# Math Help - chebyshev polynomials

1. ## 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...

2. Originally Posted by pico24
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.

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??

4. 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