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

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In summary, the conversation discusses the difficulty of finding solutions for fifth order polynomials and the limitations of using numerical methods or approximations. Some possible solutions mentioned include using analytic functions or trigonometric functions, but they may not be practical due to their complexity and length. Alternatively, Newton's method could be used but it would be time-consuming.
  • #1
jdstokes
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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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  • #2
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.
 
  • #3
That's the first time I've seen a quintuple post.
 
  • #4
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?
 
  • #5
Actually you could use Newton's method altought with such a function it would take some time.
 

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

What is the solution to the equation x + 1/(1 + x)^2 = 1/x^2?

The solution to this equation is x = -1 or x = 0.

How do you solve for x in this equation?

To solve for x, you can use algebraic manipulation and the quadratic formula to get the two solutions of x = -1 or x = 0.

What is the significance of this equation?

This equation is significant because it is a special case of the general quadratic equation. It is also an example of a rational function, which has applications in various fields of science and mathematics.

Can this equation be solved using other methods?

Yes, this equation can also be solved using graphical methods or numerical methods, such as Newton's method or the bisection method.

Are there any real-life applications of this equation?

Yes, this equation can be used to model various physical phenomena, such as the behavior of a spring or the trajectory of a projectile. It can also be used in economics and finance to model growth and decay processes.

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