Perturbation theory and asymptotics

Click For Summary
SUMMARY

The discussion focuses on finding the roots of the transcendental function f(x;a)=x^2-3ax-1-a+exp(-x/a)=0, particularly as the parameter 'a' approaches zero. The user suggests using Taylor's polynomial to approximate the exponential term, exp(-x/a), but encounters difficulties due to division by zero in the limit. The conversation emphasizes the application of perturbation theory to tackle the problem effectively.

PREREQUISITES
  • Understanding of transcendental functions
  • Familiarity with Taylor series expansion
  • Knowledge of perturbation theory
  • Basic calculus, particularly limits and continuity
NEXT STEPS
  • Study the application of Taylor series in perturbation problems
  • Research techniques for handling limits involving parameters approaching zero
  • Explore advanced perturbation methods in mathematical analysis
  • Learn about transcendental equations and their numerical solutions
USEFUL FOR

Mathematicians, physicists, and engineers dealing with perturbation theory and transcendental equations, particularly those interested in asymptotic analysis and limit behaviors.

Juggler123
Messages
80
Reaction score
0
I need to find the roots of the transcendental function,

f(x;a)=x^2-3ax-1-a+exp(-x/a)=0;

I've done many problems like this before and am fairly sure this is just a regular perturbation problem. The difficulty I'm having is with the exponential term.

Could anyone give me an idea of how to tackle this problem?

Thanks.
 
Physics news on Phys.org
Approximate the exponential with a Taylor's polynomial:
exp(-x/a)= 1- x/a+ \frac{x^2}{2a^2}+ \cdot\cdot\cdot+ (-1)^n\frac{x^n}{n!a^n}
 
Sorry I didn't make it clear in my first post, I'm finding the roots of the equation as a tends to 0. I thought about the Taylor expansion but then the terms are being divided by 0 (as a tends to zero). Is there anyway around this?

Thanks.
 

Similar threads

  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 134 ·
5
Replies
134
Views
12K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 49 ·
2
Replies
49
Views
5K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 1 ·
Replies
1
Views
4K