Многочлены Чебышёва (Chebyshev Polynomials)

Question

Chebyshev Polynomials are a family of orthogonal polynomials that pop up in all kinds of places in math, and they have a lot of quite interesting properties. One characterization of them is that they are the unique polynomials that satisfy Tn(cos(x)) = cos(n*x).

Challenge

Given an nonnegative integer n, you should output the n-th Chebyshev Polynomial. Tn(x).

Definition

The n-th Chebyshev Polynomial is given by following three term recursion:

T0(x) = 1
T1(x) = x
Tn+1(x) = 2*x*Tn(x) - Tn-1(x)

Details

If your language has a native polynomial type, you can use that one as an output, otherwise you should output a list of coefficients in ascending- or descending order, or as a string representing a polynomial.

Examples

T0(x) = 1
T1(x) = x 
T2(x) = 2x^2 - 1
T3(x) = 4x^3 - 3 x
T4(x) = 8x^4 - 8x^2 + 1
T5(x) = 16x^5 - 20x^3 + 5x
T10(x) = 512x^10 - 1280x^8 + 1120x^6 - 400x^4 + 50x^2 - 1

In the descending degree list format we'd get T3(x) = [4,0,-3,0] and in the ascending degree format we'd get T3(x) = [0,-3,0,4]

Answer 1

Mathematica, 15 bytes

#~ChebyshevT~x&

Of course, Mathematica has a builtin.

If an alternative input form is allowed (10 bytes):

ChebyshevT

takes an integer n and a variable.

Answer 2

Octave, 39 bytes

@(n)round(2^n/2*poly(cos((.5:n)/n*pi)))

Try it online!

Explanation

cos((.5:n)/n*pi) builds a vector with the roots of the polynomial, given by

poly gives the monic polynomial with those roots. Multiplying by 2^n/2 scales the coefficients as required. round makes sure that results are integer in spite of numerical precision.

Answer 3

Pari/GP, 12 bytes

Yes, a builtin. Shorter than Mathematica.

polchebyshev

Try it online!

---

Without builtin:

Pari/GP, 34 bytes

f(n)=if(n<2,x^n,2*x*f(n-1)-f(n-2))

Try it online!

Answer 4

Haskell, 62 bytes

t n|n<2=1:[0|n>0]|x<-(*2)<$>t(n-1)++[0]=zipWith(-)x$0:0:t(n-2)

Try it online!

flawr saved a byte.

Answer 5

CJam (21 bytes)

1a2,qi{0X$+2f*@.-}*;`

This is a full program: the equivalent as an anonymous block is the same length:

{1a2,@{0X$+2f*@.-}*;}

Online demo

Answer 6

Python + SymPy, 66 bytes

from sympy.abc import*
T=lambda n:n<2and x**n or 2*x*T(n-1)-T(n-2)

Try it online!

With descending coefficients representation

from sympy import*
x=symbols('x')
T=lambda n:n<2and x**n or expand(2*x*T(n-1)-T(n-2))

Try it online!

Answer 7

SageMath, 25 bytes

lambda n:chebyshev_T(n,x)

Try it online

Answer 8

MATL, 17 bytes

lFTi:"0yhEbFFh-]x

Coefficients are output in increasing order of degree.

Try it online! Or verify all test cases.

Explanation

For input n, the code applies the recursive relation n times. The two most recent polynomials are always kept on the stack. When a new polynomial is computed, the oldest is removed.

At the end, the second-last polynomial is displayed (the last polynomial is deleted), as we have done one too many iterations.

l        % Push 1
FT       % Push [0 1]. These are the first two polynomials
i:"      % Input n. Do the following n times
  0      %   Push 0
  y      %   Duplicate most recent polynomial
  h      %   Concatenate: prepends 0 to that polynomial
  E      %   Multiply coefficients by 2
  b      %   Bubble up. This moves second-most recent polynomial to top
  FF     %   Push [0 0]
  h      %   Concatenate: appends [0 0] to that polynomial
  -      %   Subtract coefficients
]        % End
x        % Delete. Implicitly display

Answer 9

Jelly, 18 bytes

Cr1µ’ßḤ0;_’’$ß$µỊ?

Try it online!

Returns a list of coefficients in ascending order.

There is another solution for 17 bytes with floating-point inaccuracies.

RḤ’÷Ḥ-*ḞÆṛæ«’µ1Ṡ?

Try it online!

