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(2012-02-15)  
A family of commuting  polynomial functions.  T_nT_p =  T_pT_n =  T_np

cos(n) is a polynomialfunction of cos(). The following relation defines a polynomial of degree nknown as theChebyshev polynomial of degree n:

cos (n)   =  T_n(cos )

The trigonometric formula  cos (n+2)  = 2 cos cos (n+1)cos n  translates into a simple recurrence relation which makes Chebyshev polynomialsvery easy to tabulate, namely:

T0(x)	=	1	Tn+2(x)   =  2 x Tn+1(x)    Tn(x)
T1(x)	=		x
T2(x)	=	-1	+	2 x2
T3(x)	=		-3 x	+	4 x3
T4(x)	=	1		8 x2	+	8 x4
T5(x)	=		5 x		20 x3	+	16 x5
T6(x)	=	-1	+	18 x2		48 x4	+	32 x6
T7(x)	=		-7 x	+	56 x3		112 x5	+	64 x7
T8 (x)	=	1		32 x2	+	160 x4		256 x6	+	128 x8

Knowing only the highest term of  T_n  and its obvious  n  distinct real zeroes, we obtain immediately  T_n  as a product of  n  factors:

If  n > 0,   then     Tn(x)   =   2 n-1		( x cos (k+½)/n )

The case  T_n(0) = (-1)ⁿ  tellssomething nice about a product of cosines.

Inverse formulas :

x0	=	T0
x1	=		T1
2 x2	=	T0	+	T2
4 x3	=		3 T1	+	T3
8 x4	=	3 T0	+	4 T2	+	T4
16 x5	=		10 T1	+	5 T3	+	T5
32 x6	=	10 T0	+	15 T2	+	6 T4	+	T6
64 x7	=		35 T1	+	21 T3	+	7 T5	+	T7
128 x8	=	35 T0	+	56 T2	+	28 T4	+	8 T6	+	T8

(2014-07-26)  A solution looking for a problem :

Chebyshev polynomials verify   T_m(T_n(x))  = T_mn(x).  This unique property makes it possible to define pairs ofclosely related functions from any pair of arithmetic functions u and v (with subexponential growth)  that areDirichlet inverses of each other,using the following symmetrical relations:

g ( x )   =		u(n) f ( Tn(x) )
f ( x )   =		v(n)  g ( Tn(x) )

:  Expand the latter right-hand-side using the definition of  g :

_{m  n} u(n) v(m) f ( T_mn (x) )  =   _{k d|k} u(d) v(k/d) f ( T_k (x) )

u and v being Dirichlet inverses,the bracket is either  1  (if k = 1)  or  0. 

This applies, in particular, when  u  is atotally multiplicativearithmetic function  [i.e., such that  u(mn)  =  u(m) u(n) for any  m & n ] in which case its Dirichlet inverse can be expressed using theMöbius function () :

v(n)   =   (n) u(n)

Using T_n(x) = x^1/n instead of Chebyshev polynomials,this pattern was used in 1859 byRiemann to linkhis (normalized) prime-counting function f =  with the celebrated jump function g = J  he obtained with  u(n) = 1/n.

(2015-12-06)  
They are denoted by the symbol  U  (simply because U comes afterT ).

They obey exactly the same second-order recurrence relation as the aboveChebychev polynomials of the first kind but the startingpoints are different:

T0(x)  =  1 T1(x)  =  x

U0(x)  =  1 U1(x)  =  2x

U0(x)	=	1	Un+2(x)   =  2 x Un+1(x)    Un(x)
U1(x)	=		2 x
U2(x)	=	-1	+	4 x2
U3(x)	=		-4 x	+	8 x3
U4(x)	=	1		12 x2	+	16 x4
U5(x)	=		6 x		32 x3	+	32 x5
U6(x)	=	-1	+	24 x2		80 x4	+	64 x6
U7(x)	=		-8 x	+	80 x3		192 x5	+	128 x7
U8 (x)	=	1		40 x2	+	240 x4		448 x6	+	256 x8

Inverse formulas :

x0	=	U0
2 x1	=		U1
4 x2	=	U0	+	U2
8 x3	=		2 U1	+	U3
16 x4	=	2 U0	+	3 U2	+	U4
32 x5	=		10 U1	+	4 U3	+	U5
64 x6	=	5 U0	+	9 U2	+	5 U4	+	U6
128 x7	=		14 U1	+	14 U3	+	6 U5	+	U7
256 x8	=	14 U0	+	28 U2	+	20 U4	+	7 U6	+	U8

(2012-02-15)  
Key to Coulombianmultipole expansion& spherical harmonics.

