My Very Own Partition...

Consider a function so constructed such that,

$Z_{s}=\cfrac { d\, \beta e^{ f(E_{ s }) } }{ d\, E_{ s } } =\beta f^{ ' }(E_{ s })e^{ f(E_{ s }) }$

where  $\beta$  is a constant,  and  $f(E_{s})$  is a function in  $E_s$  with roots  $E_{si}$  $i=1,2,3...$.  ie.

$f(E_{ s }=E_{ { si } })=0\quad \quad \quad i=1,2,3...$

then

$e^{ f(E_{ s }=E_{ { si } }) }=e^{ 0 }=1$

and,

$\left[{ \cfrac { d\, \beta e^{ f(E_{ s }) } }{ d\, E_{ s } }  }\right]_{ E_{ si } }=\beta f^{ ' }(E_{ s }=E_{ { si } }).1=Z_{ si }$

Since differentiation by product rule leaves behind one term without each root factor, we define,

$f^{ ' }(E_{ si })=p(Z_{ si }).(E_{ si }-E_{ s1 })(E_{ si }-E_{ s2 })..$

except for the factor of  $(E_{si}-E_{si})$ such that,

$f^{ ' }(E_{ si })=p(Z_{ si }).\prod \limits_{\substack{ j=1 \\ j\ne i }}^n (E_{ si }-E_{ { sj} })=\cfrac{Z_{ si }}{\beta}$

where  $p(Z_{si})$  is defined by,

$p(Z_{ si })=\cfrac { Z_{ si } }{ \beta .\prod\limits _{\substack{ j=1 \\ j\ne i }}^n (E_{ si }-E_{ { sj} }) }$

where we insist that  $E_{si}\ne E_{sj}$  for  $i\ne j$, each root of  $f(E_s)$  is unique.  $p(Z_{si})$  can also be defined as  $p(E_{si})$  as

$p'(E_{s})\prod_i(E_s-E_{si})$    is zero for all  $E_s=E_{si}$.  We will still have,

$p(E_{ si })=\cfrac { Z_{ si } }{ \beta .\prod\limits _{\substack{ j=1 \\ j\ne i }}^n (E_{ si }-E_{ { sj} }) }$

$\beta$  is defined such that,

$\int _{ 0 }^{ E_s max }{ Z_{ s } } d\, E_{ s }=Z=\left[ \beta e^{ f(E_{ s }) } \right] ^{ E_s max  }_{ 0 }=\sum\limits _{ i }^{ all\, i } Z_{ si }$

$\beta=\cfrac{Z}{\left[ e^{ f(E_{ s }) } \right] ^{ E_s max  }_{ 0 }}$

If  $e^{ f(E_{ s }max)}\rightarrow0$    for    $E_{ s }max\rightarrow \infty$

$\beta=\cfrac{Z}{0-e^{ f(0) }}=-\cfrac{Z}{e^{ f(0) }}$

at  $e^{f(E_s=0)}$.

Notice the expression for  $Z$  is always zero integrating from root to root of  $f(E_s)$.  This is because  $e^{ f(E_{ s })}$  intersect  $y=1$  repeatedly such that the area above and below the x-axis of   $e^{ f(E_{ s })}$  is equal, from root to root of  $f(E_s)$.  This means, the value of  $Z$  is determined by that part of the curve from zero to the first root of $f(E_s)$  and, the last root of  $f(E_s)$  to the tail end of  $e^{ f(E_{ s })}$  towards infinity.

If the first root is zero then  $\beta$  can be found from,

$\beta=\cfrac{Z}{0-e^{ f(E_s\,last) }}=-\cfrac{Z}{e^{ f(E_s\,last) }}=-Z$

because we know that the integration of  $Z_s$  with boundaries between roots of  $f(E_s)$  sums to zero.

No need to involve  $T$,  which ever way temperature is defined.