Question

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

11. Find the third-order Fourier approximation to \({\sin ^2}t\), without performing any integration calculations.

Short answer

The required Approximation is\({\sin ^2}t = \frac{1}{2} - \frac{1}{2}\cos 2t\).

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:

\(f\left( t \right) = {\sin ^2}t\)

Usingtrigonometric expansion, we have:

\(\begin{array}{c}{\sin ^2}t = \frac{1}{2}\left( {1 - \cos 2t} \right)\\ = \frac{1}{2} - \frac{1}{2}\cos 2t\end{array}\)

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

Therefore, the third-order expansion is:

\({\sin ^2}t = \frac{1}{2} - \frac{1}{2}\cos 2t\)

Hence, the is the required Approximation.