Let $\gamma:I\rightarrow\mathbb{R}^3$ be a curve parametrised by arclength. I want to calculate the taylor expansion using the curvature $$\kappa_\gamma:=\| \ddot\gamma \|$$ and torsion $$\tau_\gamma:=\langle \dot\eta_\gamma,b_\gamma\rangle$$
I'd be very happy, if someone could look over it and tell, if I've done following correct.
First of all, $$\Big(\dot\gamma,\frac{\ddot\gamma}{\|\ddot\gamma\|},\dot\gamma\times\frac{\ddot\gamma}{\|\ddot\gamma\|}\Big)=\Big(\dot\gamma,\eta_\gamma,b_\gamma\Big)$$
is a positive oriented orthonormal basis of $\mathbb{R}^3$.
The taylor expansion is
$$\gamma(t)=\gamma(t_0)+(t-t_0)\dot\gamma(t_0)+\frac{1}{2}(t-t_0)^2\ddot\gamma(t_0)+\frac{1}{6}(t-t_0)^3\dddot\gamma(t_0)+(t-t_0)^4 o(t-t_0)$$
I find
$$\ddot\gamma=\kappa_\gamma\eta_\gamma$$
and therefore
$$\dddot\gamma=\dot\kappa_\gamma\eta_\gamma+\kappa_\gamma\dot\eta_\gamma$$
Using the orthonormal basis I find
$$ \dddot\gamma=\langle\kappa_\gamma\dot\eta_\gamma,\dot\gamma\rangle\dot\gamma+\langle\dot\kappa_\gamma\eta_\gamma,\eta_\gamma\rangle\eta_\gamma+\langle\kappa_\gamma\dot\eta_\gamma,b_\gamma\rangle b_\gamma=-\kappa_\gamma^2\dot\gamma+\dot\kappa_\gamma\eta_\gamma+\kappa_\gamma\tau_\gamma b_\gamma $$
Therefore the taylor expansion of $\gamma$ in $t_0$ is
$$\gamma(t)=\gamma(t_0)+(t-t_0)\dot\gamma(t_0)+\frac{1}{2}(t-t_0)^2\Big(\kappa_\gamma\eta_\gamma\Big)(t_0)+\frac{1}{6}(t-t_0)^3\Big(-\kappa_\gamma^2\dot\gamma+\dot\kappa_\gamma\eta_\gamma+\kappa_\gamma\tau_\gamma b_\gamma\Big)(t_0)+(t-t_0)^4 o(t-t_0)$$
Thanks for your attention!