Repeatedly taking differences on a polynomial yields the factorial of its degree?

Consider a function such that it takes in polynomial function and creates an array of its outputs and then using that array creates another new array by calculating the absolute difference between the first $2$ values and keeps doing this until it reaches an array full of zeros.

This is much easier to show you by example.

For example take $F(x)= x^2$, the first array would be

$1,4,9,16,25,36,49,64,81$ and so on, the second would be

$3,5,7,9,11,13,15,17,19$ ( the difference between the first value and the second one)

but the third one is where it gets interesting as if we continue the pattern we would get an array filled with only $2$'s and after that it would only be zeros.

Lets do another example, $F(x)=x^3$

$1,8,27,64,125,216,343$

$7,19,37,61,91,\dotsc$ but here is the interesting part if we continue this

$12,18,24,30,\dotsc$ and once more then we get

$6,6,6,6,6,\dotsc$ after that it would just be an array of zeros

There are $2$ main observation that I made about this

Firstly, the value that is begin repeated indefinitely is equals to the factorial of the functions power. Meaning that for $F(x)=x^2$ the value being repeated is $2!$. For $F(x)=x^3$ , it's $3!$ and this is true for all polynomials (I tried it up to $x^7$, after that it got too messy)

Secondly, the value that is repeated always occurs on the $n$th iteration of the function. Meaning that for $F(x)=x^2$, we have to go through the processes $2$ times before we find the value. For $F(x)=x^3$, we have to go through it $3$ times before getting the value.

Is there any way to prove this and does this mean anything at all?

• Lookup finite differences ("an $n^{th}$ power has a constant $n^{th}$ finite difference") . Also for example this answer. – dxiv Jun 12 '17 at 3:25
• Good. You discovered a concept (through on a sequence basis) very similar to derivative. An excellent mind of discovering and inducting. Continue it if you are still young and have enough time. – BAI Jun 12 '17 at 4:45
• Welcome to Math.SE! Since one aim of the site is to collect questions and answers in forms useful for posterity, would you please consider re-titling your question to indicate its content? (Perhaps something like "Repeated differences for polynomial functions at equally-spaced inputs"...?) – Andrew D. Hwang Jun 12 '17 at 11:18
• – Simply Beautiful Art Jun 12 '17 at 17:21
• @user21820 your edit to the question is a great example of what a good title can do for a better understanding of a given topic and to promote the curiosity of the reader. – iadvd Jun 13 '17 at 0:12

Here's a fact:

• If $p(x)$ is a polynomial of degree $n$ with leading term $ax^n$ then $p(x+1)-p(x)$ is a polyomial of degree $n-1$ with leading term $a \, n \, x^{n-1}$.

(I'll prove this fact below).

Applying this fact together with an induction argument, it follows that after repeating the process $n$ times, one obtains a polynomial of degree zero whose leading term is $$a \, n \, (n-1) \ldots (2) (1) = a \, n!$$ which is just a constant having that value.

So if the original leading coefficient $a$ is equal to $1$, as it is in the specific cases $F(x)=x^n$ that you ask about, repeating the difference process $n$ times yields a constant sequence of $n!$ as you ask.

Here's a proof of the fact by applying induction (the base case $n=1$ is easy).

Assuming the induction hypothesis for polynomials of degree $\le n-1$, suppose that $$p(x) = a \, x^n + q(x)$$ where $q(x)$ is a polynomial of degree $\le n-1$.

We have $$p(x+1)-p(x) = a \, (x+1)^n - a \, x^n + \underbrace{(q(x+1)-q(x))}_{r(x)}$$ and $r(x)$ is a polynomial of degree $\le n-2$ by induction. Thus $$p(x+1)-p(x) = a \, (x^n + n \, x^{n-1} + s(x)) - a\, x^n + r(x)$$ where $s(x)$ is also a polynomial of degree $\le n-2$ (by application of the binomial theorem). Therefore $$p(x+1)-p(x) = a \, n \, x^{n-1} + (a \, s(x)+r(x))$$ which is a polynomial of degree $n-1$ with leading term as required.

What you have discovered/invented is known as the forward difference operator $D$ defined as: $\def\nn{\mathbb{N}} \def\zz{\mathbb{Z}} \def\lfrac#1#2{{\large\frac{#1}{#2}}} \def\lbinom#1#2{{\large\binom{#1}{#2}}}$

$D = ( \text{function$f$on$\zz$} \mapsto ( \text{int$n$} \mapsto f(n+1) - f(n) ) )$

Namely for any function $f$ on $\zz$ and $n \in \zz$, $D(f)(n) = f(n+1) - f(n)$.

If you think of the functions as sequences (infinite in both directions), then taking the forward difference means replacing each term with the value of the next term minus itself. What you did is essentially to repeatedly take the forward difference of the sequence of cubes:

...,-27,-8,-1, 0, 1, 8,27,...
..., 19, 7, 1, 1, 7,19,37,...
...,-12,-6, 0, 6,12,18,24,...
...,  6, 6, 6, 6, 6, 6, 6,...
...,  0, 0, 0, 0, 0, 0, 0,...
...,  0, 0, 0, 0, 0, 0, 0,...


