Interpolating a Sequence with B-spline of Order 1

Question

Let $\operatorname{rect}(x)=\chi_{\left[-\frac{1}{2}, \frac{1}{2}\right]}(x)$ be the characteristic function of the interval $\left[-\frac{1}{2}, \frac{1}{2}\right]$ and $\operatorname{rect}^{(p)}(x)=$ rect $* \ldots * \operatorname{rect}(x)$ be the $p$-times iterated convolution of $\operatorname{rect}(x)$. The function $\operatorname{rect}^{(p)}(x)$ is a piecewise polynomial function of degree $p-1$ and a prime example of $B$-spline of order $p-1$.

Given a sequence of data $a_n = \{a_1, a_2, \dots\}$, I want to construct a B-spline of order 0 to interpolate this data. The B-spline of order 0 is essentially a step function, and in my context, it corresponds to the characteristic function of the interval $\left[-\frac{1}{2}, \frac{1}{2}\right]$, denoted as $\operatorname{rect}(x)$.

The B-spline of order 0, in my case, is the characteristic function $\operatorname{rect}(x)$, defined as:

$$ \operatorname{rect}(x) = \begin{cases} 1 & \text{if } -\frac{1}{2} \leq x \leq \frac{1}{2} \\ 0 & \text{otherwise} \end{cases} $$

To interpolate the given sequence $a_n$, I can use the step function $\operatorname{rect}(x)$. The idea is to place a step of height $a_n$ at each corresponding point $x_n$ on the x-axis.

The interpolated function can be expressed as a sum of shifted and scaled step functions:

$$ f(x) = \sum_{n} a_n \cdot \operatorname{rect}\left(\frac{x - x_n}{h}\right) $$

where $h$ is the width of the steps (in my case, $h = 1$ since the interval of $\operatorname{rect}(x)$ is $\left[-\frac{1}{2}, \frac{1}{2}\right]$). In particular, I can choose $x_n=n$ obtaining:

$$ f(x) = \sum_{n} a_n \cdot \operatorname{rect}\left(\frac{x - n}{h}\right) $$

This function $f(x)$ will interpolate the given data sequence $a_n$ using a B-spline of order 0.

For example, if my data sequence is $a_n = \{2, 1, 3\}$, and the corresponding positions on the x-axis are $1, 2, 3$, then the interpolated function would be:

$$ f(x) = 2 \cdot \operatorname{rect}(x - 1) + 1 \cdot \operatorname{rect}(x - 2) + 3 \cdot \operatorname{rect}(x -3) $$

This function will provide a step-wise interpolation of the data using B-spline of order $0$.

Hence, I've been working on interpolating a sequence of data $a_n = \{a_1, a_2, \dots\}$ using a B-spline of order 0, where the B-spline is essentially a step function corresponding to the characteristic function of the interval $\left[-\frac{1}{2}, \frac{1}{2}\right]$.

Now, I am interested in extending this interpolation to a B-spline of order 1. However, I'm finding it challenging to apply the same approach to obtain the interpolation with a B-spline of order 1, as $\operatorname{rect}^2$ would theoretically be a triangle function.

Could anyone guide me on how to construct the function $f(x)$ from the sequence $a_n$ by interpolating it with a B-spline of order 1? The earlier approach involving steps and convolutions seems less straightforward in this case.

I appreciate any insights or alternative methods you may suggest.

Thank you!

Answer 1

If you need a piecewise continuous linear interpolation, then an elementary way to do that is by making

$$ f_k(x_k,y_k,x) = \left(y_k+\frac{y_{k+1}-y_k}{x_{k+1}-x_k}(x-x_k)\right)\left(\theta(x-x_k)-\theta(x-x_{k+1})\right) $$

where $\theta(\cdot)$ is the Heaviside unit step function, and $x_k,y_x$ is the data set to interpolate, and thus

$$ f(x) = \sum_k f_k(x_k,y_k,x) $$

Follows a MATHEMATICA script producing this.

n = 10;
Clear[x, linterp]
SeedRandom[2]
yk = RandomReal[{-1, 1}, n];
xk = N@Subdivide[0, 5, n - 1];

linterp[xk_, yk_, x_] := Module[{n = Length[xk], fx}, 
  fx = Sum[(yk[[k]] + (yk[[k + 1]] - yk[[k]])/(xk[[k + 1]] - xk[[k]]) (x - xk[[k]])) (UnitStep[x - xk[[k]]] - UnitStep[x - xk[[k + 1]]]), {k, 1, n - 1}];
Return[fx]
]


fx = linterp[xk, yk, x];
gr0 = ListPlot[Transpose[{xk, yk}], PlotStyle -> {Red, PointSize[0.02]}]
gr1 = Plot[fx, {x, xk[[1]], xk[[n]]}];
Show[gr1, gr0]