Given the Euclidean vector space
\left(\mathcal P_{\leq 2}, \langle \ , \ \rangle \right) with
\displaystyle \langle p, q \rangle = \int_{-1}^{1} p(x)q(x) \; dx
and the polynomials {\color{red}p}, \ {\color{blue}q} \; : [-1,1] \to \mathbb R
mit {\color{red}p(x) = Ax^2 + Bx +C} and
q(x) = {\color{blue}d}x + 1.
For which {\color{blue}d} are {\color{red}p} and {\color{blue}q} orthogonal?
{\color{blue}d} =
(-3*C-A)/B
We are looking for {\color{blue}d} such that
\displaystyle 0= \langle {\color{red}p}, {\color{blue}q} \rangle =
\int_{-1}^{1} {\color{red}p(x)}\ {\color{blue}q(x)} \; dx
.
With the given {\color{red}p}, \ {\color{blue}q} we have
\displaystyle
\int_{-1}^{1} {\color{red}\left(Ax^2 + Bx +C\right)}\
({\color{blue}d}x + 1) \; dx =
\left(negParens(fractionReduce(A,4)) {\color{blue}d} x^4 +
\left(fractionReduce(B,3){\color{blue}d} + fractionReduce(A,3)\right)x^3 +
\left(fractionReduce(C,2){\color{blue}d} + fractionReduce(B,2)\right)x^2 + Cx \right)\biggl|_{-1}^{1} =
fractionReduce(2*B,3){\color{blue}d} + fractionReduce(2*A+6*C,3).
We solve
fractionReduce(2*B,3){\color{blue}d} + fractionReduce(2*A+6*C,3) =0 for {\color{blue}d} and conclude {\color{blue}d} = fractionReduce(-3*C-A,B).