0
$\begingroup$

In Caruana / Einav (2008), there is the following polynomial of $q_i$ and $q_j$ (Eq. 14 on page 995):

$$ 0 = \frac{1}{2\theta} \left(B_t + 2 D_t q_i + F_t q_j \right)^2 + \frac{1}{\theta}\left( C_t + E_t q_j + F_t q_i\right)(B_t + 2 D_t q_j + F_t q_i) + (A'_t + B'_t q_i + C'_t q_j + D'_t q_i^2 + E'_t q_j^2 + F'_t q_i q_j) $$

where $A,B,C,D,E$ and $F$ are functions of $t$.

By arguing that '...this has to be satisfied for all values of $q_i$ and $q_j$, all of its six coefficients (which are functions of $t$) have to equal zero' and reversing signs for the derivatives, they obtain that (Eq. 15):

$$ \begin{pmatrix} A' \\ B' \\C'\\D'\\E'\\F'\end{pmatrix} = \frac{1}{\theta} \begin{pmatrix} B^2/2 + BC \\2BD+BF+CF\\BF+2BE+2CD\\ 2D^2+F^2\\F^2/2+4DE \\4DF+2EF\end{pmatrix}$$

I can see how one obtains $A'$ by setting $q_i = q_j = 0$, but cannot see where the other derivatves come from. How are these obtained?

1 Answers 1

1

A polynomial is the zero polynomial iff all the coefficients are zero. To get equation involving $B'_t$ you should perhaps write the equation comparing coefficient of $q_i$. Generally the conditions coefficients of $q_iq_j$, $q_i^2q_j$ etc should lead to the equations (i.e Eq.15 in your question). Nothing terribly difficult, bit tiresome perhaps.