I recently came up with this theorem:

For any complex polynomial $$P$$ degree $$n$$:

$$\sum\limits_{k=0}^{n+1}(-1)^k\binom{n+1}{k}P(a+kb) = 0\quad \forall a,b \in\mathbb{C}$$

Basically, if $$P$$ is quadratic, $$P(a) - 3P(a+b) + 3P(a+2b) - P(a+3b) = 0$$ (inputs of $$P$$ are consecutive terms of any arithmetic sequence). This can be generalized to any other degrees.

• Has this been discovered? If yes, what's the formal name for this phenomenon?
• Is it significant/Are there important consequences of this being true?
• Can this be generalized to non-polynomials?
• I do not know if this have a name, but your theorem comes as a linear combination of the special case $P(x) = x^m$ where $0 \leq m \leq n$, and for the monomial case, I believe it is well known. I am still searching for a more suitable reference, but this page explains a bit: mathworld.wolfram.com/FiniteDifference.html – Hw Chu Sep 26 '18 at 0:49
• For $b=1$, this follows from Theorem 1 and Theorem 2 in my post math.stackexchange.com/questions/1379172/… . (My argument is more or less the same as Steven Stadnicki's.) To get the case of general $b$, you can either generalize my argument, or you can apply my argument to the polynomial $Q$ (also of degree $n$) defined by $Q\left(x\right) = P \left(a+xb\right)$. – darij grinberg Sep 26 '18 at 2:02
• My answer here has a reference to an article by Gould that has a very extensive list of references on this result, which goes back to Euler. – Chappers Sep 26 '18 at 15:03
• What specific field of math is this? I want to learn it. – clathratus Oct 2 '18 at 3:22
• @clathratus I'm not sure, tbh. I guess it's algebra, since it involves polynomials (something you learn in Algebra class at high school), and also because most techniques I used to prove this are algebraic manipulations (i.e., divide both sides by something, etc.). I hope someone else can give you a definite answer. – Felix Fourcolor Oct 2 '18 at 5:02

In brief: this is well-known, but definitely important.

It's easiest to write this in terms of the finite difference operator $$\Delta$$: $$\Delta P(x)=P(x+1)-P(x)$$. You use $$P(x+b)$$ instead of $$P(x+1)$$, but it's easy to see that these two things are equivalent; to keep things consistent with your notation, I'll write $$\Delta_b$$ for your operator.

The most important feature of the $$\Delta_b$$ operator is how it affects the degree of a polynomial:

Theorem: for any nonconstant polynomial $$P(x)$$, the degree of $$\Delta_b P(x)$$ is one less than the degree of $$P(x)$$.

Proof outline: Note that the degree of $$\Delta_b P(x)$$ is no greater than the degree of $$P(x)$$. Now, write $$P(x) = a_dx^d+Q(x)$$, where $$Q(x)$$ is a polynomial of degree $$d-1$$ or less. Then $$P(x+b) =a_d(x+b)^d+Q(x+b)$$, so $$\Delta_b P(x) = a_d\left((x+b)^d-x^d\right)+\Delta_b Q(x)$$; by the binomial theorem $$(x+b)^d=x^d+{d\choose 1}bx^{d-1}+\ldots$$, so $$(x+b)^d-x^d={d\choose 1}bx^{d-1}+\ldots$$ is a polynomial of degree at most $$d-1$$, and thus $$\Delta_bP(x)$$ is the sum of two polynomials of degree at most $$d-1$$ (namely, $$a_d\left((x+b)^d-x^d\right)$$ and $$\Delta_b Q(x)$$), so it's of degree at most $$d-1$$ itself.

(It's slightly more challenging to prove that the degree of $$\Delta_bP(x)$$ is exactly $$d-1$$ when $$b \neq 0$$, but this can also be shown.)

Why does this matter? Because it can be shown by induction that your sum is exactly the result of applying the $$\Delta_b$$ operator $$d+1$$ times, where $$d$$ is the degree of the polynomial; since each application of $$\Delta_b$$ reduces the degree by one, then $$(\Delta_b)^dP(x)$$ is a polynomial of degree zero — a constant — and thus $$(\Delta_b)^{d+1}P(x)$$ will be identically zero. This is exactly your identity.

Now, you may know that the derivative of a polynomial of degree $$d$$ is also a polynomial of degree $$d-1$$. It turns out that this isn't a coincidence; $$\Delta$$ is very similar to a derivative in many ways, with the Newton polynomials $${x\choose d}=\frac1{d!}x(x-1)(x-2)\cdots(x-d)$$ playing the role of the monomial $$x^d$$ with respect to the derivative. For more details, I suggest starting with Wikipedia's page on finite difference calculus.

