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The most common way to represent a polynomial is writing it as a linear combination of monomials, i.e., powers of the variable. For example, the polynomial \$p(x) = x^3 + 2x^2 + x + 1\$ is a linear combination of the monomials \$x^3\$, \$x^2\$, \$x\$, and \$1\$, with coefficients \$1\$, \$2\$, \$1\$, and \$1\$ respectively.

However, the monomial basis is not the only possible basis for polynomials. Another interesting basis, useful in combinatorics and some other areas of mathematics, is the falling factorial basis. The falling factorial of degree \$k\$ is defined as:

$$(x)_k = x (x - 1) (x - 2) \ldots (x - k + 1) = \prod_{i=0}^{k-1} (x - i).$$

In particular, \$(x)_0 = 1\$.

Any polynomial \$p(x)\$ of degree \$n\$ can be uniquely expressed as a linear combination of falling factorials \$(x)_0, (x)_1, \ldots, (x)_n\$. For example, the polynomial \$p(x) = x^3 + x^2 + x + 1\$ can be expressed in the falling factorial basis as:

$$p(x) = (x)_3 + 4 (x)_2 + 3 (x)_1 + (x)_0 = x (x - 1) (x - 2) + 4 x (x - 1) + 3 x + 1.$$

A possible way to find the coefficients in the falling factorial basis is using the Stirling numbers of the second kind. In fact,

$$x^n = \sum_{k=0}^{n} S(n, k) (x)_k,$$

where \$S(n, k)\$ is the Stirling number of the second kind, which counts the number of ways to partition a set of \$n\$ items into \$k\$ non-empty subsets.

Task

In this challenge, you are given a polynomial \$p(x)\$, and you need to find its representation in the falling factorial basis. Specifically, you need to find the coefficients \$c_0, c_1, \ldots, c_n\$ such that:

$$p(x) = c_n (x)_n + c_{n-1} (x)_{n-1} + \ldots + c_1 (x)_1 + c_0 (x)_0.$$

You may take the input polynomial \$p(x)\$ in any reasonable format. For example, the polynomial \$x^4 - 4x^3 + 5x^2 - 2x\$ may be represented as:

  • a list of coefficients, in descending order: [1,-4,5,-2,0];
  • a list of coefficients, in ascending order:[0,-2,5,-4,1];
  • a string representation of the polynomial, with a chosen variable, e.g., "x^4-4*x^3+5*x^2-2*x";
  • a built-in polynomial object, e.g., x^4-4*x^3+5*x^2-2*x in PARI/GP.

You may take the degree \$n\$ or the number of coefficients \$n + 1\$ as an additional input.

You may output the list of coefficients \$[c_n, c_{n-1}, \ldots, c_1, c_0]\$ in either ascending or descending order. You may also take an additional input \$k\$, and output the \$k\$-th coefficient \$c_k\$ only.

This is , so the shortest code in bytes wins.

Test cases

Here the input polynomials are represented as lists of coefficients in descending order. The output are also in descending order.

[1] -> [1]                                  # 1 => (x)_0
[1, 1] -> [1, 1]                            # x + 1 => (x)_1 + (x)_0
[1, 0, 0] -> [1, 1, 0]                      # x^2 => (x)_2 + (x)_1
[1, 2, 1] -> [1, 3, 1]                      # x^2 + 2 * x + 1 => (x)_2 + 3 * (x)_1 + (x)_0
[1, 1, 1, 1] -> [1, 4, 3, 1]                # x^3 + x^2 + x + 1 => (x)_3 + 4 * (x)_2 + 3 * (x)_1 + (x)_0
[1, 2, 3, 4] -> [1, 5, 6, 4]                # x^3 + 2 * x^2 + 3 * x + 4 => (x)_3 + 5 * (x)_2 + 6 * (x)_1 + 4 * (x)_0
[1, -1, 1, -1, 1] -> [1, 5, 5, 0, 1]        # x^4 - x^3 + x^2 - x + 1 => (x)_4 + 5 * (x)_3 + 5 * (x)_2 + (x)_0
[1, 0, -2, 0, 1] -> [1, 6, 5, -1, 1]        # x^4 - 2 * x^2 + 1 => (x)_4 + 6 * (x)_3 + 5 * (x)_2 - (x)_1 + (x)_0
[5, 4, 3, 2, 1] -> [5, 34, 50, 14, 1]       # 5 * x^4 + 4 * x^3 + 3 * x^2 + 2 * x + 1 => 5 * (x)_4 + 34 * (x)_3 + 50 * (x)_2 + 14 * (x)_1 + (x)_0
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  • 1
    \$\begingroup\$ I think [1, 0, -2, 0, 1] -> [1, 6, 5, -1, 1] \$\endgroup\$ Commented yesterday
  • \$\begingroup\$ @att Thanks. Fixed. \$\endgroup\$ Commented yesterday

