One of the super-powers of linear algebra is that it can be applied everywhere. In the next question, you will use linear algebra to solve a non-linear problem. Given a list of n points (y) Rwith distinct a, values, it is sometimes necessary to find a polynomial curve p(x) Pn 1(R) passing through all the points. One method to do so, uses linear algebra as follows. (We will learn another method, called Lagrange interpolation, later in the course.) Consider an arbitrary polynomial p(x) = ao + a₁x + a₂x²+...+ an-1" We consider the coefficients as unknown. Each point (₂) gives a linear relation among the coefficients. The linear system is given: n-1 ao + a₁x₁ + a₂x² + + an 1x1 ao + ₁x2 + ₂x²+...+ an-12-1 Y₁ Y2 ao+a₁x₁ + a₂x²/1 + + an 1 x n 1 Yn 1. Use this method to find a quadratic polynomial passing through the points (0, 3), (1, 2), (2, 3). 2. Use this method to find a cubic passing through through the points (0,5), (1,9), (2, 11), (3,5). 3. Consider the example of the three points laying on a line, such as: (0,0), (1, 1), (2, 2). What happens to your linear system in this case? In general, what happens to this method if the points lie on a curve of lower degree than necessary?
One of the super-powers of linear algebra is that it can be applied everywhere. In the next question, you will use linear algebra to solve a non-linear problem. Given a list of n points (y) Rwith distinct a, values, it is sometimes necessary to find a polynomial curve p(x) Pn 1(R) passing through all the points. One method to do so, uses linear algebra as follows. (We will learn another method, called Lagrange interpolation, later in the course.) Consider an arbitrary polynomial p(x) = ao + a₁x + a₂x²+...+ an-1" We consider the coefficients as unknown. Each point (₂) gives a linear relation among the coefficients. The linear system is given: n-1 ao + a₁x₁ + a₂x² + + an 1x1 ao + ₁x2 + ₂x²+...+ an-12-1 Y₁ Y2 ao+a₁x₁ + a₂x²/1 + + an 1 x n 1 Yn 1. Use this method to find a quadratic polynomial passing through the points (0, 3), (1, 2), (2, 3). 2. Use this method to find a cubic passing through through the points (0,5), (1,9), (2, 11), (3,5). 3. Consider the example of the three points laying on a line, such as: (0,0), (1, 1), (2, 2). What happens to your linear system in this case? In general, what happens to this method if the points lie on a curve of lower degree than necessary?
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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Transcribed Image Text:One of the super-powers of linear algebra is that it can be applied everywhere. In the next question, you will use
linear algebra to solve a non-linear problem.
Given a list of n points (x,y) R with distinct a, values, it is sometimes necessary to find a polynomial curve
p(x) Pn 1(R) passing through all the points. One method to do so, uses linear algebra as follows. (We will learn
another method, called Lagrange interpolation, later in the course.)
Consider an arbitrary polynomial p(x) = ao + a₁x + a₂x²+...+ an-1" We consider the coefficients as
unknown. Each point (U) gives a linear relation among the coefficients. The linear system is given:
ao + a₁x₁ + a₂x² + ... + an_1x₁
ao + a₁x2 + a₂x² +
n-1
+ an 1x2
1
ao+an + ax² + + an 1 xn
=
Yı
Y2
Yn
1. Use this method to find a quadratic polynomial passing through the points (0, 3), (1, 2), (2, 3).
2. Use this method to find a cubic passing through through the points (0,5), (1,9), (2, 11), (3,5).
3. Consider the example of the three points laying on a line, such as: (0,0), (1, 1), (2, 2). What happens to your
linear system in this case? In general, what happens to this method if the points lie on a curve of lower degree
than necessary?
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