The force of interest, $\delta$, is defined as the nominal rate of interest payable continuously: \[\lim_{p\rightarrow\infty}i^{(p)}=\delta.\]
As the relationship between the effective and the nominal rate is given by: \[1+i=\left(1+\frac{i^{(p)}}{p}\right)^p\] then in the limit as $p\rightarrow\infty$ \begin{align} &1+i=\lim_{p\rightarrow\infty}\left(1+\frac{\delta}{p}\right)^p=e^\delta\\ &1+i=e^\delta. \end{align}
Re-arranging the above formula:
\begin{align} &i=e^\delta-1\\ &\ln(1+i)=\delta \end{align}
where
What is the force of interest if the nominal interest rate is $5\%$ per annum payable monthly?
\begin{align} \ln \left(1+\frac{0.05}{12}\right)^{12}&=\delta\\ \Rightarrow\delta&=0.04989612178\\ \Rightarrow\delta&\approx 4.99\% \end{align}
What is the interest rate if the force of interest is $10\%$?
\begin{align} i&=e^{0.1}-1\\ \Rightarrow i&=0.1051709181\\ \Rightarrow i&\approx 10.52\% \end{align}