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I have a Lagrangian

$$ L = \underbrace{\dot{x}^2 + \frac{1}{2}\dot{\theta}^2 + \dot{x}\dot{\theta}\cos\theta}_{\text{kinetic energy}} \underbrace{ - \frac{3}{4}x^2 + \cos\theta}_{\text{-ve potential energy}}$$

I wish to approximate the Lagrangian at the equilibrium, $x = 0 ,\theta = \pi$

I know:

$$L \approx \frac{1}{2}\delta{\bf\dot{q}}\cdot M({\bf q_*})\delta{\bf\dot{q}} - U({\bf q_*}) - \frac{1}{2}\delta {\bf q} \cdot K \delta {\bf q}$$

Where $$\bf q = q_* + \delta q$$ Are the generalised coordinates.

My issue is specifically with the potential energy. My working is as follows.

$$U = \frac{3}{4}x^2 - \cos\theta$$ We take the Taylor expansion of cos and ignore terms with order higher than 2. $$U = \frac{3}{4}x^2 - (1 - \frac{\theta^2}{2} + \ldots)$$

$$\frac{1}{2} {\bf q} \cdot K {\bf q} + c = \frac{3}{4}x^2 - 1 + \frac{\theta^2}{2} $$

So if my working is correct,

$$K = \begin{pmatrix} \frac{3}{2} & 0 \\ 0 & 1 \end{pmatrix}, c = -1$$

However, if I then use this K in the approximation of the Lagrangian, I do not get what I am required.

Specifically, I need to show the spproximate Lagrangian takes the form $$ L = (\delta\dot{x})^2 + \frac{1}{2}(\delta\dot{\theta})^2 + (\delta\dot{x})(\delta\dot{\theta}) - \frac{3}{4}(\delta x)^2 + \frac{1}{2}(\delta\theta)^2 + c$$

I appear to get the $\delta\theta$ term with the wrong sign.

Any help would be appreciated, thanks.

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