Science:Math Exam Resources/Courses/MATH307/December 2010/Question 04 (a)

MATH307 December 2010

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Question 04 (a)

In this question we are once again given 4 points (x1, y1), (x2, y2), (x3, y3) and (x4, y4) in the plane. This time we want to find a quadratic function $\displaystyle q(x)=ax^{2}+bx+c$ that comes closest to going through the points by doing a least squares fit. (a) The least squares equation you need to solve to find the coefficients a, b and c has the form ATAa = ATb. Write down expressions for A, a and b.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
Write the conditions that q(x) goes through all four points in matrix notation. The values xi and yi are known, and the values a, b and c are unknown.

Checking a solution serves two purposes: helping you if, after having used the hint, you still are stuck on the problem; or if you have solved the problem and would like to check your work. If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work. If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.

Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. If the quadratic polynomial could fit all four points, then these conditions would be satisfied: ${\begin{aligned}y_{1}&=ax_{1}^{2}+bx_{1}+c\\y_{2}&=ax_{2}^{2}+bx_{2}+c\\y_{3}&=ax_{3}^{2}+bx_{3}+c\\y_{4}&=ax_{4}^{2}+bx_{4}+c\end{aligned}}$ this is equivalent to: ${\begin{aligned}{\begin{pmatrix}x_{1}^{2}&x_{1}&1\\x_{2}^{2}&x_{2}&1\\x_{3}^{2}&x_{3}&1\\x_{4}^{2}&x_{4}&1\end{pmatrix}}{\begin{pmatrix}a\\b\\c\end{pmatrix}}&={\begin{pmatrix}y_{1}\\y_{2}\\y_{3}\\y_{4}\end{pmatrix}}\\A\mathbf {a} &=\mathbf {b} \end{aligned}}$. However, we only have 3 unknowns and 4 equations, hence we ask for the best solution (in the least square sense), which is found by solving $A^{T}A\mathbf {a} =A^{T}\mathbf {b}$. Therefore, $A={\begin{pmatrix}x_{1}^{2}&x_{1}&1\\x_{2}^{2}&x_{2}&1\\x_{3}^{2}&x_{3}&1\\x_{4}^{2}&x_{4}&1\end{pmatrix}},\quad \mathbf {a} ={\begin{pmatrix}a\\b\\c\end{pmatrix}},\quad \mathbf {b} ={\begin{pmatrix}y_{1}\\y_{2}\\y_{3}\\y_{4}\end{pmatrix}}$.

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