TheLegendre polynomials (A008316) are recursively defined by:

	P0(x)	=	1	;	Pn(x)   =   (2-1/n) x Pn-1(x)  (1-1/n) Pn-2(x)
	P1(x)	=		x
2	P2 (x)	=	-1	+	3 x2
2	P3 (x)	=		-3 x	+	5 x3
8	P4 (x)	=	3		30 x2	+	35 x4
8	P5 (x)	=		15 x		70 x3	+	63 x5
16	P6 (x)	=	-5	+	105 x2		315 x4	+	231 x6
16	P7 (x)	=		-35 x	+	315 x3		693 x5	+	429 x7
128	P8 (x)	=	35		1260 x2	+	6930 x4		12012 x6	+	6435 x8

They are linked to the expressions of spherical harmonics in terms of the colatitude  [0,[  and the longitude   (modulo2).

(2012-02-16)  
Radial part of the solution of theSchrödinger equationfor hydrogenoids.

Laguerre's equation  is a second-order linear differential equation:

x y''  +  (1-x) y'  +  n y   =   0

It has non-singular solutions only when  n  is a non-negative integer. In that case,  a solution is  L_n(n),  the Laguerre polynomial of order ngiven by:

Sorin is credited for the following generalized Laguerre equation :

x y''  +  (+1-x) y'  +  n y   =   0

This is satisfied by theLaguerre function,  defined by:

L	) n	=				n+ n-p		(-x)p p!
			n=1

Because of the way binomial coefficients  vanish, a polynomial  (a finite sum)  called associated Laguerre polynomial is so obtained when  n  is a non-negative integer. Otherwise,  the above is a divergent series  which is Borel-summable.

Ordinary Laguerre polynomials  correspond to the special case  = 0.

(2012-02-18)  
Eigenstates of the quantum harmonic oscillator.

The above are more popular than the simpler modified Hermite polynomials  He_n  which can be defined via: H_n(x)   =   2^n/2 He_n(2^½x)

(2014-12-07)  

The reverse Bessel polynomials  tabulated belowappear in the transfer functions ofBessel-Thomson filters

(2013-04-24)  

Like Hermite polynomials  and Euler polynomials,  the sequenceof  Bernoulli polynomials  start with some nonzero constant polynomial (namely, 1)  and subsequentlyverify the Appell property, which is to say:

dB_n(x) / dx   =   n B_n-1(x)

This relation becomes a recursive definition if the successive constants of integration  are given as a prescribed sequence:

B_n   =   B_n(0)

The polynomials can be expressed in terms of that sequence of numbers:

Bn (x)   =

n

n k

Bk   xn-k

On the two competing definitions of Bernoulli numbers :

With the convention adopted in Numericana (i.e.,  B₁ = ½)  we have:

Bn = Bn(1)

Authors who posit that  B₁ = -½   specify instead that B_n = B_n(0). Note that those two conventions only differ in the case  n = 1.

The Bernoulli polynomials are not affected by the choice of convention for Bernoulli numbers. Neither are relations between those polynomials,  like:

B_n (1 - x)   =   (-1)ⁿ B_n(x)
B_n (1 + x)   =   B_n(x)  +  n x^n-1

Besides the aforementioned case  n = 1,  B_n  vanishes for odd values of  n.

By the vonStaudt-Clausen theorem (1840)  the denominator of B_2n  is the product of all primes  p  forwhich  p-1  divides  2n.

(2019-11-24)  

After deriving explicit formulas up to  p = 17, Johann Faulhaber observed that,   if  p = 2q+1  is odd, then the sum of the p-th powers of the integers from 0 to n  isa polynomial of degree  q+1  in the variable  x = n(n+1)/2. A related expression holds for a nonzero even  p,  namely:

	n	k 2q+1	=	Fq+1(x)
If  q > 0,  then:	n	k 2q	=	n+½ 2q+1		d dx	Fq+1	(x)

That result was proved in full generality by Carl Jacobi,  in 1834.

