Elementary solutions to f(z+1)-f(z)=g(z) in entire functions - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T06:32:16Z http://mathoverflow.net/feeds/question/4434 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/4434/elementary-solutions-to-fz1-fzgz-in-entire-functions Elementary solutions to f(z+1)-f(z)=g(z) in entire functions Boris Bukh 2009-11-06T20:50:11Z 2009-11-07T18:05:04Z <p>Let g(z) be an entire function of a complex variable z. Does there exist an entire function f(z) such that f(z+1)-f(z)=g(z)? As I learned several years back, the answer to this <a href="http://www.math.niu.edu/~rusin/known-math/01_incoming/coho_func" rel="nofollow">is apparently 'yes'</a>, but I have not felt satisfied with the proof because it goes beyond my expertise.</p> <p>I tried to find f using the power series expansion of g, for that works when g is a polynomial. But the results of partial inversions kept diverging. Representing g as an integral via Cauchy's formula, and doing inversion inside the integration led to similar problems. Perhaps I am overly optimistic, but a question this elementary should have equally elementary solution. Is there such a solution? If not, is there a reason to expect that no simple and elementary solution should exist?</p> http://mathoverflow.net/questions/4434/elementary-solutions-to-fz1-fzgz-in-entire-functions/4437#4437 Answer by Michael Lugo for Elementary solutions to f(z+1)-f(z)=g(z) in entire functions Michael Lugo 2009-11-06T21:14:04Z 2009-11-06T21:14:04Z <p>I suspect you want a constructive solution; if someone gives you g, you want to be able to say what f is.</p> <p>Write Df(z) for f(z+1) - f(z). We want to solve Df(z) = g(z). Of course this only determines f up to a constant.</p> <p>Let g(z) = (z)<sub>k</sub> = z(z-1)(z-2)...(z-k+1). Then f(z) = (z)<sub>(k+1)</sub>/(k+1) satisfies Df(z) = g(z). (The falling power (z)<sub>k</sub> is to finite differences what z<sup>k</sup> is to ordinary differential calculus.)</p> <p>We can use this to "integrate" (i. e. given g, find f) the ordinary powers z<sup>k</sup> by using the fact that ordinary powers can be written as linear combinations of falling powers. The coefficients of the basis change are <a href="http://en.wikipedia.org/wiki/Stirling_number" rel="nofollow">Stirling numbers</a>. So, for example, </p> <p>g(z) = z<sup>3</sup> = (z)<sub>3</sub> + 3(z)<sub>2</sub> + (z)<sub>1</sub></p> <p>and so Df(z) = g(z) where f is</p> <p>(z)<sub>4</sub>/4 + (z)<sub>3</sub> + (z)<sub>2</sub>/2 = z<sup>2</sup> (z-1)<sup>2</sup>/4. In general, given g(z) = z<sup>k</sup> one gets </p> <p>f(z) = &sum;<sub>k=1</sub><sup>n</sup> S(n,k) (n)<sub>k+1</sub>/(k+1)</p> <p>where S(n,k) is a Stirling number of the second kind. But these polynomials are not that nice, and I'm not sure how one would use this to find, say, the solution to Df(z) = exp(z).</p> http://mathoverflow.net/questions/4434/elementary-solutions-to-fz1-fzgz-in-entire-functions/4438#4438 Answer by Andrew Critch for Elementary solutions to f(z+1)-f(z)=g(z) in entire functions Andrew Critch 2009-11-06T21:15:04Z 2009-11-06T21:15:04Z <p>An answer to part of your question: it makes sense that a power series approach isn't so straightforward, since substitution of formal power series only makes formal sense when you're plugging in something with zero constant term, not something like 1+x. In particular, a sequence of polynomials p(x) converging to a power series f(x) in the Krull topology (i.e. eventually agreeing coefficients) doesn't imply that p(1+x) converges to f(1+x) in the Krull topology, even if f(1+x) happens to make sense due to f being the power series of an entire function.