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Is there an interesting or useful necessary and sufficient condition for a rational function $f \in \mathbb{C}(X)$ to be expressible as the difference $g(X+1)-g(X)$ of another rational function $g \in \mathbb{C}(X)$?

As an example, every polynomial $f$ is expressible in such a way since we have the Faulhaber's formula.

Alternately, we can observe that $(X+1)^n - X^n = \sum\limits_{k=0}^{n-1}{n\choose k}X^k$ and so given a polynomial $f(X) = a_nX^n + \ldots + a_1X + a_0$ we may solve the system of equations obtained by equating the $k$-th terms of $f$ with those of $\sum\limits_{k=0}^{n}\left(\sum\limits_{i=k+1}^{n+1}b_i{i \choose k}\right)X^k$. The associated matrix consists of the upper triangular matrix with entries $A_{ij} = {j \choose i-1}, 1 \leq i \leq j \leq n+1$, and its determinant is clearly nonzero since it is the product of the diagonal entries ${i \choose i-1} = i$, so the determinant is $(n+1)!$.

The solution for the $b_i$, $1 \leq i \leq n+1$ will give a polynomial $g(X) = b_{n+1}X^{n+1} + \ldots b_1X + b_0$ such that $f(X) = g(X+1)-g(X)$ ($b_0$ is arbitrary).

Another example is the rational function $f(X) = \frac{1}{X(X+1)}$ where $g(X) = -\frac{1}{X}$.

As a non-example, consider the function $f(X) = \frac{1}{X}$. Suppose there exist polynomials $p(X), q(X) \in \mathbb{C}[X]$ such that $\frac{p(X+1)}{q(X+1)}-\frac{p(X)}{q(X)} = \frac{1}{X}$. Assume moreover that $p$ and $q$ are coprime after dividing by any common factors.

Then we have $Xp(X+1)q(X)-Xp(X)q(X+1)=q(X+1)q(X)$ so that $q(X)$ divides $Xq(X+1)$ and $q(X+1)$ divides $Xq(X)$. This means we have polynomials $\alpha(X), \beta(X)$ such that $$Xq(X+1) = \alpha(X)q(X)$$ $$Xq(X) = \beta(X)q(X+1)$$

By degree considerations, we have $\deg \alpha = \deg \beta = 1$.

Then $X^2q(X) = \alpha(X)\beta(X)q(X)$ which means $X^2 = \alpha(X)\beta(X)$ and finally this forces $\alpha(X) = cX$ and $\beta(X) = \frac{1}{c}X$ for some $c \neq 0$. Substituting in the equations above we obtain $q(X+1) = cq(X)$ which is only possible if $q$ is constant because any zero would lead to infinitely many zeros.

But this is impossible because $\frac{p(X)}{q(X)}$ would then be a polynomial, and its forward difference cannot equal $\frac{1}{X}$.

I believe this proof can generalize to other cases where $f(X) = \frac{1}{g(X)}$ where $g$ is irreducible (that would just be the linear functions since $\mathbb{C}$ is algebraically closed).

But I have not found a "nice" condition on $f$ which could be necessary and sufficient. The continuous analogue of the problem is easy, we know that $f$ has a rational antiderivative if and only if its partial fraction decomposition does not contain denominators of degree $1$.

