Question
Asked Nov 1, 2019
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From Numerical Analysis 8th edition by Richard Burden. This is #1 from Section 8.1

1. Compute the linear least squares polynomial for the data of Example 2 (repeated below).
Yi
2
1
0.00
1.0000
2
0.25 1.2840
3 0.50 1.6487
4 0.75 2.1170
5 1.00 2.7183
Note: this can be done effectively by hand - special case tabular method or with matrices.
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1. Compute the linear least squares polynomial for the data of Example 2 (repeated below). Yi 2 1 0.00 1.0000 2 0.25 1.2840 3 0.50 1.6487 4 0.75 2.1170 5 1.00 2.7183 Note: this can be done effectively by hand - special case tabular method or with matrices.

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Expert Answer

Step 1

The data is given as

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Xi yi 1 0.00 1.0000 2 0.25 1.2840 0.50 1.6487 4 0.75 2.1170 1.00 2.7183 Consider the linear least square polynomial as y=a+bx

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Step 2

On substituting the data points in the function we obtain the system as follows.  

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0 1 1 1.2840 1 0.25 1.6487 1 0.5 b 1 0.75 2.1170 2.7183 1 Consider the residual vectorr as follows 1 p 1 (1 0 1.2840 p-0.25q р 1.6487 -p-0.5q 1 0.25 1.2840 r =1.6487 1 0.5 1 0.75 2.1170 p-0.75q 2.1170 2.7183- р- 2.7183 1 1

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Step 3

Now,

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2 Obtain 2 f (1-p)1.2840-p-025q) +(1.6487-p-0.5q)+(2.1170-p-0.75q) +(2.7183-p-q) _ Differentiatefpartially withp and q and equate them to 0 = 8.7681-5p-2.5q = 0 др = 5.4515 - 2.5p-1.875q 0 8.7681 5 The new system is obtained as 2.5 р which has a solution 5.4515 2.5 1.875

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