Lagrange Interpolation Formula Calculation Of Y(6)

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In the realm of numerical analysis, Lagrange's interpolation formula stands as a powerful technique for approximating the value of a function at a specific point, given a set of data points. This method is particularly useful when the function's explicit form is unknown, but we have a collection of x and y values. In this article, we will delve into the application of Lagrange's interpolation formula to find the value of y at x = 6, based on the provided dataset: X: 3, 5, 7, 9, 11 and Y: 6, 24, 58, 108, 174. Our discussion will encompass the theoretical underpinnings of the formula, a step-by-step application to the given problem, and a detailed explanation of the calculations involved. This approach will not only solve the problem at hand but also provide a comprehensive understanding of how to employ Lagrange's interpolation in similar scenarios. Whether you are a student grappling with numerical methods or a professional seeking practical tools for data analysis, this guide aims to offer clarity and precision in using this essential mathematical technique. The strength of Lagrange's method lies in its ability to construct a polynomial that passes through each given data point, allowing for accurate estimations within the range of the provided data. This makes it an indispensable tool in fields ranging from engineering to finance, where data interpolation is a common requirement. Let's embark on this exploration to unravel the intricacies of Lagrange's interpolation and its practical application.

Understanding Lagrange's Interpolation Formula

Before we dive into the specifics of calculating y(6), let’s first understand the Lagrange's interpolation formula itself. This formula constructs a polynomial that passes through a given set of points. Suppose we have n + 1 data points (x₀, y₀), (x₁, y₁), ..., (xₙ, yₙ). The Lagrange interpolation polynomial P(x) is given by:

P(x) = Σᵢ<binary data, 1 bytes><binary data, 1 bytes><binary data, 1 bytes>₀ⁿ yᵢ Lᵢ(x)

where Lᵢ(x) are the Lagrange basis polynomials, defined as:

Lᵢ(x) = ∏₀≤ⱼ≤ₙ, ⱼ≠i (x - xⱼ) / (xᵢ - xⱼ)

The essence of Lagrange's interpolation lies in creating a weighted sum of the y-values, where the weights are the Lagrange basis polynomials. Each Lᵢ(x) is constructed such that it equals 1 when x = xᵢ and 0 when x equals any other data point's x-value. This property ensures that the interpolation polynomial P(x) passes exactly through each of the given data points. The formula effectively constructs a polynomial that smoothly connects the data points, providing a reliable estimate of the function's value at any point within the range of the given data. Understanding this foundation is crucial for applying the method correctly and interpreting the results effectively. The beauty of Lagrange's interpolation is its simplicity and directness; it avoids the need to solve a system of equations, which is often required in other interpolation methods. By carefully constructing the basis polynomials, we can accurately approximate the function's behavior between the known data points, making it an invaluable tool in various scientific and engineering applications. The formula may seem complex at first glance, but breaking it down into its components—the basis polynomials and the weighted sum—reveals its elegant and practical nature.

Step-by-Step Calculation of y(6) Using Lagrange's Interpolation

Now, let's apply the Lagrange's interpolation formula to our specific problem: finding y(6) given the data points (3, 6), (5, 24), (7, 58), (9, 108), and (11, 174). We have five data points, so n = 4. We will calculate the Lagrange basis polynomials L₀(x), L₁(x), L₂(x), L₃(x), and L₄(x) and then use them to find y(6).

  1. Calculate L₀(x):

    L₀(x) = [(x - 5)(x - 7)(x - 9)(x - 11)] / [(3 - 5)(3 - 7)(3 - 9)(3 - 11)]

    L₀(x) = [(x - 5)(x - 7)(x - 9)(x - 11)] / [(-2)(-4)(-6)(-8)]

    L₀(x) = [(x - 5)(x - 7)(x - 9)(x - 11)] / 384

  2. Calculate L₁(x):

    L₁(x) = [(x - 3)(x - 7)(x - 9)(x - 11)] / [(5 - 3)(5 - 7)(5 - 9)(5 - 11)]

    L₁(x) = [(x - 3)(x - 7)(x - 9)(x - 11)] / [(2)(-2)(-4)(-6)]

    L₁(x) = [(x - 3)(x - 7)(x - 9)(x - 11)] / 96

  3. Calculate L₂(x):

    L₂(x) = [(x - 3)(x - 5)(x - 9)(x - 11)] / [(7 - 3)(7 - 5)(7 - 9)(7 - 11)]

    L₂(x) = [(x - 3)(x - 5)(x - 9)(x - 11)] / [(4)(2)(-2)(-4)]

    L₂(x) = [(x - 3)(x - 5)(x - 9)(x - 11)] / 64

  4. Calculate L₃(x):

    L₃(x) = [(x - 3)(x - 5)(x - 7)(x - 11)] / [(9 - 3)(9 - 5)(9 - 7)(9 - 11)]

    L₃(x) = [(x - 3)(x - 5)(x - 7)(x - 11)] / [(6)(4)(2)(-2)]

    L₃(x) = [(x - 3)(x - 5)(x - 7)(x - 11)] / (-96)

