xs are not spaced equally.
Uses of Lagrange Interpolation Theorem – In science, a complicated function needs a lot of time and energy to be solved. This makes experiments difficult to run. In order to create a slightly less complex version of the original function, the interpolation method comes in use.
Conclusion
The Lagrange theorem generalizes the well-established mathematical facts like a line is uniquely determined by 2 points, 3 points uniquely determine the graph of a quadratic polynomial, and so on. There is a caveat here i.e.; the points must have different x coordinates. The image enlargement technique uses principles of the Lagrange theorem in trying to describe the tendency of image data by using interpolation polynomials to estimate unknown data. This helps in image enlargement.
The theorem can be expressed in a mathematical formula as shown here:
\[P (x) = \sum i = 1_{n}\sum i = 1_{n}P_{i}P_{i}(x)\]
Yi and it applies to all values of x, whether they are equally spaced or not.
FAQs on Lagrange Interpolation Theorem
1. What is the Lagrange Interpolation Theorem in numerical analysis?
The Lagrange Interpolation Theorem provides a straightforward method for finding a unique polynomial of the lowest possible degree that passes exactly through a given set of data points. In essence, if you have a set of distinct points (x₀, y₀), (x₁, y₁), ..., (xₙ, yₙ), this theorem helps construct a polynomial P(x) such that P(xᵢ) = yᵢ for all points. It is a fundamental concept in numerical approximation and data fitting.
2. What is the formula for Lagrange Interpolation?
The formula for the Lagrange interpolating polynomial P(x) for a set of n+1 data points (x₀, y₀), (x₁, y₁), ..., (xₙ, yₙ) is given by the sum of Lagrange basis polynomials:
P(x) = Σ (from j=0 to n) [yⱼ * Lⱼ(x)]
Where each Lagrange basis polynomial Lⱼ(x) is defined as:
Lⱼ(x) = Π (from i=0 to n, where i≠j) [(x - xᵢ) / (xⱼ - xᵢ)]
Each Lⱼ(x) has the unique property that it equals 1 at x = xⱼ and 0 at all other data points xᵢ.
3. What is the difference between interpolation and extrapolation?
The key difference lies in the range where you are estimating a value. Interpolation is the process of estimating a value that lies within the range of your known data points. For example, if you have data for x=1 and x=5, interpolating would mean finding a value for x=3. In contrast, extrapolation is the process of estimating a value that lies outside the range of your known data points, such as finding a value for x=7. Interpolation is generally considered more reliable than extrapolation.
4. Why is Lagrange interpolation an important method in mathematics?
Lagrange interpolation is important for several reasons in applied Maths and numerical methods. Its primary importance stems from:
- Function Approximation: It allows us to approximate complicated functions with simpler polynomials, which are easier to analyse and integrate.
- Proof of Existence: The formula itself serves as a constructive proof that a unique polynomial of degree at most 'n' exists for any n+1 points.
- Foundation for Other Methods: It forms the theoretical basis for other numerical techniques, such as numerical integration formulas like the Newton-Cotes formulas.
5. What are the main limitations or disadvantages of using the Lagrange interpolation method?
Despite its utility, the Lagrange method has significant limitations. The primary disadvantages are:
- Computational Inefficiency: If a new data point is added, the entire set of basis polynomials must be recalculated from scratch, making it computationally expensive for growing datasets.
- Runge's Phenomenon: For a high number of equally spaced points, the resulting polynomial can exhibit large oscillations near the endpoints of the interval, leading to poor approximation between the points.
- Numerical Instability: The method can be sensitive to the choice of data points, potentially leading to large errors in the resulting polynomial.
6. How does the Lagrange method differ from Newton's interpolation method?
Both methods yield the same unique interpolating polynomial, but they differ in their computational approach. The main difference is that Newton's interpolation method is recursive. It uses a divided differences table, which allows for the easy addition of new data points without recomputing the entire formula. In contrast, the Lagrange method requires a complete recalculation if a new point is introduced. Therefore, Newton's method is generally more efficient for practical applications where data may be added incrementally.
7. Can you provide a simple example of finding a value using Lagrange interpolation?
Certainly. Suppose we have two points: (1, 3) and (4, 9). We want to find the linear polynomial that passes through them. Here, x₀=1, y₀=3, and x₁=4, y₁=9.
The Lagrange basis polynomials are:
L₀(x) = (x - x₁) / (x₀ - x₁) = (x - 4) / (1 - 4) = -(x - 4)/3
L₁(x) = (x - x₀) / (x₁ - x₀) = (x - 1) / (4 - 1) = (x - 1)/3
The interpolating polynomial P(x) is:
P(x) = y₀L₀(x) + y₁L₁(x) = 3 * [-(x - 4)/3] + 9 * [(x - 1)/3]
P(x) = -(x - 4) + 3(x - 1)
P(x) = -x + 4 + 3x - 3
P(x) = 2x + 1
This is the unique line passing through (1, 3) and (4, 9).
