# Cubic B-Spline (Basic Spline) Non-Zero Domain

Why does the cubic B-spline (Basic Spline) takes on the value zero outside the interval $[x_i, x_{i+4}]$? Specifically, why is the lower value of the interval $x_i$?

The cubic B-spline is defined as:

$$B_i(x) = (x_{i+4} - x_i) f_x [x_i, x_{i+1}, x_{i+2}, x_{i+3}, x_{i+4}]$$

$f_x[x_i, x_{i+1}, x_{i+2}, x_{i+3}, x_{i+4}]$ is the 4-th order divided difference of $f_x(t) = (t-x)_+^3$. Note that if $x<x_i$, then still $f_x[x_i, x_{i+1}]$ will be non-zero.

Another formula for the cubic B-spline is

$$B_i(x) = (x_{i+4} - x_i) \sum_{j=i}^{j=i+4} \frac{(x_j-x)_+^3}{\psi'_i(x_j)}$$

where $\psi'_n(x_i) = (x_i - x_0) (x_i - x_1) \ldots (x_i-x_{i-1}) (x_i - x_{i+1}) \ldots (x-x_n)$

Note that if I use the second formula, I can see easily why if $x > x_{i+4}$, then the value of the B-spline is zero. (It is assumed $x_{i+4}> x_{i+3} > \ldots > x_i$). But still, if $x < x_i$, then the value of $B_i(x)$ is not zero.

• The cubic $B$-spline is intended to be built up of four cubic pieces within its support interval, and zero otherwise. You might wish to peer at de Boor's treatment here. – J. M. isn't a mathematician Sep 18 '11 at 21:31
• By the way: you do realize why $(t)_+=0$ if $t$ is negative, right? Given the ordering of the points, can you see why your sum in your second B-spline formula becomes zero? – J. M. isn't a mathematician Sep 19 '11 at 2:08
• @J.M.: I am aware of the notation $(t)_+$, and I see why the second B-spline formula becomes zero only when $x>x_{i+4}$. I cannot see why it becomes zero when $x<x_i$, because I believe all the terms inside the $()_+$ will be positive, and therefore, all of the summation terms will be positive. – jrand Sep 19 '11 at 14:51

The original question was: why does the value of the B-spline take on zero when $x<x_i$?

Note the following definition for the value of the b-spline, equation (3.7.33) in Atkinson:

$$B_i(x) = (x_{i+4} - x_i) f_x[x_i, x_{i+1}, x_{i+2}, x_{i+3}, x_{i+4}] (3.7.33)$$

Note that we know: $x_i < x_{i+1} < x_{i+2} < x_{i+3} < x_{i+4}$ and $f_x(t) = (t-x)_+^3$. From equation (3.7.12), we know the relation between the m-th divided difference and the m-th deriviative:

$$f[x_0, x_1, \ldots, x_m] = \frac{f^{(m)} (\xi)}{m!} (3.7.12)$$ and $\min\{x_0, x_1, \ldots, x_m\} \leq \xi \leq \max\{x_0, x_1, \ldots, x_m\}$.

Combining (3.7.33) and (3.7.12), we get for the 4-th divided difference,

$$f_x[x_i, x_{i+1}, x_{i+2}, x_{i+3}, x_{i+4}] = \frac{f_x^{(4)} (\xi)}{4!}$$ and $\xi \geq x_i$. If we know $\xi \geq x_i$ and (the original assumption), $x<x_i$, then we combine these two facts as: $x < \xi$.

If $x < \xi$, then in the expression $f_x(\xi) = (\xi-x)_+^3$, we get a "regular" 3-rd degree polynomial. If we take the fourth derivative of a 3-rd degree polynomial, we get zero. Then, $B_i(x) = 0$ when $x < x_i$.

Credit goes to my university professor who helped me to see this.