Laguerre Polynomials

Question

Laguerre polynomials are nontrivial solutions to Laguerre's equation, a second-order linear differential equation: \$xy''+(1-x)y'+ny=0\$. For a given value of \$n\$, the solution, \$y\$, is named \$L_n(x)\$. To avoid trivial solutions, the polynomials are non-constant except for \$n=0\$.

The polynomials can be found without calculus using recursion:

\$L_0(x)=1\$

\$L_1(x)=1-x\$

\$L_{k+1}(x)=\frac{(2k+1-x)L_k(x)-kL_{k-1}(x)}{k+1}\$

Summation can be used to the same end:

\$L_n(x)=\sum\limits_{k=0}^{n}{n\choose k}\frac{(-1)^k}{k!}x^k\$

\$L_n(x)=\sum\limits_{i=0}^n\prod\limits_{k=1}^i\frac{-(n-k+1)x}{k^2}\$

The first Laguerre polynomials are as follows:

Coefficients can be found here.

The Challenge

Given a nonnegative integer \$n\$ and a real number \$x\$, find \$L_n(x)\$.

Rules

• This is code-golf so the shortest answer in bytes wins.

• Assume only valid input will be given.

• Error should be under one ten-thousandth (±0.0001) for the test cases.

Test Cases

Here, \$n\$ is the first number and \$x\$ is the second.

In: 1 2
Out: -1

In: 3 1.416
Out: -0.71360922

In: 4 8.6
Out: −7.63726667

In: 6 -2.1
Out: 91.86123261

Answer 1

Python 2, 53 bytes

f=lambda n,x:n<1or((2*n-1-x)*f(n-1,x)-~-n*f(n-2,x))/n

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Answer 2

Wolfram Language (Mathematica), 9 bytes

LaguerreL

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Answer 3

Jelly,  11  10 bytes

ŻṚṀc÷!ƲḅN}

A dyadic Link accepting \$n\$ on the left and \$x\$ on the right which yields \$L_n(x)\$.

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How?

This makes the observation that
\$L_n(x)=\sum\limits_{k=0}^{n}{n\choose k}\frac{(-1)^k}{k!}x^k=\sum\limits_{k=0}^{n}{(-x)^k}\frac{n\choose k}{k!}\$
which is the evaluation of a base \$-x\$ number with n+1 digits of the form \$\frac{n\choose k}{k!}\$.

ŻṚṀc÷!ƲḅN} - Link: n, x
Ż          - zero-range (n) -> [0, 1, 2, ..., n]
 Ṛ         - reverse -> [n, ..., 2, 1, 0]
      Ʋ    - last four links as a monad - f(I=that):
  Ṁ        -   maximum -> n
   c       -   {that} binomial (I) -> [nCn, ..., nC2, nC1, nC0]
     !     -   {I} factorial -> [n!, ..., 2!, 1!, 0!]
    ÷      -   division -> [nCn÷n!, ..., nC2÷2!, nC0÷0!]
        N} - negate right argument -> -x
       ḅ   - convert from base (-x) -> -xⁿnCn÷n!+...+-x²nC2÷2!+-x¹nC1÷1!+-x°nC0÷0!

Answer 4

MATL, 5 bytes

_1iZh

Inputs are \$n\$, then \$x\$. Try it online! Or verify all test cases.

How it works

This uses the equivalence of Laguerre polynomials and the (confluent) hypergeometric function:

\$ L_n(x) = {} _1F_1(-n,1,x) \$

_    % Implicit input: n. Negate
1    % Push 1
i    % Input: x
Zh   % Hypergeometric function. Implicit output

Answer 5

JavaScript (ES6),  48 42  41 bytes

Expects (x)(n). May output true instead of 1.

x=>g=k=>k<1||((x-k---k)*g(k)+k*g(k-1))/~k

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Answer 6

Python 3.8 (pre-release), 66 bytes

L=lambda n,x:((2*n-1-x)*L(d:=n-1,x)-d*L(n-2,x))/n if n>1else 1-n*x

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Direct implementation of the recursive algorithm, with one interesting part: L(1,x) and L(0,x) can be combined as L(n,x)=1-n*x.

Could save 2 bytes using L=lambda n,x:n>1and((2*n-1-x)*L(d:=n-1,x)-d*L(n-2,x))/n or 1-n*x, but L(n) is not necessarily zero.

Answer 7

APL (Dyalog Unicode), 16 bytes

1⊥⍨0,⎕×(-÷⌽×⌽)⍳⎕

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A full program that takes n and x from two separate lines of stdin.

