Let $F=\{X \subset \Bbb R^2\; \vert \; \text{card} \, X\in \Bbb N\; \land \forall_{(x_i,y_i)\in X}\, i\neq j\to x_i\neq x_j\}$. Let $P$ be the set of all polynomials from $\Bbb R$ to $\Bbb R$ of any degree. There exists a mapping $M:F\to P$ whose definition is as follows: Let $A\in F$, then $M(A)\in P$ is the unique polynomial of smallest possible degree whose graph goes through every point in A. Let $f:\Bbb{R}\to\Bbb{R}$ be an arbitrary analytic function. Let $\{ a\}_n$ be an arbitrary sequence in $\Bbb R$ with no repeats. Consider the sequence of sets recursively defined as follows: $S_1=\{ (a_1,f(a_1))\}$, $S_n=S_{n-1}\, \cup \, \{(a_n, f(a_n))\}$. $\forall_{n\in \Bbb N}\, S_n\in F$. This naturally arises a sequence of polynomials $b_n(x)=M(S_n)$. However, if you chose a different $\{a\}_n$, that would lead to a different sequence of sets $\{S\}_n$, which would lead to a different sequence of polynomials $\{b\}_n$. How would I go about proving the criteria on $\{a\}_n$ so that the limit of the associated $\{b\}_n$ is $f$ similarly to how the limit of the sequence of Taylor polynomials for an analytic function $f$ is $f$? Perhaps there are even certain further criteria on $f$ (maybe on the domain?) to make it work for all $\{a\}_n$? For example, if we restrict the domain of $f$ to be a compact subset of $\Bbb R$ lets call $D$, then by the Weierstrass approximation theorem, we can approximate the function as close as we like with a polynomial. Would that mean any $\{a\}_n\subset D$ would work? Another example is let $f(x) = \text{sin}\, x$, sequence 1 be $a_n = 2\pi n$, and sequence 2 be $a_n=0.01\pi n$. Using the above algorithm with $f$ and sequence 1, the limit of $b_n$ is the zero function, but using sequence 2, the limit of $b_n$ seems to approach the sine wave. EDIT: The previous statement is not proven, only conjectured. See the counterexample below in the answer section by Tyler J. This shows that there must be some important constraints on $\{a\}_n$ so that the "information" or "essence" of $f$ is captured by the points in $S_n$. What are the constraints?

  • $\begingroup$ Did you try writing the formulas of the mentioned Newton polynomials ? $\endgroup$ – reuns Mar 29 '17 at 23:37

If the sequence $\{a_n\}_{n\in\mathbb{N}}$ has an accumulation point, it will yield $f$, as you can apply the principle of permanence to the difference between $f$ and the limit of the polynomials, assuming they converge (edit: further, assuming the limit is analytic!).

However, your above example demonstrates that this condition is sufficient, but not necessary.

edit: Actually, your example reveals some other interesting information! Consider $g(x) = \sin(201x)$ and $f(x) = \sin(x)$. On the sequence $a_n = 0.01\pi n$, $g(a_n) = f(a_n)$. Therefore, if we try to find a limit of interpolating polynomials for $g(x)$ using this sequence, we'll recover the same limiting polynomial as with $f(x)$ (which I'll trust is $f(x)$ itself, as you claim).

  • $\begingroup$ I have a computer program that will calculate $M(A)$ for any $A\in F$. I manually tested the example of $f(x)=\text{sin} \, x$ and $a_n=0.01\pi n$ with the first 30 points of the sequence and the output polynomials seemed to approach the Maclaurin series for sin. Your counter example of $g(x)=\text{sin} \, (201x)$ is so interesting though. It intersects $f(x)$ at every x value in the sequence. Evidently, even that sequence isn't a sufficient sampling of the sine wave for the interpolating polynomials. I wonder how to define a rigorous sampling sequence given a certain function. $\endgroup$ – Tyler Alkatraz Randall Mar 30 '17 at 23:35
  • $\begingroup$ My suspicion is that it is necessary and sufficient, without a priori information about $f$ aside from analyticity, for the sequence to have an accumulation point. $\endgroup$ – Tyler J Mar 31 '17 at 2:08

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