(2013-04-24)  

E_n(x)

Relation to the Sequence of Euler Numbers :

Euler numbers can be expressed in terms of the above Euler polynomials:

E_n   =   2ⁿ E_n(½)

The Euler numbers of odd index vanish. The signs of even-indexed Euler numbers alternate.

(2021-07-15)   .
M_n(x)  is the coefficient of  tⁿ /n!  in the expansion  of  (1+t)^x/ (1-t)^x

Mittag-Leffler polynomials were first discussed under that name in 1940,  by Harry Bateman (1882-1946).

They obey the same binomial formula  as ordinary powers:

(2021-07-20)  
The binomial polynomials form a group under umbral composition.

(2020-06-02)  
Irreducible divisors of   x ⁿ 1   over the rationals.

The  n^th cyclotomic polynomial _n is the unique  monic polynomial dividing x ^k 1   for  k = n  but not for any lesser value of  k.

When n > 1, _n  is palindromic. If  n  has at most two distinct odd prime factors,  then the coefficients of _n  stay within  {-1,0,1}. That holds for n < 105;  the first product of three distinct odd primes (Adolph Migotti, 1883). Those coefficients can be arbitrarily large (Issai Schur, 1931). Furthermore, any  given integer occurs as a coefficient of some cyclotomic polynomial  (Jiro Suzuki,1987).

_n  is an irreducible polynomial over the rationals, whose degree is equal to the Euler totient   (n). That nontrivial fact is due to Carl F. Gauss.

The following definition also holds for  n = 0 (as an empty product  is 1).

n (x)   =			x exp( i 2k / n)

For n > 0,  the cyclotomic polynomial _n can thus be defined as the unique monicpolynomial whose roots are the primitive  n^th  roots of unity.

As with any multiplicative function,  the (rarely used)  value of    at zero is  (0) = 0  (as its one-line definition  implies) which confirms ₀ = 1.

₀  from Maple®  (albeit with due apologies).

The following factorization yields as many factors as there are divisors of n:

xn 1   =		k(x)
	The following interesting equation involves the Möbius function  :  n(x)   =     ( xn/k 1 ) (k) (2020-06-03)   Nontrivial factors of   (p x2) p 1  when  p  is an odd prime. A polynomial  Pp  can be defined for which the following identity holds, which provides a nontrivial factorization  of some special integers: ( p x2) p (-1)m   =  ( p x2 (-1)m)  Pp(-x)  Pp(x) Here,  p = 2m+1  is an odd prime (see Sophie Germain identity  for p=2). Pp (x)   =   Ap( p x2)  +  (p x) Bp( p x2)  where  Ap  and  Bp  are both palindromic monic  polynomials. Ap  hasdegree  m.   Bp  has degree m-1. Polynomials  Pp(x) = Ap(y) + (p x) Bp(y)   with   y = p x 2  Prime p  1   y  y2 y3 y4 y5 y6 y7 y8 y9 y10 y11 y12 y13 y14 y15 + p=3 A 1 1 B 1 p=5 A 1 3 1 B 1 1 + p=7 A 1 3 3 1 B 1 1 1 + p=11 A 1 5 -1 -1 5 1 B 1 1 -1 1 1 p=13 A 1 7 15 19 15 7 1 B 1 3 5 5 3 1 p=17 A 1 9 11 -5 -15 -5 11 9 1 B 1 3 1 -3 -3 1 3 1 + p=19 A 1 9 17 27 31 31 27 17 9 1 B 1 3 5 7 7 7 5 3 1 + p=23 A 1 11 9 -19 -15 25 25 -15 -19 9 11 1 B 1 3 -1 -5 1 7 1 -5 -1 3 1 p=29 A 1 15 33 13 15 57 45 19 45 57 15 13 33 15 1 B 1 5 5 1 7 11 5 5 11 7 1 5 5 1 + p=31 A 1 15 43 83 125 151 169 173 173 169 151 125 83 43 15 1 B 1 5 11 19 25 29 31 31 31 29 25 19 11 5 1 For p=31 (and x=9) this factors a nice 102-digit semiprime:   (251131+1) / 2512  =  889923919072997985238634558820908333948499157179463 × 1111413273683146858652465162019244587926917356315577 That factorization would take a long time with a general-purpose  program. For compactness, we'll give palindromic polynomials as lists of coefficientswith underlined central ones  (so the mirror endings can be freely truncated). =(1, 19, 79, 183, 285, 349, 397, 477, 579,627, 579, 477, 397, 349, 285, 183, 79, 19, 1) B37 =(1, 7, 21, 39, 53, 61, 71, 87,101,101, 87, 71, 61, 53, 39, 21, 7, 1)   A41 =(1, 21, 67, 49, 7, 35, 15, 11, -23, -65,-31, -65, -23, 11, 15, 35, 7, 49, 67, 21, 1) B41 =(1, 7, 11, 3, 3, 5, 1, 1, -9,-7,-7, -9, 1, 1, 5, 3, 3, 11, 7, 1)   A43+ =(1, 21, 81, 169, 223, 225, 213, 223, 229, 197,159,159, 197, 229, 223, 213, 225, 223, 169... B43+ =(1, 7, 19, 31, 35, 33, 33, 35, 33, 27,23, 27, 33, 35, 33, 33, 35, 31, 19, 7, 1)   A47+ = (1, 23, 65, -15, -169, -97, 179, 287, -37, -375, -149,311,311, -149, -375, -37, 287, 179... B47+ = (1, 7, 7, -15, -25, 5, 41, 25, -37, -49, 15,57, 15, -49, -37, 25, 41, 5, -25, -15, 7, 7, 1)   A53 = (1, 27, 113, 103, -155, -219, 263, 513, -59, -465, 75, 551, 93,-357,93, 551, 75, -465, -59... B53 = (1, 9, 19, -1, -35, -3, 67, 41, -51, -39, 57, 57,-31,-31, 57, 57, -39, -51, 41, 67, -3, -35, -1...   A59+ = (1, 29, 111, 55, -85, 47, 11, 53, 131, -245, 41, 103, -111, 227,-103,-103, 227, -111, 103... B59+ = (1, 9, 15, -5, -5, 9, -3, 21, -9, -25, 25, -11, 9, 19,-31,19, 9, -11, 25, -25, -9, 21, -3, 9, -5, -5...   A61 = (1,31,191,637,1541,2979,4881,7029,9125,10953,12397,13511,14379,15053,15511,15667... B61 = (1, 11, 47, 131, 281, 497, 761, 1037, 1291, 1501, 1663, 1789, 1887, 1961,2001,2001...   A67+ = (1,33,193,565,1055,1429,1599,1803,2225,2637,2617,2195,1869,1875,1865,1469,991,991... B67+ = (1, 43, 99, 155, 187, 205, 243, 301, 329, 297, 243, 225, 233, 209, 147,111, 147, 209, 233...   A71+ = (1, 35, 169, 155, -109, 233, 597, 39, 101, 445, 163, 293, 89, -203, 249, -49, -505,37,37... B71+ = (1, 11, 25, 1, -5, 63, 43, -9, 43, 37, 21, 35, -19, 1, 29, -47, -35,23,-35, -47, 29, 1, -19, 35... Those are linked to cyclotomic polynomials  via the following equality: (p x2)p 1   =   ( p x2 1) [ Ap(p x2)2  p2x2  Bp(p x2)2] As a polynomial identity,  that's equivalent  (using  y = p x2) to a simpler relation  (albeit not directly applicable to integer factorization): yp 1   =   (y 1) [ Ap(y)2 (p y) Bp(y)2]   (with  1 = (-1)(p+1)/2) (2020-10-20)   Manufacturing remarkable identities. Clearly,   x5 + x4 + x3 + x2 + x + 1      =    x3 (x2 + x + 1) + x2 + x + 1 = (x3 + 1) (x2 + x + 1) This can be used to factor   x5 + x4 + 1   : x5 + x4 + 1      =    x5 + x4 + x3 + x2 + x + 1 x (x2 + x + 1) = (x3 x + 1) (x2 + x + 1) My first quintic equation (10:28) by Steve Chow  (blackpenredpen, 2020-06-03).

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