MHB Use the inverse function theorem to estimate the change in the roots

Click For Summary
To estimate the change in the roots of the cubic polynomial p(λ) = λ^3 + a2λ^2 + a1λ + a0 when the coefficients change, the inverse function theorem can be applied. By substituting the root x1 into the polynomial and taking the total derivative with respect to x1, one can relate the changes in the coefficients to the changes in the roots. The change in coefficients, represented as Δa, is specified as 0.01 times the original coefficients. This method allows for a systematic estimation of how small perturbations in the coefficients affect the roots of the polynomial. Understanding this relationship is crucial for applying the inverse function theorem effectively.
i_a_n
Messages
78
Reaction score
0
Let $p(\lambda )=\lambda^3+a_2\lambda^2+a_1\lambda+a_0=(\lambda-x_1)(\lambda-x_2)(\lambda-x_3)$ be a cubic polynomial in 1 variable $\lambda$. Use the inverse function theorem to estimate the change in the roots $0<x_1<x_2<x_3$ if $a=(a_2,a_1,a_0)=(-6,11,-6)$ and $a$ changes by $\Delta a=0.01a$. How can I use the inverse function theorem to estimate?
 
Physics news on Phys.org
ianchenmu said:
Let $p(\lambda )=\lambda^3+a_2\lambda^2+a_1\lambda+a_0=(\lambda-x_1)(\lambda-x_2)(\lambda-x_3)$ be a cubic polynomial in 1 variable $\lambda$. Use the inverse function theorem to estimate the change in the roots $0<x_1<x_2<x_3$ if $a=(a_2,a_1,a_0)=(-6,11,-6)$ and $a$ changes by $\Delta a=0.01a$. How can I use the inverse function theorem to estimate?

You could start by filling in $x_1$ in $p(\lambda )=\lambda^3+a_2\lambda^2+a_1\lambda+a_0=0$ and taking the (total) derivative with respect to $x_1$.
 
I like Serena said:
You could start by filling in $x_1$ in $p(\lambda )=\lambda^3+a_2\lambda^2+a_1\lambda+a_0=0$ and taking the (total) derivative with respect to $x_1$.

What $\Delta a$ means? Can you give me a more complete answer? Thank you.

Who can provide me a complete answer?
 
Last edited:

Similar threads

  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 0 ·
Replies
0
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 26 ·
Replies
26
Views
4K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K