2 c c2 cos 6 arcsin x 13 poly1 d convert series sin

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Unformatted text preview: 1 we have, y2 = a0 1 α4 α2 1 − x 2 + x4 + · · · . x 2! 4! Since y2 is singular as x tends to zero and y1 does not, this indicates that the two solutions are linearly independent. 5 AMATH 351: Assignment 4 > Chebyshev Polynomial First we define Chebyshev's DE as stated in the equation with p and then find the solution. > p d 6; p := 6 (1.1) (1.1) > ode1 d 1 K x2 $ diff y x , x, x K x$diff y x , x C p2 $y x = 0; d2 d ode1 := K 2 C 1 x y x Kx y x C 36 y x = 0 2 dx dx > soln1 d dsolve ode1, y x ; soln1 := y x = _C1 sin 6 arcsin x (1.2) C _C2 cos 6 arcsin x (1.3) > poly1 d convert series sin 6 arcsin x , x = 1 , polynom ; 143 poly1 := 6 I 2 csgn I x K 1 xK1 C I 2 csgn I x K 1 xK 1 3/2 2 3861 21879 C I 2 csgn I x K 1 xK 1 5/2 C I csgn I x K 1 2x 16 64 230945 264537 K 1 7/2 C I 2 csgn I x K 1 xK 1 9/2 C I 2 csgn I x 1024 4096 K1 xK1 (1.4) (1.4) 11 / 2 > poly2 d convert series cos 6 arcsin x , x = 1 , polynom ; poly2 := 35 K 36 x K 210 x K 1 C 93 csgn I x K 1 256 2 2 K 448 x K 1 xK1 3 K 432 x K 1 4 49245 256 (1.5) (1.5) CK 5 > Order d 3; Order := 3 (1.6) (1.6) > soln2 d dsolve ode1, y x , series, x = 1 ; 143 1287 soln2 := y x = _C1 x K 1 1 C xK1 C 12 32 C _C2 1 C 36 x K 1 C 210 x K 1 > poly3 d convert Cx 3 K Cx 1 1C 143 12 2 CO xK1 K Cx C 1 xK1 2 CO xK1 3 (1.7) (1.7) 3 1287 32 K Cx 1 2 CO K 1 , polynom ; (1.8) (1.8) poly3 := > poly4 d convert xK1 > p3 d plot xK1 2 1 C 36 K C x C 210 K C x 2 C O K C x 1 1 1 2 poly4...
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This document was uploaded on 01/26/2014.

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