Explanation

Cr1µ’ßḤ0;_’’$ß$µỊ?  Input: integer n
                Ị   Insignificant - abs(n) <= 1
                    If true, n = 0 or n = 1
   µ                  Monadic chain
C                       Complement, 1-x
 r1                     Range to 1
                    Else
               µ      Monadic chain
    ’                   Decrement
     ß                  Call itself recursively
      Ḥ                 Double
       0;               Prepend 0
         _              Subtract with
            $             Monadic chain
          ’’                Decrement twice
              $           Monadic chain
             ß              Call itself recursively

Answer 10

Ruby + polynomial, 59 58+13 = 72 71 bytes

Uses the -rpolynomial flag.

f=->n{x=Polynomial.new 0,1;n<2?[1,x][n]:2*x*f[n-1]-f[n-2]}

Answer 11

J, 33 bytes

(0>.<:)2&*1:p.@;9:o._1^+:%~1+2*i.

Try it online!

Assumes that floating-point inaccuracies are acceptable and creates the emoji (0>.<:)

For 41 bytes, there is another solution that avoids floats.

(0&,1:)`(-&2((-,&0 0)~2*0&,)&$:<:)@.(>&1)

Try it online!

Answer 12

Python 2, 79 bytes

f=lambda n:n>1and[2*a-b for a,b in zip([0]+f(n-1),f(n-2)+[0,0])]or range(1-n,2)

Try it online!

Answer 13

Axiom, 40 bytes

f(n,x)==(n<2=>x^n;2*x*f(n-1,x)-f(n-2,x))

results

(9) -> for i in [0,1,2,3,4,5,10] repeat output ["f(y)",i,"=", f(i,y)]
   ["f(y)",0,"=",1]
   ["f(y)",1,"=",y]
                   2
   ["f(y)",2,"=",2y  - 1]
                   3
   ["f(y)",3,"=",4y  - 3y]
                   4     2
   ["f(y)",4,"=",8y  - 8y  + 1]
                    5      3
   ["f(y)",5,"=",16y  - 20y  + 5y]
                      10        8        6       4      2
   ["f(y)",10,"=",512y   - 1280y  + 1120y  - 400y  + 50y  - 1]
                                                               Type: Void

it is possible define one substitution law for formula in Axiom use above f() function for expansion of cos(n*x) where n is one integer

(9) -> o:=rule cos(n*%y)==f(n,cos(%y))
   (9)  cos(%y n) == 'f(n,cos(%y))
                    Type: RewriteRule(Integer,Integer,Expression Integer)
                                                              Time: 0 sec
(10) -> b:=o cos(20*x)
   (10)
                 20                18                16                14
     524288cos(x)   - 2621440cos(x)   + 5570560cos(x)   - 6553600cos(x)
   +
                  12                10               8              6
     4659200cos(x)   - 2050048cos(x)   + 549120cos(x)  - 84480cos(x)
   +
               4            2
     6600cos(x)  - 200cos(x)  + 1
                                                 Type: Expression Integer
                       Time: 0.48 (EV) + 0.02 (OT) + 0.10 (GC) = 0.60 sec

Answer 14

C# (.NET Core), 126 bytes

f=n=>n==0?new[]{1}:n==1?new[]{0,1}:new[]{0}.Concat(f(n-1)).Select((a,i)=>2*a-(i<n-1?f(n-2)[i]:0)).ToArray();

Byte count also includes:

using System.Linq;

Try it online!

The function returns polynomial as an array of coefficients in ascending order (from x^0 to x^n)

Explanation:

f = n =>                          // Create a function taking one parameter (int)
    n == 0 ? new[] { 1 } :        // If it's 0, return collection [1]
    n == 1 ? new[] { 0, 1 } :     // If it's 1, return collection [0,1] (so x + 0)
    new[] { 0 }                   // Else create new collection, starting with 0
        .Concat(f(n - 1))         // Concatenate with f(n-1), effectively multiplying polynomial by x
        .Select((a, i) => 2 * a - (i < n - 1 ? f(n - 2)[i] : 0))
                                  // Multiply everything by 2 and if possible, subtract f(n-2)
        .ToArray();               // Change collection to array so we have a nice short [] operator
                                  // Actually omitting this and using .ElementAt(i) is the same length, but this is my personal preference

Answer 15

JavaScript (ES6), 65 bytes

f=n=>n?n>1?[0,...f(n-1)].map((e,i)=>e+e-(f(n-2)[i]||0)):[0,1]:[1]

Inefficient for large n. Interesting but sadly also inefficient:

n=>[...Array(n+1)].map(g=(m=n,i)=>i<0|i>m?0:m<2?i^m^1:g(m-1,i-1)*2-g(m-2,i))

Very efficient for 68 bytes:

f=(n,a=[1],b=[0,1])=>n?f(n-1,b,[0,...b].map((e,i)=>e+e-(a[i]||0))):a

Returns an array of coefficients in ascending order.