Using the last equation we define two values for the second derivative, namely $$ s''_{i}(x_i)= f_i, $$ and $$ s''_{i}(x_{i+1})= f_{i+1}, $$ and setting up a straight line between \( f_i \) and \( f_{i+1} \) we have $$ s_i''(x) = \frac{f_i}{x_{i+1}-x_i}(x_{i+1}-x)+ \frac{f_{i+1}}{x_{i+1}-x_i}(x-x_i), $$ and integrating twice one obtains $$ s_i(x) = \frac{f_i}{6(x_{i+1}-x_i)}(x_{i+1}-x)^3+ \frac{f_{i+1}}{6(x_{i+1}-x_i)}(x-x_i)^3 +c(x-x_i)+d(x_{i+1}-x). $$