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Solution to x + 1/(1 + x)^2 = 1/x^2

by jdstokes
Tags: 1 or 1, 1 or x2, solution
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jdstokes
#1
Dec5-05, 08:07 PM
P: 527
In other words

x^5 + 2x^4 + x^3 - 2x - 1 = 0.

I am aware that fifth order polynomials are generally not analytically soluble. Are there any clever ways to at least approximately solve this equation without resorting to numerical methods or fourth order taylor approximation which does not capture the asymptotic behaviour.
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Hurkyl
#2
Dec5-05, 09:06 PM
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I am aware that fifth order polynomials are generally not analytically soluble.
That's not quite accurate:

There is no (general) expression for the roots of a polynomial of degree 5 or higher in terms of the integers, +, -, *, /, and n-th root functions. (for integer n)


There are certainly analytic solutions -- e.g. there are functions that maps 6 complex numbers to the solution to a polynomial with those numbers as coefficients, and I believe they can be made analytic on large regions.

I also think that such things can be solved in terms of sines and cosines (and arcsines and arccosines), but I don't know how much that helps, since generally sines and cosines can only be "evaluated" through numerical approximation.
devious_
#3
Dec6-05, 10:37 PM
P: 347
That's the first time I've seen a quintuple post.

jdstokes
#4
Dec7-05, 08:02 AM
P: 527
Solution to x + 1/(1 + x)^2 = 1/x^2

lol. I've heard about such solutions, they typically span hundreds of pages and are thus of little practical use. Any other thoughts on approximate solutions, or am I stuck up the proverbial creek?
Werg22
#5
Dec7-05, 09:28 PM
P: 1,520
Actually you could use Newton's method altought with such a function it would take some time.


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