In general the period of oscillation is given by $\Delta(1/B)=1/B_{2}-1/B_{1}$, where $B_{i}$ is such that $\varepsilon_{f_{i}}(B_{i})=\mu$. Therefore $\mu=\alpha\sqrt{B_{i}n_{i}}-s_{i}\mu_{\mathrm{B}}B_{i}$, where for simplicity we have noted $n_{f_{i}}=n_{i}$ and $s_{f_{i}}=s_{i}$. Using the approximation $\mu^{2}+2s_{i}\mu\mu_{\mathrm{B}}B_{i}+\mu_{\mathrm{B}}^{2}B_{i}^{2}\simeq\mu^{2}+2s_{i}\mu\mu_{\mathrm{B}}B_{i}$, which holds for typical values of magnetic field, we obtain the $B_{i}$ at which the peaks occur:

$$\frac{1}{B_{i}}=\frac{n_{i}\alpha^{2}}{\mu^{2}}-\frac{2s_{i}\mu_{\mathrm{B}}}{\mu}.\label{1/Bi}$$

We shall consider that the two peaks correspond to a fixed spin, with the oscillation being given when the LL changes by one. Thus we take $\Delta n=n_{2}-n_{1}=1$, while $s=s_1=s_2$. Therefore, from equation (\[1/Bi\]) we obtain the period $\Delta(1/B)$ and frequency $\omega=\left[\Delta(1/B)\right]^{-1}$

$$\omega_{s}=\frac{\mu^{2}}{\alpha^{2}}.\label{freq}$$

Then we can write equation (\[1/Bi\]) as $1/B_{i,s}=n_{i}/\omega_{s}+\Delta_{s}$, where $\Delta_s$ is the phase

$$\Delta_{s}=-\frac{2s_{i}\mu_{\mathrm{B}}}{\mu}.\label{phase}$$

From equations (\[freq\]) and (\[phase\]) we see that $\omega_{1}=\omega_{-1}$ and $\Delta_{1}=-\Delta_{-1}$. This means that the two sawtooth peaks have the same frequency but different phase. The peaks amplitude are obtained from equation (\[DeltaM\]). Suppose the magnetic field is increased so the last sorted position $f$ changes to $f-1$. For $\Delta n$ and $\Delta m_{P}$ we easily get $\Delta n=D/\mathcal{A}=2B/\phi$ and $\Delta m_{P}=D\mu_{\mathrm{B}}s_{f}/\mathcal{A}=2B\mu_{\mathrm{B}}s_{f}/\phi$, so in general

$$A_{s}=\frac{-\mu+s\mu_{\mathrm{B}}B}{\phi}.\label{amp}$$

We are now in position to express the two SO as an infinite series, whose amplitude, frequency and phase are given by equations (\[freq\]), (\[phase\]) and (\[amp\]). We 