This powerful abstraction makes it easy to get a lot of things. For instance, the numbers obtained here can be easily used to obtain the general formula for sum of cubes!

General method for indefinite summation

The key is that:

$D\left( \text{int$n$} \mapsto \lbinom{n}{k+1} \right) = \left( \text{int$n$} \mapsto \lbinom{n}{k} \right)$ for any $k \in \zz$.

This is to be expected because it follows directly from Pascal's triangle, especially if we define $\lbinom{n}{k}$ using the triangle.

This means that if we have any function $f$ on $\zz$ such that $f(n) = \sum_{k=0}^\infty a_k \lbinom{n}{k}$ for any $n \in \zz$, then we get the Newton series:

$D(f)(n) = \sum_{k=0}^\infty a_{k+1} \lbinom{n}{k}$ for any $n \in \zz$.

From a high-level perspective, this is the discrete version of the Taylor series, and indeed for such a function we easily see that $f(n) = \sum_{k=0}^\infty D^k(f)(0) \lbinom{n}{k}$ for any $n \in \zz$, because $\binom{0}{0} = 1$ while $\lbinom{0}{k} = 0$ for any $k \in \nn^+$.

This works for any polynomial function $f$ on $\zz$, since $D^k(f)$ is the zero function once $k$ is larger than the degree of $f$, so we can use it to immediately find the series for $(\text{int n} \mapsto n^3)$, and then just take the anti-difference by shifting the coefficients of the series the other way. The undetermined constant that appears will drop out once we perform a definite sum like if we want the sum of the first $m$ cubes.

Note also that $D^k(f)$ is the constant function with value $k!$ if $f(n) = n^k$ for every $n$. Lee Mosher has already explained this particular fact by directly proving it, but another way to see it is that the highest order term in its Newton series is $k! \lbinom{n}{k}$, because $\lbinom{n}{k}$ is the only term that can contribute the $k$-th power of $n$. Since $D$ merely shifts the coefficients, $D^k(f) = \left( \text{int$n$} \mapsto k! \lbinom{n}{0} \right)$ and we are done.

Sum of $p$ powers

For example if we want $\sum_{k=1}^{n-1} k^3$ we first find the iterated forward differences of the sequence of cubes $( n^3 )_{n \in \zz}$:

..., 0, 1, 8,27,64,...
..., 1, 7,19,37,...
..., 6,12,18,...
..., 6, 6,...
..., 0,...


So we immediately get $n^3 = 0 \binom{n}{0} + 1 \binom{n}{1} + 6 \binom{n}{2} + 6 \binom{n}{3}$ and hence $\sum_{k=0}^{n-1} k^3 = 0 \binom{n}{1} + 1 \binom{n}{2} + 6 \binom{n}{3} + 6 \binom{n}{4} = \lfrac{n(n-1)}{2} \Big( 1 + \lfrac{6(n-2)}{3} \big( 1 + \lfrac{n-3}{4} \big) \Big) = \Big( \lfrac{n(n-1)}{2} \Big)^2$.

Computation efficiency

This is far more efficient than the usual method (namely by taking summation on both sides of $(n+1)^3-n^3 = 3n^2+3n+1$ and telescoping) because the series using binomial coefficients is easy to manipulate and easy to compute. For sum of $p$-powers we only need $O(p^2)$ arithmetic operations to find the forward-differences and then $O(p^2)$ more to simplify the series form into a standard polynomial form. In contrast, the other method requires $O(p^3)$ arithmetic operations.

Indefinite summation of non-polynomials

Also, for a wide class of non-polynomial functions, we can still compute the indefinite sum without using the series, by using the discrete analogue to integration by parts, here called summation by parts.

To derive it, simply check that $D(f \times g)(n) = f(n+1) g(n+1) - f(n) g(n) = f(n+1) D(g)(n) - D(f)(n) g(n)$ and so we get the product rule:

$D(f \times g) = R(f) \times D(g) + D(f) \times g$

where $R$ is the right-shift operator defined as:

$R = ( \text{function$f$on$\zz$} \mapsto ( \text{int$n$} \mapsto f(n+1) ) )$

Namely for any function $f$ on $\zz$ and $n \in Z$, $R(f)(n) = f(n+1)$.

For convenience we also define the summation operator:

$S = ( \text{function$f$on$\zz$} \mapsto ( \text{int$n$} \mapsto \sum_{k=0}^{n-1} f(k) ) )$

Then we have the important property that $DS(f) = f$ for any function $f$ on $\zz$, analogous to the fundamental theorem of calculus.

Now by substituting $f$ with $S(f)$ into the product rule and taking summation on both sides we get summation by parts:

$S( f \times g ) = S(f) \times g - S( R(S(f)) \times D(g) ) + c$ for some constant function $c$ on $\zz$.

Example indefinite sum

Using this we can easily compute things like $\sum_{k=1}^n k^3 3^k$ by applying it three times, each time reducing the degree of the polynomial part. There are other ways to achieve this using differentiation, but this method is purely discrete.