Thread Subject: Polynomial roots

Hello,

I am trying to find roots of a polynomial of the form:

(1-d*x^(-1))*(1-conj(d)*x)-b^2=0

d is complex number,
b is real number,

The main problem is how to make Matlab interpret properly orders of
the polynomial - some of them are negative some positive.

Thanks for any help.

Pawel

Pawel <prulikowski@removegmail.com> wrote:
> Hello,
>
> I am trying to find roots of a polynomial of the form:
>
> (1-d*x^(-1))*(1-conj(d)*x)-b^2=0
>
> d is complex number,
> b is real number,
>
> The main problem is how to make Matlab interpret properly orders of
> the polynomial - some of them are negative some positive.

Your equation is equivalent to:

                           2 2 2
                        d x + (- d + b - 1) x + d
                      - ---------------------------- = 0
                                     x

then your answer is: roots([ d, -d^2+b^2-1, d ])
or analitically solved:

                4 2 2 4 2 2 2
          SQRT(d + (- 2 b - 2) d + b - 2 b + 1) - d + b - 1
    x = - --------------------------------------------------------
                                    2 d

                 4 2 2 4 2 2 2
           SQRT(d + (- 2 b - 2) d + b - 2 b + 1) + d - b + 1
    x = --------------------------------------------------------
                                     2 d

Hi Pawel,

you schould write the function in Matrixform
like this:

x^3 -3ix^2 +ix -1
c=[1 -3i i -1]

to get the roots of this function use the roots command

roots(c)

ans =

  -0.0020 +16.6079i
  -0.0024 -13.6079i
   0.0044 + 0.0000i

The next MATLAB function, which I remember is the fzero command.

Hope this example help, let me know.

Best regards

Sebastian Nowoisky

 Pawel wrote:
>
>
> Hello,
>
> I am trying to find roots of a polynomial of the form:
>
> (1-d*x^(-1))*(1-conj(d)*x)-b^2=0
>
> d is complex number,
> b is real number,
>
> The main problem is how to make Matlab interpret properly orders of
> the polynomial - some of them are negative some positive.
>
> Thanks for any help.
>
> Pawel

Pawel wrote:
>
>
> Hello,
>
> I am trying to find roots of a polynomial of the form:
>
> (1-d*x^(-1))*(1-conj(d)*x)-b^2=0
>
> d is complex number,
> b is real number,
>
> The main problem is how to make Matlab interpret properly orders of
> the polynomial - some of them are negative some positive.
>
> Thanks for any help.
>
> Pawel
  
This is not a polynomial as roots
can understand it.

However, why not multiply by x?
Then it is of a proper form for
both conv and roots to manipulate.

 (x-d)*(1-conj(d)*x) - x*b^2 = 0

Thus,

 p = conv([1,-d],[[-conj(d),1] - [0,b^2,0];

 roots(p)

Your original post had a product of
n terms of this form, so multiply by
x^n.

HTH,
John D'Errico

pisz_na.mirek wrote:
>
>
> Pawel <prulikowski@removegmail.com> wrote:
>> Hello,
>>
>> I am trying to find roots of a polynomial of the form:
>>
>> (1-d*x^(-1))*(1-conj(d)*x)-b^2=0
>>
>> d is complex number,
>> b is real number,
>>
>> The main problem is how to make Matlab interpret properly
orders
> of
>> the polynomial - some of them are negative some positive.
>
> Your equation is equivalent to:
>
> 2 2 2
> d x + (- d + b - 1) x + d
> - ---------------------------- = 0
> x
>
> then your answer is: roots([ d, -d^2+b^2-1, d ])
> or analitically solved:
>
> 4 2 2 4 2 2 2
> SQRT(d + (- 2 b - 2) d + b - 2 b + 1) - d + b - 1
> x = - --------------------------------------------------------
> 2 d
>
> 4 2 2 4 2 2 2
> SQRT(d + (- 2 b - 2) d + b - 2 b + 1) + d - b + 1
> x = --------------------------------------------------------
> 2 d
>
>

Hello Mirek,

Thanks a lot for all your suggestions - the problem is slightly more
complicated as the actual, proper equation looks like this:

_N
||[(1-dk*x^(-1))*(1-conj(dk)*x)]-b^2=0
k=1

How would you handle this problem ?

Regards

Pawel

Pawel <prulikowski@removegmail.com> wrote:
>
> Hello Mirek,
>
> Thanks a lot for all your suggestions - the problem is slightly more
> complicated as the actual, proper equation looks like this:
>
> _N
> ||[(1-dk*x^(-1))*(1-conj(dk)*x)]-b^2=0
> k=1
>
> How would you handle this problem ?


Is -b^2 outside of Pi (multiplication)? If so then your complex term is
equivalent to
                               2 2
                           dk x + (- dk - 1) x + dk
(%o125) - --------------------------
                                       x

You can remove denominator by multipling your equation by x^N. So matlab
solution is

p=1;
for k=1:N
  p=conv(p,[ -d(k), d(k)^2+1, -d(k) ]);
end;
p(end-N)=p(end-N)+(-b^2); % add -b^2*x^N
roots(p)

In article <ef40263.3@webcrossing.raydaftYaTP>, Pawel <prulikowski@REMOVEgmail.com> wrote:

> pisz_na.mirek wrote:
> >
> >
> > Pawel <prulikowski@removegmail.com> wrote:
> >> Hello,
> >>
> >> I am trying to find roots of a polynomial of the form:
> >>
> >> (1-d*x^(-1))*(1-conj(d)*x)-b^2=0
> >>
> >> d is complex number,
> >> b is real number,
> >>
> >> The main problem is how to make Matlab interpret properly
> orders
> > of
> >> the polynomial - some of them are negative some positive.
> >
> > Your equation is equivalent to:
> >
> > 2 2 2
> > d x + (- d + b - 1) x + d
> > - ---------------------------- = 0
> > x
> >
> > then your answer is: roots([ d, -d^2+b^2-1, d ])
> > or analitically solved:
> >
> > 4 2 2 4 2 2 2
> > SQRT(d + (- 2 b - 2) d + b - 2 b + 1) - d + b - 1
> > x = - --------------------------------------------------------
> > 2 d
> >
> > 4 2 2 4 2 2 2
> > SQRT(d + (- 2 b - 2) d + b - 2 b + 1) + d - b + 1
> > x = --------------------------------------------------------
> > 2 d
> >
> >
>
> Hello Mirek,
>
> Thanks a lot for all your suggestions - the problem is slightly more
> complicated as the actual, proper equation looks like this:
>
> _N
> ||[(1-dk*x^(-1))*(1-conj(dk)*x)]-b^2=0
> k=1
>
> How would you handle this problem ?
>
> Regards
>
> Pawel

All of us have suggested essentially the
same thing.

Multiply by x^N.

Then use roots.

What is the problem?

John


--
The best material model of a cat is another, or preferably the same, cat.
A. Rosenblueth, Philosophy of Science, 1945

Those who can't laugh at themselves leave the job to others.
Anonymous