10 Answers 10

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Wolfram Language (Mathematica), 28 bytes

\!\(\_{x,#}#2\)/#!/.x->0&

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Input [k, p], where p is a polynomial in terms of x. Outputs \$c_k\$.

is \[DifferenceDelta]. This character has code point U+F4A4; the documentation incorrectly claims that it has code point U+2206. That character, ∆, is actually the similar-looking \[Laplacian] (undocumented).

Returns \$\frac{\Delta^k p(0)}{k!}\$, where \$\Delta^k p\$ is the \$k\$th "discrete derivative" of \$p\$ defined by \begin{align*} \Delta^0 p(x)&=p(x),\\ \Delta^{n+1}p(x)&=\Delta^n p(x+1)-\Delta^n p(x). \end{align*}

(This is analogous to obtaining the coefficients of the usual power series representation via differentiation, where the coefficient of the \$x^k\$ term is \$\frac{D^k p(0)}{k!}\$.)

Wolfram Language (Mathematica), 27 bytes

0~Range~#~StirlingS2~#2.#3&

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Input [n, k, p], where n is the degree of the polynomial and p is an ascending list of coefficients . Outputs \$c_k\$.

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4
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R, 126 114 107 bytes

\(v,`^`=\(n,k)`if`(n<1,k<1,k*(n-1)^k+(n-1)^(k-1))*(k>=0))lapply(l<-(seq(v)-1),\(k)sum(v*sapply(l,\(n)n^k)))

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Jelly, 26 bytes

cŻ$Ø-ṁ×Ɗׯ*¥SA:!{
J’çþ`æ×@

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This is embarassingly long.

Takes and outputs the coefficients in little-endian order.

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Jelly, 14 bytes

ḅJ;@ṪW;JI÷¥\Ḣ€

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Input a list of coefficients in descending order. Output a list of coefficients in ascending order.

Port of my Mathematica solution, since I was sad no one tried that approach.

I don't know Jelly. I spent half an hour clicking around the language spec trying to figure out how to do things.

ḅ               evaluate p at
 J                1..n+1
  ;@Ṫ           prepend the constant term (= p(0))
     W;J   \    for i=1..n, keeping intermediate values,
        I÷¥       take successive differences, divide by i
            Ḣ€  first element of each
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3
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Python3, 498 bytes

from itertools import*
def R(a):
 d={}
 for i in a:
  for j in i:d[j]=d.get(j,0)+i[j]
 return d
def x(k):
 if k==0:return{0:1}
 q=[{1:1,0:-i}for i in range(k)]
 while len(q)>1:
  a,b,*q=q;d={}
  for j in a:
   for J in b:d[j+J]=d.get(j+J,0)+a[j]*b[J]
  q=[d]+q
 return q[0]
def f(p):
 P=[x(i)for i in range(len(p))]
 for E in count():
  for c in product(*[range(E)for _ in p]):
   u=R([{i:C*B[i]for i in B}for C,B in zip(c,P)])
   if all(u.get(k,0)==p.get(k,0)for k in range(max(*u,*p)+1)):return c

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1
  • \$\begingroup\$ 454 bytes \$\endgroup\$ Commented 2 hours ago
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Maple, 47 bytes

(p,k)->add(p[i]*Stirling2(i-1,k),i=1..nops(p));

Takes a list p in ascending order and non-negative integer k, and outputs the kth coefficient, using @Arnauld's formula.

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3
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Charcoal, 26 bytes

ＦＬθ⊞υＥθ⎇ι↨…§υ⊖ιλι¬λＩＥυΣ×θι

Attempt This Online! Link is to verbose version of code. I/O is a list of coefficients in ascending order. Explanation: Uses @Arnauld's formula along with dynamic programming to calculate the Stirling numbers of the second kind.

ＦＬθ

Loop for \$ k \$ from \$ 0 \$ to \$ n \$.

⊞υＥθ⎇ι↨…§υ⊖ιλι¬λ

Calculate \$ S(n, k) = S(n - 1, k - 1) + k S(n - 2, k - 1) + k^2 S(n - 3, k - 1) + \dots \$ as this is a more convenient way to generate a transposed array S[k][n].

ＩＥυΣ×θι

Output the dot product of each row of S with the input.