</p> http://mathoverflow.net/questions/4434/elementary-solutions-to-fz1-fzgz-in-entire-functions/4450#4450 Answer by Dan Piponi for Elementary solutions to f(z+1)-f(z)=g(z) in entire functions Dan Piponi 2009-11-06T21:53:06Z 2009-11-06T21:53:06Z <p>An informal approach is to write the finite difference operator as exp(D)-1 where D is d/dx. We're then trying to evaluate 1/(exp(D)-1) f(x).</p> <p>For example, consider the case f(x) = xcos(x).</p> <p>Write cos(x)=(exp(ix)+exp(-ix))/(2i)</p> <p>Quantum physicists make much use of the 'theorem': f(D) exp(ax) = exp(ax) f(a+D)</p> <p>So 1/(exp(D)-1) (xexp(ix)) = exp(ix) 1/(exp(D+i)-1) x</p> <p>1/(exp(D+i)-1) can be expanded as a power series in D. As we're applying it to x, only the constant and D^1 terms survive.</p> <p>Reassembling and simplifying gives g(x) = ((1+x)(cos(x)-cos(1+x))+cos(1+x))/(4sin(1/2)^2)</p> <p>Amazingly, Mathematica simplifies g(x+1)-g(x) to xcos(x) so it's probably correct. No doubt an analyst could make these kinds of shenanigans with D (which I believe go back to Heaviside) rigorous.</p> http://mathoverflow.net/questions/4434/elementary-solutions-to-fz1-fzgz-in-entire-functions/4461#4461 Answer by S. Carnahan for Elementary solutions to f(z+1)-f(z)=g(z) in entire functions S. Carnahan 2009-11-06T23:24:26Z 2009-11-06T23:24:26Z <p>I haven't thought through the necessary estimates, but it seems like you can use uniform approximation by polynomials on compact sets. For any disc of radius N, there is a polynomial g<sub>N</sub> that differs from g on that disc by at most, say, 1/N or something, and a polynomial f<sub>N</sub> that satisfies f<sub>N</sub>(z+1)-f<sub>N</sub>(z) = g<sub>N</sub>(z). You can also demand bounds on how the derivatives of g<sub>N</sub> differ from those of g. The hard part is putting bounds on how much f<sub>N</sub> varies (on a disc of radius N-1) if you vary g<sub>N</sub> by a small amount.</p> <p>Anyway, if you get past that part, you can let N go to infinity.</p> http://mathoverflow.net/questions/4434/elementary-solutions-to-fz1-fzgz-in-entire-functions/4494#4494 Answer by moonface for Elementary solutions to f(z+1)-f(z)=g(z) in entire functions moonface 2009-11-07T05:18:26Z 2009-11-07T05:18:26Z <p>First, you can find f smooth with this property. Now you wish to replace f by f+h, with h periodic, so that f+h is holomorphic. To do this, you need to be able to solve <img src="http://latex.mathoverflow.net/png?%5Cbar%7Bd%7D" alt="\bar{d}" title="" />h = some fixed smooth periodic function. </p> <p>I think you can solve this by expanding the fixed periodic function into Fourier series and solving term-by-term. I suspect this is pretty close to (a translation of) the solution suggested in the link you gave. </p> http://mathoverflow.net/questions/4434/elementary-solutions-to-fz1-fzgz-in-entire-functions/4542#4542 Answer by fedja for Elementary solutions to f(z+1)-f(z)=g(z) in entire functions fedja 2009-11-07T18:05:04Z 2009-11-07T18:05:04Z <p>It took me some time to find a solution that satisfies both requirements:</p> <p>a) If should be based on the power series expansion</p> <p>b) It should use no tools heavier than contour integration.</p> <p>So, let $g(z)=\sum a_ k z^k$. We know that $a_ k$ decay faster than any geometric progression. We want analytic functions $F_ k(z)$ such that $F_ k(z+1)-F_ k(z)=z^k$ and $F_ k(z)$ grows not faster than a geometric progression as a function of $k$ in every disk. Then $f=\sum a_ k F_ k$ is what we want. So, just choose any odd multiples $r_ k\in(k,k+2\pi)$ of $\pi$ and put $$F_ k(z)=\frac{k!}{2\pi i}\oint_ {|z|=r_ k}\frac {e^{z\zeta}}{e^\zeta-1}\frac{d\zeta}{\zeta^{k+1}}$$<br /> The key is that $|e^\zeta-1|\ge \frac 12$ on the circle and $r_ k^k\ge k!$, so $|F_ k(z)|\le 2e^{|z|(k+2\pi)}$ on the plane.</p>