MHB First order corrections to a polynomial equation

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To approximate solutions to a polynomial equation with a small perturbation, the first order correction can be found using the Taylor expansion. By defining g(x) = f(x) + e, where e is small, and identifying x0 as a root of f(x), the correction is derived from the equation 0 = g(x) = f(x0) + e + (x-x0)f'(x0). This leads to the first order correction being (x-x0) = -e / f'(x0). This method resembles the first iteration of the Newton-Raphson algorithm, which can identify new roots unless f' is zero in the interval. The discussion highlights the application of Taylor expansion in solving perturbed polynomial equations.
topsquark
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I've seen this done in a video but I can no longer find the video! :(

What I would like to do is approximate the solutions to a polynomial equation in terms of a small perturbation. For example, say we have y = f(x) and know the corresponding zeros exactly. How would I go about finding first order corrections to the zeros if we have y = f(x) + e (e is small)?

Any help would be appreciated.

-Dan
 
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topsquark said:
I've seen this done in a video but I can no longer find the video! :(

What I would like to do is approximate the solutions to a polynomial equation in terms of a small perturbation. For example, say we have y = f(x) and know the corresponding zeros exactly. How would I go about finding first order corrections to the zeros if we have y = f(x) + e (e is small)?

Any help would be appreciated.

-Dan

Hey Dan!

Let g(x)=f(x)+e, and let x0 be a root of f(x).
Then we're looking for an x near x0 such that g(x)=0.
The first order Taylor expansion is: g(x) = g(x0) + (x-x0)g'(x0)
It follows at the first order that 0 = g(x) = f(x0) + e + (x-x0)f'(x0).
And therefore the first order correction is (x-x0) = -e / f'(x0).

This would actually be the first iteration of a Newton-Raphson algorithm, that will find the new zero unless f' is zero somewhere between the old and the new root.
In that case we can use f'' to see if the zero disappeared, or if there are 2 new roots at each side.

Is this perchance what you had in mind?
 
I like Serena said:
Hey Dan!

Let g(x)=f(x)+e, and let x0 be a root of f(x).
Then we're looking for an x near x0 such that g(x)=0.
The first order Taylor expansion is: g(x) = g(x0) + (x-x0)g'(x0)
It follows at the first order that 0 = g(x) = f(x0) + e + (x-x0)f'(x0).
And therefore the first order correction is (x-x0) = -e / f'(x0).

This would actually be the first iteration of a Newton-Raphson algorithm, that will find the new zero unless f' is zero between the old and the new root.
In that case we can use f'' to see if the zero disappeared, or if there are 2 new roots at each side.

Is this perchance what you had in mind?
That's the one! Thanks. I knew Taylor was probably in there somewhere but I couldn't figure out how to apply it.

-Dan
 

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