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3
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JavaScript (ES6), 77 bytes

Expects an array of coefficients in ascending order. Returns an array in the same format.

a=>a.map((_,k)=>a.reduce((t,v,n)=>t+v*(s=n=>n?s(--n,k--)+ ++k*s(n):!k)(n),0))

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Method

This is based on the formula suggested in the challenge.

Given the coefficients \$a_0\$, \$a_1\$, ... \$a_N\$, we compute:

$$b_k=\sum_{n=k}^N a_nS(n,k)=\sum_{n=0}^N a_nS(n,k)$$

where \$S(n,k)\$ is the Stirling number of the second kind with the usual convention \$S(n,k)=0\$ for \$n<k\$.

Commented

a =>                 // a[] = coefficients in ascending order
a.map((_, k) =>      // for each coefficient at index k in a[]:
  a.reduce(          //   for each coefficient v at index n in a[],
    (t, v, n) =>     //   with t as the accumulator:
    t +              //     add to t ...
    v * (            //     ... v multiplied by S(n, k)
                     //     actually implemented as s(n) where s is ...
      s = n =>       //     ... a recursive function taking only n explicitly:
      n ?            //       if n is not 0:
        s(--n, k--)  //         compute S(n - 1, k - 1)
        + ++k * s(n) //         + k * S(n - 1, k)
      :              //       else:
        !k           //         return 1 if k = 0, or 0 otherwise
    )(n),            //     initial call to s
    0                //     start with t = 0
  )                  //   end of reduce()
)                    // end of map()
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3
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Jelly, 16 bytes

Ḣ;J€×J+ŻƲ{\×"⁸SƲ

A monadic Link that accepts a list of integers, the monomial coefficients in ascending order and yields the falling factorial coefficients in ascending order.

Try it online! Or see the test-suite (showing I/O in descending order).

Note: a constant polynomial (an input of length one) yields an additional, redundant zero coefficient at \$x_1\$, I'm guessing that's acceptable, if not Ḣ;ȧJ€×J+ŻƲ{\×"⁸SƲ$ would work for \$18\$ bytes.

How?

Effectively employs the recurrence relation of Stirling numbers of the second kind, \$S(n, k)\$:

$$S(n+1, k) = k S(n, k) + S(n, k-1) \space \vert \space 0 \lt k \lt n$$ $$S(n, 0) = S(0, k) = 0$$ $$S(n, n) = 1$$

Ḣ;J€×J+ŻƲ{\×"⁸SƲ - Link: list of integers, C
Ḣ                - behead C -> removes the constant term, N, from C
               Ʋ - last four links as a monad - f(Rest)
  J€             -   range of length of each -> [[1], [1], ..., [1]]
          \      -   cumulative reduce by:
        Ʋ{       -     last four links as a monad - f(Current (initially [1])):
     J           -       range of length -> [1..length(Current)]
    ×            -       {Current} multiplied by {that} (vectorises) -> M
       Ż         -       prefix {Current} with a zero
      +          -       {M} add {[0]+Current} (vectorises) -> next Current
           ×"⁸   -   {that} zipwise multiplied by {Rest}
              S  -   column-wise sum
 ;               - {N} concatenate {that}
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Nekomata, 9 bytes

xbᶦ∆haxF/

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A port of @att's Jelly answer, which in turn is based on @att's Mathematica answer. So make sure to upvote those answers as well!

Takes input as a list of coefficients in descending order. Outputs a list of coefficients in ascending order.

xbᶦ∆haxF/     Takes [5,4,3,2,1] as an example
x           Enumerate
              [5,4,3,2,1] -> [5,4,3,2,1], [0,1,2,3,4]
 b          Convert to base, automatically vectorized
              [5,4,3,2,1], [0,1,2,3,4] -> [1,15,129,547,1593]
              1 is [5,4,3,2,1] to base 0, 15 is to base 1, etc.
  ᶦ∆        Take differences zero or more times until failure
              [1,15,129,547,1593] -> [14,114,418,1046]
              [14,114,418,1046] -> [100,304,628]
              [100,304,628] -> [204,324]
              [204,324] -> [120]
              [120] -> []
    h       Head
              [1,15,129,547,1593] -> 1
              [14,114,418,1046] -> 14
              [100,304,628] -> 100
              [204,324] -> 204
              [120] -> 120
     a      All possible results
              [1,14,100,204,120]
      x     Enumerate
              [1,14,100,204,120] -> [1,14,100,204,120], [0,1,2,3,4]
       F    Factorial
              [0,1,2,3,4] -> [1,1,2,6,24]
        /   Divide
              [1,14,100,204,120], [1,1,2,6,24] -> [1,14,50,34,5]
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