Chapter 5.4, Problem 35E

Suppose you wish to fit a function of the form $f\left(t\right)=c+p\sin \left(t\right)+q\cos \left(t\right)$ to a given continuous function $g\left(t\right)$ on the closed interval from 0 to $2\pi$ . One approach is to choose n equally spaced points a between 0 and $2\pi \left[a_i=i\cdot \left(2\pi /n\right),\text{for}i=1,\mathrm{...},n,\text{say}\right]$ . We can fit a function
$f_n\left(t\right)=c_n+p_n\sin \left(t\right)+q_n\cos \left(t\right)$ to the data points $\left(a_i,g\left(a_i\right)\right)$ , for $i=1,\mathrm{...},n$ . Now examine what happens to the coefficients $c_n,p_n,q_n$ of $f_n\left(t\right)$ as n approaches infinity.

To find $f_n\left(t\right)$ , we make an attempt to solve the equations
$f_n\left(a_i\right)=g\left(a_i\right),$ for $i=1,\mathrm{...},n,$
or
$\left|\begin{array}{l}{c}_n+p_n\sin \left(a_1\right)+q_n\cos \left(a_1\right)=g\left(a_1\right)\\ c_n+p_n\sin \left(a_2\right)+q_n\cos \left(a_2\right)=g\left(a_2\right)\\ ⋮⋮\\ c_n+p_n\sin \left(a_n\right)+q_n\cos \left(a_n\right)=g\left(a_{,}\right)\end{array}\right|$ ,
or
$A_n\left[\begin{array}{l}{c}_n\\ p_n\\ q_n\end{array}\right]={\stackrel{\to }{b}}_n$ ,
where
$A_n=\left[\begin{array}{ccc}1& \sin \left(a_1\right)& \cos \left(a_1\right)\\ 1& \sin \left(a_2\right)& \cos \left(a_2\right)\\ ⋮& ⋮& ⋮\\ 1& \sin \left(a_2\right)& \cos \left(a_n\right)\end{array}\right],{\stackrel{\to }{b}}_n=\left[\begin{array}{l}g\left(a_1\right)\\ g\left(a_2\right)\\ ⋮\\ g\left(a_n\right)\end{array}\right]$
a. Find the entries of the matrix $A_n^T{A}_n$ and the components of the vector $A_n^T{\stackrel{\to }{b}}_n$ .
b. Find $\underset{n\to \infty }{\lim }\left(\frac{2\pi }{n}{A}_n^T{A}_n\right)$ and $\underset{n\to \infty }{\lim }\left(\frac{2\pi }{n}{A}_n^T\stackrel{\to }{b}\right)$ .
Hint: Interpret the entries of the matrix $\left(2\pi /n\right)A_n^T{A}_n$ and thecomponents of the vector $\left(2\pi /n\right)A^T\stackrel{\to }{b}$ as Riemann sums. Then the limits are the corresponding Riemann integrals. Evaluate as many integrals as you can. Note that $\underset{n\to \infty }{\lim }\left(\frac{2\pi }{n}{A}_n^T{A}_n\right)$ is a diagonal matrix
c. Find
$\begin{array}{l}\underset{n\to \infty }{\lim }\left[\begin{array}{l}{c}_n\\ p_n\\ q_n\end{array}\right]=\underset{n\to \infty }{\lim }{\left(A_n^T{A}_n\right)}^{-1}{A}_n^T{\stackrel{\to }{b}}_n\\ =\underset{n\to \infty }{\lim }\left[{\left(\frac{2\pi }{n}{A}_n^T{A}_n\right)}^{-1}\left(\frac{2\pi }{n}{A}_n^T{\stackrel{\to }{b}}_n\right)\right]\\ ={\left[\underset{n\to \infty }{\lim }\left(\frac{2\pi }{n}{A}_n^T{A}_n\right)\right]}^{-1}\underset{n\to \infty }{\lim }\left(\frac{2\pi }{n}{A}_n^T{\stackrel{\to }{b}}_n\right)\end{array}$
The resulting vector $\left[\begin{array}{l}c\\ p\\ q\end{array}\right]$ gives you the coefficient of the desired function
$f\left(t\right)=\underset{n\to \infty }{\lim }{f}_n\left(t\right)$ .
Write $f\left(t\right)$ . The function $f\left(t\right)$ is called the first Fourier approximation of $g\left(t\right)$ . The Fourier approximation satisfies a “continuous” least-squares condition, an idea we will make more precise in the next section.