Roots of a sixth order polynomial in symbolic form

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RAN 2021-3-18

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评论： Bruno Luong 2021-4-6

采纳的回答： Bruno Luong

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Hello all,

I have a sixth order polynomial in symbolic form:

eqn1=a1*s^6 + a2*s^5 + a3*s^4 a4*s^3 + a5*s^2 +a6*s + a7 == 0

There is no way of finding analytical roots of the polynomial in symbolic form.

After substituting values of co-efficients, and using "root" function, I end up with six roots (complex and real) in numerical form.

Each of these numerical root is a function of the coefficients and that is clear. Now I would like to extract this function of co-efficient to sort of make it more analytic.

Any idea how to do it?

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Answer 1

Bruno Luong 2021-3-19

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编辑：Bruno Luong 2021-3-19

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"Ok let me elaborate a bit further. I have a system with a sixth order denominator polynomial, like i mentioned in the question. I want to evaluate the settling time, which is now dependent on the coefficient of the polynomial. The aim is to find out which coefficients of the polynomial have the most influence on the settling time. To that end, I found the poles of the denominator and time constant associated with each pole numerically. So now if I want to increase or decrease the settling time, i do not know which coefficient to change. What I would like to do is figure out which coefficient has most effect on settling time and change accourdingly. Is that possible?"

If you want to have the sensitivity of the roots with respect to the coefficiensts, then you can just make the derivative of the equation

P(s) = 0;

P is polynomial associates with a(:).

with respect to the coefficient ak:=a(k), you'll find out that

ds/dak = s.^(length(a)-k)./(polyval(polyder(a),s));

So now you know how to change the coefficients to make the setting time changes, at least locally, close to the current polyomial.

You might also be interested in function MATLAB stock function poly() that gives the coefficient from the roots.

Check the formula for a random third order polynomial:

a=rand(1,4);
for k=1:length(a)
    step=1e-3;
    ap=a;
    ap(k)=ap(k)+step;
    s=roots(a);
    t=roots(ap);
    dsdak_num = (t-s)/eps % finite difference
    dsdak = s.^(length(a)-k)./(polyval(polyder(a),s)) 
end

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RAN 2021-4-6

@Bruno Luong: In my case, all coefficients, i.e a1=f(x1,x2), a2=f(x1,x2)...etc. are a function of two variables x1 and x2 which I have the freedom to play with. Unfortunately I cannot disclose all the details. So if i take a derivative of d (roots)/da1, by changing x1 or/and x2 then other coefficients change as well. Hence I come back to the same question as to which coefficient affected the most. Unfortunately, in this case there is no possible way of finding the sensitivity of roots by changing a single co-efficient while keeping the rest of the co-efficients unchanged. Your answer could have worked if all co-efficients were not dependent on each other.

Bruno Luong 2021-4-6

If your decision parameters are (x1,x2) and not (a1,a2) then you might use the derivative chain rule to compute the sensitivity to the decision parameter.

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Answer 2

Walter Roberson 2021-3-18

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编辑：Walter Roberson 2021-3-18

"Each of these numerical root is a function of the coefficients and that is clear."

However, the roots cannot reliably be expressed as algebraic functions of the coefficients: that is, as expressions involving only addition, subtraction, multiplication, division, integer powers, integer roots.

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Walter Roberson 2021-3-18

Do you mean to say that each numerical root is not a strict function of coefficients?

In mathematics, a function is a mapping, in this case

which ultimately can be defined by listing out all of the possibilities. There does not need to be any formula or rule involved, and there does not need to be "continuity" in any sense. If you had a "black box" that you could input 7 inputs, and it always gave the same output for those inputs, then the "black box" acts like a function, no matter whether it processes electronically, or by carefully balanced weights and springs, or by the different possible inputs being physical pieces with different magnetic patterns. Discrete inputs mapping to discrete outputs repeatably is all that is needed for function .

This naturally gets very expensive to list out as the number of possible inputs increases, so people tend to turn to formulas to calculate the outputs. Formulas in general do not need to be continuous: it is entirely acceptable to say "this formula applies to negative numbers except the negative integers, with it being undefined at negative integers" (such as the Γ function)

Polynomials are one kind of formula. Ratios of polynomials extend the possibilities. But there are things that cannot be defined in terms of finite polynomials (or ratios of polynomials) -- for example sin() and exp() can be expressed in polynomial-like terms but require infinite series. The Taylor theorem says you can construct an approximating infinite polynomial from any infinitely-continuous formula. There are plenty of formulas that you could truncate a Taylor series to a finite number of terms to get a finite relative error bound approximation of the true formula, but any periodic function requires infinite terms. Every polynomial goes to +/- infinity as x approaches infinity, so Taylor series cannot be used to construct a finite approximation of a bounded function.

There are some common functions that cannot be expressed as polynomials. I mentioned sin() earlier. The general formula for length of a curve is another.

It is incorrect to say that in such cases that the results are not "strict functions" of the coefficients, but you cannot express the results using "algebraic numbers" (addition, subtraction, multiplication, division, integer powers, radicals).

RAN 2021-3-19

编辑：Walter Roberson 2021-3-19

Hi Walter,

Thank you for the detailed answer You did comment on a question asked many years ago, that there could be a solution for 5th order in special cases ( https://uk.mathworks.com/matlabcentral/answers/106964-roots-of-quintic-equation-using-matlab ). Please can you comment on these cases?

Also I think there exist a general solution to quintic which was a recent development.

https://math.stackexchange.com/questions/1555743/how-do-you-solve-5th-degree-polynomials

https://mathworld.wolfram.com/QuinticEquation.html

Specifically reducing the 5th order to Bring-Jerrard form. Is there any sort of implementation of this in Matlab?

Walter Roberson 2021-3-19

some quintics can be factored into pieces that are individually of degree no more than 4; roots can be found for the fragments individually. For example quintics that have constant coefficient 0 can be factored .

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