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 is apparently 'yes', but I have not felt satisfied with the proof because it goes beyond my expertise.

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?

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    $\begingroup$ As a general rule, elementary questions need not have elementary solutions. That's never been a useful principle in mathematics. $\endgroup$ – Qiaochu Yuan Nov 6 '09 at 21:15
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    $\begingroup$ In particular, insisting that if f(z+1) - f(z) = g(z) and g is elementary, then f must be elementary is analogous to insisting that the integral of an elementary function must be elementary. $\endgroup$ – Michael Lugo Nov 6 '09 at 21:17
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    $\begingroup$ Hoping for an elementary solution is akin to Occam's razor: one should strive for the simplest explanation that explains the phenomenon at hand. Sometimes it happens there is such an explanation, sometimes not. If not, the reasons for its non-existence are instructive. $\endgroup$ – Boris Bukh Nov 7 '09 at 14:00

It took me some time to find a solution that satisfies both requirements:

a) If should be based on the power series expansion

b) It should use no tools heavier than contour integration.

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 $$ F_ k(z)=\frac{k!}{2\pi i}\oint_ {|z|=r_ k}\frac {e^{z\zeta}}{e^\zeta-1}\frac{d\zeta}{\zeta^{k+1}} $$
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.

  • $\begingroup$ That is beautiful! $\endgroup$ – Boris Bukh Nov 7 '09 at 18:25
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    $\begingroup$ I agree, this is a very neat solution. Note that the Bernoulli polynomial $B_k(z)$ satisfies the equation $B_k(z+1)-B_k(z) = kz^{k-1}$, and you can work out fedja's integral explicitly to show that it's taking the Bernoulli polynomial and shifting it by a periodic function to zero out the low-order Fourier modes on the interval [0,1] (which have the most significant contribution). However, if we insist on that approach (without the contour integral), it seems much trickier to argue that the sum will converge in the end. $\endgroup$ – Darsh Ranjan Nov 10 '09 at 9:23
  • $\begingroup$ @DarshRanjan, still, it is very helpful to know that actually we can just take $F_k(z)=\frac1{k+1}B_{k+1}(z)$ here $\endgroup$ – მამუკა ჯიბლაძე May 20 '17 at 7:21

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).

For example, consider the case f(x) = xcos(x).

Write cos(x)=(exp(ix)+exp(-ix))/(2i)

Quantum physicists make much use of the 'theorem': f(D) exp(ax) = exp(ax) f(a+D)

So 1/(exp(D)-1) (xexp(ix)) = exp(ix) 1/(exp(D+i)-1) x

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.

Reassembling and simplifying gives g(x) = ((1+x)(cos(x)-cos(1+x))+cos(1+x))/(4sin(1/2)^2)

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.


I suspect you want a constructive solution; if someone gives you g, you want to be able to say what f is.

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.

Let g(z) = (z)k = z(z-1)(z-2)...(z-k+1). Then f(z) = (z)(k+1)/(k+1) satisfies Df(z) = g(z). (The falling power (z)k is to finite differences what zk is to ordinary differential calculus.)

We can use this to "integrate" (i. e. given g, find f) the ordinary powers zk by using the fact that ordinary powers can be written as linear combinations of falling powers. The coefficients of the basis change are Stirling numbers. So, for example,

g(z) = z3 = (z)3 + 3(z)2 + (z)1

and so Df(z) = g(z) where f is

(z)4/4 + (z)3 + (z)2/2 = z2 (z-1)2/4. In general, given g(z) = zk one gets

f(z) = ∑k=1n S(n,k) (n)k+1/(k+1)

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).

  • $\begingroup$ Actually, Df(z) = g(z) determines f up to addition of an arbitrary function of period 1, i. e., a solution of Df(z) = 0. $\endgroup$ – Darsh Ranjan Nov 7 '09 at 2:07
  • $\begingroup$ As explained in my question, the problem with this approach is convergence. Andrew Critch in his answer explained why this approach is bound to be troubled. $\endgroup$ – Boris Bukh Nov 7 '09 at 10:36
  • $\begingroup$ Darsh: that's a good catch. I went too far with the analogy to integration. Boris: that explains why theoretically one might expect trouble, but I was going through with it anyway and hoping that things might work out formally anyway. $\endgroup$ – Michael Lugo Nov 7 '09 at 17:55

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.


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 $\bar{d}\, h =$ some fixed smooth periodic function.

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.


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 gN that differs from g on that disc by at most, say, 1/N or something, and a polynomial fN that satisfies fN(z+1)-fN(z) = gN(z). You can also demand bounds on how the derivatives of gN differ from those of g. The hard part is putting bounds on how much fN varies (on a disc of radius N-1) if you vary gN by a small amount.

Anyway, if you get past that part, you can let N go to infinity.


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