Conference rusure::math

>   If one has an even polynomial e(z) (with real coefficients) with imaginary
>   roots, is it always possible to add an odd polynomial o(z), also with
>   real coefficients, to form a polynomial with positive roots?

Sure. Pick any odd polynomial o(x) with real positive roots and degree > 
degree(e). Then o(x) - e(x) is what you seek.

        Re .1
        
>> Sure. Pick any odd polynomial o(x) with real positive roots and degree > 
>> degree(e). Then o(x) - e(x) is what you seek.
        
        No, as o(x) - e(x) isn't odd.  Besides, I think .0 should
        have included the restriction not to change the degree of
        the result.
        
        Dan

    .0 says can you "add an odd polynomial o(z)... to form a
    polynomial...". 
    
    To avoid confusion for readers of the notesfile, shouldn't we just say 
    e(z) + o(z) rather than o(z) - e(z), since e(z) + (-o(z)) will have the
    same roots as o(z) - e(z), and -o(z) will also be an odd polynomial
    with real positive roots and same degree as o(z)?  Some readers are
    likely to think that the problem wasn't read carefully, rather than
    seing the two polynomials as equivalent for the purposes of this
    problem.
    
    Also, the way the problem is stated, e(z) + o(z) doesn't have to be
    odd or even, it just has to have real coefficients and positive roots.
    Am I wrong?  Am I missing something?
    
    Jon

>   If one has an even polynomial e(z) (with real coefficients) with imaginary
>   roots, is it always possible to add an odd polynomial o(z), also with
>   real coefficients, to form a polynomial with positive roots?

    Dan is right, I forgot to restrict the degree of o(z)
    to be less than e(z)...

    - Jim

	Let r_j be the roots of p, all positive reals.   Then

	e(z) = 0

=>	prod (z-r_j) = - prod(-z-r_j)

=>	prod (z-r_j)/(-z-r_j) = -1

=>	prod |(z-r_j)|/|-z-r_j| = 1

	Now suppose that Re(z) > 0.   Then |z-r_j| < |-z-r_j| since r_j is +ve
for all j.   Thus the left hand side is < 1.   Similarly, if Re(z) < 0, the
left hand side is > 1.   Hence Re(z) = 0.

Regards,
Andrew.

	re .3,
        
>>    Also, the way the problem is stated, e(z) + o(z) doesn't have to be
>>    odd or even, it just has to have real coefficients and positive roots.
        
        That's correct.
        
        Dan

>   If one has an even polynomial e(z) (with real coefficients) with imaginary
>   roots, is it always possible to add an odd polynomial o(z), also with
>   real coefficients, to form a polynomial with positive roots?

	No.   Suppose that we have a polynomial p(x) = e(x) + o(x), where the
zeros of p are r_j > 0, and the degree of p is n.   Examine:

	g(v) = prod (iv-r_j)/(-iv-r_j) as v runs from 0 to infinity.

	When g(0) = 1, g(v) -> (-1)^n as v -> infinity, and every time
g(v) = 1, you have a root of o(x), every time g(v) = -1, you have a root of
e(x).   Thus the roots of e(x) alternate with those of o(x).   By counting
the number of times that g(v) wraps round the origin as v goes from 0 to
infinity, it's easy to see that all the roots of e(x) and o(x) are
distinct.

	So in particular, if e(x) has a repeated root, we know that it cannot
= p(x) + p(-x), where p has real positive roots.

	But in fact, there are much stronger constraints on e(x).   
d� log(g(v))/dv� < 0, which means the zeros of e(x) are getting further
and further apart as v increases.

	Also, the fact that the zeros of e and o must alternate means that
there is no point in looking for an o of arbitarily high degree.   o can
have degree at most one greater than that of e.

Regards,
Andrew.

	Let me pick my own nits:

>	d� log(g(v))/dv� < 0

	The lhs is imaginary.   What I meant was obviously:

	d� -i*log(g(v))/dv� < 0

	Curious I should make that slip given the discussion in the next
note...

A