Q35E Suppose you wish to fit a functi... [FREE SOLUTION]

Chapter 5: Q35E (page 247)

Suppose you wish to fit a function of the form $f (t) = c + p s i n (t) + q c o s (t)$ to a given continuous function $g (t)$ on the closed interval from $0$ to $2 蟺$ . One approach is to choose $n$ equally spaced points $a_{i}$ between $0$ and $2 蟺$ [ $a_{i} = i 鈰� (\frac{2 蟺}{n})$ , for $i = 1, . .., n$ , say]. We can fit a function $f_{n} (t) = c_{n} + p_{n} s i n (t) + q_{n} c o s (t)$ to the data points , $(a_{i}, g (a_{i}))$ for $i = 1, . .., n$ . Now examine what happens to the coefficient $c_{n}, p_{n}, q_{n}$ of $f_{n} (t)$ as $n$ approaches infinity.
To find $f_{n} (t)$ , we make an attempt to solve the equations $f_{n} (a_{i}) = g (a_{i})$ , for $i = 1, . .., n$
orlocalid="1668166660598" $| \begin{matrix} c_{n} + p_{n} s i n (a_{1}) + q_{n} c o s (a_{1}) = g (a_{1}) \\ c_{n} + p_{n} s i n (a_{2}) + q_{n} c o s (a_{2}) = g (a_{2}) \\ 鈰� \\ c_{n} + p_{n} s i n (a_{n}) + q_{n} c o s (a_{n}) = g (a_{n}) \end{matrix} |$ or $A_{n} [\begin{matrix} c_{n} \\ p_{n} \\ q_{n} \end{matrix}] = {\overset{鈫�}{b}}_{n}$ where $A_{n} = [\begin{matrix} 1 & s i n (a_{1}) & c o s (a_{1}) \\ 1 & s i n (a_{2}) & c o s (a_{2}) \\ 鈰� & 鈰� & 鈰� \\ 1 & s i n (a_{n}) & c o s (a_{n}) \end{matrix}]$ , ${\overset{鈫�}{b}}_{n} = [\begin{matrix} g (a_{1}) \\ g (a_{2}) \\ 鈰� \\ g (a_{n}) \end{matrix}]$
(a)Find the entries of the matrix $A_{n}^{T} A_{n}$ and the components of the vector $A_{n}^{T} {\overset{鈫�}{b}}_{n}$ .
(b)Find $\underset{n 鈫� 鈭�}{l i m} (\frac{2 蟺}{n} A_{n}^{T} A_{n})$ androle="math" localid="1668167052477" $\lim_{n 鈫� 鈭�} (\frac{2 蟺}{n} A_{n}^{T} \overset{鈫�}{b})$ . Hint: Interpret the entries of the matrix $\frac{2 蟺}{n} A_{n}^{T} A_{n}$ and the components of the vector $\frac{2 蟺}{n} A_{n}^{T} \overset{鈫�}{b}$ as Riemann sums. Then the limits are the corresponding Riemann integrals. Evaluate as many integrals as you can. Note that i $\underset{n 鈫� 鈭�}{l i m} (\frac{2 蟺}{n} A_{n}^{T} A_{n})$ s a diagonal matrix.
(c) Find
$\begin{matrix} \underset{n 鈫� 鈭�}{l i m} [\begin{matrix} c_{n} \\ p_{n} \\ q_{n} \end{matrix}] = \underset{n 鈫� 鈭�}{l i m} {(A_{n}^{T} A_{n})}^{鈭� 1} A_{n}^{T} {\overset{鈫�}{b}}_{n} \\ = \underset{n 鈫� 鈭�}{l i m} [{(\frac{2 蟺}{n} A_{n}^{T} A_{n})}^{鈭� 1} (\frac{2 蟺}{n} A_{n}^{T} {\overset{鈫�}{b}}_{n})] \\ = {[\underset{n 鈫� 鈭�}{l i m} (\frac{2 蟺}{n} A_{n}^{T} A_{n})]}^{鈭� 1} \underset{n 鈫� 鈭�}{l i m} (\frac{2 蟺}{n} A_{n}^{T} {\overset{鈫�}{b}}_{n}) . \end{matrix}$
The resulting vector $[\begin{matrix} c \\ p \\ q \end{matrix}]$ gives you the coefficient of the desired function $f (t) = \underset{n 鈫� 鈭�}{l i m} f_{n} (t)$ .
Write $f (t)$ . The function $f (t)$ is called the first Fourier approximation of $g (t)$ . The Fourier approximation satisfies a 鈥渃ontinuous鈥� least-squares condition, an idea we will make more precise in the next section.

