Its an answer but i cant read handwritten . please type it for me. i give you postive rate. ## Deriving the Polynomial Expression using Lagrange Interpolation### Polynomial Expression\[ P(x) = \sum_{i=1}^n L_i(x) y_i, \]where \[ L_i(x) = \prod_{\substack{j=1 \\ j \neq i}}^n \frac{x - x_j}{x_i - x_j}. \]### Task: Show that $\sum_{i=1}^n L_i(x) = 1$ holds in general.### Proof:We are given $ n $ distinct real values $ x_1, x_2, \ldots, x_n $ and $ n $ real values $ y_1, y_2, \ldots, y_n $ (not necessarily distinct). There is a unique polynomial $ P $ with real coefficients satisfying $ P(x_i) = y_i $ for $ i \in \{1, 2, \ldots, n\} $ such that the degree of $ P \leq n-1 $.#### Definition of $ L_i(x) $:\[ L_i(x) = \prod_{\substack{j=1 \\ j \neq i}}^n \frac{x - x_j}{x_i - x_j} \]Thus,\[ L_i(x) = \frac{(x - x_1)(x - x_2) \ldots (x - x_{i-1})(x - x_{i+1}) \ldots (x - x_n)}{(x_i - x_1)(x_i - x_2) \ldots (x_i - x_{i-1})(x_i - x_{i+1}) \ldots (x_i - x_n)} \]#### Simplification:Then,\[ L_i(x_i) = 1 \text{ and } L_i(x_j) = 0 \text{ for } i \neq j \]Similarly, construct polynomials $ L_2, L_3, \ldots, L_n $ such that $ L_j(x_j) = 1 $ and $ L_j(x_i) = 0 $ for all $ i \neq j $.### Writing $ L_i(x) $ in another form:\[ L_i(x) = \frac{f(x)}{(x **Lagrange Interpolation Polynomial**The provided image shows a step in the process of constructing a polynomial using Lagrange Interpolation. Lagrange Interpolation is a method for finding the polynomial of least degree that passes through a given set of points.The text in the image reads:---Then $ p(x) = \sum_{i=1}^{n} L_i(x) y_i $ is a polynomial with real coefficients satisfying $ P(x_i) = y_i $. This therefore proves that $ \sum_{i=1}^{n} L_i(x) = 1 $---### Explanation:- $ p(x) = \sum_{i=1}^{n} L_i(x) y_i $: This equation represents the polynomial $ p(x) $ as a sum of terms, where each term is the product of $ L_i(x) $ and $ y_i $.- $ L_i(x) $: These are the Lagrange basis polynomials, which are constructed such that $ L_i(x_j) = 0 $ for $ j \neq i $ and $ L_i(x_i) = 1 $.- $ y_i $: These are the values (or y-coordinates) corresponding to the given x-coordinates $ x_i $.The polynomial $ p(x) $ is constructed to ensure that it passes through the given points $(x_1, y_1), (x_2, y_2), ..., (x_n, y_n)$.Finally, the proof states that the sum of all Lagrange basis polynomials $ L_i(x) $ is equal to 1:\[ \sum_{i=1}^{n} L_i(x) = 1 \]The square symbol ✓ denotes the end of the proof, a common notation in mathematical writings to indicate that the theorem is proven conclusively.This explanation is foundational in understanding how to interpolate a set of points using a polynomial in numerical analysis or computational mathematics. ### Graphs and DiagramsIn the context of Lagrange Interpolation, a graph would typically show:- The given data points $(x_i, y_i)$ on a coordinate plane.- The polynomial curve passing through these points.- The basis polynomials $ L_i(x) $ individually plotted to show how they contribute to the final polynomial $ p(x) $.These visual aids help illustrate

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Description: Its an answer but i cant read handwritten . please type it for me. i give you postive rate. ## Deriving the Polynomial Expression using Lagrange Interpolation### Polynomial Expression\[ P(x) = \sum_{i=1}^n L_i(x) y_i, \]where \[ L_i(x) = \prod_{\subs

Answered: PG) = ≤ Li (x) yi

Math

Advanced Math

PG) = ≤ Li (x) yi

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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