Chapter 10.3, Problem 1E

a.

To determine

To find: the best order 2 least squares approximation to the data using the basic function $1and\cos 2\pi t$

Explanation of Solution

Given information:

The initial conditions that are given are,

here is the best order 2 least squares approximation to the data using the basic function $1and\cos 2\pi t$

Calculation:

As it’s known that by using DFT equation here is the best order 2 least squares approximation to the data using the basic function $1and\cos 2\pi t$

According to Theorem 10.11, the order 2 least square approximation results from dropping all but the first two terms in the trigonometric interpolating function 

  $P_4(t)=0+0\cos 2\pi t+\sin 2\pi t+0\cos 2\pi t$ .

Therefore, the approximating function is  $P_2(t)=0$ .

b.

To determine

To find: the best order 2 least squares approximation to the data using the basic function $1and\cos 2\pi t$

Explanation of Solution

Given information:

The initial conditions that are given are,

here is the best order 2 least squares approximation to the data using the basic function $1and\cos 2\pi t$

Calculation:

As it’s known that by using DFT equation, here is the best order 2 least squares approximation to the data using the basic function $1and\cos 2\pi t$

.Similar to (a).

Dropping all but the first two terms from the trigonometric interpolating function

  $P_4(t)=\cos 2\pi t+\sin 2\pi t$  yields  $P_2(t)=\cos 2\pi t$ .

c.

To determine

To find: the best order 2 least squares approximation to the data using the basic function $1and\cos 2\pi t$

Explanation of Solution

Given information:

The initial conditions that are given are, here is the best order 2 least squares approximation to the data using the basic function $1and\cos 2\pi t$

Calculation:

As it’s known that by using DFT equation, here is the best order 2 least squares approximation to the data using the basic function $1and\cos 2\pi t$

Similar to (a). Dropping all but the first two terms from the trigonometric interpolating function

  $P_4(t)=-\cos 4\pi t$ yields  $P_2(t)=0$ .