Let f be a meromorphic non-entire function in the plane, and suppose that for every k ≥ 0, the derivative f(k) has only real zeros. We have proved that then f(az + b) = P(z)/Q(z) for some real numbers a and b where a ≠ 0, where Q(Z) = zn or Q(z) = (z2 + 1)n, n is a positive integer, and P is a polynomial with only real zeros such that deg P ≤ deg Q + 1 ; or f(az + b) = C(z -i)-n or f(az + b) = C(z - α)/(z -i) where a is real and C is a non-zero complex constant. In this paper we provide part of the proof of this theorem, by obtaining the following result. Let f be given by f(z) = g(z)/(z2 + 1)n where g is a real entire function of finite order with g(i)g(-i) ≠ 0 and n is a positive integer. If f, f′, and f″ have only real zeros then g is a polynomial of degree at most 2n + 1. Conversely, if f is of this form where p is a polynomial of degree at most 2n with only real zeros, then f(k) has only real zeros for all k ≥ 0. If the degree of g is 2n + 1 then f(k) has only real zeros for all k ≥ 0 if, and only if, f and f′ have only real zeros.
|Original language||English (US)|
|Number of pages||72|
|Journal||Annales Academiae Scientiarum Fennicae Mathematica|
|State||Published - Dec 1 1998|
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