Tob Ernack
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    Not enough for an answer, but: I believe $f(X) = X^{-2}$ is also not the forward difference of a rational function, and in fact no $X^{-k}$ is; an argument can be built around the fact that if we have rational functions $f(), g()\in\mathbb{C}(x)$ with $g(X)=\Delta f(X)$, then $f(X)\in \mathbb{Q}(X)\Leftrightarrow g(X)\in\mathbb{Q}(X)$ (up to a constant); but then $\lim_{X\to\infty} g(X)\in\mathbb{Q}$ if it exists, but this conflicts with $\sum_{n=1}^\infty f(n)=\frac{\pi^2}6\not\in\mathbb{Q}$. – Steven Stadnicki Mar 22 '18 at 19:05
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    A very minor quibble about your interesting question: given any property $\phi$, there is always a necessary and sufficient condition for $\phi$, e.g., $\phi$ itself. What you are asking for is an interesting or useful necessary and sufficient condition, I think. – Rob Arthan Mar 22 '18 at 19:10
  • @RobArthan indeed, I am looking for "useful" or "interesting" conditions, but I don't know how to precisely define what that should be... Perhaps a condition which can be checked with a sufficiently fast deterministic algorithm given $f$. But I do not wish to add arbitrary restrictions. – Tob Ernack Mar 22 '18 at 19:11
  • @TobErnack: so just say "interesting" or "useful" in your question. Everyone will no what you mean and pedants like me won't complain. $\ddot{\smile}$. – Rob Arthan Mar 22 '18 at 19:12
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    Another proof that $\frac 1 X$ is a non-example: if ${\frac 1 X}=\Delta g$, then $\frac{1}{X^2}=\Delta (-g')$ with $-g'\in \Bbb C(X)$. But this can't be by @StevenStadnicki 's argument. – Arnaud Mortier Mar 22 '18 at 19:23
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    I think you can show that if $z_0$ is a singularity of $f$ then $f$ must have a singularity at $z_0+n$ for some non-zero integer $n.$ That eliminates a lot of your functions. – Thomas Andrews Mar 22 '18 at 19:29
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    Boring but probably accurate answer to the original question: over $\mathbb{C}(x)$, any rational function $f(X)$ can be written as a polynomial plus a sum of terms $a_{n,t}(X-t)^{-n}$. Then there exists a $g(X)$ with $f(X)=\Delta g(X)$ iff for each degree $n$, the corresponding condition holds on the set ${a_{n,t}}$, and I think this could easily be written purely combinatorially. – Steven Stadnicki Mar 22 '18 at 19:32
  • (i.e., there are no 'spurious' examples; everything can be read out from the decomposition of $f()$.) – Steven Stadnicki Mar 22 '18 at 19:33
  • @StevenStadnicki nice argument for $X^{-2n}$. As a minor issue, we do not have proofs that $\zeta(2n+1)$ is irrational except for $\zeta(3)$, I believe. – Tob Ernack Mar 22 '18 at 19:35
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    @TobErnack Again, the derivative argument takes care of the odd powers if you have a proof for even powers. – Arnaud Mortier Mar 22 '18 at 19:47
  • @ArnaudMortier indeed I see it now, thanks. – Tob Ernack Mar 22 '18 at 19:50
  • What about $f(x) = 1, g(x) = x$? – jim Mar 22 '18 at 20:40

2 Answers2

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Every rational fraction $f\in \mathbb{C}(X)$ can be written uniquely as a sum $$f(X) = P(X) + R(X)$$ where $P$ is a polynomial and $R$ is the polar part. Now the polar part is uniquely expressed as a finite sum $$R(X)= \sum_{k\ge 1, \alpha \in \mathbb{C}}\frac{c_{k,\alpha}}{(X-\alpha)^k}$$

The map $g(X) \mapsto g(X+1) -g(X)$ takes polynomial (polar) part to polynomial (polar) part. So $f$ is in the image of this map if and if only if its polar part is. It is easy to see that the necessary and sufficient condition is $$\sum_{n\in \mathbb{Z}} c_{k,\alpha +n}=0$$ for all $k\ge 1$ and $\alpha \in \mathbb{C}$. (finitely many linear conditions).

ADDED: Consider now the case $f\in \mathbb{Q}(X)$. The decomposition in partial fraction of $f$ happens in a finite (Galois) extension $K$ of $\mathbb{Q}$. Assume that there exists $g\in \mathbb{C}(X)$ so that $f(X) = g(X+1) - g(X)$. Since the linear conditions above are satisfied then we can find $g\in K(X)$ so that $f(X) = g(X+1)-g(X)$. Now average all the equalities $f(X) =g^{\sigma}(X+1)-g^{\sigma}(X)$ for all $\sigma \in Gal(K/\mathbb{Q})$. We get $\bar g\in \mathbb{Q}(X)$ so that $f(X) = \bar g(X+1) - \bar g(X)$.

