Re: question about zeroes of polynomials




In article <42C2DDD7.7040503@xxxxxxxxxxxxxxxxx>,
Christophe <christop@xxxxxxxxxxxxxxxxx> writes:
>dear all,
>
>I have the following problem. We have a polynomial Q(x) of degree 3+N in x:
>
>Q(x):= x^(3+N)-rho*3*x^(2+N)+rho*3*x^(1+N)-rho*x^N + rho*(x-1)^3*P(x)
>
>with P(x) any polynomial of degree N-1 and rho a real number. If N=2
>for instance, P(x) looks like P(x)=a*x+b, and we obtain that Q is of
>degree 5. If N=1, P(x) is just a nonzero constant. If N=0, P(x)=0.
>
>Now it is easy to see that if rho=1, 1 is a triple root of Q (3 of the
>3+N roots are equal to 1). The question I have now is what happens with
>these three special roots if we perturb rho a little bit towards zero.
>So if rho is not 1 but 1-delta (with delta real, positive and
>approximately 0, so f.i.0.0000001). Of course, these three roots that
>were originally 1 might become complex then, such as
>1.0000010201193+0.1846304081e-5*I.
>
>The thing I would really like to prove is that at least one of these
>three roots (that were 1 if rho was 1), becomes larger than 1 in
>absolute value (modulus) if rho is slightly perturbed in the way
>described above.
>
>I thought it should be possible to prove this using some kind of
>perturbation on rho or the roots (where we write f.i. rho=1+delta and
>can then neglect some higher order terms in delta) but I do not seem to
>get this kind of formulation right.
>
>Can anyone help me out?
>
>Thank you so much,
>
>Christophe

write rho=1+eps. then you can develop the zeroes near one of the form 1+deltaz
with deltaz a (complex) power
series in the powers of eps^(1/3), one of them real of course. now the coefficient
in front of eps^{1/3} will tell you whether you are right. this will clearly
depend also on the coeffs of P(x).
the book of Wilkinson: rounding erros in algebraic processes has a nice account
for this problem, showing how to proceed and giving impressive examples.
the general framework is "algebraic functions", a branch in the theory of
functions of one complex argument.
hth
peter
.



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