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Chebyshev Polynomial The Chebyshev Polynomials of the first kind

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Chebyshev Polynomial The Chebyshev Polynomials of the first kind Powered By Docstoc
					Chebyshev Polynomial

The Chebyshev Polynomials of the first kind denoted by                           n(x)         =   ; n= 0,1,2,… are

Solutions of the chebyshev’s differential equation:-



       To show that                 n(x)   is a solution of the chebyshev’s differential equation :

         n(x)   =                                ; n= 0,1,2,…
         `n(x) =                                     )

         ``n(x) =                                    )+                               )


        Now : (1-           )                        )+                                   )

        = -                                  +                      )–



        Hence        n(x)   is a solution of shebyshevi differential equation

        n(x) can be written as :
         n(x) =               (1-                                 (1-        )2 + ….
        To verify equation (1) , we use that
          n(x) =              )
        Let               , so cos
        , then n(x) = cos n

        Now by DeMoivre’s theorem :
                         n
        (cos               = cos n + i sin n
                                                                        n
        So that n(x) = cos n
        Applying the binomial theorem on (cos                                        )2
        We get :
                         n         2
        (cos               =         +                                      (i sin                 (i sin )2 +
                                    (i sin )3 +                    (i sin )4 + ……
        From this we get that:
                           n                             n
                             =                                                (i sin )2 +              (i sin )4
                                    and            =            =1-
              n(x)   =          -                (1- ) +         (1-         )2 ….
       Now : By equation (1) we can list the first few chebyshev’s polynomials of the first kind
       :-
       So , To find 0(x) = = 1.
       To find 4(x) Put n = 4 in equation (1) we get :-
         4(x)   =   -                    (1-     )+               (1-       )2
                =       –                 (1-    ) + (1-2x2+x4)
             = – 6x2 + 6x4 + 1 – 2x2 + x4
             = 8x4 – 8x2 + 1
       And so on To find the rest:
                             2             3
        1(x) = x , 2(x) = 2x -1 , 3(x) = 4x -3x
                   5       3
        5(x) = 16x – 20x + 5x
                   6       4      2
        6(x) = 32x – 48x + 18x - 1
                    7        5      3
        7(x) = 64 x - 112x + 56 x – 7x



       Note:
        1(x) is a polynomial of order (1)

        2(x) is a polynomial of order (2)



       In general           k(x)   is a polynomial of order (k).

       ** Generating function for n(x):
       The generating function for n(x) is given by :
                    =                     n(x)



       ** special values of the chebyshev polynomials n(x) :
       (i)      n(1) =                 =         =
       (ii)      n(-1) =                   =        =        .
       (iii)        2n(0)        =                                      =                 =            =    .
       (iv)      2n+1(0) =                        =                   =
        =0
       In fact we can also user equation (1) to verify these results.
       We can also verify the above four result by using the generation function for                                     n(x):

       (i)          n(1):     substitute x = 1 in                                =            n(x)

                                     -
                                                              n
                                   =                  n(1)t       =>             ==                   tn

                                                                   n
                            =>            =              n(1)t          , but         =          tn

So :         tn =                        tn , Now equating the Coefficients of tn we get                   n(x)   = 1.
If we set x = 0 in the generating function we get :-

         =               n(0)        , But              =

Hence :                                                       n(0)


Equating the Coefficient, of tn , once when n is even and once when n is odd , we get :

 2n(0)   = (-1)n ,          2n+1(0)   =0

              (ii)           n(-1)   = (-1)n : to verify this set x = -1 in the generating function we get:
                                        =                n(-1)

                                            =                   =>       =           n(-1)

                                 But             =
                                                                             n(-1)                                     we get
                                 :
                                     n(-1)   = (-1)n


Rule :       n(-x)   = (-1)n     n(x)


Proof : to prove this rule we using equation (1) and substituting –x we get :

 n(-x)   = (-x)n -          (-x)n-2(1-(-x)n-2) +              (-x)n-4(1-(-x) 2)2 - …..

= (-1)n (x)n -           (-1)n-2 xn-2 (1-x2) +                (-1)n-4 xn-4(1-x2)2 - ….

= (-1)n [(x)n -          (-1)-2 xn-2 (1-x2) +                 (-1)-4 xn-4(1-x2)2 - ….]

= (-1)n [(x)n -          xn-2 (1-x2) +                xn-4(1-x2)2 - ….]

So    n(-x)    = (-1)n      n(x)


*** Recurrence Relations for                         n(x):-


(i)   n+1(x)    – 2x     n(x)   +     n-1(x)     =0

(ii)(1-x2)       n(x)   = -n x       n(x)   +n       n-1(x)


To verify these recurrence relations, we use

  n(x)   =                            , So Let                   = , so that cos              and   n(x)   = cos(n ).

No to prove (i) :

 n+1(x)      – 2x    n(x)   +   n-1(x)   = cos((n+1) )-2cos cos(n ) + cos((n-1) )
= cos(n               ) – 2cos cos(n ) + cos(n

But cos(n           + ) = cos(n )cos( ) – sin(n )sin( )

and cos(n                           cos(n )cos( ) + sin(n )sin( )

So : cos(n      ) – 2cos cos(n ) + cos(n                                                        = cos(n )cos( ) – sin(n )sin( ) –
2cos cos(n ) + cos(n )cos( ) + sin(n )sin( ) = 0

So :        n+1(x)   – 2x          n(x)   +    n-1(x)   = 0 which recurrence relation (i)

To prove recurrence relation (ii) we start with the left hand side:

(1-x2)      n(x)    = (1-x2)               n(x)



= (1-cos2                 cos(n ).            = (sin2 ) (-n sin(n )) . (               )

So (1-x2)          n(x)   = n sin                       ). ---------------- (a)

Now : the Right hand side:

-n x     n(x)   +n        n-1(x)   = -n cos          cos(n ) + n cos ((n-1) )

= -n cos        cos (n ) + n cos                   cos (n ) + -n x         n(x)   +n   n-1(x)


So -n x      n(x)   +n       n-1(x)       = -n x     n(x)   +n    n-1(x)   .-------------------(b)

From (a) and (b) we get recurrence relation (ii):

(1-x2)      n(x)    = -n x         n(x)   +n      n-1(x)


Examples : Evaluate (a)                       2(x)   , (b)       2(x)


If   0(x)   =1 ,     1(x)   = x.

       (a) To evaluate 2(x) we use recurrence relation (i):-
            2(x) - 2x 1(x) + 0(x) = 0
                     2
            2(x) = 2x – 1

       (b) To find 2(x) we use recurrence relation (ii) :-
           (1-x2) 2(x) = -2 x 2(x) + 2 1(x) = -2x(2x2-1)+2x = -4x3+2x + 2x = 4x(1-x2)
            So       2(x)    =                  = 4x.
**** Orthogonality:

                      dx =



Examples:
(i)                   dx =                 ((since n=m=0))

(ii)                  dx =               ((since n=m=2))

(iii)                    dx =            ((since n           ))

(iv)                              dx = 0
Since 32x7-56x5+28x3-3x =         3(x)     4(x)

So                      dx = 0 ((since n          ))


In addition :
 3(x): Polynomial of order 3.

 3(x) Polynomial of order 4.

        3(x)   3(x)                                    AND        odd = 0

				
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posted:2/26/2012
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