Question

Let \(B\) be the basis of \({{\mathop{\rm P}\nolimits} _3}\) consisting of the Hermite polynomials in Exercise 21, and let \(p\left( t \right) = 7 - 12t - 8{t^2} + 12{t^3}\). Find the coordinate vector of \({\mathop{\rm p}\nolimits} \) relative to \(B\).

Short answer

The coordinate vector of \({\mathop{\rm p}\nolimits} \) relative to \(B\) is \({\left[ {\mathop{\rm p}\nolimits} \right]_B} = \left[ {\begin{array}{*{20}{c}}3\\3\\{ - 2}\\{\frac{3}{2}}\end{array}} \right]\).

Step-by-step solution

Definition of the coordinate vector of x

Suppose\(B = \left\{ {{{\mathop{\rm b}\nolimits} _1},...,{{\mathop{\rm b}\nolimits} _n}} \right\}\)is a basis for\(V\)and x is in\(V\). Thecoordinatesof\({\mathop{\rm x}\nolimits} \)relative to the basis \(B\)(or the\(B\)-coordinates of x) are the weights \({c_1},...,{c_n}\) such that \({\mathop{\rm x}\nolimits} = {c_1}{b_1} + ... + {c_n}{b_n}\).

Determine the coordinate vector of \({\mathop{\rm p}\nolimits} \) relative to \(B\)

The coordinate vector of\(p\left( t \right) = 7 - 12t - 8{t^2} + 12{t^3}\)with respect to\(B\)is as shown below:

\({c_1}\left( 1 \right) + {c_2}\left( {2t} \right) + {c_3}\left( { - 2 + 4{t^2}} \right) + {c_4}\left( { - 12t + 8{t^3}} \right) = 7 - 12t - 8{t^2} + 12{t^3}\)

Equate the coefficient of\(t\)to produce the system of the equation as shown below:

\(\begin{aligned} {c_1}\,\,\,\,\,\,\,\,\,\, - 2{c_3}\,\,\,\,\,\,\,\,\,\,\,\,\, &= 7\\\,\,\,\,\,\,\,2{c_2}\,\,\,\,\,\,\,\,\,\, - 12{c_4} &= - 12\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,4{c_3}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, &= - 8\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,8{c_4} &= 12\end{aligned}\)

By solving the system of the equation, you get\({c_1} = 3,{c_2} = 3,{c_3} = - 2,{c_4} = \frac{3}{2}\). Therefore,\({\left[ {\mathop{\rm p}\nolimits} \right]_B} = \left[ {\begin{array}{*{20}{c}}3\\3\\{ - 2}\\{\frac{3}{2}}\end{array}} \right]\).

Thus, the coordinate vector of \({\mathop{\rm p}\nolimits} \) relative to \(B\) is \({\left[ {\mathop{\rm p}\nolimits} \right]_B} = \left[ {\begin{array}{*{20}{c}}3\\3\\{ - 2}\\{\frac{3}{2}}\end{array}} \right]\).