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I am finishing my Ph.D. dissertation in engineering and I would like to show a simple proof. I am having troubles formalizing my ideas into a proof though. I think in a mathematics paper this concept is obvious, but I think a simple proof will go a long way in an engineering paper.

I need to show that a thin-plate spline RBF approximation: \begin{equation} \mathcal{P}_f(\boldsymbol{x}) = \sum_{k=1}^N c_k r_k^2 \log (r_k) + P\left(\boldsymbol{x}\right) \end{equation}

is real analytic, where $r_k=||\boldsymbol{x}-\boldsymbol{x_k}||$, $c_k$'s are constant, $\boldsymbol{x}_k$ are constant training nodes and $P\left(\boldsymbol{x}\right)$ is a polynomial in the coordinates of $\boldsymbol{x}$, e.g. for $\boldsymbol{x}\in\mathbb{R}^2$ , $P\left(\boldsymbol{x}\right)=d_0 + d_1x_1 +d_2x_2$, where the $d$'s are the coefficients and $x_1$ and $x_2$ are the coordinates of $\boldsymbol{x}$.

I know that the sums, products, and compositions of real analytic functions are real analytic , that the Logarithm is real analytic on $\left(0,\infty\right)$, and that any real polynomial is real analytic.

Would it be appropriate to do something along the following:

Theorem 1: $\mathcal{P}_f$ is a real analytic function on $\left(0,\infty\right)$

Axiom 1: the sums, products, and compositions of real analytic functions are real analytic

Axiom 2: the Logarithm is real analytic on $\left(0,\infty\right)$

Axiom 3: any real polynomial is real analytic.

...

Vlad
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Dr. Papa
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  • I don't think your description of $P_f$ makes sense. It looks as if you have a collection of numbers $x_k$ (or points $x_k$). But then $r$ is defined to be the distance from $x$ (what's x?) to one of them. Perhaps what you meant is this: $$ P_f(x) = \sum_{k=1}^N c_k r_k^2 \log(r_k) + P(x) $$ where $P(x)$ is a polynomial in the coordinates of $x$, $r_k$ denotes $| x - x_k |$, and the $c_k$s are constants.

    It's not clear whether the $x_k$ are points in space or numbers. In the RBF stuff I've seen, polynomial $P$ is typically linear/affine in the coordinates of $x$. Can you clarify?

     – John Hughes Dec 18 '13 at 22:25
  • You are exactly right. I've updated my question to clarify things. – Dr. Papa Dec 18 '13 at 23:01

1 Answers1

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Well, the proof would look something like this:

To show $P_f$ analytic, we observe that $P$ is analytic (it's just an affine function!), so all that remains is to show that the summation is analytic. We do so by showing that each term is. To do this, we need only show that the function $g(x) = x^2 \log |x|$ is analytic. (Indeed, this is both sufficient and necessary, as the case where $N = 1$ and $x_1 = 0$ shows.)

The only problem remaining is that I'm not at all certain that the function $g$ really is analytic. I look at $g(1+x) = (1+x)^2 \log(|1 + x|)$, and for positive $x$, the Taylor series for that (around x = 0) is the product of the series for the two factors, i.e., it's $$ (1 + 2x + x^2) * (x - \frac{x^2}{2} + \frac{x^3}{3} - \ldots) = \\ (x - \frac{x^2}{2} + \frac{x^3}{3} - \ldots) + 2x((x - \frac{x^2}{2} + \frac{x^3}{3} - \ldots))+ x^2((x - \frac{x^2}{2} + \frac{x^3}{3} - \ldots)). $$ The $n+1$th term of that series (after the first few) has coefficient $c_{n+1} = \pm (\frac{1}{n+1} - \frac{2}{n} + \frac{1}{n-1}) = \pm \frac{2}{n(n - 1)(n+1)}$. If you apply the ratio test to try to find the radius of convergence, you end up with $$ \rho = x \frac{n (n-1) (n+1)}{(n+1) (n) (n+2)}= x \frac{ (n-1) }{(n+2)} $$ which in the limit looks like just $x$. So you get a radius of convergence of just 1. That doesn't sound like a real analytic function to me.

I may have done the computation wrong, of course. But in general, I wouldn't expect this result to be real-analytic, and my computation seems to suggest it's not.

John Hughes
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  • hmmm. I'll need to look at what you have here a little longer to understand what you are doing, but why do you say "I woudn't expect this result to be real-analytic"? I have read that the product of real-analytic functions are real-analytic and that the logarithm is real-analytic on $(0,\infty)$. So, isn't this just the product of the logarithm and a polynomial $x^2$? Do I have some wrong information or am I missing something? – Dr. Papa Dec 19 '13 at 06:01
  • I thought you meant that it was real analytic on its whole domain, which would include being expressible by a convergent power series at the points $x_i$. That would require that $x^2 \log(|x|)$ be expressible by a convergent power series around $x = 0$. Writing down that power series would take me lots of work with L'hopital's rule; then I'd have to check convergence, too.

    If you just want to claim that $P_f$ is real analytic away from ${x_1, x_2, \ldots, x_N}$, that's easy: on the complement of that set, each of the terms is analytic (by your arguments), so the whole thing is.

     – John Hughes Dec 19 '13 at 12:40