91Ó°ÊÓ

The Inner Productfor any two arbitrary functions is given by:

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

The given orthogonal polynomial is:

\[{p_3}\left( t \right) = \frac{5}{6}{t^3} - \frac{{17}}{6}t\]

From example 2, we have:

\[\begin{array}{l}{p_0}\left( t \right) = 1\\{p_1}\left( t \right) = t\\{p_2}\left( t \right) = {t^2} - 2\\g = \left( {3,5,5,4,3} \right)\end{array}\]

Now, the cubic trend function can be given as:

\(\begin{array}{c}\hat p = \frac{{\left\langle {g,{p_0}} \right\rangle }}{{\left\langle {{p_0},{p_0}} \right\rangle }}{p_0} + \frac{{\left\langle {g,{p_1}} \right\rangle }}{{\left\langle {{p_1},{p_1}} \right\rangle }}{p_1} + \frac{{\left\langle {g,{p_2}} \right\rangle }}{{\left\langle {{p_2},{p_2}} \right\rangle }}{p_2} + \frac{{\left\langle {g,{p_3}} \right\rangle }}{{\left\langle {{p_3},{p_3}} \right\rangle }}{p_3}\\ = \frac{{20}}{5}\left( 1 \right) - \frac{1}{{10}}t - \frac{7}{{14}}\left( {{t^2} - 2} \right) + \frac{2}{{10}}\left( {\frac{5}{6}{t^3} - \frac{{17}}{6}t} \right)\\ = 5 - \frac{2}{3}t - \frac{1}{2}{t^2} + \frac{1}{6}{t^3}\end{array}\)

Hence, this is the required answer.