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Nth degree polynomial

  1. Dec 9, 2013 #1
    How would I go about solving an Nth degree polynomial function such that N>=5?

    Ax[itex]^{n}[/itex]+Bx[itex]^{n-1}[/itex]+.........+Z = 0
     
  2. jcsd
  3. Dec 9, 2013 #2

    SteamKing

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    Iteration, unless you stumble on some kind of factorization.
     
  4. Dec 9, 2013 #3
    That might take a while with this
    u(k) = (900-3k)r[itex]^{k-1}[/itex]
    s(n) = Σ[itex]_{k=1...n}[/itex]u(k)
    find r for which s(5000) = -600,000,000,000

    But ok no problem, I guess I will write a program to iterate through all possible values of r.
     
  5. Dec 9, 2013 #4
    You would use an algorithm like Newton's method to find the roots. Testing all possible values of r is both impossible and also not particularly efficient.
     
  6. Dec 9, 2013 #5
    I made some assumptions but I got the algorithm working and I got r
     
  7. Dec 9, 2013 #6
    The Galois group of a general complex polynomial of degree ##n\geq 5## is not solvable. Thus, you won't be able to come up with something like the quadratic formula for ##n\geq 5##. You have to find other ways to find roots.
     
  8. Dec 10, 2013 #7
    Given:

    $$ w(z)=a_0+a_1 z+a_2 z^2+\cdots+a_nz^n$$

    by Laurent's Expansion Theorem applied to algebraic functions, we can always write down an explicit expression (just like the quadratic formula) for the roots in terms of complex-contour integrals:

    $$
    w(z)=b\prod_{j=1}^N \left(z-\frac{1}{2k_j\pi i}\mathop{\oint} \frac{z_{j,k}(\zeta)}{\zeta}d\zeta\right)^{k}
    $$

    However, the integral is in general not solvable in terms of elementary functions so must be evaluated numerically.
     
    Last edited: Dec 10, 2013
  9. Dec 10, 2013 #8
    You know I meant "in terms of radicals, powers, addition, and subtraction," right? :tongue:

    That's actually REALLY cool to me. I've never thought of that, so I've learned something new today. :approve:
     
  10. Dec 10, 2013 #9
    We need to try one in case the reader is not sure how to do this. Let's take:

    $$f(z,w)=z-w(w-1)^2(w-2)^3(w-(1+i))^4=0$$

    And in order to extract the fourth-order root, we will integrate over the 4-cycle branch corresponding to this root, the ##w_{4,1}## branch, along an ##8\pi## circular path say midway the distance to the nearest singular point of ##f(z,w)## which in this case is about 0.026 so let's let ##z=0.01 e^{it}## Then we have:

    $$\frac{dw}{dt}=-\frac{\frac{df}{dz}}{\frac{df}{dw}}\frac{dz}{dt}=g(z,w)$$

    Then we solve the IVP:

    $$\frac{dw}{dt}=g(z,w),\quad w(0)=w_{4,1}(0.01)$$

    where ##w_{4,1}(0.01)## means we are starting the integration on one of the determinations of this 4-cycle branch..

    We integrate that numerically and obtain the (numeric) function ##w_{4,1}##.

    Then we would integrate:

    $$\frac{1}{8\pi i}\oint \frac{w_{4,1}(t)}{z(t)}z'(t) dt$$

    and when I solve the IVP for each determination of that branch, and solve that integral numerically, I get four values very close to ##1+i##.

    Now, if I've explained that well, the reader can now write Mathematica code to extract the third-order root. It should only take about 10 lines of code. You would of course have to find which of the 10 determinations of the function correspond to the 3-cycle branch but you can just integrate over all of them; three results should be very close to 2.
     
    Last edited: Dec 10, 2013
  11. Dec 10, 2013 #10
    There is a Master's Thesis (in PDF format) from Oxford University posted at the following link:

    http://eprints.maths.ox.ac.uk/16/1/mekwi.pdf

    I think it does a pretty good job providing a general overview (and easily readable) of the various iterative methods available for solving polynomial equations.

    Forming a Companion Matrix for the polynomial and solving for its eigenvalues (which are the roots of the original equation) is a very robust technique. However, it is not computationally efficient; it is better to use a method specifically designed to solve for the roots of a polynomial equation.

    The Jenkins-Traub and Laguerre methods are two algorithms popular for solving for the roots of polynomial equations. There are several more; a search should reveal them.

    Do you need to write a program to do this task?
    Or do you just need it done, regardless of the technique?

    In addition to the commercial software that can do this (e.g. - MATLAB, Maple, Mathematica, etc), there are also several free tools available online that can perform this task. For example, the following Javascript program uses the Jenkins-Traub algorithm to solve for the roots of a polynomial up to degree 100:

    http://www.akiti.ca/PolyRootRe.html [Broken]

    Hope this helps.
     
    Last edited by a moderator: May 6, 2017
  12. Dec 10, 2013 #11
    What I did was just figure out what r = 1 was, then r = 10 was. Then I found out the answer was between, so I binary searched through the decimals until I got to a precise enough r. I was hoping to come up with a true maths solution. But I guess that was good enough. r = 1.002322108633
     
  13. Dec 10, 2013 #12

    Office_Shredder

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    That gets you one answer, but there is one for each degree (possibly complex). Depending on the context of the question you might need to do more work.
     
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