fixed $\mu$, we can obtain the function $\delta$ by the extrema condition $\partial M_{T}/\partial x=0$, where $x=1/B$. Now, neglecting the spin splitting, equation (\[MT\]) becomes

$$M_{T}=-\frac{2\mu}{\pi\phi}\arctan\left[\cot\left(\pi\omega x+\sum_{n}\pi f_{n}\right)\right],\label{MTSP}$$

where we defined $f_{n}\equiv\left[1+e^{\beta\mu\left(x-x_{n}\right)/2x}\right]^{-1}$. Then, the extrema condition $\partial M_{T}/\partial x=0$ can be written as

$$1=\frac{\beta\mu}{8\omega x^{2}}\sum_{n}x_{n}\textrm{sech}^{2}\left[\frac{\beta\mu\left(x-x_{n}\right)}{4x}\right].\label{x extreme}$$

The $x=1/B$ that satisfy equation (\[x extreme\]) give all the new maxima and minima in the MO at $T\neq0$. Notice that at zero temperature $\beta\rightarrow\infty$ and equation (\[x extreme\]) implies $x\rightarrow x_{n}$, as expected. In general, for a peak at $x_{l}=l/\omega$, to obtain its shift it is sufficient to consider only the $n$ close to $l$ (say $l-2<n<l+2$) in the summation of equation (\[x extreme\]). The new extrema will be located at $x=x_{l}-p\delta_{l}$, where $\delta_{l}$ is the corresponding shift and $p=1(-1)$ for the maxima (minima). In this way we can rewrite equation (\[x extreme\]) as a function of $\delta_{l}$ and the considered peak $x_{l}$:

$$1 =\frac{\beta\mu}{8\omega\left(x_{l}-p\delta_{l}\right)^{2}}
 \sum_{n}x_{n}\textrm{sech}^{2}\left[\frac{\beta\mu\left(x_{l}-x_{n}-p\delta_{l}\right)}{4\left(x_{l}-p\delta_{l}\right)}\right].\label{delta extreme}$$

![\[fig7\]Graphical solution of equation (\[delta extreme\]) for different LL $l$ and temperatures (a) $T=5$ K, (b) $T=10$ K. All cases correspond to $\mu=250$ meV and maxima shift, so $p=1$ in equation (\[delta extreme\]).](fig7.eps)

In figure \[fig7\] it is show the graphical solution of equation (\[delta extreme\]) 