Short Answer

Expert verified

(a) The entries of matrix $A_{n}^{T} A_{n}$ are

$A_{n}^{T} A_{n} = [\begin{matrix} n & {鈭�}_{i = 1}^{n} \sin (a_{i}) & {鈭�}_{i = 1}^{n} \cos (a_{i}) \\ {鈭�}_{i = 1}^{n} \sin (a_{i}) & {鈭�}_{i = 1}^{n} \sin^{2} (a_{i}) & {鈭�}_{i = 1}^{n} \sin (a_{i}) \cos (a_{i}) \\ {鈭�}_{i = 1}^{n} \cos (a_{i}) & {鈭�}_{i = 1}^{n} \sin (a_{i}) \cos (a_{i}) & {鈭�}_{i = 1}^{n} \cos^{2} (a_{i}) \end{matrix}]$ and the components of the vector

$A_{n}^{T} \overset{鈫�}{b}$ are $A_{n}^{T} \overset{鈫�}{b} = [\begin{matrix} {鈭�}_{i = 1}^{n} g (a_{i}) \\ {鈭�}_{i = 1}^{n} g (a_{i}) \sin (a_{i}) \\ {鈭�}_{i = 1}^{n} g (a_{i}) \cos (a_{i}) \end{matrix}]$

(b) The values of limits are $\lim_{n 鈫� 鈭�} (\frac{2 蟺}{n} A_{n}^{T} A_{n}) = [\begin{matrix} 2 蟺 & 0 & 0 \\ 0 & 蟺 & 0 \\ 0 & 0 & 蟺 \end{matrix}]$ and

$\lim_{n 鈫� 鈭�} (\frac{2 蟺}{n} A_{n}^{T} \overset{鈫�}{b}) = [\begin{matrix} {鈭�}_{0}^{2 蟺} g (t) d t \\ {鈭�}_{0}^{2 蟺} g (t) \sin t d t \\ {鈭�}_{0}^{2 蟺} g (t) \cos t d t \end{matrix}]$

Step by step solution

Determine the transpose of a matrix

The transpose of a matrix is a matrix obtained when the rows and columns of a matrix are interchanged.

Find the entries of AnTAn

The given matrix is $A_{n} = [\begin{matrix} 1 & \sin (a_{1}) & \cos (a_{1}) \\ 1 & \sin (a_{2}) & \cos (a_{2}) \\ 鈰� & 鈰� & 鈰� \\ 1 & \sin (a_{n}) & \cos (a_{n}) \end{matrix}]$ . So, its transpose is

$A_{n}^{T} = [\begin{matrix} 1 & 1 & 鈰� & 1 \\ \sin (a_{1}) & \sin (a_{2}) & 鈰� & \sin (a_{n}) \\ \cos (a_{1}) & \cos (a_{2}) & 鈰� & \cos (a_{n}) \end{matrix}]$

Here, the dot product of $A_{n}^{T}$ and $A_{n}^{}$ is equal to the matrix $A_{n}^{T} A_{n}^{}$ . So, $i j^{th}$ entry of $i^{th}$ row of $A_{n}^{T}$ and $j^{th}$ column of $A_{n}$ .

Therefore, the entries of the matrix $A_{n}^{T} A_{n}$ is given below:

Similarly, the components of $A_{n}^{T} \overset{鈫�}{b}$ are given below:

$\begin{matrix} A_{n}^{T} \overset{鈫�}{b} = A_{n}^{T} = [\begin{matrix} 1 & 1 & 鈰� & 1 \\ \sin (a_{1}) & \sin (a_{2}) & 鈰� & \sin (a_{n}) \\ \cos (a_{1}) & \cos (a_{2}) & 鈰� & \cos (a_{n}) \end{matrix}] [\begin{matrix} g (a_{1}) \\ g (a_{2}) \\ 鈰� \\ g (a_{n}) \end{matrix}] \\ = [\begin{matrix} {鈭�}_{i = 1}^{n} g (a_{i}) \\ {鈭�}_{i = 1}^{n} g (a_{i}) \sin (a_{i}) \\ {鈭�}_{i = 1}^{n} g (a_{i}) \cos (a_{i}) \end{matrix}] \end{matrix}$

Thus, the entries of matrix $A_{n}^{T} A_{n}$ are

and the components of the vector are $A_{n}^{T} \overset{鈫�}{b} = [\begin{matrix} {鈭�}_{i = 1}^{n} g (a_{i}) \\ {鈭�}_{i = 1}^{n} g (a_{i}) \sin (a_{i}) \\ {鈭�}_{i = 1}^{n} g (a_{i}) \cos (a_{i}) \end{matrix}]$

Find the limits

Here, the limit $\lim_{n 鈫� 鈭�} \frac{2 蟺}{n} A_{n}^{T} A_{n}$ can be represented as Riemann sums. So, the entries of the matrix can be obtained as follows:

