Question

In Exercises 5-14, the space is \(C\left[ {0,2\pi } \right]\) with inner product (6).

12. Find the third-order Fourier approximation to \({\cos ^3}t\), without performing any integration calculations.

Short answer

The required Approximation is \({\cos ^3}t = \frac{3}{4}\cos t + \frac{1}{4}\cos 3t\).

Step-by-step solution

Inner Product

The Inner Productfor any two arbitrary functions is given by:

\(\left\langle {f,g} \right\rangle = \int_0^{2\pi } {f\left( t \right)g\left( t \right)dt} \)

Find the third-order Approximation

As per the question, we have:

\(f\left( t \right) = {\cos ^3}t\)

Usingtrigonometric expansion, we have:

\(\begin{array}{c}{\cos ^3}t = \frac{1}{4}\left( {3\cos t + \cos 3t} \right)\\ = \frac{3}{4}\cos t + \frac{1}{4}\cos 3t\end{array}\)

Since this expansion itself is a trigonometric polynomial of the order less than 3.

Therefore, the third-order expansion is:

\({\cos ^3}t = \frac{3}{4}\cos t + \frac{1}{4}\cos 3t\)

Hence, the is the required Approximation.