Using Reme's Algorithm to Find q2(x) for ex on [-1,1]

[squenshl]

Homework Statement

Find q₂(x) for f(x) = e^x on [-1,1] using Reme's second algorithm.

Homework Equations

The Attempt at a Solution

For the first iteration:
Step a of the algorithm gives a₀ = 0.989141, a₁ = 1.130864, a₂ = 0.553940 and E = 0.0443369. The question I am asking is how do I get these values from using step a of the algorithm before I can carry on with the algorithm.

 

[lurflurf]

This is not a clearly stated problem, but I know what you are trying to do because it is a common practice problem.
we desire a minimax quadratic approximation to exp(x) on [-1,1]
a₀+a₁ x+a₂ x²
such that the maximal norm
||-e^x+a₀+a₁ x+a₂ x²||
is minimized
we begin by chosing n+2 points (with n=2 because it is quadratic) in [-1,1]
often Chebyshev points are chosen so we have {-1,-.5,.5,1}
we solve the linear system
a₀+a₁ (-1)+a₂ (-1)²-E=e^-1
a₀+a₁ (-.5)+a₂ (-.5)²+E=e^-.5
a₀+a₁ (.5)+a₂ (.5)²-E=e^.5
a₀+a₁ (1)+a₂ (1)²+E=e¹
for a₀,a₁,a₂,E
and find
a₀=.98914107
a₁=1.1308643
a₂=.55393956
E=.044336860

 

[squenshl]

Cheers heaps.
Now for step 2.
z₀ = -1.0, z₁ = -0.438621, z₂ = 0.560939, z₃ = 1.0
For these z_i: f(z₀) - q(z₀) = -0.0443369
f(z₁) - q(z₁) = 0.0452334
f(z₂) - q(z₂) = -0.0452334
f(z₃) - q(z₃) = 0.0443369
Thanks again.

 

[squenshl]

I think I got it. Firstly I solve f(z) - q₂(z) = 0 using Newton-Raphson method to get z₁ & z₂ where z₀ = -1.0 and z₃ = 1.0 always stay the same getting the numbers required.

 

[lurflurf]

Do you mean solve f'(z) - q'₂(z) = 0 ?
Newton-Raphson method is often used, but other methods may also be used.
Keep steping until your abscissa and error are as saccurate as desired.

 

[batik]

How to find zi?