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 giveninner productis:

\[\begin{array}{c}\left\langle {p,q} \right\rangle = p\left( { - 5} \right)q\left( { - 5} \right) + p\left( { - 3} \right)q\left( { - 3} \right) + p\left( { - 1} \right)q\left( { - 1} \right)\\ + p\left( 1 \right)q\left( 1 \right) + p\left( 3 \right)q\left( 3 \right) + p\left( 5 \right)q\left( 5 \right)\end{array}\]

For the\[{p_2}\], we have:

\[\begin{array}{l}{p_0}\left( t \right) = 1\\{p_1}\left( t \right) = t\end{array}\]

Now, the trend function can be given as:

\(\begin{array}{c}{p_2}\left( t \right) = {t^2} - \frac{{\left\langle {{t^2},t} \right\rangle }}{{\left\langle {t,t} \right\rangle }}1 - \frac{{\left\langle {{t^2},t} \right\rangle }}{{\left\langle {t,t} \right\rangle }}t\\ = {t^2} - \frac{{35}}{3}\end{array}\)

Scaling by\(\frac{3}{8}\), we have:

\({p_2}\left( t \right) = \frac{3}{8}{t^2} - \frac{{35}}{8}\)

Hence, this is the required proof.

The givendatais:

\[\left( { - 5,1} \right),\left( { - 3,1} \right),\left( { - 1,4} \right),\left( {1,4} \right),\left( {3,6} \right),\left( {5,8} \right)\]

For the\[{p_2}\], we have:

\[\begin{array}{l}{p_0}\left( t \right) = 1\\{p_1}\left( t \right) = t\\g = \left( {1,1,4,4,6,8} \right)\end{array}\]

Now, the trend function can be given as:

\(\begin{array}{c}\hat p = \frac{{\left\langle {g,{p_0}} \right\rangle }}{{\left\langle {{p_0},{p_0}} \right\rangle }}{p_0} + \frac{{\left\langle {g,{p_1}} \right\rangle }}{{\left\langle {{p_1},{p_1}} \right\rangle }}{p_1} + \frac{{\left\langle {g,{p_2}} \right\rangle }}{{\left\langle {{p_2},{p_2}} \right\rangle }}{p_2}\\ = \frac{{24}}{6}\left( 1 \right) + \frac{{50}}{{70}}t + \frac{6}{{24}}\left( {\frac{3}{8}{t^2} - \frac{{35}}{8}} \right)\\ = \frac{{59}}{{16}} + \frac{5}{7}t + \frac{3}{{112}}{t^2}\end{array}\)

Hence, this is the required answer.