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Mathematics question (solving nth degree polynomials)

Orsorum

Lifer
Let's say you have an nth-degree polynomial f(t).

f(t) = 1-dt+nt^2-mt^3+pt^4-mt^5+nt^6-dt^6+t^7

Is there a regular formula to find the roots of the equation in terms of its coefficients? (e.g. for an equation of degree 2, f(t) = 1-dt+t^2, t = d+- ((root) d^2 -4)/2) In doing so, we also want to restrict the roots to ones that are less than or equal to 1.

Thoughts? Or am I going to have to go through case by case and check every root? 🙁
 
There are no formulae to generically solve polynomials of greater than fourth order, it has also been proven (by Galois IIRC) that such a formula cannot exist.

Your best bet would probably be to use numerical methods to find the roots.
 
Haircut is right. In addition, it has been shown that there is no formula for degree five (or greater). You can find the cubic and quartic formulas in a CRC manual or a good math book (although they are VERY cumbersome to use).

You CAN show that if there are any rational roots, they are + or - the ratios of the factors of the first and last terms. This is trivial in your case (where they are both 1.... so +/- 1 is the only possibility).

Note that you are not guaranteed ANY roots for polynomials of even degree--but you are guaranteed at least one root (although it may be outside your -1<t<1 range) for odd polynomials.

For your particular problem, I have no idea. Its been a LONG time since I've taken a "formal" algebraic class..... we engineers (even us mechanics nuts) are always solving things in a set method. Best of luck.

 
That is what I suspected. I looked around for a while on google a few months ago; looks like number by number is the best I can do. 🙁

Thanks!
 
*sigh* Well, two hours and fifty, sixty combinations of integers later, I have some answers.

When examining the polynomial f(t) = 1-dt+mt^2-nt^3+mt^4-dt^5+t^6, we notice that it has roots <= |1| for combinations of d, m, and n as follows:

Around -6 <= d <= 6, the roots are symmetric around zero , where d and n are symmetric and m is identical to its counterpart i.e. there are roots at (d, m, n) = (-3, 6, -7) and (3, 6, 7), (-5, 11, -14) and (5, 11, 14), etc. The actual list is considerably bigger (54 combinations so far).

I am still stuck on the few roots directly around d=0. It should be symmetric so that there are roots at d= (0, 2, 0) and (0, -2, 0), (0, 1, -1) and (0, -1, 1), and so forth. But it's not. Bah! There is symmetry at (0, 1, 0) and (0, -1, 0), but that's the only one. Three anomalies left to figure out. 🙁
 
Ok, I havea question..

What in God's name would a real world application for this be? 😛
 
Originally posted by: Eli
Ok, I havea question..

What in God's name would a real world application for this be? 😛


these things arise when one is trying to design equipment. Another area is statistics specifically market research.
 
Originally posted by: whitecloak
Originally posted by: Eli
Ok, I havea question..

What in God's name would a real world application for this be? 😛


these things arise when one is trying to design equipment. Another area is statistics specifically market research.
Design equipment eh?

hmmm......

🙁
 
Good lord, I can't give a presentation on this! (he just suggested that maybe I should try to give a short talk on it to some undergrads or someone)... scary!

We may have the discrepancy at d=0 figured out... it's still symmetric, there just aren't two competing values of d... so you have to basically split the roots down the middle to see the symmetry. The starting point is d=0, n=-1, m=0.
 
w00000000000000000000000000000000t

The upper and lower bounds for the values correspond to Pascal's triangle. I don't know what it means, but it's pretty damn cool either way.
 
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