In fact, we can also prove the converse (and this answers the question about generalizing to non-polynomials in the negative). I'll work in terms of $$\Delta$$, rather than $$\Delta_b$$, but again all the results generalize readily.

Note that $$\Delta^n P(x)$$ only depends on the values of $$P(x+i)$$ for $$i$$ an integer between $$0$$ and $$n$$; thus, a function can take arbitrary values for $$0\lt x\lt1$$ and still satisfy the identity; we can't say much about general points. However, it does constrain the values at integers:

Theorem: suppose that $$\Delta^{d+1}f(x)\equiv 0$$ identically. Then there exists a polynomial $$P(x)$$ of degree $$d$$ such that $$f(n)=P(n)$$ for all integers $$n$$.

The proof works by induction. For simplicity's sake, I'll consider all functions as being on $$\mathbb{Z}$$ now, and not consider non-integer values at all. Note first of all that if $$\Delta f(x)=g(x)$$, then $$f(n)=f(0)+\sum_{i=0}^{n-1}g(i)$$. (Proof by induction: the case $$n=1$$ is true by definition, since $$g(0)=\Delta f(0)=f(1)-f(0)$$ implies that $$f(1)=f(0)+g(0)$$. Now, assuming it's true for $$n=k$$, at $$n=k+1$$ we have $$f(k+1)=f(k)+g(k)$$ $$=f(0)+\sum_{i=0}^{k-1}g(i)+g(k)$$ $$=f(0)+\sum_{i=0}^kg(i)$$.) In particular, if $$\Delta f(x)\equiv 0$$ identically, then $$f(n)=f(0)$$ for all integers $$n$$; $$f()$$ is constant on $$\mathbb{Z}$$.

This gives us the base case for our induction; to induct we just need to show that if $$\Delta f(x)$$ is a polynomial of degree $$d$$, then $$f(x)$$ is polynomial of degree $$d+1$$. But suppose for concreteness that $$\Delta f(x)=P(x)=\sum_{i=0}^da_ix^i$$. Then $$f(n)=f(0)+\sum_{k=0}^{n-1}P(k)$$ $$=f(0)+\sum_{k=0}^{n-1}\left(\sum_{i=0}^da_ik^i\right)$$ $$=f(0)+\sum_{i=0}^da_i\left(\sum_{k=0}^{n-1}k^i\right)$$. Now, for each $$i$$ the sum $$\sum_{k=0}^{n-1}k^i$$ in parentheses in this last expression is known to be a polynomial of degree $$i+1$$ (see e.g. https://en.wikipedia.org/wiki/Faulhaber%27s_formula ), so the whole expression is a polynomial of degree $$d+1$$, as was to be proved.

• "this is well-known, but definitely important." Did you mean this particular theorem, or the idea of finite difference in general? – Felix Fourcolor Sep 26 '18 at 4:47
• @FelixFourcolor Sort of both - the idea of finite differences in the broad, but also particularly this consequence that the difference operator behaves like a derivative especially with respect to polynomials (and that therefore the n+1st difference is zero, which is your theorem). It's used, for instance, in building up interpolating polynomials. – Steven Stadnicki Sep 26 '18 at 5:20
• Would you happen to know why I got those down-votes on my answer below with no explanation added? It seems to me that I have a perfectly valid formal power series proof. Maybe you can advise here. – Marko Riedel Sep 26 '18 at 15:04
• @MarkoRiedel I'm afraid I don't; I wasn't one of the downvoters and it looks broadly sensible to me. It might just be a matter of a gap in levels between the question and the proof, but I wouldn't expect that to garner quite so many downvotes in and of itself. – Steven Stadnicki Sep 26 '18 at 15:39
• @FelixFourcolor I'll try to amend my answer to address that; the short version is that your conditions impose no restrictions on the function's values from e.g. $0$ to $b$, but force the function to be 'polynomial at multiples of $b$'. I'll add the proof of that (along with a clearer definition) in my answer. – Steven Stadnicki Oct 2 '18 at 21:25

What you wrote is the finite difference operator of order $$n+1$$, acting on $$P$$, $$\sum\limits_{k=0}^{n+1}(-1)^k\binom{n+1}{k}P(a+kb) =\Delta^{n+1} P(a+kb).$$

Notice that by linearity it suffices for the property to hold for all monomials $$k^m, m\le n$$ and it is easily explained by the fact that the first order difference of a polynomial is a polynomial of degree one less.

