Lagrange Interpolation: Solved Examples & Easy Explanations

Hey everyone! Ever stumbled upon a set of data points and wished you could predict what happens between them? Well, that's where Lagrange interpolation swoops in, making you the hero of the day! It's a fantastic technique used in numerical analysis to find a polynomial that passes through a given set of points. Think of it as connecting the dots, but with a mathematical twist. This method is named after the brilliant mathematician Joseph-Louis Lagrange, and it's super handy when you want to estimate values within a range where you have data. We're going to dive deep into this cool concept, breaking it down into easy-to-understand chunks, with plenty of solved examples to boot. Whether you're a student scratching your head over a problem or just curious about how math can solve real-world puzzles, you're in the right place. Let’s get started and unravel the magic of Lagrange interpolation together, shall we?

Grasping the Basics of Lagrange Interpolation

Alright, let’s start with the basics, shall we? Lagrange interpolation is all about finding a polynomial function that fits a given set of data points perfectly. Imagine you have a few points scattered on a graph, and you want to draw a smooth curve that passes exactly through each of those points. That's essentially what Lagrange interpolation does! This method is particularly useful when you have a set of data points and need to estimate the value of the function at a point that wasn't originally measured. It's used in various fields, from engineering and physics to computer science and finance. The beauty of this method lies in its ability to provide a unique polynomial for any given set of data points, ensuring that you can interpolate values with a high degree of accuracy. The formula might look a little intimidating at first, but don't worry, we'll break it down into easy pieces. With a bit of practice, you’ll be interpolating like a pro in no time, guys!

Here’s the gist:

• Data Points: You start with a set of data points, each consisting of an x-value and a corresponding y-value. These are your knowns.

• The Goal: The aim is to find a polynomial, often denoted as P(x), that passes through all these points.

• The Formula: The Lagrange interpolation formula is the key. It looks something like this:

  P(x) = ∑ yᵢ * Lᵢ(x), where Lᵢ(x) = ∏ (x - xⱼ) / (xᵢ - xⱼ)*, for j = 1 to n, j ≠ i

  • Where:
    • xᵢ and yᵢ are your data points.
    • Lᵢ(x) are the Lagrange basis polynomials, the secret sauce that makes it all work.
    • n is the number of data points.

Breaking Down the Formula

Okay, let's decode this formula, because, let's be honest, it looks a bit scary at first glance. But don't worry, we'll go through it step by step. Firstly, the Lagrange interpolation formula is a sum. You're adding up a bunch of terms to get your final polynomial. Each term in the sum is the product of yᵢ (the y-value of a data point) and Lᵢ(x) (the Lagrange basis polynomial). The Lagrange basis polynomial is a function that has a value of 1 at xᵢ and 0 at all other xⱼ values. This clever construction ensures that the resulting polynomial P(x) passes through all the data points. The formula for Lᵢ(x) is a product. You multiply a series of fractions together. The numerator of each fraction is (x - xⱼ), where x is the point at which you want to interpolate, and xⱼ is the x-value of another data point. The denominator is (xᵢ - xⱼ). This means for each data point, you're creating a polynomial that is 1 at that point and 0 at all others. By summing these scaled polynomials, you ensure the final curve passes through all the given data points. So, the main idea is to construct a polynomial that fits your data exactly, using these basis polynomials as building blocks. It is not that complex, trust me!

Solved Example 1: Interpolating with Three Points

Time to put theory into practice, yeah? Let's work through a solved example to see how Lagrange interpolation works in action. Imagine you're given three data points: (1, 3), (2, 1), and (3, 4). Our goal is to find the polynomial P(x) that passes through these points. We'll use the Lagrange interpolation formula to do this. Remember, the formula is: P(x) = ∑ yᵢ * Lᵢ(x).

Step-by-Step Solution

1. Identify the Data:
   • We have: x₁ = 1, y₁ = 3; x₂ = 2, y₂ = 1; x₃ = 3, y₃ = 4
2. Calculate the Lagrange Basis Polynomials:
   • Let's find L₁(x), L₂(x), and L₃(x).
     • L₁(x) = ((x - x₂) / (x₁ - x₂)) * ((x - x₃) / (x₁ - x₃)) = ((x - 2) / (1 - 2)) * ((x - 3) / (1 - 3)) = ((x - 2) / -1) * ((x - 3) / -2) = (1/2) * (x - 2) * (x - 3)*
     • L₂(x) = ((x - x₁) / (x₂ - x₁)) * ((x - x₃) / (x₂ - x₃)) = ((x - 1) / (2 - 1)) * ((x - 3) / (2 - 3)) = (x - 1) * ((x - 3) / -1) = -(x - 1) * (x - 3)*
     • L₃(x) = ((x - x₁) / (x₃ - x₁)) * ((x - x₂) / (x₃ - x₂)) = ((x - 1) / (3 - 1)) * ((x - 2) / (3 - 2)) = (1/2) * (x - 1) * (x - 2)*
3. Apply the Lagrange Interpolation Formula:
   • P(x) = y₁ * L₁(x) + y₂ * L₂(x) + y₃ * L₃(x)
   • P(x) = 3 * (1/2) * (x - 2) * (x - 3) + 1 * [-(x - 1) * (x - 3)] + 4 * (1/2) * (x - 1) * (x - 2)
4. Simplify and Find P(x):
   • P(x) = (3/2)(x² - 5x + 6) - (x² - 4x + 3) + 2(x² - 3x + 2)
   • P(x) = (3/2)x² - (15/2)x + 9 - x² + 4x - 3 + 2x² - 6x + 4
   • P(x) = (3/2 - 1 + 2)x² + (-15/2 + 4 - 6)x + (9 - 3 + 4)
   • P(x) = (5/2)x² - (19/2)x + 10

