Question

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

9. Find the third-order Fourier approximation to \(f\left( t \right) = 2\pi - t\).

Short answer

The required Approximation is\(f\left( t \right) = \pi + 2\sin t + \sin 2t + \frac{2}{3}\sin 3t\).

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:

\(f\left( t \right) = 2\pi - t\)

Using theinner productrule, we have the Fourier Transformation as:

\(\frac{{{a_0}}}{2} = \frac{1}{2}\left\{ {\frac{1}{\pi }\int_0^{2\pi } {\left( {2\pi - t} \right)dt} } \right\} = \pi \)

Similarly,

\({a_k} = \frac{1}{\pi }\int_0^{2\pi } {\left( {2\pi - t} \right)\cos ktdt} = 0\)

And,

\({b_k} = \frac{1}{\pi }\int_0^{2\pi } {\left( {2\pi - t} \right)\sin ktdt} = \frac{2}{k}\)

Hence,therequired Approximation at k=3 is\(f\left( t \right) = \pi + 2\sin t + \sin 2t + \frac{2}{3}\sin 3t\).