Undetermined coefficients - deriving formula

Click For Summary
SUMMARY

The discussion focuses on deriving a formula for the integral approximation \(\int_{a}^{b} f(x) dx \approx c_{0} f(a) + c_{1} f(b) + c_{2} f'(a) + c_{3} f'(b)\) that is exact for polynomials of the highest degree possible. The user applies a change of variable \(y = x - a\) and reformulates the integral, leading to \(\int_{0}^{b-a} f(y) dy \approx c_{0} f(0) + c_{1} f(b-a) + c_{2} f'(0) + c_{3} f'(b-a)\). Clarification is sought regarding the conditions for the formula's accuracy with respect to polynomial degrees.

PREREQUISITES
  • Understanding of integral calculus and approximation methods
  • Familiarity with polynomial functions and their derivatives
  • Knowledge of change of variables in integrals
  • Basic concepts of numerical integration techniques
NEXT STEPS
  • Research the method of undetermined coefficients in numerical integration
  • Learn about polynomial interpolation and its applications
  • Study the derivation of Simpson's Rule for integral approximation
  • Explore the concept of Taylor series and its relevance to approximation
USEFUL FOR

Students in calculus courses, educators teaching numerical methods, and anyone interested in the mathematical foundations of integral approximation techniques.

muso07
Messages
53
Reaction score
0

Homework Statement


Derive a formula of the form \int\stackrel{b}{a}f(x)dx \approx c_{0}f(a)+c_{1}f(b)+c_{2}f'(a)+c_{3}f'(b) that is exact for polynomials of the highest degree possible.

Apply a change of variable: y=x-a

Homework Equations


The Attempt at a Solution


I don't get the "highest degree possible" thing. Like, if it's exact for quadratics, I know it has to be exact for f(x)=1,x,x2.

But applying the change of variable, I got \int\stackrel{b-a}{0}f(y)dy \approx c_{0}f(0)+c_{1}f(b-a)+c_{2}f'(0)+c_{3}f'(b-a)

Any help much appreciated. :)
 
Last edited:
Physics news on Phys.org
Whoops, that's meant to be integral from a to b for the first one and 0 to b-a for the second one.
 
Can anyone even point me in the right direction?
 

Similar threads

  • · Replies 21 ·
Replies
21
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 15 ·
Replies
15
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 3 ·
Replies
3
Views
1K
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K