Chruser
2017-08-17, 07:31 AM
Perhaps you can assist me with some intuition. I'm looking for the number of zeros of f_{m,n}(z) : = \frac{\mathrm{d}^m}{\mathrm{d}z^m}\left[\left(\frac{P(z)}{Q(z)}\right)^n\right], where P(z) and Q(z) are nonconstant polynomials with no common zeros, and m,\,n are nonnegative integers. As an example, the case P(z) = 2(z-1)^4(z-3)(z-7),\:Q(z) = (z+1)(z-2),\:m=0,\:n=5 yields f_{0,5}(z) = \frac{32(z-1)^{20}(z-3)^5(z-7)^5}{(z+1)^5(z-2)^5}, which has thirty zeros (the zeros of the polynomial in the numerator).
For some convenient notation, let p : = \deg{P(z)},\: q : = \deg{Q(z)} and Q(z) = A(z-z_1)(z-z_2)\dots (z-z_q), where A is a complex number. I tried to decompose the rational function f_{m,n}(z) in terms of its poles as follows:
f_{m,n}(z) = H_{m,n}(z) + \frac{C_{z_1,1}}{(z-z_1)^{m+1}} + \frac{C_{z_1,2}}{(z-z_1)^{m+2}} + \dots + \frac{C_{z_1,n}}{(z-z_1)^{m+n}} + \frac{C_{z_2,1}}{(z-z_2)^{m+1}} + \frac{C_{z_2,2}}{(z-z_2)^{m+2}} + \dots + \frac{C_{z_q,n}}{(z-z_q)^{m+n}}.\:\: ( 1 )
Here, the C_{z_i,j} are complex numbers, and H_{m,n}(z) is a polynomial of degree \max\{(p-q)n-m,\,0\}. Specifically, if p\ge q and 0\le m\le (p-q)n (which is precisely when H_{m,n}(z) does not vanish), we can write the terms in the right-hand side of equation ( 1 ) as a fraction with a common denominator, such that the polynomial H_{m,n}(z)[(z-z_1)(z-z_2)\dots(z-z_q)]^{m+n} dominates the degree in its numerator. Hence, f_{m,n}(z) has (p-q)n-m+q(m+n) = pn + (q-1)m zeros in this case.
There seem to be two more distinct cases, but I'm not sure how to prove what the number of zeros is in them. The answer in those (remaining) cases should (probably, based on my numerical experiments) be
pn + (q-1)m if p < q,
and
q(m+n)-(m+1), if p \ge q and m > (p-q)n.
Any ideas?
For some convenient notation, let p : = \deg{P(z)},\: q : = \deg{Q(z)} and Q(z) = A(z-z_1)(z-z_2)\dots (z-z_q), where A is a complex number. I tried to decompose the rational function f_{m,n}(z) in terms of its poles as follows:
f_{m,n}(z) = H_{m,n}(z) + \frac{C_{z_1,1}}{(z-z_1)^{m+1}} + \frac{C_{z_1,2}}{(z-z_1)^{m+2}} + \dots + \frac{C_{z_1,n}}{(z-z_1)^{m+n}} + \frac{C_{z_2,1}}{(z-z_2)^{m+1}} + \frac{C_{z_2,2}}{(z-z_2)^{m+2}} + \dots + \frac{C_{z_q,n}}{(z-z_q)^{m+n}}.\:\: ( 1 )
Here, the C_{z_i,j} are complex numbers, and H_{m,n}(z) is a polynomial of degree \max\{(p-q)n-m,\,0\}. Specifically, if p\ge q and 0\le m\le (p-q)n (which is precisely when H_{m,n}(z) does not vanish), we can write the terms in the right-hand side of equation ( 1 ) as a fraction with a common denominator, such that the polynomial H_{m,n}(z)[(z-z_1)(z-z_2)\dots(z-z_q)]^{m+n} dominates the degree in its numerator. Hence, f_{m,n}(z) has (p-q)n-m+q(m+n) = pn + (q-1)m zeros in this case.
There seem to be two more distinct cases, but I'm not sure how to prove what the number of zeros is in them. The answer in those (remaining) cases should (probably, based on my numerical experiments) be
pn + (q-1)m if p < q,
and
q(m+n)-(m+1), if p \ge q and m > (p-q)n.
Any ideas?