How to Find Interpolation Polynomial Using Lagrange Formula
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 Explained with Formula
1. What is Lagrange Interpolation Theorem?
The Lagrange Interpolation Theorem states that through n + 1 distinct data points, there exists a unique polynomial of degree at most n that passes exactly through those points. If the points are (x₀, y₀), (x₁, y₁), ..., (xₙ, yₙ) with distinct x-values, then the interpolating polynomial is constructed as a linear combination of special basis polynomials. This theorem is fundamental in polynomial interpolation, numerical analysis, and curve fitting.
2. What is the formula for Lagrange interpolation?
The Lagrange interpolation formula is given by P(x) = Σ (yᵢ Lᵢ(x)), where Lᵢ(x) are the Lagrange basis polynomials. It can be written as:
- P(x) = Σi=0 to n yᵢ Lᵢ(x)
- Lᵢ(x) = ∏j≠i (x − xⱼ)/(xᵢ − xⱼ)
Each Lᵢ(x) equals 1 at x = xᵢ and 0 at all other given data points, ensuring the polynomial passes through every point exactly.
3. How do you construct a Lagrange interpolating polynomial step by step?
To construct a Lagrange interpolating polynomial, compute each basis polynomial and combine them using the given y-values.
- Step 1: Identify the given data points (x₀, y₀), …, (xₙ, yₙ).
- Step 2: Compute each Lᵢ(x) using Lᵢ(x) = ∏(x − xⱼ)/(xᵢ − xⱼ), j ≠ i.
- Step 3: Multiply each Lᵢ(x) by its corresponding yᵢ.
- Step 4: Add all terms to obtain P(x).
The resulting polynomial is the required interpolation polynomial.
4. Can you give a simple example of Lagrange interpolation?
Yes, for points (1,2) and (3,4), the Lagrange interpolating polynomial is P(x) = x + 1.
- L₀(x) = (x − 3)/(1 − 3) = (x − 3)/(-2)
- L₁(x) = (x − 1)/(3 − 1) = (x − 1)/2
- P(x) = 2·L₀(x) + 4·L₁(x)
After simplification, P(x) = x + 1, which passes through both given data points exactly.
5. Why is Lagrange interpolation polynomial unique?
The Lagrange interpolation polynomial is unique because there is exactly one polynomial of degree ≤ n that passes through n + 1 distinct points.
- If two such polynomials existed, their difference would be a polynomial of degree ≤ n.
- This difference would have n + 1 distinct zeros.
- A non-zero polynomial of degree ≤ n cannot have more than n zeros.
Hence, the interpolating polynomial must be unique.
6. What are Lagrange basis polynomials?
The Lagrange basis polynomials Lᵢ(x) are special polynomials that equal 1 at xᵢ and 0 at all other given x-values.
- Formula: Lᵢ(x) = ∏j≠i (x − xⱼ)/(xᵢ − xⱼ)
- Degree of each Lᵢ(x) is n.
- They form the building blocks of the interpolation polynomial.
These basis functions ensure accurate polynomial interpolation through all data points.
7. What is the degree of the Lagrange interpolating polynomial?
The degree of the Lagrange interpolating polynomial is at most n when constructed from n + 1 data points.
- If all points lie on a lower-degree curve, the degree may be less.
- Otherwise, the maximum possible degree is exactly n.
This property follows directly from the Lagrange Interpolation Theorem in numerical methods.
8. What is the difference between Lagrange and Newton interpolation?
The main difference is that Lagrange interpolation uses basis polynomials directly, while Newton interpolation uses divided differences and a recursive form.
- Lagrange form: Direct formula using Lᵢ(x).
- Newton form: Uses divided difference table.
- Newton form is easier to update when new data points are added.
Both methods produce the same unique interpolating polynomial but differ in computational approach.
9. Where is Lagrange interpolation used in real life?
Lagrange interpolation is used in numerical analysis, data fitting, and scientific computing to estimate unknown values between known data points.
- Curve fitting in engineering experiments
- Computer graphics and animation
- Signal processing and numerical simulations
- Estimating missing data values
It is especially useful when an exact polynomial passing through given data points is required.
10. What are common mistakes in using Lagrange interpolation?
Common mistakes in Lagrange interpolation include calculation errors and using repeated x-values.
- Using non-distinct x-values (violates theorem conditions).
- Incorrect computation of denominators (xᵢ − xⱼ).
- Forgetting to simplify the final polynomial.
- Applying high-degree interpolation leading to oscillation (Runge’s phenomenon).
Careful algebra and verifying distinct data points help avoid errors in polynomial interpolation.
