Continuous differentiable spline or function resembling floor

Question

I'd need any (real-valued) function (whatever meets the following description at least approximately) continuous and thrice differentiable everywhere (or twice if 3 not possible), with the following features.

Seen "by far, it would be resembling the floor function $\lfloor x \rfloor $, exactly equal to $j$ at each positive integer $j$, but then slightly increasing on the "horizontal" part and finally more suddenly raise to the next integer, so approximately doing as follows. Let $\epsilon$ and $\alpha$ two arbitrarily small positive numbers given by the user. The function should be:

$j$ at any integer $j$ (I need only $j$ > 2, no need to consider negative integers)

increasing from $j$ to $j +\alpha$ near the the "horizontal part" of the floor, say approximately over $(j, j + 1 - \epsilon$].

suddenly increasing (or non-decreasing), from $j + \alpha$ to $(j + 1)$, "near the step" of a floor function [say approximately over $[j + 1 - \epsilon, j+1$).

These pieces should be "joined" in such a way that the function is continuous and differentiable. (Any function with the above characteristics will do: of course, the simpler the better: I need it for an iterative program)

PS. added later

If useful, I found this paper: http://math.arizona.edu/~shankar/projects/TermPaper_Yilu.pdf

which has something which perhaps may come close (see pag. 10/11). But it is missing the increasing "horizontal" part.

Answer 1

You can build the function you need from an alternating sequence of linear pieces and polynomial pieces.

Over the interval $(j-1, j - \epsilon]$, you just use a straight line that increases from $j-1$ to $j -1+\alpha$. Call it's slope $m$, which we can calculate from $\epsilon$ and $\alpha$. In fact, $m = \alpha/(1-\epsilon$.

Similarly, over the interval $(j, j +1 - \epsilon]$, you just use a straight line that increases from $j$ to $j +\alpha$.

Now we need to fit in a polynomial piece $p$ that provides a steep upwards ramp over the interval $(j - \epsilon, j]$. The requirements are: $$p(j - \epsilon) = j-1+\alpha \quad ; \quad p(j) = j$$ $$p'(j - \epsilon) = m \quad ; \quad p'(j) = m$$ A cubic polynomial has four coefficients, so you will be able to satisfy these four conditions. You could just assume $p(x) = ax^3 + bx^2 + cx + d$, write down the four equations arising from the four conditions, and solve for $a$, $b$, $c$, $d$. This is the brute force approach; there are cleverer approaches using Hermite interpolation techniques, which give you the required curve immediately, without the need to solve a system of linear equations. See this article, for example. Applying the Hermite cubic formulae, we get $$ p(t) = (m - \epsilon)(2t^3 - 3t^2) + mt + j - \epsilon \qquad (0 \le t \le 1) $$ where $t = (x - j + \epsilon)/\epsilon$. You only have to do this once. Thereafter, you can get other polynomial filler pieces just by shifting this basic one to the left or right.

The curve constructed this way will be only once differentiable -- there will be discontinuities in the second derivative at all the joints. If you want a curve that's twice differentiable, you let $p$ be a quintic polynomial, and you add two more conditions that force it to mate nicely with the adjacent linear pieces: $$ p''(j - \epsilon) = 0 \quad ; \quad p''(j) = 0 $$ To get a curve that's thrice differentiable, use a polynomial of degree 7, and so on.

Answer 2

For your requirement, the traditional spline-computing algorithm will yield $f(x)=x$.

So you'll also need to define a "midpoint" between every two integers in your domain.

For example: $f(2)=2,f(2.5)=2.1,f(3)=3,f(3.5)=3.1,f(4)=4,f(4.5)=4.1$.

Here is a piece of Python code for any given number of points $(x_0,y_0),(x_1,y_1),...,(x_N,y_N)$:

class Point:
    def __init__(self,x,y):
        self.x = 1.0*x
        self.y = 1.0*y

def Spline(points):
    N   = len(points)-1
    w   =     [(points[i+1].x-points[i].x)      for i in range(0,N)]
    h   =     [(points[i+1].y-points[i].y)/w[i] for i in range(0,N)]
    ftt = [0]+[3*(h[i+1]-h[i])/(w[i+1]+w[i])    for i in range(0,N-1)]+[0]
    A   =     [(ftt[i+1]-ftt[i])/(6*w[i])       for i in range(0,N)]
    B   =     [ftt[i]/2                         for i in range(0,N)]
    C   =     [h[i]-w[i]*(ftt[i+1]+2*ftt[i])/6  for i in range(0,N)]
    D   =     [points[i].y                      for i in range(0,N)]
    return A,B,C,D

def PrintSpline(points,A,B,C,D):
    for i in range(0,len(points)-1):
        func = str(points[i].x)+' <= x <= '+str(points[i+1].x)+' : f(x) = '
        components = []
        if A[i]:
            components.append(str(A[i])+'(x-'+str(points[i].x)+')^3')
        if B[i]:
            components.append(str(B[i])+'(x-'+str(points[i].x)+')^2')
        if C[i]:
            components.append(str(C[i])+'(x-'+str(points[i].x)+')')
        if D[i]:
            components.append(str(D[i]))
        if components:
            func += components[0]
            for i in range (1,len(components)):
                if components[i][0] == '-':
                    func += ' - '+components[i][1:]
                else:
                    func += ' + '+components[i]
            print func
        else:
            print func+'0'

Here is how you can print the Spline function of the $6$ points in the example above:

points = [
    Point(2,2),Point(2.5,2.1),
    Point(3,3),Point(3.5,3.1),
    Point(4,4),Point(4.5,4.1),
]
A,B,C,D = Spline(points)
PrintSpline(points,A,B,C,D)

Here is Wolfram's graphic illustration (sorry about the missing piece $4\leq{x}\leq4.5$):

You can change the characteristics of the "horizontal" parts by "playing" with the midpoint values.