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I have a generic function: $g(s,u)$. Now I want to have a local approximation near the point $(s^{\star}, u^{\star})$ in the quadratic form $s^{T} Q s + u^{T} R u$ to apply an optimal control algorithm named LQR. I have to apply a second order Taylor expansion which will result in $g(s^{\star}, u^{\star}) + g_s(s^{\star}, u^{\star})(x-x^{\star}) + g_u(s^{\star}, u^{\star})(u-u^{\star}) + \frac{1}{2} \left[g_{ss}(s^{\star}, u^{\star})(s-s^{\star})^2 + g_{su}(s^{\star}, u^{\star})2(s-s^{\star})(u-u^{\star}) + g_{uu}(s^{\star}, u^{\star})(u-u^{\star})^2 \right]$

Now my question is which partial derivatives form the matrices $Q$ and $R$.

I know how to derive this matrices for an example with one variable, then $\frac{1}{2} x^T Q x + x^T q + q_0$ can be written as $\begin{bmatrix}s \\ 1\end{bmatrix}^T \begin{bmatrix}Q & q \\ q^T & q_0\end{bmatrix} \begin{bmatrix}s \\ 1\end{bmatrix}$ which is a quadratic form. But how to do this with two variables?

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    Rephrased the question here: http://math.stackexchange.com/questions/198088/quadratic-approximation-of-a-cost-function-with-a-taylor-expansion2012-09-17

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What you're doing makes no sense. I assume that $g : \mathbb{R}^n \times \mathbb{R}^m \to \mathbb{R}$. The local approximation of $g$ at $(s^*,u^*)$ should not be a quadratic form

$ g (s, u) \approx s^T Q s + u^T R u$

but rather a quadratic polynomial, the 2nd order Taylor approximation of $g$, which is

$g (s,u) \approx g (s^*, u^*) + q^T (s - s^*) + r^T (u - u^*) + (s - s^*)^T Q (s - s^*) + (u - u^*)^T R (u - u ^*)$

If $g (s^*, u^*) = 0$, $q = 0_n$, and $r = 0_m$, then $(s^*, u^*)$ is a local minimum and you obtain a quadratic form as a local approximation

$g (s,u) \approx (s - s^*)^T Q (s - s^*) + (u - u^*)^T R (u - u ^*)$

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    Yes, $g$ is my cost function for which I want to obtain a quadratic approximation. Is $Q$ equal to $g_{ss}$? And $R$ to $g_{uu}$? And what to do with $g_{su}$?2012-09-13