# Deriving a tridiagonal system for cubic spline interpolation

Can anyone explain how $B_{i-1} = 1/4$ and $B_{i+1} = 1/4$ were chosen in line 6 of the picture, just above the matrix?

I'm trying to understand cubic splines but this result seems like it came out of nowhere.

For equally spaced nodes, the cubic splines consist of sums of several copies of translated and scaled copies of the cardinal spline $$B(t) = \begin{cases} 1 - \frac34 t^2(2-|t|), \qquad & |t|\le 1 \\ {(2-|t|)^3}/{4} ,\qquad & 1\le |t|\le 2 \\ 0 \qquad & \text{otherwise} \end{cases}$$ The cardinal spline is designed to have value $1$ at $0$, to be zero outside of $[-2,2]$, and to be $C^2$ smooth as a cubic spline ought to be. These properties determine it uniquely.
Observe that in addition to $B(0)=1$, we have nonzero values at neighboring integers: $B(\pm 1) = 1/4$. This is where $1/4$ in that formula comes from: evaluating a basis spline at a node that neighbors its maximum. $$B_{i-1}(x_i) = B_{i+1}(x_i)=\frac14$$