The Result

So, the polynomial P(x) = (5/2)x² - (19/2)x + 10 passes through the points (1, 3), (2, 1), and (3, 4). You can plug in the x-values of your original points to verify that they correspond to the y-values. This means our Lagrange interpolation successfully found the polynomial that fits our data perfectly! Amazing, right? Now you know how to use Lagrange interpolation for three data points. Isn't that great?

Solved Example 2: More Points, More Fun!

Let’s crank things up a notch with another solved example! This time, we'll interpolate with a few more data points to solidify your understanding. Suppose we have the following data points: (0, 1), (1, 3), (2, 2), and (3, 0). Our objective remains the same: find the polynomial P(x) that accurately represents these points. This example gives you practice with a slightly more complex calculation, showing how to handle more data points using the same fundamental principles. Ready to give it a shot?

Following the Steps

1. Define the Data:
   • x₁ = 0, y₁ = 1; x₂ = 1, y₂ = 3; x₃ = 2, y₃ = 2; x₄ = 3, y₄ = 0
2. Calculate Lagrange Basis Polynomials:
   • L₁(x) = ((x - 1) / (0 - 1)) * ((x - 2) / (0 - 2)) * ((x - 3) / (0 - 3)) = (-1)(x - 1) * (x - 2) * (x - 3) / (-6) = (1/6)(x - 1)(x - 2)(x - 3)
   • L₂(x) = ((x - 0) / (1 - 0)) * ((x - 2) / (1 - 2)) * ((x - 3) / (1 - 3)) = (x) * ((x - 2) / -1) * ((x - 3) / -2) = (1/2)x(x - 2)(x - 3)
   • L₃(x) = ((x - 0) / (2 - 0)) * ((x - 1) / (2 - 1)) * ((x - 3) / (2 - 3)) = (1/2)x * (x - 1) * (x - 3) / (-1) = (-1/2)x(x - 1)(x - 3)
   • L₄(x) = ((x - 0) / (3 - 0)) * ((x - 1) / (3 - 1)) * ((x - 2) / (3 - 2)) = (1/3)x * (x - 1) * (x - 2) = (1/6)2x(x - 1)(x - 2)
3. Apply the Lagrange Interpolation Formula:
   • P(x) = y₁ * L₁(x) + y₂ * L₂(x) + y₃ * L₃(x) + y₄ * L₄(x)
   • P(x) = 1 * (1/6)(x - 1)(x - 2)(x - 3) + 3 * (1/2)x(x - 2)(x - 3) + 2 * (-1/2)x(x - 1)(x - 3) + 0 * (1/6)x(x - 1)(x - 2)
4. Simplify to Find P(x):
   • P(x) = (1/6)(x³ - 6x² + 11x - 6) + (3/2)(x³ - 5x² + 6x) - x(x² - 4x + 3) + 0
   • P(x) = (1/6)x³ - x² + (11/6)x - 1 + (3/2)x³ - (15/2)x² + 9x - x³ + 4x² - 3x
   • P(x) = (1/6 + 3/2 - 1)x³ + (-1 - 15/2 + 4)x² + (11/6 + 9 - 3)x - 1
   • P(x) = (1/3)x³ - (9/2)x² + (41/6)x - 1

The Final Result

Thus, the polynomial P(x) = (1/3)x³ - (9/2)x² + (41/6)x - 1 accurately passes through the data points (0, 1), (1, 3), (2, 2), and (3, 0). You can check by plugging in the x-values and confirming the resulting y-values. And that’s it! You've successfully used Lagrange interpolation on another set of data points. Great job, guys! This method allows you to find a polynomial that fits these points perfectly, and it's a fundamental concept in numerical analysis. You’re doing amazing!

Practical Applications of Lagrange Interpolation

Now that you've got the hang of the method, you might be wondering,