How it works

1⊥⍨0,⎕×(-÷⌽×⌽)⍳⎕
              ⍳⎕  ⍝ Take n and generate 1..n
       (-÷⌽×⌽)    ⍝ Compute i÷(n+1-i)^2 for i←1..n
   0,⎕×           ⍝ Multiply x to each and prepend 0, call it B
1⊥⍨               ⍝ Convert all ones from base B to single number

The mixed base conversion looks like this:

1..n:                ... n-3          n-2          n-1          1
B:            0      ... (n-3)x/4^2   (n-2)x/3^2   (n-1)x/2^2   nx
digits:       1      ... 1            1            1            1
digit values: x^n/n! ... (nC3 x^3/3!) (nC2 x^2/2!) (nC1 x^1/1!) (nC0 x^0/0!)

It is essentially a fancy way to write the sum of product scan over 1, nx, (n-1)x/2^2, (n-2)x/3^2, .... This happens to be shorter than a more straightforward -x-base conversion (evaluating a polynomial at -x):

APL (Dyalog Unicode), 18 bytes

(-⎕)⊥⌽1,(!÷⍨⊢!≢)⍳⎕

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Answer 8

Python 3.8 (pre-release), 61 bytes

L=lambda k,x:k<1or[1-x,L(w:=k-1,x)*(k+w-x)-L(k-2,x)*w][k>1]/k

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Answer 9

JavaScript (Node.js), 36 bytes

x=>(i=0,g=n=>n?1-x*n/++i/i*g(n-1):1)

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Just convert the formula to this, and use recursive:

$$ L_n(x) = \sum_{i=0}^n\prod_{k=1}^i\frac{-(n-k+1)x}{k^2} $$

Answer 10

J, ³⁷ 20 bytes

-5 thanks to @Bubbler

Calculates the polynomial adapted from the summation formula and uses J's p. operator to calculate that polynomial with a given x.

(p.-)~i.((!]/)%!)@,]

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J, 45 byte

Alternative Recursive function.

1:`-@.[~ ::((>:@]%~($:*[-~1+2*])-]*($:<:))<:)

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How it works

We define a hook (fg), which is x f (g n). f is (p.-)~ so it will be evaluated as ((i.((!]/)%!)@,]) n) p. (- x).

(p.-)~i.((!]/)%!)@,]
      i.         @,] enumerate 3 -> 0 1 2, append 3 -> 0 1 2 3, …
         (!]/)       3 over i
              %      divided by
               !     !i
   -                 negate x
 p.                  apply -x to the polynomial expressed in J as
                     1 3 1.5 0.166667, so 1-3(-x)+1.5(-x)^2+0.16(-x)^3

Answer 11

Pari/GP, 39 bytes

Using the formula \$L_n(x)=\sum_{k=0}^n \binom{n}{k}\frac{(-1)^k}{k!} x^k\$.

l(n,x)=sum(k=0,n,n!*(-x)^k/(n-k)!/k!^2)

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---

Pari/GP, 45 bytes

Using the generating function \$\sum_{n=0}^\infty x^n L_n(t)= \frac{1}{1-x} e^{-xt/(1-x)}\$.

l(n,t)=Vec(exp(-x*t/(1-x)+O(x^n++))/(1-x))[n]

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Answer 12

Charcoal, 29 bytes

⊞υ¹ＦＮ⊞υ×⌈υＬυＩ↨Ｅυ∕⌈υ×ιＸ§⮌υκ²±Ｎ

Try it online! Link is to verbose version of code. Uses a slightly modified version of the summation given in the question. Explanation:

⊞υ¹ＦＮ⊞υ×⌈υＬυ

Calculate the factorials from \$0!\$ to \$n!\$.

Ｉ↨Ｅυ∕⌈υ×ιＸ§⮌υκ²±Ｎ

For each index \$i\$ from \$0\$ to \$n\$ calculate \$\frac{n!}{i!(n-i)!^2}\$ and then perform base conversion from base \$-x\$ which multiplies each term by \$(-1)^{n-i}x^{n-i}\$ and takes the sum.

If we set \$k=n-i\$ we see that we calculate \$\sum\limits_{k=0}^{n}{\frac{n!(-1)^k}{(n-k)!k!^2}x^k}=\sum\limits_{k=0}^{n}{n\choose k}\frac{(-1)^k}{k!}x^k\$ as required.