ADDED:

The linear necessary and sufficient conditions follow from the following combinatorial lemma: for a sequence $(a_n)_{n\in \mathbb{Z}}$ of finite support there exists a sequence of finite support $b= (b_n)_{n\in \mathbb{Z}}$ so that $a= \Delta b$ if and only if $\sum a_n =0$. Moreover, in that case $b$ is uniquely defined. This is the discrete analogue of the result: a function with compact support on $\mathbb{R}$ has an antiderivative with compact support if and only if its total integral is $0$. In our case we want $b_{n+1}-b_n =a_n$ for all $n$. The unique solution with finite support is given by $b_n=\sum_{k<n}a_k$ for all $n$.

ADDED: As @Steven Stadnicki: noticed, if a solution $g$ exists, it is unique up to a constant. Moreover, if $f$ has coefficients in a subfield $K$, so can $g$ be taken with coefficients in $K(X)$.

We give now some results in particular cases:

  1. Assume that $f(X) = \frac{P(X)}{Q(X)}$ where $Q$ has simple integral roots and $\deg Q \ge \deg P +2$. Then $f$ is a difference. Indeed, it is enough to check that the sum of residues of $f$ is $0$, and this follows from Cauchy's formula.

  2. Assume that $\deg f =-1$. Then $f$ is not a difference. This can be seen in two ways: one, since the sum of residues is not $0$; another way is to notice that $\deg f = \deg g -1$, but $g$ is at most of degree $-1$ ( basically we look at the rate of decrease at $\infty$.

ADDED:

If the equation $f(X)= h(X+2)-h(X)$ has a solution $h$, then the equation $f(X)=g(X+1)-g(X)$ has a solution $g(X)= h(X+1)+h(X)$.

orangeskid
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  • Yep, that's basically what I got, only maybe expressed cleaner. But it unclear what you mean by "polar." – Thomas Andrews Mar 22 '18 at 20:58
  • @Thomas Andrews: it is really the remainder over the divisor. and then do partial fractions. the poles of the functions show up. – orangeskid Mar 22 '18 at 21:04
  • @Thomas Andrews: Still it's not a very useful criterion in that one needs to find the decomposition into partial fractions. For instance, it's easy to see directly for $f(X) = \frac{1}{X(X+1)\ldots (X+m)}$, but its parial fraction decomposition is a bit more involved. – orangeskid Mar 22 '18 at 21:08
  • @Thomas Andrews: i wonder how it works for fields like $\mathbb{Q} $ – orangeskid Mar 22 '18 at 21:36
  • I was going to also ask about $\mathbb{Q}$ but since the partial fraction decomposition becomes more complicated I preferred to keep it at $\mathbb{C}$ in the question. – Tob Ernack Mar 22 '18 at 22:06
  • Also just to make sure, the $c_{k,\alpha}$ here are polynomials of degree less than $k$ right? I think this system of linear conditions is a satisfying criterion, since it is easy to verify given the partial fraction decomposition of $f$. – Tob Ernack Mar 22 '18 at 23:12
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    @Tob Ernack: the $c_{k, \alpha}$ are complex numbers (constants). Btw, I added a part that says that for a fraction with rational coefficients, the existence of a $g$ with complex coefficients implies the existence of $g$ with rational coefficients. – orangeskid Mar 22 '18 at 23:16
  • I think this answer is correct, but I vigorously disagree with 'it is easy to see' - I think that part (that there can't be any 'accidents') is the crux of the problem, and it's far from trivial. – Steven Stadnicki Mar 22 '18 at 23:51
  • @Steven Stadnicki: It is consequence of the following basic lemma: for a sequence $(a_n){n\in \mathbb{Z}}$ of finite support there exists a sequence of finite support $b= (b_n){n\in \mathbb{Z}}$ so that $a= \delta b$ if and only if $\sum a_n =0$ – orangeskid Mar 23 '18 at 00:06
  • @orangeskid I think the combinatorial piece is pretty easy; what's less clear (even though I think it's true and maybe even straightforward) is that you can't have 'accidents' where e.g. $\frac1{X+1}-\frac2{(X+2)^2}+\frac1{X+3}$ turns out to be a difference of two functions. – Steven Stadnicki Mar 23 '18 at 01:02
  • @Steven Stadnicki: I think one considers a fixed power and coefficients at a string of locations differing by integers. All the action is confined to such a place. Then use lemma. Btw, it's nice to notice that $P(X)/Q(X)$ is a difference if all the roots of $Q$ are integral and simple and $\deg P + 2 \le \deg Q$. see the material added., – orangeskid Mar 23 '18 at 01:55
  • @Steven Stadnicki: I was trying using Wolfram Alpha to find an $f(X)$ rational that is not a difference, so that the sum $\sum_{n\ge 1}f(n)$ is rational, but no success...Wolfram Alpha either cannot evaluate symbolically or gives (plausible) irrationals. – orangeskid Mar 23 '18 at 02:01
  • If I understand correctly, the lemma is applied by noting that if $c_{k,\alpha}$ is the coefficient of $(X-\alpha)^{-k}$ in the expansion of $f$, and $f = \Delta g$, then we must have $c_{k,\alpha} = b_{k,\alpha+1}-b_{k,\alpha}$ where $b_{k,\alpha}$ is the analogous coefficient for $g$. Then $(c_{k,\alpha+n})$ and $(b_{k,\alpha+n})$ have finite support (finitely many singularities) so the lemma applies. Is that correct? – Tob Ernack Mar 23 '18 at 03:55
  • @Tob Ernack: Yes, exactly. Moreover, the solutions for $b_{k,\alpha}$ is unique. – orangeskid Mar 23 '18 at 08:55
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If $f(z)$ has a singularity at $0$ (WLOG) then we must have either $g(z)$ having a singularity at $0$ or at $1$ (or both.)

