Is possible to construct infinitely differentiable functions that interpolate through arbitrary points, the way polynomial splines do? If so, do they have a name and is there an algorithm for constructing the curves.
I suppose a infinite degree polynomial spline might qualify but I was hoping there might be something more like a bump function.
I definitely am a novice when it comes to this stuff, so please do not hesitate to correct me if I misusing any terminology.
On
Yes - if you have a finite list of points on the plane that pass the vertical line test, you can make a smooth (infinitely differentiable) function that goes through all of them.
This can actually be done somewhat explicitly, using the formulas at [1]. Define:
$$ f(x) = \begin{cases} 0 & x \leq 0 \\ e^{-1/x}& x > 0 \\ \end{cases}. $$
It can be proved that this is a smooth function. Then define
$$ g(x) = \frac{f(x)}{f(x)+f(1-x)}. $$
The function $g(x)$ is smooth, defined for all real $x$, and has the property that $g(x) = 0$ for all $x \leq 0$ while $g(x) = 1$ for all $x \geq 1$. By adding together shifted and scaled copies of $g(x)$, you can make a smooth function that goes through any finite list of points that pass the vertical line test.
1: http://en.wikipedia.org/wiki/Non-analytic_smooth_function
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It's a bit disingenuous to say polynomial splines are infinitely differentiable; a spline with nth order pieces can only have at most C^{n-1} continuity. That being said, a different basis set (e.g. the Fourier basis or rational functions) would in principle be infinitely differentiable, but might be a very "wiggly" way to "connect the dots" so to speak. Polynomial interpolation, whatever method you use, will also suffer from this "wiggliness".
Why would you need your approximant to be infinitely differentiable anyway? Unless you're going to be doing something special I don't see it being a necessity.
On
Along with all the other nice answers, I'd like to also mention Exporational B-Splines (ERBSs) which, as I understand it, are a direct generalization of B-Splines to the infinite degree case:
https://www2.isye.gatech.edu/~brani/wp/ExpoSplines.pdf
They have a number of nice properties that are like the "smooth" versions of b-splines
The archetypical version of this is
$$ \frac{\int_{-1}^{x} e^{-\frac{t^2}{1-t^2}} dt}{\int_{-1}^{1} e^{-\frac{t^2}{1-t^2}} dt} $$
but the paper features an extension by several parameters to modify the shape in complicated ways.
Further extensions to the equivalent of NURBS (Non-Uniform Exporational B-Splines, NUERBS) or with knot multiplicities that are not infinite (Generalized Exporational B-Splines, GERBS) also exist
and yet another approach might involve the Fabius function which can be defined through a recursive integration scheme where you keep integrating over an interval and then rescaling the result so it doesn't explode.
If you are curious about that, see:
Recursive Integration over Piecewise Polynomials: Closed form?
For n data points, there is a unique polynomial of degree n−1 that passes through the points; one way to get it is the Lagrange polynomial.
Another class of interpolating functions is the thin plate spline, which is typically used for 2D image warping, but can easily be applied to 1D interpolation as well, which is a strictly simpler problem. The Wikipedia article on polyharmonic splines, of which the thin plate spline is a special case, has examples of other infinitely differentiable splines.
I believe that in principle, you can choose any basis function whose translations are linearly independent, and find an interpolating function as a sum of translations and scalings of the basis. The above are essentially examples of particular basis functions people have used because of their other nice properties.
Edit: Sorry, thin plate splines and other examples in the article, apart from the Gaussian, are infinitely differentiable almost everywhere. However, if you really want infinite differentiability everywhere, you can use any infinitely differentiable function as basis instead, such as the Gaussian, or $x^2/\sqrt{x^2+1}$.