Use economisation to find linear approximation to x^2-x-1?

Question

I've been given the solution to this question... It uses chebychev, and you get:

$1/2(2x^2-1)-2x-1/2$

So the Chebyshev economisation polynomial is $-2T1 -1/2 T0$

I can see the logic in how this question has been answered, but would really appriciate it if someone could give me a method with steps on how to calculate a common question!

Thanks in advance

Answer 1

Recalling $$ T_0(x) = 1 \qquad T_1(x) = x \qquad T_2(x) = 2x^2-1 $$ and using method of indeterminate coefficients: $$ x^2 - x -1 = a_2 T_2(x) + a_1 T_1(x) + a_0 T_0(x) = 2 a_2 x^2 + a_1 x + (a_0-a_2) $$ Equating polynomial coefficients we get the system $2 a_2 = 1$, $a_1 = -1$ and $a_0 - a_2 = -1$, which readily gives $a_2 = \tfrac{1}{2}$, $a_1 =-1$ and $a_0 =-\tfrac{1}{2}$.

First order economization hence: $$ \mathcal{Econ}_1\left(x^2-x-1\right) = -T_1(x) - \tfrac{1}{2} T_0(x) = -x - \frac{1}{2} $$ Representing graphically in Mathematica: