How is this 'root stability' differential equation derived?

Click For Summary
SUMMARY

The discussion focuses on the derivation of the differential equation related to the sensitivity of polynomial roots to coefficient errors, specifically in the context of Wilkinson's polynomial. The equation describes how the rate of change of the roots, denoted as \(\frac{d\beta_j}{dt}\), is determined by the error function \(c(\beta_j)\) and the derivative of the original polynomial \(p'(\beta_j)\). The derivation involves perturbing the polynomial and applying differentiation with respect to the error scaling parameter \(t\), ultimately leading to the first-order approximation \(\frac{d\beta_j}{dt} = -\frac{c(\alpha_j)}{p'(\alpha_j)}\).

PREREQUISITES
  • Understanding of polynomial functions and their roots
  • Familiarity with differential calculus and differentiation techniques
  • Knowledge of Taylor series expansion
  • Basic concepts of perturbation theory in mathematics
NEXT STEPS
  • Study the derivation of Wilkinson's polynomial and its implications on root stability
  • Learn about perturbation methods in polynomial root analysis
  • Explore Taylor series and their applications in approximating functions
  • Investigate sensitivity analysis in numerical methods for polynomials
USEFUL FOR

Mathematicians, numerical analysts, and students studying polynomial root sensitivity and stability analysis will benefit from this discussion.

tomizzo
Messages
113
Reaction score
2
I'm currently studying the sensitivity of polynomial roots as a function of coefficient errors. Essentially, small coefficient errors of high order polynomials can lead to dramatic errors in root locations.

Referring to the Wilkinson polynomial wikipedia page right here,[/PLAIN] you can see that there is a differential equation listed under the 'stability analysis' section. This derivative explains how the rate of change of the roots (with respect to some error scaling parameter 't') equals the error evaluated at the polynomial error function divided by the original polynomials derivative... That was a mouthful..

So my question: How is this differential equation derived? Maybe I'm just rusty on my calculus, but would someone be willing to demonstrate how it was derived given the information from the original problem statement?

Thanks,
 
Last edited by a moderator:
Physics news on Phys.org
I think the notation in the article is confusing; \alpha_j is used both for the root of original polynomial and a function which gives the root of the perturbed polynomial. These really need to be given different symbols.

Define p(x) = \prod_j (x - \alpha_j). We perturb this to q(x,t) = p(x) + tc(x) = \prod_j (x - \beta_j(t)). Now by definition q(\beta_j(t),t) = p(\beta_j(t)) + tc(\beta_j(t)) = 0 for every t, so we can differentiate with respect to t to obtain 0 = p&#039;(\beta_j)\frac{d\beta_j}{dt} + c(\beta_j) + tc&#039;(\beta_j)\frac{d\beta_j}{dt} which we rearrange to obtain <br /> \frac{d\beta_j}{dt} = -\frac{c(\beta_j)}{p&#039;(\beta_j) + tc&#039;(\beta_j)}. The right hand side can now be expanded as a Taylor series about t = 0, and as \beta_j(0) = \alpha_j this yields to first order <br /> \frac{d\beta_j}{dt} = -\frac{c(\alpha_j)}{p&#039;(\alpha_j)} as required.
 
  • Like
Likes   Reactions: The Bill

Similar threads

  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 65 ·
3
Replies
65
Views
8K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 9 ·
Replies
9
Views
7K