All velocities in this problem are assumed to be aligned in the $\mathrm{x}$-direction. The rapidity $\phi$ of a particle or frame with respect to a frame $\mathrm{S}$ is defined as
$$
\tanh \phi=\beta=\frac{v}{c}
$$
where $v$ is the velocity of the particle or frame with respect to $\mathrm{S} \cdot \tanh \phi=$ $\frac{e^\phi-e^{-\phi}}{e^\phi+e^{-\phi}}$ is the hyperbolic tangent function. Show that if a particle has rapidity $\phi_1$ with respect to frame 1 and frame 1 has rapidity $\phi_2$ with respect to frame 2 , the particle has rapidity $\phi_1+\phi_2$ with respect to frame 2 . It may be useful to know that $\tanh \left(\phi_1+\phi_2\right)=\frac{\tanh \phi_1+\tanh \phi_2}{1+\tanh \phi_1 \tanh \phi_2}$.
Now, consider a particle which travels at a velocity $v_1$ with respect to frame $S_1$, which travels at velocity $v_2$ with respect to frame $S_2$, which travels at velocity $v_3$ with respect to frame $S_3$, and so on until frame $S_{n-1}$ which travels at velocity $v_n$ with respect to frame $S_n$. All of these velocities are aligned. Show that the velocity of the particle in frame $S_n$ is
$$
u=c \cdot \frac{\prod_{i=1}^N\left(1+\beta_i\right)-\prod_{i=1}^N\left(1-\beta_i\right)}{\prod_{i=1}^N\left(1+\beta_i\right)+\prod_{i=1}^N\left(1-\beta_i\right)}
$$
where $\beta_i=\frac{v_i}{c}$.