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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 |

language: | English |

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