Algebraic Substitution: Solving F(r) Using r=r_+(1+\rho^2)

  • Thread starter Thread starter cristo
  • Start date Start date
  • Tags Tags
    Substitution
Click For Summary
SUMMARY

The discussion focuses on the algebraic substitution of the function F(r) = (r - r_+)(r - r_-)/r^2 using the substitution r = r_+(1 + ρ^2). The resulting expression simplifies to F = (ρ^2[r_+(1 + ρ^2) - r_-])/(r_+^2(1 + ρ^2)^2). Participants confirm that the final form can be expressed as (r_+ρ^2(r_+ - r_-)/r_+^2) * [1 + O(ρ^2)], indicating the use of a Taylor expansion for approximation. The discussion highlights the importance of recognizing Taylor expansions in algebraic manipulations.

PREREQUISITES
  • Understanding of algebraic functions and substitutions
  • Familiarity with Taylor series expansions
  • Knowledge of mathematical notation and simplification techniques
  • Basic calculus concepts related to limits and approximations
NEXT STEPS
  • Study Taylor series and their applications in function approximation
  • Explore algebraic manipulation techniques in calculus
  • Learn about asymptotic analysis and big O notation
  • Review advanced algebraic functions and their properties
USEFUL FOR

Students and professionals in mathematics, particularly those studying algebraic functions and calculus, as well as anyone interested in function approximation techniques.

cristo
Staff Emeritus
Science Advisor
Messages
8,145
Reaction score
75

Homework Statement



I have this function [tex]F(r)=\frac{(r-r_+)(r-r_-)}{r^2}[/tex] and I want to make the subsitution [itex]r=r_+(1+\rho^2)[/itex].

Homework Equations



None.

The Attempt at a Solution



So, I sub in, to obtain [tex]F=\frac{[r_+(1+\rho^2)-r_-][r_+(1+\rho^2)-r_-]}{r_+^2(1+\rho^2)^2}=\frac{\rho^2[r_+(1+\rho^2)-r_-]}{r_+^2(1+\rho^2)^2}[/tex].

Now, the solutions say that this is equal to [tex]\frac{r_+\rho^2(r_+-r_-)}{r_+^2}\cdot [1+O(\rho^2)][/tex], however I cannot, for the life of me, see how to get this from my line above! Can anyone help?
 
Physics news on Phys.org
Looks like a Taylor expansion to me, but I could be wrong...
 
Hootenanny said:
Looks like a Taylor expansion to me, but I could be wrong...

Yup, you're correct. I never spot things like that. Thanks! :smile:
 

Similar threads

Replies
11
Views
2K
  • · Replies 7 ·
Replies
7
Views
3K
Replies
4
Views
2K
  • · Replies 2 ·
Replies
2
Views
5K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
2
Views
2K
  • · Replies 6 ·
Replies
6
Views
3K