Considering a Perturbed Function F_eps(x)
Date: 11/20/2009 at 09:59:34 From: Rhoan Subject: f(x) = x^5 - 300x^2 - 126x + 5005 which has a root a=5 Let eps denote a small number. Consider the perturbed function F_eps(x) = f(x) + epsx^5 = (1 + eps)x^5 - 300x^2 - 126x + 5005 Let a(eps) denote the perturbed root of F_eps(x) = 0 corresponding to a(0) = 5. Estimate a(eps) - 5. I do not have anything to show for this problem. I am not sure how to start working it as I do not understand perturbed functions. I have tried to research it, but I am still not clear on its meaning.
Date: 11/27/2009 at 19:55:45 From: Doctor Vogler Subject: Re: f(x) = x^5 - 300x^2 - 126x + 5005 which has a root a=5 Hi Rhoan, Thanks for writing to Dr Math. That's a good question. Basically, they are asking you to define a function implicitly using F_eps(x). Specifically, consider F_eps(x) = f(x) + eps*x^5 = (1 + eps)(x - a(eps))(x - b(eps)) ... (x - e(eps)), where a(0) = 5. If you estimate (as in Taylor's Theorem) a(eps) = a(0) + eps*a'(0) + (smaller stuff on the order of eps^2), then you only have to compute a'(0). My first thought is to differentiate (with respect to epsilon, not x) both sides of the equation, which results in f(x) + eps*x^5 = (1 + eps)(x - a(eps))(x - b(eps))...(x - e(eps)) Substitute eps = 0, and then try to simplify the right side of the equation. I find that this doesn't work out so well, since you can't quite find values for b'(0) and so on. But it does work out more nicely if you only pull out one factor, instead of separating into five. That leaves an equation like f(x) + eps*x^5 = (x - a(eps))((1 + eps)x^4 + b(eps)x^3 + c(eps)x^2 + d(eps)x + e(eps)) Then do as I suggested: Differentiate both sides of the equation with respect to eps, not x; substitute eps = 0; and finally try to simplify the right side of the equation. You might find that it simplifies very nicely when you substitute x = 5. If you have any questions about this or need more help, please write back and show me what you have been able to do, and I will try to offer further suggestions. - Doctor Vogler, The Math Forum http://mathforum.org/dr.math/
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