If $0$ is a singularity of $g$ then since $f(-1)=g(0)-g(-1)$, either $f$ has a singularity at $-1$ or $g$ does (or both.) If $g$ does, then we can keep moving down and find that either $g$ has a singularity at $-k$ or $f$ has singularities at $0$ and $-k.$ So for some $k> 0$, we must have $f(z)$ with singularities at $0,-k$ and $g$ has a singularity at $-1,\dots,-(k-1).$ We have to be able to stop at some point because $f$ must have finitely many singularities.

If $1$ is a singularity of $g$, then either $2$ is a singularity of $g$ or $1$ is a singularity of $f$. We then move up or down, and find that $0$ and $k$ must be a singularity of $f$ for some integer $k>0.$

This means that if $f$ has a singularity at $z_0$ then there must be an integer $n\neq 0$ such that $f$ has a singularity at $z_0+n.$

An Algorithm

Here is an algorithm that doesn't require you to find the exact roots.

The general strategy: Each time through the loop, we will reduce the problem to another $f$ with fewer singularities.

Write $f(x)=\frac{p(x)}{q(x)}$ with $\deg p<\deg q,$ and $p,q$ are relatively prime. (Any polynomial part can be handled separately.) Compute $q_n(x)=\gcd(q(x),q(x+n))$ for $n=1,2,\dots$ until either:

  1. $q_n(x)\neq 1,$ or
  2. $n$ is greater than the diameter of the set of (complex) zeros of $q(x).$

If 2. is reached, and $f\neq 0,$ then there is no $g$. (We can at least compute an upper bound for the diameter, so we only have to check finitely many $n.$)

Otherwise, we've found a $q_n(x)\neq 1.$ Write $q(x)=a(x)b(x)$ where $q_n(x)\mid a(x)\mid q_n^k(x)$ for some $k$ and and $\gcd(a(x),b(x))=1.$ Basically, let $a(x)=\gcd(q(x),q_n^k(x))$ for some large $k,$ so that $a(x)$ has all repetitions of the roots $z_0$ of $q$ such that $z_0+n$ is also a root.

We know that $b(x)$ is non-constant, because any root with a maximal real part is not a root of $q_n(x).$

Then we can write:

$$f(x)=\frac{p(x)}{q(x)}=\frac{c(x)}{a(x)} + \frac{d(x)}{b(x)},$$ using standard partial fraction approaches.