$$(k+1)^m-k^m=k^m+mk^{m-1}+\cdots-k^m.$$

Illustration ($$n=3$$):

$$\Delta^4 k^m=((4^m-3^m)-(3^m-2^m))-((3^m-2^m)-(2^m-1^m)) \\-((3^m-2^m)-(2^m-1^m))-((2^4-1^m)-(1^4-0^m)) \\=4^m-4\cdot3^m+6\cdot2^m-4\cdot1^m+0^m.$$ and $$\begin{matrix} 1&&1&&1&&1&&1 \\&0&&0&&0&&0 \\&&0&&0&&0 \\&&&0&&0 \\&&&&0 \end{matrix}$$

$$\begin{matrix} 0&&1&&2&&3&&4 \\&1&&1&&1&&1 \\&&0&&0&&0 \\&&&0&&0 \\&&&&0 \end{matrix}$$

$$\begin{matrix} 0&&1&&4&&9&&16 \\&1&&3&&5&&7 \\&&2&&2&&2 \\&&&0&&0 \\&&&&0 \end{matrix}$$

$$\begin{matrix} 0&&1&&8&&27&&64 \\&1&&7&&19&&37 \\&&6&&12&&18 \\&&&6&&6 \\&&&&0 \end{matrix}$$

Final remark:

On can show that $$\Delta_{n+1}k^{n+1}=(-1)^nn!$$, so that for a polynomial of degree $$n+1$$ the sum is

$$(-1)^nn!p_{n+1}b^{n+1},$$ independently of $$a$$.

• You don't need to include "illustrations", as I was the one who discovered this phenomenon myself, I must have known it. Instead, you could talk more about your final remark and maybe a sketch of proof for it. For now, your answer brings very limited new information to the table. That's the reason for my downvote. – Felix Fourcolor Oct 4 '18 at 6:20
• @FelixFourcolor: posting such a naive question made me think you needed simple explanations. Sorry to have disturbed you. – Yves Daoust Oct 4 '18 at 7:11

This result is known and was demonstrated by a student named Ruiz.

Here's the reference :

Sebastián Martín Ruiz, An Algebraic Identity Leading to Wilson's Theorem, The Mathematical Gazette, 80 (489) 579-582 (Nov. 1996).

You accesss to it at JSTOR : http://www.jstor.org/stable/3618534

This is not a new result. What you wrote is the formula for the $$n$$-th order forward difference and can be found even on Wikipedia.

I will show a representation formula which is valid for $$C^n(\mathbb R)$$ functions, not just polynomials. The same can be found also here.

The key point is that finite differences and derivatives commute: $$D\Delta_h=\Delta_hD$$.

For $$f\in C^1(\mathbb R)$$ you can compute $$\frac1h\Delta_h[f](x) = \frac{f(x+h)-f(x)}h = \frac1h \int_0^h D[f](x+x_1) \,dx_1$$

For $$f\in C^n(\mathbb R)$$, iterating the above formula, you get $$\begin{split} \frac1{h^n}\Delta_h^n[f](x) &= \frac{\frac1{h^{n-1}}\Delta_h^{n-1}[f](x+h) - \frac1{h^{n-1}}\Delta_h^{n-1}[f](x)} {h} \\ &= \frac1h \int_0^h \frac1{h^{n-1}}D\Delta_h^{n-1}[f](x+x_1) \,dx_1 \\ &= \frac1{h^2} \int_0^h \int_0^h \frac1{h^{n-1}} D^2\Delta_h^{n-2}[f](x+x_1+x_2) \,dx_1dx_2 \\ &= \,\cdots \\ &= \frac1{h^n} \int_0^h\dotsi\int_0^h D^n[f](x+x_1+\dotsb+x_n) \,dx_1\dotsm dx_n . \end{split}$$

The last integral is a weighted average of $$D^n[f]$$ over the segment $$[x,x+nh]$$. More precisely, let $$X_1,\dotsc,X_n\sim\mathrm{Uniform}(0,h)$$ be i.i.d. and $$S=X_1+\dotsb+X_n$$. Then the previous expression can be viewed as $$\frac1{h^n}\Delta_h^n[f](x) = \mathbb{E}\left[D^n[f](S)\right]$$

The mean value theorem for integrals applies and tells us that there exists $$x^*\in(x,x+nh)$$ such that $$\frac1{h^n}\Delta_h^n[f](x) = D^n[f](x^*).$$

Regarding your proposition, it follows trivially from the fact that if $$P$$ is a polynomial of degree $$n$$, then $$D^{n+1}[P]=0$$ identically, hence also $$\Delta_h^{n+1}[P]=0$$.

I can recommend reading at least Finite difference, Indefinite sum and Newton polynomial to learn about related material.