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Math Help - chebyshev polynomials

  1. #1
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    chebyshev polynomials

    I am having trouble with this question (gap in my notes, always happens)

    The nth Chebyshev polynomial Tn(x) is defined by Tn(x) = cos(n cos^-1(x)) for -1<x<1 and n = 0,1,2,3,...
    Assume these polynomials satisfy the recurrence relation

    xTn(x) = 1/2Tn+1(x) + 1/2Tn-1(x) for n =1,2,3,4,...

    a) Express the polynomial p(x) = -2x^4 +3x^3 -2x^2 +x -2 in the form
    c0T0(x) + c1T1(x) +...+c4T4(x) for suitable constants c0,c1,...,c4.
    (Hint: x^k = x.x^k-1)

    b) Econimise p(x) first to a degree three polynomial and then to a degree two polynomial.

    c) For each of the approximations in part (b), give an upper bound for the total error and state the range of values of x for which this bound is valid.


    I dont know where to start with part a) but if I can get help with it I should have no problems with parts b) & c)

    Thanks for any help...
    Last edited by pico24; May 11th 2010 at 11:11 AM.
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  2. #2
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    Quote Originally Posted by pico24 View Post
    The nth Chebyshev polynomial Tn(x) is defined by Tn(x) = cos(n cos^-1(x)) for -1<x<1 and n = 0,1,2,3,...
    Assume these polynomials satisfy the recurrence relation

    xTn(x) = 1/2Tn+1(x) + 1/2Tn-1(x) for n =1,2,3,4,...

    a) Express the polynomial p(x) = -2x^4 +3x^3 -2x^2 +x -2 in the form
    c0T0(x) + c1T1(x) +...+c4T4(x) for suitable constants c0,c1,...,c4.
    (Hint: x^k = x.x^k-1)
    From the definition T_n(x) = \cos(n \cos^{-1}(x)) you can check that T_0(x) = 1 and T_1(x) = x.

    Then use the hint and the recurrence relation, like this: x^2 = x.x = x.T_1(x) = \tfrac12(T_2(x) + T_0(x)) = \tfrac12(T_2(x)+1).

    Continuing in the same way,

    \begin{aligned}x^3 = x.x^2 = x.\tfrac12(T_2(x)+1) &= \tfrac12(xT_2(x) + x) \\ &= \tfrac12\bigl(\tfrac12(T_3(x)+T_1(x))+T_1(x)\bigr) = \tfrac14(T_3(x) + 3T_1(x)), \end{aligned}

    and similarly x^4 = \ldots (you do that one). Once you know how to express each power of x in terms of Chebyshev polynomials, you can plug those expressions into the formula for p(x) to express that in terms of Chebyshev polynomials.
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  3. #3
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    thanks for reply

    is x^4= x.x^3 =
    [1/2 (1/4T4(x) + 3/4T2(x)) + T2(x)]
    = 1/8T4(x) + 3/8T2(x) + 1/2T2(x)
    = 1/8T4(x) + 7/8T2(x)?

    How to I find c0,c1,...,c4

    Is p(x) = -2(value for x^4) + 3(value x^3....etc correct??
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  4. #4
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    x.x^3= 1/8T4(x) + 1/2T2(x) + 3/8T0(x) = x^4

    p(x)= -2[1/8T4(x)+1/2T2(x)+3/8T0(x)] + 3[1/4T3(x)+3/4T1(x)] - 2[1/2T2(x)+1/2T0(x)] + T1(x) - 2T0(x)

    = T0(2-1-6/8) + T1(1+9/4) + T2(-1-1) + T3(3/4) + T4(-1/4)

    p(x) = 1/4T0 + 13/4T1 - 2T2 + 3/4T3 - 1/4T4


    For part b)
    Economise deg 3
    q(x) = 1/4T0 + 13/4T1 - 2T2 + 3/4T3
    error Abs{p(x) - q(x)} < abs(1/4)

    Economise deg 2
    r(x) = 1/4T0 + 13/4T1 - 2T2
    error Abs{q(x) - r(x)} < abs(3/4)

    Part c)

    total error abs[p(x) - q(x)] + [q(x) - r(x)] = abs(1/4 + 3/4) = 1
    for max xE[-1,1]

    Can you tell me is this correct?

    Thanks
    Last edited by pico24; May 11th 2010 at 05:12 PM.
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