0 & 0 & a_3 & b \\
0 & 0 & 0 & a_4 
\end{array}\right)}\\
& = &
\left(\begin{array}{cccc}
f(a_1) & 0 & 0 & 0 \\
0 & f(a_2) & \Delta f(a_2,a_3)(c) & \Delta^2 f(a_2,a_3,a_4)(c,b) \\
0 & 0 & f(a_3) & \Delta f(a_3,a_4)(b) \\
0 & 0 & 0 & f(a_4) 
\end{array}\right).\quad\quad\end{aligned}$$ Applying Proposition \[prop:3.1\] to $a=\left(\begin{array}{cc}
a_1 & 0\\
0 & a_2
\end{array}\right),c=\left(\begin{array}{cc}
a_3 & b\\
0 & a_4
\end{array}\right)$ and $b=\left(\begin{array}{cc}
0 & 0\\
c & 0
\end{array}\right)$ under the form of provides an estimate for $\Delta^2 f(a_2,a_3,a_4)(c,b)$. As the size of the formula in question becomes quite large, we shall split it. We have $$\begin{aligned}
\Im f\left(\begin{array}{cc}
a_3 & b\\
0 & a_4
\end{array}\right)^{-1} 
& = & \left(\begin{array}{cc}
\Im f(a_3) & \frac{\Delta f(a_3,a_4)(b)}{2i}\\
\frac{\Delta f(a_3,a_4)(b)^*}{-2i} & f(a_4)
\end{array}\right)^{-1}\\
& = & \left(\begin{array}{cc}
e_{11} & e_{12}\\
e_{21} & e_{22}
\end{array}\right),\end{aligned}$$ where $$\begin{aligned}
e_{11}&=&\left(\Im f(a_3)-\frac{\Delta f(a_3,a_4)(b)(\Im f(a_4))^{-1}\Delta f(a_3,a_4)(b)^*}{4}\right)^{-1}\\
e_{12}&=&\left(\Im f(a_3)-\frac{\Delta f(a_3,a_4)(b)(\Im f(a_4))^{-1}\Delta f(a_3,a_4)(b)^*}{4}
\right)^{-1}\\
& & \mbox{}\times\frac{\Delta f(a_3,a_4)(b)}{-2i}(\Im f(a_4))^{-1}\\
e_{21}&=&(\Im f(a_4))^{-1}\frac{\Delta f(a_3,a_4)(b)^*}{2i}\\
& & \mbox{}\times\left(\Im f(a_3)-\frac{\Delta f(a_3,a_4)(b)(\Im f(a_4))^{-1}\Delta f(a_3,a_4)(b)^*}{4}
\right)^{-1}=e_{12}^*\\
e_{22}&=&\left(\Im f(a_4)-\frac{\Delta f(a_3,a_4)(b)^*(\Im f(a_3))^{-1}\Delta f(a_3,a_4)(b)}{4}\right)^{-1}.\end{aligned}$$ Thus, in the left-hand side of preserves only one nonzero element, in the position $22$ (lower right corner), namely $$\begin{aligned}
\lefteqn{\Delta 