2. Consider the data points (1,2), (2,3), (3,7). (a) Find a degree 2 polynomial passing through the three data points. (b) Find a degree 1 polynomial f(x) = ao + a1 with minimum error E-1 y; - f(x;)[². [2]
2. Consider the data points (1,2), (2,3), (3,7). (a) Find a degree 2 polynomial passing through the three data points. (b) Find a degree 1 polynomial f(x) = ao + a1 with minimum error E-1 y; - f(x;)[². [2]
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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Question
This is a solution from my professor about how to solve 2a and 2b on the problems listed. Can you please break them down into much smaller steps because I feel like it goes really fast (especially 2b)?
![2
By the computation in the last paragraph, we have Q = [u,u2] and R =
0.
%3D
%3D
||x2|]
2. Consider the data points (1, 2), (2,3), (3, 7).
(a) Find a degree 2 polynomial passing through the three data points.
(b) Find a degree 1 polynomial f(x) = ao + a1x with minimum error
E-1 ly; - f(x;)/².
1
1 1||bo
Solution. (a) Suppose g(x) = bo + b1x + b2x² fits the data. Then 3=1
2 4b1
%3D
1
39
b2
Solving the system, we see that (bo,b1,b2) = (4,–3.5, 1.5). So, g(x) = 4- 3.5x + 1.5x².
%3D
[1
1
(b) By the least square theory, we let B == | 1 2 and b = |3 and solve BT Bâu = BTb.
1
%3D
%3D
7
We see that = (ao,a1) = (-1,2.5). So, f (x) = -1 + 2.5x is the best fit for the data.
%3D
%3D
3.
3. Find a SVD for A =2
3
and find a unit vector x eR? such that Ax has the maximum length.
%3D
-2
17
Solution. We consider B = AT A
8.
det(B - AI) = (17 - A)2 - 82 = (25 - A)(9 – ). Then
%3D
|
8.
17
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Transcribed Image Text:2
By the computation in the last paragraph, we have Q = [u,u2] and R =
0.
%3D
%3D
||x2|]
2. Consider the data points (1, 2), (2,3), (3, 7).
(a) Find a degree 2 polynomial passing through the three data points.
(b) Find a degree 1 polynomial f(x) = ao + a1x with minimum error
E-1 ly; - f(x;)/².
1
1 1||bo
Solution. (a) Suppose g(x) = bo + b1x + b2x² fits the data. Then 3=1
2 4b1
%3D
1
39
b2
Solving the system, we see that (bo,b1,b2) = (4,–3.5, 1.5). So, g(x) = 4- 3.5x + 1.5x².
%3D
[1
1
(b) By the least square theory, we let B == | 1 2 and b = |3 and solve BT Bâu = BTb.
1
%3D
%3D
7
We see that = (ao,a1) = (-1,2.5). So, f (x) = -1 + 2.5x is the best fit for the data.
%3D
%3D
3.
3. Find a SVD for A =2
3
and find a unit vector x eR? such that Ax has the maximum length.
%3D
-2
17
Solution. We consider B = AT A
8.
det(B - AI) = (17 - A)2 - 82 = (25 - A)(9 – ). Then
%3D
|
8.
17
2)
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