Mollie Wise

2022-11-23

Let B be the basis ${\mathbb{P}}_{3}$ consisting of the Hermite polynomials $1,2t,-2+4{t}^{2}$ and $-12t+8{t}^{3}$; and let $p\left(t\right)=1+16{t}^{2}-8{t}^{3}$. Find the coordinate vector of p relative to B.
$\left[p{\right]}_{B}=\left[\begin{array}{c}-7\\ -6\\ -4\\ -1\end{array}\right]$

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Expert

Let $p\left(t\right)=a.1+b\left(2t\right)+c\left(-2+4{t}^{2}\right)+d\left(-12t+8{t}^{3}\right)\phantom{\rule{0ex}{0ex}}=\left(8d\right){t}^{3}+\left(4c\right){t}^{2}+\left(2b-12d\right)t+\left(a-2c\right)\phantom{\rule{0ex}{0ex}}⇒-8{t}^{3}+16{t}^{2}+1=\left(8d\right){t}^{3}+\left(4c\right){t}^{2}+\left(2b-12d\right)t+\left(a-2c\right)\phantom{\rule{0ex}{0ex}}⇒\left\{\begin{array}{l}8d=-8\\ 4c=16\\ 2b-12d=0\\ a-2c=1\end{array}⇒\left\{\begin{array}{l}d=-1\\ c=4\\ b=-6\\ a=9\end{array}$
So $\left[p{\right]}_{B}=\left[\begin{array}{c}9\\ -6\\ 4\\ -1\end{array}\right]$

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