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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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](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F190bb1c4-42bb-47b5-853f-a418d6d98c12%2F16c9f4a7-7f6c-4f97-960e-c73ff9282b9d%2Fartzf6l_processed.jpeg&w=3840&q=75)
Transcribed Image Text:## 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 Diagrams
In 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](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F190bb1c4-42bb-47b5-853f-a418d6d98c12%2F16c9f4a7-7f6c-4f97-960e-c73ff9282b9d%2Fnp9gmvp_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**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 Diagrams
In 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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