  5. Calculate L₄(x):

    L₄(x) = [(x - 3)(x - 5)(x - 7)(x - 9)] / [(11 - 3)(11 - 5)(11 - 7)(11 - 9)]

    L₄(x) = [(x - 3)(x - 5)(x - 7)(x - 9)] / [(8)(6)(4)(2)]

    L₄(x) = [(x - 3)(x - 5)(x - 7)(x - 9)] / 384

With the basis polynomials calculated, the next crucial step is to substitute x = 6 into each of these polynomials. This will give us the weights Lᵢ(6) that we need to combine with the corresponding y-values. The calculations involve carefully plugging in the value of 6 and simplifying the expressions. This step is vital for accurately determining the contribution of each data point to the final interpolated value. It's a process that requires attention to detail and precision in arithmetic to ensure the correct weights are obtained. These weights will then be used in the final summation to compute the value of y(6). The accuracy of the interpolated value hinges on the correctness of these intermediate calculations, making this a pivotal stage in the application of Lagrange's interpolation formula. The methodical approach in computing each Lᵢ(6) will pave the way for a reliable estimate of the function's value at the desired point.

Substituting x = 6 into Basis Polynomials and Final Calculation

Now, substitute x = 6 into each Lᵢ(x):

  1. L₀(6):

    L₀(6) = [(6 - 5)(6 - 7)(6 - 9)(6 - 11)] / 384 = [(1)(-1)(-3)(-5)] / 384 = 15 / 384

  2. L₁(6):

    L₁(6) = [(6 - 3)(6 - 7)(6 - 9)(6 - 11)] / 96 = [(3)(-1)(-3)(-5)] / 96 = -45 / 96

  3. L₂(6):

    L₂(6) = [(6 - 3)(6 - 5)(6 - 9)(6 - 11)] / 64 = [(3)(1)(-3)(-5)] / 64 = 45 / 64

  4. L₃(6):

    L₃(6) = [(6 - 3)(6 - 5)(6 - 7)(6 - 11)] / (-96) = [(3)(1)(-1)(-5)] / (-96) = 15 / (-96) = -15 / 96

  5. L₄(6):

    L₄(6) = [(6 - 3)(6 - 5)(6 - 7)(6 - 9)] / 384 = [(3)(1)(-1)(-3)] / 384 = 9 / 384

Now, we use the Lagrange's interpolation formula:

y(6) = Σᵢ<binary data, 1 bytes><binary data, 1 bytes><binary data, 1 bytes>₀⁴ yᵢ Lᵢ(6)

y(6) = (6 * 15/384) + (24 * -45/96) + (58 * 45/64) + (108 * -15/96) + (174 * 9/384)

y(6) = 90/384 - 1080/96 + 2610/64 - 1620/96 + 1566/384

To simplify, let's find a common denominator, which is 384:

y(6) = (90 - 4320 + 15660 - 6480 + 1566) / 384

y(6) = 6416 / 384

y(6) = 36

Therefore, using Lagrange's interpolation formula, we find that y(6) = 36. This result is obtained by meticulously applying the formula, calculating the basis polynomials, substituting the value of x, and finally, combining the weighted y-values. The process exemplifies the power and precision of numerical methods in estimating function values when an explicit function is not available. The final value of 36 represents the interpolated estimate of y at x = 6, based on the given dataset. This calculation not only answers the specific question but also underscores the broader applicability of Lagrange's interpolation in various scientific and engineering contexts, where data analysis and estimation are paramount. The accuracy of this method, as demonstrated in this example, makes it an indispensable tool for practitioners and researchers alike.

Conclusion

In conclusion, we have successfully applied Lagrange's interpolation formula to find y(6) for the given data set. The step-by-step approach, from understanding the formula to calculating the basis polynomials and substituting values, demonstrates the practical application of this powerful interpolation technique. The result, y(6) = 36, provides an accurate estimate based on the provided data points. Lagrange's interpolation is a valuable tool in numerical analysis, enabling us to approximate function values when a direct equation is not available. This method is widely used in various fields, including engineering, computer graphics, and data science, for its ability to fit a polynomial through a set of points accurately. The detailed process outlined in this article serves as a comprehensive guide for anyone looking to apply Lagrange's interpolation in their work or studies. The formula's strength lies in its ability to handle a wide range of data sets, making it a versatile choice for interpolation tasks. Furthermore, the clarity of the steps involved ensures that even those new to numerical methods can grasp the concept and apply it effectively. The success of this calculation highlights the importance of mastering such techniques for problem-solving in various domains. Ultimately, Lagrange's interpolation is more than just a formula; it's a gateway to understanding and utilizing the power of numerical methods in real-world applications. The insights gained from this exercise underscore the significance of mathematical tools in data analysis and the broader scientific landscape. As we continue to generate and analyze data in ever-increasing volumes, the ability to interpolate accurately and efficiently will remain a critical skill for researchers and practitioners across disciplines.

Keywords: Lagrange's interpolation formula, numerical analysis, data interpolation, polynomial approximation, y(6) calculation