Given a Taylor series is there any way to determine the (max) number of zeros that this series can have? Some Taylor series eg: $1+x-x^2-x^3+....$ have no (real) zeros. However, do the same for $x-x^{3}/3!+x^{5}/5!-...$ and one gets infinite (real) zeros
Is there a way to figure out which series will have zeros and which will not and for those that do is there a way to predict the greatest number of said zeros?
Any help/suggestions most welcome. 
 A: For smooth functions $f$, the derivatives at a single point don't tell you much:


*

*the set of zeroes is of course a closed set

*the zero'th derivative tells you whether $f(t)=0$

*if $f^{(n)}(t)\neq 0$ for some $n>0$ then $t$ has a neighbourhood where $f$ has no zeroes, except possibly at $t$


Any possibility allowed by these conditions can be constructed using bump functions.
If you have bounds on the derivatives over the whole range, more could be said - there are several question like this on this site.
A function is real-analytic if the Taylor series converges everywhere to the value of $f.$ There is then a tautological answer: the function is determined by the Taylor series, so the zeroes are determined by the Taylor series. Even without looking at the Taylor series, the Weierstrass factorization theorem says that a set is the zero set of a non-constant analytic function if and only if all its points are isolated.
A sufficient condition for having a finite number of zeroes is the the function is analytic and a Taylor series satisfies a linear recurrence. This implies that the function is rational. This is not a necessary condition, for example $f(z)=\exp(z).$
