In this build method the force of mortality is a piecewise linear function of age. The level is set to match the change in the survival fraction across an age interval and the slope is set to match person-years lived. Let there be a set of survival fractions \(\sfrac_i\) and person-years lived \begin{equation} \ssf_i=\int_0^{x_i} \sfrac(a)da \end{equation} at ages \(x_i\) for \(i=0,\ldots,n-1\). The force of mortality at age \(x\in[x_i,x_{i+1})\) takes the form \begin{equation} \force(x) = \frac{\alpha}{\dx}+\slope\dx \left(A-\frac{1}{2}\right), \end{equation} where \begin{equation} \alpha=\log\left(\frac{\sfrac_i}{\sfrac_{i+1}}\right),\quad A=\frac{x-x_i}{x_{i+1}-x_i},\quad \dx=x_{i+1}-x_i\\ . \end{equation} Let \begin{equation} \dssf=\ssf_{i+1}-\ssf_i \end{equation} then the parameter \begin{equation} z=\dx\sqrt{|\slope|/2},\quad \end{equation} is determined by the condition \begin{equation} \label{eqn:dssf} \frac{\dssf}{\dx}= \frac{1}{z} \left[\sfrac_{i+1}E_\slopesign\left(\slopesign\frac{\alpha}{2z}+\frac{z}{2}\right)- \sfrac_i E_\slopesign\left(\slopesign\frac{\alpha}{2z}-\frac{z}{2}\right)\right] \end{equation} where \begin{equation} \omega=\left\{\begin{array}{rl} 1, & \dssf/\dx > (\sfrac_i-\sfrac_{i+1})/\alpha\\ -1, & \dssf/\dx <(\sfrac_i-\sfrac_{i+1})/\alpha \end{array} \right. \end{equation} and \begin{equation} E_\slopesign(x) = e^{\slopesign x^2} \int_0^x e^{-\slopesign t^2} dt \end{equation} \(E_\slopesign(x)\) is related to the error function for \(\slopesign =1\) and the Dawson function for \(\slopesign=-1\). After solving for \(z\) the survival fraction and number of person-years lived for any \(x\in [x_i,x_{i+1})\) are given by the expressions \begin{equation} \sfrac(x) = \sfrac_i \exp\left(-\alpha A+\slopesign z^2 A(1-A)\right) \end{equation} and \begin{equation} \ssf(x)=\ssf_i+\frac{\dx}{z} \left[ \sfrac(x) E_\slopesign\left(\slopesign\frac{\alpha}{2z}+\left(A-\frac{1}{2}\right)z\right) -\sfrac_i E_\slopesign\left(\slopesign\frac{\alpha}{2z}-\frac{z}{2}\right)\right] \end{equation}
Unlike other build methods \(\hermiteforce\) requires both a frequency measure of mortality (death rate, death probability, survival fraction) and a duration measure (person-years lived, person-years lived after, life expectancy or separation factor). This extra information is necessary when one wants to reproduce both survival fraction and life expectancy data in a published life table or when one wants to convert death rates into death probabilities using a specified set of separation factors. The duration measure data is given in a RESULTS object. The name of the results object must be given in the build method extrainfo string.
If \(z\) is small then Equation\(~\ref{eqn:dssf}\) can be solved using the expansion \begin{equation} 2\frac{\dssf}{\dx}=\sqrt{\sfrac_i . \sfrac_{i+1}}\sum_{n\ge 0} \frac{(\omega z^2/4)^n}{n!}\ifunc{n}{\frac{\alpha}{2}} \end{equation} where \begin{equation} \ifunc{n}{x} := \int_{-1}^1 (1-t^2)^n e^{xt} dt . \end{equation} Rearranging terms gives the iteration \begin{equation} \label{eqn:zstart} \frac{\omega z^2}{4} \leftarrow \frac{1}{\sqrt{\sfrac_i \sfrac_{i+1}}} \frac{2\frac{\dssf}{\dx}-\sqrt{\sfrac_i\sfrac_{i+1}}\ifunc{0}{\frac{\alpha}{2}}}{% \sum_{n\ge 1} \frac{(\omega z^2/4)^{n-1}}{n!}% \ifunc{n}{\frac{\alpha}{2}}} \end{equation} which converges quickly for small \(z\) from the starting point \(z=0\). If \(z\) is not small then the above iteration can be used to set a starting point for a solution of Equation\(~\ref{eqn:dssf}\) using Newton. Derivatives can be found without much additional computation using the identity \begin{equation} E_\slopesign '(x)=2\slopesign x E_\slopesign (x)+ 1 . \end{equation}
The iteration\(~(\ref{eqn:zstart})\) requires \(\ifunc{n}{x}\) for \(n=0,1,\ldots,\nmax\) for some suitably large \(\nmax\). If \(\ifunc{\nmax-1}{x}\) and \(\ifunc{\nmax}{x}\) are known then remaining values can be calculated using the stable downwards recurrence \begin{equation} \label{eqn:ifunc_recd} \ifunc{n}{x}=\left(1+\frac{1}{2(n+1)}\right) \ifunc{n+1}{x}+\frac{x^2}{4(n+1)(n+2)}\ifunc{n+2}{x} \end{equation} Let \(\nifunc{n}{x}\) be the normalized function \begin{equation} \nifunc{n}{x}=\frac{\ifunc{n}{x}}{\ifunc{N}{x}} \end{equation} The \(\nifunc{n}{x}\)'s also satisfy the recurrence \((\ref{eqn:ifunc_recd})\) because it is linear. By definition \begin{equation} \nifunc{\nmax}{x}=1 . \end{equation} And equation~(\ref{eqn:ifunc_recd}) can be used to prove the continued fraction representation \begin{equation} \nifunc{\nmax-1}{x}^{-1}=\frac{1}{% 1+\frac{1}{2\nmax}+\frac{\frac{x^2}{4\nmax(\nmax+1)}}{% 1+\frac{1}{2(\nmax+1)}+\frac{\frac{x^2}{4(\nmax+1)(\nmax+2)}}{% 1+\ldots} }} \end{equation} which can be evaluated efficiently using Lentz. With these two starting conditions recurrence can be used to generate all \(\nifunc{n}{x}\) for \(n=\nmax-2,...,0\). Then \begin{equation} \ifunc{n}{x}=\ifunc{\nmax}{x}\nifunc{n}{x} \end{equation} where \(\ifunc{\nmax}{x}\) is determined by the known value at \(n=0\) \begin{equation} \ifunc{0}{x}=2 \frac{\sinh(x)}{x}. \end{equation}