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

  • Thread starter Thread starter squenshl
  • Start date Start date
  • Tags Tags
    Algorithm
Click For Summary

Homework Help Overview

The discussion revolves around finding the quadratic approximation \( q_2(x) \) for the function \( f(x) = e^x \) on the interval \([-1, 1]\) using Reme's second algorithm. Participants are exploring the steps involved in applying the algorithm and the necessary calculations to achieve the desired approximation.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to clarify how to derive specific coefficients from the first step of Reme's algorithm. Other participants discuss the minimax quadratic approximation and the choice of points for the approximation. There is also mention of using the Newton-Raphson method to find roots related to the approximation.

Discussion Status

Participants are actively engaging with the problem, sharing their calculations and methods. Some guidance has been offered regarding the use of the Newton-Raphson method, and there is an ongoing exploration of how to find the necessary points \( z_i \) for the next steps in the algorithm.

Contextual Notes

There is a mention of using Chebyshev points for the approximation, and the discussion includes specific values and errors associated with the approximation process. The clarity of the problem statement has been noted as a potential issue.

squenshl
Messages
468
Reaction score
4

Homework Statement


Find q2(x) for f(x) = ex 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.
 
Physics news on Phys.org
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
||-ex+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
 
Cheers heaps.
Now for step 2.
z0 = -1.0, z1 = -0.438621, z2 = 0.560939, z3 = 1.0
For these zi: 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.
 
Last edited:
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.
 
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.
 
How to find zi?
 

Similar threads

Replies
1
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 47 ·
2
Replies
47
Views
5K
  • · Replies 7 ·
Replies
7
Views
4K
Replies
2
Views
1K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
3
Views
3K