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In Kuhl and Giardina's 1982 paper, it derives the coefficients, $a_n$ and $b_n$, by equating the two definitions of $\dot{x}(t)$:

$$ \begin{align*} \dot{x}(t) &= \sum_{n=1}^\infty \alpha_n \cos\frac{2n \pi t}{T} + \beta_n \sin\frac{2n \pi t}{T} \\ \dot{x}(t) &= \sum_{n=1}^\infty -\frac{2n\pi}{T} a_n \sin\frac{2n \pi t}{T} + \frac{2n\pi}{T} b_n \cos\frac{2n \pi t}{T} \end{align*} $$ Where $$ \alpha_n = \frac{2}{T} \sum_{p=1}^K \frac{\Delta x_p}{\Delta t_p} \left( \sin\frac{2n \pi t_p}{T} - \sin\frac{2n \pi t_{p-1}}{T} \right) \\ \beta_n = -\frac{2}{T} \sum_{p=1}^K \frac{\Delta x_p}{\Delta t_p} \left( \cos\frac{2n \pi t_p}{T} - \cos\frac{2n \pi t_{p-1}}{T} \right) $$ Their result is $$ a_n = \frac{T}{2n^2 \pi^2} \sum_{p=1}^K \frac{\Delta x_p}{\Delta t_p} \left( \cos\frac{2n \pi t_p}{T} - \cos\frac{2n \pi t_{p-1}}{T} \right)\\ b_n = \frac{T}{2n^2 \pi^2} \sum_{p=1}^K \frac{\Delta x_p}{\Delta t_p} \left( \sin\frac{2n \pi t_p}{T} - \sin\frac{2n \pi t_{p-1}}{T} \right) $$ However, when I tried to derive them, I get: $$ \frac{2n\pi}{T} b_n = \alpha_n = \frac{2}{T} \sum_{p=1}^K \frac{\Delta x_p}{\Delta t_p} \left( \sin\frac{2n \pi t_p}{T} - \sin\frac{2n \pi t_{p-1}}{T} \right)\\ b_n = \frac{1}{n\pi} \sum_{p=1}^K \frac{\Delta x_p}{\Delta t_p} \left( \sin\frac{2n \pi t_p}{T} - \sin\frac{2n \pi t_{p-1}}{T} \right) $$ and similar for $a_n$. The results are off exactly by $\frac{2n\pi}{T}$, the coefficient before $b_n$ in the second definition. So maybe I missed something in my derivation.

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