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It is often stated that a single-valued analytic function f(z) is uniquely and completely determined if (1) it is analytic at all points of a convergent sequence of points in the complex plane and at their limit point and (2) one is given the points of the sequence and the values of f(z) at each of these points.

Let z(1),z(2),...,z(n)... be a convergent sequence of complex numbers which are strictly decreasing in absolute value as n increases, and whose limit point is zero. Let f(z) be analytic at all the points of this sequence and at their limit point. Supposing that for each positive integer i one is given z(i) and f(z(i)). Does there then always exist a unique power series P(z) centered at zero such that (3) the radius of convergence R of P(z) is positive (or infinite) and (4) if k is any positive integer for which the absolute value of z(k) is less than R, P(z(k))=f(z(k))?

If such a unique power series exists, how do we obtain its coefficients from the data we are given? One can set up an equation for these unknown coefficients involving two infinite column vectors and an infinite Vandermonde matrix. The rows of the matrix are all of the form 1,z(j),z(j)^2,z(j)^3...where j is a positive integer. But I do not know what conditions are needed to insure that such matrices have a unique inverse.

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I took the liberty of fixing your paragraph breaks –  Yemon Choi Nov 9 '12 at 20:27
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Obviously, you need, at a minimum, that the sequence $f(z_i)$ converges, otherwise, it's hopeless. Building on Steve's answer below, the sequence of pairs $\bigl(z_i,f(z_i)\bigr)$ will have to have the property that the higher differences converge as well. Even that won't be enough to get everything you want because, obviously, you can throw away any finite number of the 'data points' $\bigl(z_i,f(z_i)\bigr)$ and it won't affect the limits, so if a solution exists for the remaining data and has large enough $R$, then you'll get a contradiction if the missing data don't match the function $f$. –  Robert Bryant Nov 9 '12 at 21:10
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4 Answers 4

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If I am not misunderstanding your question, you read off the power series pretty much directly from the given data. You know f(0). You also know f'(0) by using the definition of the derivative. The higher derivatives can all be determined by using higher order difference equations http://en.wikipedia.org/wiki/Finite_difference#Higher-order_differences. Since the function is analytic the taylor series you get this way does converge on some disk...

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Steve, I believe you have the right approach to this problem. However one must still find a definition of the nth derivative of f(z) at z=0 as the limit of a formula involving finite differences which requires no information other than the data we have. Suppose z(i) is a real number for each positive integral i. Many of the approximations to the nth derivative of f(z) at z=0 require that we know f(z) at values of z between z(k) and z(k+1)-for some integers k-and we do not have this information. –  Garabed Gulbenkian Nov 11 '12 at 19:07
    
Here is the recipe for f''(0) which you should be able to generalize. Take 3 points within a distance of $\delta$ from 0. Find the unique quadratic passing through all 3 points. Half the lead term of this quadratic is an approx of the second derivative. Form a sequence of such approximations as $/delta$ goes to 0. The limit of this sequence is the second derivative. –  Steve Nov 11 '12 at 20:23
    
It is sort of sad that such basic observations are not usually given to our calculus students, or complex analysis students for that matter. It was many years after learning these subjects, and feeling somewhat dissatisfied with my state of knowledge, that I finally figured these things out for myself. This plus analyticity is, in my opinion, the real reason for the identity theorem. –  Steve Nov 11 '12 at 20:25
    
Thanks, Steve. That was exactly what I was looking for. So, to approximate the nth derivative we can use the n+1 points z(k),z(k+1) ...z(k+n) together with the values of f(z) at these points and let k approach infinity. To obtain the leading coefficient of the nth degree interpolating polynomial we have just to solve a system of n+1 linear equations whose matrix is a (finite) Vandermonde matrix -which always has a unique inverse. So it seems that these matrices are useful after all. –  Garabed Gulbenkian Nov 13 '12 at 15:25
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For "most" choices of the values $f(z_i)$, there will be no holomorphic function with the given values. The reason if there is such a function then it is determined by any infinite subsequence of the given data. In other words, suppose that some values $f(z_i)$ are the values of a holomorphic $f$, and suppose that you try to change some of these values, while keeping infinitely many of them (including the value at the limit point) the same. Then the only function that could fit these new data would be the same $f$, since it's determined by the infinitely many unchanged values. So the new data, if they differed at all from the old, would not be the values of a holomorphic function.

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To say Robert Bryant's explanation in another way: assuming (as in the OP) that the $f(z_i)$ are already the values of a function $f$ analytic in a nbd $U$ of $0$, there is a unique power series $P$ such that $P(z(k))=f(z(k))$ for large enough $k$, and not for any positive integer $k$ for which $|z(k)|$ is less than the radius of convergence of $P$, that is $z(k)\in U\cap B(0,R)$. The reason is that the latter set is not connected, and we can't apply the principle of isolated zeros therein, not even if we known that $U$ is connected. Also note that changing finitely many $f(z_i)$ always produces the values of an analytic function on a nbd $U'$ of $0$, but of course $U\cap U'$ will be disconnected (the simplest way is to take as $U'$ a disjoint union of small nbd's of $0$ and of finitely many $z(i)$'s).

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As mentioned, if you know that your points come from sampling a holomorphic function then you can use limits of finite differences to compute all of the derivatives at the limit point. Or do it step by step, compute $f(0)=\lim_{z\rightarrow0}f(z)$, then replace all points $(z,y)$ with $(z,(y-f(0))/z)$ and continue.

However, without assuming you're starting with a holomorphic function there is very little you can say using limits. For example you can take $f(z)=\exp(-z^2)$ and take $x_i$ converging to 0 from the right, and you'd get the power series expansion at $0$: $f(z)=0+0z+0z^2+\cdots$. Even worse, you could add a tiny random error to each sample so as to not effect any of the limits, but still mess up the function.

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