Giulio

Il codice è

`\begin{equation}
\begin{split}
D=&-1+\\
&|\frac{\left(1-\frac{1}{4} n (1-p)-\frac{1}{16} (2-n)^2 (1+p)-\frac{1}{4} \left(1+\frac{n^2}{4}\right) (1+p)\right) \text{Log}\left[\frac{1}{2} \left(1-\frac{1}{4} n (1-p)-\frac{1}{16} (2-n)^2 (1+p)-\frac{1}{4} \left(1+\frac{n^2}{4}\right) (1+p)\right)\right]}{2 \text{Log}[2]}\\
&-\frac{\left(-1+\frac{1}{4} n (1-p)+\frac{1}{16} (2-n)^2 (1+p)+\frac{1}{4} \left(1+\frac{n^2}{4}\right) (1+p)\right) \text{Log}\left[\frac{1}{2} \left(1-\frac{1}{4} n (1-p)-\frac{1}{16} (2-n)^2 (1+p)-\frac{1}{4} \left(1+\frac{n^2}{4}\right) (1+p)\right)\right]}{2 \text{Log}[2]}\\
&-\frac{\left(1-\frac{1}{4} n (1-p)+\frac{1}{16} (2-n)^2 (1+p)-\frac{1}{4} \left(1+\frac{n^2}{4}\right) (1+p)\right) \text{Log}\left[\frac{1}{2} \left(1-\frac{1}{4} n (1-p)+\frac{1}{16} (2-n)^2 (1+p)-\frac{1}{4} \left(1+\frac{n^2}{4}\right) (1+p)\right)\right]}{2 \text{Log}[2]}\\
&-\frac{\left(1+\frac{1}{4} n (1-p)-\frac{1}{16} (2-n)^2 (1+p)+\frac{1}{4} \left(1+\frac{n^2}{4}\right) (1+p)\right) \text{Log}\left[\frac{1}{2} \left(1+\frac{1}{4} n (1-p)-\frac{1}{16} (2-n)^2 (1+p)+\frac{1}{4} \left(1+\frac{n^2}{4}\right) (1+p)\right)\right]}{2 \text{Log}[2]}\\
&-\frac{\left(1-\sqrt{\frac{1}{4} (2-n)^2 p^2+\left(-\frac{1}{4} n (1-p)+\frac{1}{16} (2-n)^2 (1+p)-\frac{1}{4} \left(1+\frac{n^2}{4}\right) (1+p)\right)^2}\right) \text{Log}\left[\frac{1}{4} \left(1-\sqrt{\frac{1}{4} (2-n)^2 p^2+\left(-\frac{1}{4} n (1-p)+\frac{1}{16} (2-n)^2 (1+p)-\frac{1}{4} \left(1+\frac{n^2}{4}\right) (1+p)\right)^2}\right)\right]}{2 \text{Log}[2]}\\
&+\frac{1}{2 \text{Log}[2]}\left(\frac{1}{4} n (1-p)+\frac{1}{16} (2-n)^2 (1+p)+\frac{1}{4} \left(1+\frac{n^2}{4}\right) (1+p)-\sqrt{\frac{1}{4} (2-n)^2 p^2+\left(-\frac{1}{4} n (1-p)+\frac{1}{16} (2-n)^2 (1+p)-\frac{1}{4} \left(1+\frac{n^2}{4}\right) (1+p)\right)^2}\right) \text{Log}\left[\frac{1}{2} \left(\frac{1}{4} n (1-p)+\frac{1}{16} (2-n)^2 (1+p)+\frac{1}{4} \left(1+\frac{n^2}{4}\right) (1+p)-\sqrt{\frac{1}{4} (2-n)^2 p^2+\left(-\frac{1}{4} n (1-p)+\frac{1}{16} (2-n)^2 (1+p)-\frac{1}{4} \left(1+\frac{n^2}{4}\right) (1+p)\right)^2}\right)\right]+\frac{\left(-1-\sqrt{\frac{1}{4} (2-n)^2 p^2+\left(-\frac{1}{4} n (1-p)+\frac{1}{16} (2-n)^2 (1+p)-\frac{1}{4} \left(1+\frac{n^2}{4}\right) (1+p)\right)^2}\right) \text{Log}\left[\frac{1}{4} \left(1+\sqrt{\frac{1}{4} (2-n)^2 p^2+\left(-\frac{1}{4} n (1-p)+\frac{1}{16} (2-n)^2 (1+p)-\frac{1}{4} \left(1+\frac{n^2}{4}\right) (1+p)\right)^2}\right)\right]}{4 \text{Log}[2]}\\
&-\frac{\left(1+\sqrt{\frac{1}{4} (2-n)^2 p^2+\left(-\frac{1}{4} n (1-p)+\frac{1}{16} (2-n)^2 (1+p)-\frac{1}{4} \left(1+\frac{n^2}{4}\right) (1+p)\right)^2}\right) \text{Log}\left[\frac{1}{4} \left(1+\sqrt{\frac{1}{4} (2-n)^2 p^2+\left(-\frac{1}{4} n (1-p)+\frac{1}{16} (2-n)^2 (1+p)-\frac{1}{4} \left(1+\frac{n^2}{4}\right) (1+p)\right)^2}\right)\right]}{4 \text{Log}[2]}\\
&+\frac{1}{2 \text{Log}[2]}\left(\frac{1}{4} n (1-p)+\frac{1}{16} (2-n)^2 (1+p)+\frac{1}{4} \left(1+\frac{n^2}{4}\right) (1+p)+\sqrt{\frac{1}{4} (2-n)^2 p^2+\left(-\frac{1}{4} n (1-p)+\frac{1}{16} (2-n)^2 (1+p)-\frac{1}{4} \left(1+\frac{n^2}{4}\right) (1+p)\right)^2}\right) \text{Log}\left[\frac{1}{2} \left(\frac{1}{4} n (1-p)+\frac{1}{16} (2-n)^2 (1+p)+\frac{1}{4} \left(1+\frac{n^2}{4}\right) (1+p)+\sqrt{\frac{1}{4} (2-n)^2 p^2+\left(-\frac{1}{4} n (1-p)+\frac{1}{16} (2-n)^2 (1+p)-\frac{1}{4} \left(1+\frac{n^2}{4}\right) (1+p)\right)^2}\right)\right]
\end{split}
\end{equation}`