Let $$\begin{align}f_1(x)&=\frac{c(x-n)}{a(x-n)}+\frac{d(x)}{b(x)}. \end{align}$$

Note that if $h(x)=\frac{c(x)}{a(x)},$ then $h(x)-h(x-n)=\Delta(h(x-1)+h(x-2)+\cdots h(x-n))$, so $f_1$ and $f$ differ by an element in the range of $\Delta.$

So $f_1$ is in the range of $\Delta$ if and only if $f$ is.

If $f_1=0$, we are done.

Otherwise we apply the process to $f_1.$

We know that $f_1$ has a smaller set of singularities - we have removed the singularities $z_0$ such that $z_0+n$ is also a singularity, and potentially added in more repetitions of the singularities at $z_0+n.$ What this means is that this process must eventually stop.

For example, if $$f(x)=\frac{3x^3+16x^2+28x+12}{(x+3)(x+2)^2x^2}$$ then $q_1(x)=x+3,$ and you get:

$$f(x)=\frac{-1}{x+3}+\frac{x^3+4x^2+8x+4}{(x+2)^2x^2}$$

Then you get $$f_1(x)=\frac{-1}{x+2}+\frac{x^3+4x^2+8x+4}{(x+2)^2x^2}=\frac{2x^2+8x+4}{(x+2)^2x^2}.$$

For this $f_1$, we get $q_2(x)=(x+2)^2$ and we solve:

$$f_1(x)=\frac{-(x+3)}{(x+2)^2} + \frac{x+1}{x^2}.$$

Then we get:

$$f_2(x)=\frac{-(x+1)}{x^2}+\frac{x+1}{x^2}=0$$

So our resulting expression is:

$$\begin{align}f(x)&=\left(\frac{-1}{x+3}-\frac{-1}{x+2}\right)+\left(\frac{-(x+3)}{(x+2)^2}+\frac{x+1}{x^2}\right)\\ &=-\Delta\left(\frac{1}{x+2} + \frac{x+2}{(x+1)^2}+\frac{x+1}{x^2}\right) \end{align}$$

In this, I am obviously using the knowledge of the roots to do the work, you need not. All of the steps above can be computed without the roots. You can compute the $q_n,$ then the $a,b$, then $c,d$ all using standard operations.

For example, if the singularities of $f$ were at $0,1,2\pm i,3\pm i,5\pm i$, then your first $a(x)$ would be found with $a(x)$ having roots $0$ and $2\pm i.$ $f_1$ would have remaining singularities a subset of $\{1,3\pm i,5\pm i\}.$ The next step would find $n=2$ and remove singularities at $3\pm i.$ Then we'd find out if $f_2=0.$ If it isn't, the next attempt to find an $n$ will fail.

This algorithm also works in any $k(x)$ where $k\subseteq \mathbb C$ is a field, because all the algorithmic steps work inside the field.

In particular, if $f(x)\in k(x)$ has a $g(x)\in\mathbb C(x)$ then it has a solution in $k(x).$

Thomas Andrews
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  • Thank you! So if I understand the first part of your answer, starting with a singularity of $f$ at $z_0$, by continuing the recurrence and having finitely many singularities for $g$, we are forced to have a second singularity for $f$ at some $z_0 \pm n$. Then if $f$ has two singularities spaced out at $n$ (and none further apart) we can reduce their distance by adding a term of the form $g(z+1)-g(z)$ where $g(z) = A/(z-z_0-n)$ which moves the second singularity at $z_0+n-1$. – Tob Ernack Mar 22 '18 at 21:58
  • Then we can use induction to find a linear combination of the $g_i(z+1)-g_i(z)$ which leaves us with either a polynomial or a rational function with a single unpaired singularity. And the first case leads to a solution of the recurrence, the second case leads to impossibility. – Tob Ernack Mar 22 '18 at 22:00
  • Ah in my first comment I should really have said $g(z) = p(z)/(z-z_0-n)^k$ (as in your answer) since the singularities might have order greater than $1$. – Tob Ernack Mar 22 '18 at 22:09