How to find roots of a polynomial with fractional orders?

I want to find the roots of the following fractional order polynomial:

P=s^2.1+2*s^1.2+1;

Using the solve() function (solve(s^2.1+2*s^1.2+1)), Matlab finds some roots but how does Matlab compute them? Whats the algorithm?

Answer 1

[Roger Stafford]

I don't know how 'solve' does it, but here's how I would proceed. Define the variable t = s^(3/10). Then your equation can be rewritten as

 t^7+2*t^4+1 = 0

which will have seven roots, (allowing complex numbers.) For each of these, the equation

 s^3-t^10 = 0

should have three roots for s, making a total of 21 possible roots, though there could be duplications among them.

[Bilal]

Thank you so much, it seems a good idea.. I will carefully try it and write back the results..

[Bilal]

I tried your advice and it worked thanks so much.. I found 21 roots as you told. I put them in the fractional order equation and one of them (and its complex conjugate) made the equation zero. Now i wonder, after finding these 21 roots, how i can find the right roots without trial and error?

[Roger Stafford]

The problem here is the way matlab evaluates a fractional power, particularly the fractional power of a negative or complex value. For example, the quantity s^(21/10) in principle ought to have ten possible values as given in terms of the tenth-roots of unity, but matlab gives only one of these. (Actually it would be a nuisance if it did otherwise.) For each of your 21 s-roots there should be one value of s^(21/10) among those ten that will make s^(21/10)+2*s^(12/10)+1 equal to zero (if s^(12/10) is also chosen properly.) That is, all 21 of the roots you found should be roots if the evaluation of your expression is done appropriately.

However, to give a straightforward answer to your question, the best way I know of for your purposes is simply to try all 21 possible roots with matlab and see which one or ones satisfy it.

[Walter Roberson]

solve(s^(sym(21)/sym(10))==1)

will lead to all the roots, but solve(s^(21/10)==1) will only lead to 1.

[Roger Stafford]

To expound at greater length on my statement yesterday as to the possible ways of evaluating, say, s^(12/10), suppose s is equal to -1. Then s^(1/10) can be any one of the ten values, exp(i*k*pi/10), for k = 1,3,5,...,19, and hence s^(12/10) = (s^(1/10))^12 can be any one of the five values, exp(i*k*pi/5), for k = 0,2,4,6,8.

Answer 2

[bahar beladi]

i have problem with this too:(