Solving an Equation 5th degree

Hi Guys, After i have done some research i need still a little bit explanation and help. I hope that sb can help me.

I have an Equation of 5th degree. I solved this equation with the command solve and i get 5 Solution with roots(...,z)

My Questions:

1. Is there any posibility to get an explicit Solution (like MaxDegree, 5) --> I think not.

2. How can i solve these roots without symbolic Toolbox ? I used vpa() but i don*t want to use the symbolic Toolbox.

3. I think that maybe with an numerical approximation these roots could be solved.Is that right ? Are there any numerical function from Matlab which could be used for that? or which are special for this problem ?

4.Is there any way how i could solve the roots without Symbolic Toolbox or Numeric approximation ? Maybe with a function of Matlab or an self created solver ?

Answer 1

[James Tursa]

5th order polynomials cannot be solved symbolically in general. E.g., see this:

https://en.wikipedia.org/wiki/Algebraic_solution

You could get all of the solutions numerically with the roots() function. You could also write your own code to get the roots, but I don't know why you would want to.

EDIT

If you don't want to use the roots() function for some reason, you could pick some method to get one of the guaranteed real roots (e.g. even a simple bisection method will work), then do a polynomial divide using this real root to reduce the 5th order polynomial to 4th order, then calculate the remaining roots explicitly using known formulae. E.g., see general formula for roots of a quartic here:

https://en.wikipedia.org/wiki/Quartic_function

That being said, I have no idea how much time this will save, or how well/poorly behaved the explicit formulae will be numerically for the specific coefficients you will be using vs the eigenvalue method used by roots().

Answer 2

[IlPadrino]

Thanks JamesTursa for the answer. Do you think maybe it is possible to solve the 5th order polynomial with setting constrains ? For Example i am only interested in Solution which are positive and real.

The problem is that an explicit Solution would ideal:/ The reason is the runtime... If i have an explicit with parameters... i just have to add the parameters to the explicit Solution and i can calculate my result. If i want to do it numerical i have the problem hat if i am changing the parameters i have always to run the approximation again :/ Also i am not allowed to use the symbolic Toolbox in my final skript

[James Tursa]

How much are the coefficients changing? Perhaps you could use some type of iterative method if the roots don't change much from the previous solution.

[Walter Roberson]

Explicit solutions in polynomial extended with square roots is not possible in general.

[IlPadrino]

@James Tursa only one of the parameters is changing ... and only step wise...means only increasing by one or 0.1... do you know an iterative method which is using this benefit ?

[Walter Roberson]

https://math.stackexchange.com/questions/309178/polynomial-root-finding

and

http://mathworld.wolfram.com/HalleysIrrationalFormula.html

[John D'Errico]

You CANNOT have a symbolic, explicit solution. No matter how badly you want it, the mathematically impossible is still just impossible.

Constraints are irrelevant. You cannot just decide that you want only to solve for the real roots analytically. There is no simple way around the mathematical facts.

Anyway, roots is not that slow. In fact, it is pretty fast, certainly so compared to any code that you would write yourself.