Answer 13

Japt -x, ^{28 27} 26 bytes

ò@l *VpX /Xl ²*JpX /(U-X l

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Japt, ^{30 29} 28 bytes

ò x@l *VpX /Xl ²*JpX /(U-X l

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Explanation

ò x@l *VpX /Xl ²*JpX /(U-X l
ò                               // Create a array [0, 1, ..., U]
  x                             // sum the array after mapping through
   @                            // Function(X)
    l                           //    U!
      *VpX                      //    times V ** X
           /Xl ²                //    divided by X! ** 2
                *JpX            //    times (-1) ** X
                     /(U-X l    //    divided by (U - X)!

• U is the first input
• V is the second input
• ** represents exponentiation
• ! represents factorial

Answer 14

C (gcc), 91 bytes

i;k;float f(n,x)float x;{float p,s=0;for(i=++n;k=i--;s+=p)for(p=1;--k;)p*=(k-n)*x/k/k;x=s;}

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Straighforward implementation of polynomial expansion. Slightly golfed less

i;k;
float f(n,x)float x;{
  float p,s=0;
  for(i=++n;k=i--;s+=p)
    for(p=1;--k;)
      p*=(k-n)*x/k/k;
  x=s;
}

Answer 15

Fortran (GFortran), 69 68 bytes

read*,n,a
print*,sum([(product([((j-n-1)*a/j/j,j=1,i)]),i=0,n)])
end

-1 byte thanks to @ceilingcat

The program reads in an implicit integer n and real a. Summation and product operations are performed using arrays (initialized using implicit loops) with the intrinsics sum() and product().

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Answer 16

05AB1E, 16 11 bytes

DÝR©c®!/I(β

Port of @JonathanAllan's Jelly answer, so make sure to upvote him as well!

Try it online or verify all test cases.

Original 16 bytes answer:

1λèN·<I-₁*N<₂*-N/

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Explanation:

D                 # Duplicate the first (implicit) input-integer `n`
 Ý                # Pop one, and push a list in the range [0,n]
  R               # Reverse it to range [n,0]
   ©              # Store this list in variable `®` (without popping)
    c             # Get the binomial coefficient of `n` and each value in this list
     ®!           # Push list `®` again, and get the factorial of each
       /          # Divide the values at the same positions of the two lists
        I(        # Push the second input `x`, and negate it
          β       # Convert the list from base-(-x) to a single decimal value
                  # (which is output implicitly as result)

---

Uses the recursive formula:

$$a(n)=\frac{a(n-1)\times(2n-1-x)-(n-1)\times a(n-2)}{n}$$

 λ                # Create a recursive environment
  è               # to get the 0-based n'th value afterwards
                  # (where `n` is the first implicit input)
                  # (which will be output implicitly as result at the end)
1                 # Starting with a(-1)=0 and a(0)=1,
                  # and for every other a(N), we'll:
                  #  (implicitly push a(N-1))
   N·             #  Push `N` doubled
     <            #  Decrease it by 1
      I-          #  Decrease it by the second input `x`
        *         #  Multiply it by the implicit a(N-1)
         N<       #  Push `N`-1
           ₂*     #  Multiply it by a(N-2)
             -    #  Decrease the a(N-1)*(2N-1-x) by this (N-1)*a(N-2)
              N/  #  And divide it by `N`: (a(N-1)*(2N-1-x)-(N-1)*a(N-2))/N

Answer 17

Scala, 81 bytes

Use the recursive formula of Laguerre Polynomials.

---

Golfed version. Try it online!

def f(n:Int,x:Double):Double={if(n<1)1 else((2*n-1-x)*f(n-1,x)-(n-1)*f(n-2,x))/n}

Ungolfed version. Try it online!

object Main {
  def f(n: Int, x: Double): Double = {
    if(n < 1) 1
    else ((2*n - 1 - x) * f(n - 1, x) - (n - 1) * f(n - 2, x)) / n
  }

  def main(args: Array[String]): Unit = {
    println(f(1, 2.0))
    println(f(3, 1.416))
    println(f(4, 8.6))
    println(f(6, -2.1))
  }
}

Answer 18

HTML, 1 Byte, Invalidated by newest edit.

0

\$y=0\$ is a solution to the equation stated in the question for all \$x\$ and \$n\$.

Answer 19

Jelly, 0 Bytes, invalidated by newest edit

Try it online! Note that \$y=0\$ is a solution for all \$x\$ and \$n\$. Port of my HTML answer.

Thanks to Jonathan Allen for pointing out this port, and also that this might be a polyglot.