But since identically

${\displaystyle {\frac {\overline {(\epsilon _{p}-{\overline {\epsilon }}_{p})^{0}}}{{\overline {\epsilon }}_{p}{}^{0}}}=1,\quad {\frac {\overline {(\epsilon _{p}-{\overline {\epsilon }}_{p})^{1}}}{{\overline {\epsilon }}_{p}{}^{1}}}=0}$

the value of the corresponding expression for any index will be independent of ${\displaystyle \Theta }$ and the formula reduces to

${\displaystyle {\overline {{\bigg (}{\frac {\epsilon _{p}-{\overline {\epsilon }}_{p}}{{\overline {\epsilon }}_{p}}}{\bigg )}^{h+1}}}={\frac {2h}{n}}{\overline {{\bigg (}{\frac {\epsilon _{p}-{\overline {\epsilon }}_{p}}{{\overline {\epsilon }}_{p}}}{\bigg )}^{h}}}+{\frac {2h}{n}}{\overline {{\bigg (}{\frac {\epsilon _{p}-{\overline {\epsilon }}_{p}}{{\overline {\epsilon }}_{p}}}{\bigg )}^{h-1}}}}$

(235)

we have therefore

${\displaystyle {\overline {{\bigg (}{\frac {\epsilon _{p}-{\overline {\epsilon }}_{p}}{{\overline {\epsilon }}_{p}}}{\bigg )}^{0}}}=1,\quad {\overline {{\bigg (}{\frac {\epsilon _{p}-{\overline {\epsilon }}_{p}}{{\overline {\epsilon }}_{p}}}{\bigg )}^{3}}}={\frac {8}{n^{2}}},}$

${\displaystyle {\overline {{\bigg (}{\frac {\epsilon _{p}-{\overline {\epsilon }}_{p}}{{\overline {\epsilon }}_{p}}}{\bigg )}^{1}}}=0,\quad {\overline {{\bigg (}{\frac {\epsilon _{p}-{\overline {\epsilon }}_{p}}{{\overline {\epsilon }}_{p}}}{\bigg )}^{4}}}={\frac {48}{n^{3}}}+{\frac {12}{n^{2}}},}$

${\displaystyle {\overline {{\bigg (}{\frac {\epsilon _{p}-{\overline {\epsilon }}_{p}}{{\overline {\epsilon }}_{p}}}{\bigg )}^{2}}}={\frac {2}{n}},\quad \mathrm {etc.} }$[1]

It will be observed that when ${\displaystyle \psi }$ or ${\displaystyle {\overline {\epsilon }}}$ is given as function of ${\displaystyle \Theta }$, all averages of the form ${\displaystyle {\overline {\epsilon ^{h}}}}$ or ${\displaystyle {\overline {(\epsilon -{\overline {\epsilon }})^{h}}}}$ are thereby deter-

^[1]

1. ↑ In the case discussed in the preceding foot-notes we get easily

   ${\displaystyle {\overline {(\epsilon _{q}-{\overline {\epsilon }}_{q})^{h}}}={\overline {(\epsilon _{p}-{\overline {\epsilon }}_{p})^{h}}},}$

   and

   ${\displaystyle {\overline {{\bigg (}{\frac {\epsilon _{q}-{\overline {\epsilon }}_{q}}{{\overline {\epsilon }}_{q}-\epsilon _{a}}}{\bigg )}^{h}}}={\overline {{\bigg (}{\frac {\epsilon _{p}-{\overline {\epsilon }}_{p}}{{\overline {\epsilon }}_{p}}}{\bigg )}^{h}}}.}$

   For the total energy we have in this case

   ${\displaystyle {\overline {{\bigg (}{\frac {\epsilon -{\overline {\epsilon }}}{{\overline {\epsilon }}-\epsilon _{a}}}{\bigg )}^{h+1}}}={\frac {h}{n}}{\overline {{\bigg (}{\frac {\epsilon -{\overline {\epsilon }}}{{\overline {\epsilon }}-\epsilon _{a}}}{\bigg )}^{h}}}+{\overline {{\frac {h}{n}}{\bigg (}{\frac {\epsilon -{\overline {\epsilon }}}{{\overline {\epsilon }}-\epsilon _{a}}}{\bigg )}^{h-1}}}.}$

   ${\displaystyle {\overline {{\bigg (}{\frac {\epsilon -{\overline {\epsilon }}}{{\overline {\epsilon }}-\epsilon _{a}}}{\bigg )}^{2}}}={\frac {1}{n}}.\quad {\overline {{\bigg (}{\frac {\epsilon -{\overline {\epsilon }}}{{\overline {\epsilon }}-\epsilon _{a}}}{\bigg )}^{4}}}={\frac {3}{n^{2}}}+{\frac {6}{n^{3}}},}$

   ${\displaystyle {\overline {{\bigg (}{\frac {\epsilon -{\overline {\epsilon }}}{{\overline {\epsilon }}-\epsilon _{a}}}{\bigg )}^{3}}}={\frac {2}{n^{2}}},\quad \mathrm {etc.} }$