# A General Statistical Physics Framework for Assignment Problems
## Appendix H Properties of the General Function \(G(x,a,b)\) and Its Derivatives
### \(g(x,a,b)\) and Its Derivative \(g^{\prime}(x,a,b)\)  
Note that  
\[g(x,a,b)= \frac{(b-a+1)}{e^{(b-a+1)x}-1}+\frac{1}{1-e^{x}}+b,\] \[g^{\prime}(x,a,b)= -\frac{(b-a+1)^{2}e^{(b-a+1)x}}{(e^{(b-a+1)x}-1)^{2}}+\frac{e^{x} }{(e^{x}-1)^{2}},\]  
\(g\) and its derivative \(g^{\prime}\) are defined when \(e^{(b-a+1)x}-1\
eq 0\) and \(e^{x}-1\
eq 0\), i.e., when \(x\
eq 0\). They can be extended by continuity at 0:  
\[g(0,a,b)= \frac{a+b}{2},\] \[g^{\prime}(0,a,b)= \frac{(b-a+1)^{2}-1}{12}.\]  
We have \(g^{\prime}(0,a,b)>0\) as \(b-a+1>1\). In addition,  
\[\lim_{x\to-\infty}g(x,a,b)=a,\] \[\lim_{x\to+\infty}g(x,a,b)=b,\] \[\lim_{x\to-\infty}g^{\prime}(x,a,b)=0,\] \[\lim_{x\to+\infty}g^{\prime}(x,a,b)=0.\]  
Figure 11 shows a few examples of \(g\) and \(g^{\prime}\) for different values of \(a\) and \(b\).  
We now prove property 1 of Theorem A3.  
Proof.: Let us define \(m=b-a+1\). As \(a<b\), we have \(m>1\). Let \(x\) be a real number; then,  
\[g^{\prime}(x,a,b)= -\frac{m^{2}e^{mx}}{(e^{mx}-1)^{2}}+\frac{e^{x}}{(e^{x}-1)^{2}},\]  
which we can rewrite as  
\[g^{\prime}(x,a,b)\] \[= \frac{1}{4}\bigg{(}-\frac{m^{2}}{\sinh^{2}(\frac{mx}{2})}+\frac{1 }{\sinh^{2}(\frac{x}{2})}\bigg{)}\] \[= \frac{1}{4}\bigg{(}\frac{\sinh^{2}(\frac{mx}{2})-m^{2}\sinh^{2}( \frac{x}{2})}{\sinh^{2}(\frac{mx}{2})\sinh^{2}(\frac{x}{2})}\bigg{)}\] \[= \frac{1}{4}\bigg{(}\frac{(\sinh(\frac{mx}{2})-m\sinh(\frac{x}{2} ))(\sinh(\frac{mx}{2})+m\sinh(\frac{x}{2}))}{\sinh^{2}(\frac{mx}{2})\sinh^{2}( \frac{x}{2})}\bigg{)}.\]  
From this formulation, it is clear that \(g^{\prime}(x,a,b)\) is even. As \(g^{\prime}(0,a,b)\) is strictly positive (see above), we only need to consider \(x>0\). Let us define \(y=x/2\). For \(y>0\), the denominator of \(g^{\prime}(x,a,b)\) and the term \(\sinh(my)+m\sinh(y)\) are strictly positive. The only remaining term is \(\sinh(my)-m\sinh(y)\), which we call \(f(y)\). We have  
\[f^{\prime}(y) = m(\cosh(my)-\cosh(y))\] \[= 2m\sinh((m+1)y)\sinh((m-1)y).\]  
As \(m>1\), \(\sinh((m+1)y)\) and \(\sinh((m-1)y)\) are strictly positive when \(y>0\): \(f(y)\) is monotonically increasing for \(y\geq 0\). As \(f(0)=0\), we obtain \(f(y)>0\) for \(y>0\). Therefore \(g^{\prime}(x,a,b)>0\) for all \(x>0\), and since it is an even function and \(g^{\prime}(0,a,b)>0\), we find that \(\forall x\in\mathbf{R}\quad g^{\prime}(x,a,b)>0\).  
_Appendix H.3. The Function \(l(x,a,b)=G(x,a,b)-xg(x,a,b)\)_  
Recall that  
\[l(x,a,b)=G(x,a,b)-xg(x,a,b).\]  
We nowprove property 2 of Theorem A3.  
Proof.: We prove first that \(l(x,a,b)\) is even. Note that  
\[G(-x,a,b) = \ln\bigg{(}\frac{e^{-ax}-e^{-(b+1)x}}{1-e^{-x}}\bigg{)}\] \[= \ln\bigg{(}\frac{e^{(b+1)x}-e^{ax}}{e^{x}-1}\bigg{)}+\ln\bigg{(} \frac{e^{x}}{e^{(b+1)x}e^{ax}}\bigg{)}\] \[= -bx-ax+G(x,a,b),\]  
and  
\[g(-x,a,b) = \frac{(b-a+1)}{e^{-(b-a+1)x}-1}+\frac{1}{1-e^{-x}}+b\] \[= \frac{(b-a+1)e^{(b-a+1)x}}{1-e^{(b-a+1)x}}+\frac{e^{x}}{e^{x}-1}+b\] \[= a+b-g(x,a,b).\]Therefore,  
\[l(-x,a,b) = G(-x,a,b)+xg(-x,a,b)\] \[= G(x,a,b)-xg(x,a,b)\] \[= l(x,a,b).\]  
Let now \(x\) be a positive real number. Then,  
\[l^{\prime}(x,a,b) = G^{\prime}(x,a,b)-g(x,a,b)-xg^{\prime}(x,a,b).\] \[= -xg^{\prime}(x,a,b)\]  
We know that \(g^{\prime}(x,a,b)\) is positive for all real numbers \(x\) (property 1 of Theorem A3); therefore, \(l^{\prime}(x,a,b)\) is negative on \(\mathbb{R}^{+}\) and \(l(x,a,b)\) is monotonic decreasing on this domain. Since:  
\[l(0,a,b) = \ln(b-a+1),\] \[\lim_{x\to+\infty}l(x,a,b) = 0,\]  
we have \(0\leq l(x,a,b)\leq\ln(b-a+1)\) for \(x\in\mathbb{R}^{+}\), and since \(l(x,a,b)\) is even, \(0\leq l(x,a,b)\leq\ln(b-a+1)\) for \(x\in\mathbb{R}\), which concludes the proof of property 2 of Theorem A3.