# How does the cubic spline basis exactly look like

By the definition

(1) $S(x)=\begin{cases} S_0 = a_0x^3 + b_0x^2 + c_0x + d_0, & \text{if }t_0\le x\le t_1\\ .....\\ S_{k} = a_kx^3 + b_kx^2 + c_kx + d_k, & \text{if }t_{k-1}\le x\le t_k\\ \end{cases}$

Using the necessary conditions which ensure that $S(x)$ is twice differentiable at the knots $t_0,...,t_k$ yields:

(2) $\ S_i(x) = \frac{z_i}{6h_i}(t_{i+1}-x)^3 + \frac{z_{i+1}}{6h_i}(x-t_{i})^3 + (\frac{y_{i+1}}{h_i}-\frac{z_{i+1}h_i}{6})(x-t_i) + (\frac{y_{i}}{h_i}-\frac{z_{i}h_i}{6})(t_{i+1}-x)$

where $z_i = S_i''(t_i)$, $y_i = S_i(t_i)$ and $h_i = t_{i+1}-t_i$.

According to Wood - Generalized Additive Models an Introduction with R (page 145, 146), one can rewrite (1) and (2) to

(3) $S(X) = a_j^-(x)\beta_j + a_j^+(x)\beta_{j+1} + c_j^-(x)\boldsymbol{F}_j\boldsymbol{\beta} + c_j^+(x)\boldsymbol{F}_{j+1}\boldsymbol{\beta}$

Since this is just a reformulation of (2) I'm really interested in how the exact formulation of $b_i(x)$ will look like since

$S(x) = \sum_{i=1}^k b_i(x)\beta_i$

is actually used to fit the data.