# zeros of exponential polynomials

Let $\exp[n;z]$ denote the $n$th Taylor polynomial for the exponential function.

In the 1920's Szegő initiated the study of the asymptotic properties of the zeros (rescaled by dividing by $n$) of this family of polynomials and one consequence of his results is that they can approach arbitrarily closely to the imaginary axis. This prompts the following question:

Is it possible for $\exp[n;z]$ to have a root which lies precisely on the imaginary axis?

• Any such $z$ would have to have $\operatorname{Im}(z)>1$, because $\operatorname{Im}(z)\leq 1$ would imply $\cos[n,z]>1$, where $\cos[n,z]$ is the $n$th Taylor polynomial of $\cos(z)$. Jun 3, 2013 at 18:25
• @DouglasB.Staple One can also give larger lower bounds based on the error estimate for Taylor approximation of $\sin$ and $\cos$. As long as both approximations have error $<1/\sqrt{2}$, they cannot vanish at the same point. (Lower order approximations can be inspected directly.) This way, a computer program can produce an increasing sequence of lower bounds for the hypothetical common root of $\cos[n,z]$ and $\sin[n,z]$. Jun 4, 2013 at 2:09
• Thank you @user79365, that is an interesting analysis. Jun 4, 2013 at 3:25
• MathOverflow: Zeros of exponential polynomials Jan 26, 2018 at 0:20

This is equivalent to asking if there is a simultaneous real zero of the two polynomials $\cos[n,z]$ and $\sin[n,z]$. But for any $n$, one of these two polynomials is the derivative of the other, so they are only simultaneously zero at a repeated root of the higher-degree one.
So the question is equivalent to asking if the Taylor polynomial centered at 0 of $\sin$ or $\cos$ ever has a repeated real root.