## Linear Algebra and Its Applications, Exercise 3.4.25

Exercise 3.4.25. Given $y = x^2$ over the interval $-1 \le x \le 1$ what is the closest line $C + Dx$ to the parabola formed by $y$?

Answer: This amounts to finding a least-squares solution to the equation $\begin{bmatrix} 1&x \end{bmatrix} \begin{bmatrix} C \\ D \end{bmatrix} = y$, where the entries 1, $x$, and $y = x^2$ are understood as functions of $x$ over the interval -1 to 1 (as opposed to being scalar values).

Interpreting the traditional least squares equation $A^TAx = A^Tb$ in this context, here the matrix $A = \begin{bmatrix} 1&x \end{bmatrix}$ and we have

$A^TA = \begin{bmatrix} 1 \\ x \end{bmatrix} \begin{bmatrix} 1&x \end{bmatrix} = \begin{bmatrix} (1, 1)&(1, x) \\ (x, 1)&(x, x) \end{bmatrix}$

where the entries of $A^TA$ are the dot products of the functions, i.e., the integrals of their products over the interval -1 to 1.

We then have

$(1, 1) = \int_{-1}^1 1 \cdot 1 \;\mathrm{d}x = 2$

$(1, x) = (x, 1) = \int_{-1}^1 1 \cdot x \;\mathrm{d}x = \left( \frac{1}{2}x^2 \right) \;\big|_{-1}^1 = \frac{1}{2} \cdot 1^2 - \frac{1}{2} \cdot (-1)^2 = \frac{1}{2} - \frac{1}{2} = 0$

$(x, x) = \int_{-1}^1 x^2 \;\mathrm{d}x = \left( \frac{1}{3}x^3 \right) \;\big|_{-1}^1 = \frac{1}{3} \cdot 1^3 - \frac{1}{3} \cdot (-1)^3 = \frac{1}{3} + \frac{1}{3} = \frac{2}{3}$

so that

$A^TA = \begin{bmatrix} (1, 1)&(1, x) \\ (x, 1)&(x, x) \end{bmatrix} = \begin{bmatrix} 2&0 \\ 0&\frac{2}{3} \end{bmatrix}$

Continuing the interpretation of the least squares equation $A^TAx = A^Tb$ in this context, the role of $b$ is played by the function $y = x^2$, and we have

$A^Ty = \begin{bmatrix} 1 \\ x \end{bmatrix} x^2 = \begin{bmatrix} (1,x^2) \\ (x, x^2) \end{bmatrix}$

where again the entries are dot products of the functions. From above we have

$(1, x^2) = \int_{-1}^1 1 \cdot x^2 \;\mathrm{d}x = \frac{2}{3}$

and from previous exercises we have

$(x, x^2) = \int_{-1}^1 x \cdot x^2 \;\mathrm{d}x = \int_{-1}^1 x^3 \;\mathrm{d}x = 0$

so that

$A^Ty = \begin{bmatrix} \frac{2}{3} \\ 0 \end{bmatrix}$

To get the least squares solution $\bar{C} + \bar{D}x$ we then have

$\begin{bmatrix} 2&0 \\ 0&\frac{2}{3} \end{bmatrix} \begin{bmatrix} \bar{C} \\ \bar{D} \end{bmatrix} = \begin{bmatrix} \frac{2}{3} \\ 0 \end{bmatrix}$

From the second equation we have $\bar{D} = 0$. From the first equation we have $2\bar{C} = \frac{2}{3}$ or $C = \frac{1}{3}$.

The line of best fit to the parabola $y = x^2$ over the interval $-1 \le x \le 1$ is therefore the horizontal line with $y$-intercept of $\frac{1}{3}$.

NOTE: This continues a series of posts containing worked out exercises from the (out of print) book Linear Algebra and Its Applications, Third Edition by Gilbert Strang.

If you find these posts useful I encourage you to also check out the more current Linear Algebra and Its Applications, Fourth Edition, Dr Strang’s introductory textbook Introduction to Linear Algebra, Fifth Edition and the accompanying free online course, and Dr Strang’s other books.

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