$\lim_{n 鈫� 鈭�} \frac{2 蟺}{n} {鈭�}_{i = 1}^{n} \sin (a_{i}) = {鈭�}_{0}^{2 蟺} \sin t d t \lim_{n 鈫� 鈭�} \frac{2 蟺}{n} {鈭�}_{i = 1}^{n} \cos (a_{i}) = {鈭�}_{0}^{2 蟺} \cos t d t \lim_{n 鈫� 鈭�} \frac{2 蟺}{n} {鈭�}_{i = 1}^{n} \sin (a_{i}) \cos (a_{i}) = {鈭�}_{0}^{2 蟺} \sin t \cos t d t$

Therefore, the limit $\lim_{n 鈫� 鈭�} \frac{2 蟺}{n} A_{n}^{T} A_{n}$ is shown below:

$\begin{matrix} \lim_{n 鈫� 鈭�} (\frac{2 蟺}{n} A_{n}^{T} A_{n}) = [\begin{matrix} 2 蟺 & {鈭�}_{0}^{2 蟺} \sin t d t & {鈭�}_{0}^{2 蟺} \cos t d t \\ {鈭�}_{0}^{2 蟺} \sin t d t & {鈭�}_{0}^{2 蟺} \sin^{2} t d t & {鈭�}_{0}^{2 蟺} \sin t \cos t d t \\ {鈭�}_{0}^{2 蟺} \cos t d t & {鈭�}_{0}^{2 蟺} \sin t \cos t d t & {鈭�}_{0}^{2 蟺} \cos^{2} t d t \end{matrix}] \\ [\begin{matrix} 2 蟺 & 0 & 0 \\ 0 & 蟺 & 0 \\ 0 & 0 & 蟺 \end{matrix}] \end{matrix}$

In the same way, the value of $\lim_{n 鈫� 鈭�} \frac{2 蟺}{n} A_{n}^{T} \overset{鈫�}{b}$ is:

$\lim_{n 鈫� 鈭�} \frac{2 蟺}{n} A_{n}^{T} \overset{鈫�}{b} = [\begin{matrix} {鈭�}_{0}^{2 蟺} g (t) d t \\ {鈭�}_{0}^{2 蟺} g (t) \sin t d t \\ {鈭�}_{0}^{2 蟺} g (t) \cos t d t \end{matrix}]$

Thus the values of limits are

$\lim_{n 鈫� 鈭�} (\frac{2 蟺}{n} A_{n}^{T} A_{n}) = [\begin{matrix} 2 蟺 & 0 & 0 \\ 0 & 蟺 & 0 \\ 0 & 0 & 蟺 \end{matrix}]$

and

$\lim_{n 鈫� 鈭�} (\frac{2 蟺}{n} A_{n}^{T} A_{n}) = [\begin{matrix} {鈭�}_{0}^{2 蟺} g (t) d t \\ {鈭�}_{0}^{2 蟺} g (t) \sin t d t \\ {鈭�}_{0}^{2 蟺} g (t) \cos t d t \end{matrix}]$

Find the function

Find the required limit $\lim_{n 鈫� 鈭�} [\begin{matrix} c_{n} \\ p_{n} \\ q_{n} \end{matrix}]$ as follows:

$\begin{matrix} \lim_{n 鈫� 鈭�} [\begin{matrix} c_{n} \\ p_{n} \\ q_{n} \end{matrix}] = {[\lim_{n 鈫� 鈭�} (\frac{2 蟺}{n} A_{n}^{T} A_{n})]}^{鈭� 1} \lim_{n 鈫� 鈭�} (\frac{2 蟺}{n} A_{n}^{T} {\overset{鈫�}{b}}_{n}) \\ = {[\begin{matrix} 2 蟺 & 0 & 0 \\ 0 & 蟺 & 0 \\ 0 & 0 & 蟺 \end{matrix}]}^{鈭� 1} [\begin{matrix} {鈭�}_{0}^{2 蟺} g (t) d t \\ {鈭�}_{0}^{2 蟺} g (t) \sin t d t \\ {鈭�}_{0}^{2 蟺} g (t) \cos t d t \end{matrix}] \\ = [\begin{matrix} \frac{1}{2 蟺} {鈭�}_{0}^{2 蟺} g (t) d t \\ \frac{1}{蟺} {鈭�}_{0}^{2 蟺} g (t) \sin t d t \\ \frac{1}{蟺} {鈭�}_{0}^{2 蟺} g (t) \cos t d t \end{matrix}] \\ = [\begin{matrix} c \\ p \\ q \end{matrix}] \end{matrix}$

Hence, the above calculation implies that $f (t) = c + p \sin t + q \cos t$

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Short Answer

Step by step solution

Determine the transpose of a matrix

Find the entries of AnTAn

Find the limits

Find the function

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