$f(x)=\sum_{n=-\infty}^\infty e^{nx} \longrightarrow f(x) = \infty\text{ for all }x\text{ values}$
\frac{y f''(x)}{1!}=\sum_{n=-\infty}^\infty \frac {y n^2 e^{nx}}{1!}
$\frac{y^2 f^{(4)}(x)}{2!}=\sum_{n=-\infty}^\infty \frac {y^2 n^4 e^{nx}}{2!}$
$\frac{y^3 f^{(6)}(x)}{3!}=\sum_{n=-\infty}^\infty \frac {y^3 n^6 e^{nx}}{3!}$
$ \vdots $
{g(x,y)=f(x)+\frac{y.f''(x)}{1!}+\frac{y^2 f^{(4)}(x)}{2!}+\cdots=\sum_{n=-\infty}^\infty e^{nx+n^2 y}} \text{ if }y=x/2
$ g(x,x/2)=\sum_{n=-\infty}^\infty e^{\frac{x[(n+1)^2 - 1]}{2}}$
$ g(x,x/2)=e^{-x/2}\cdot\sum_{n=-\infty}^\infty e^{\frac{x(n+1)^2}{2}}=e^{-x/2}\cdot\sum_{n=-\infty}^\infty e^{\frac{xn^2}{2}}$
this result has values for all $x<0$
My question:
How could we get such result although we produced $g(x,x/2)$ from $f(x)$ that doesn't have any defined values for $x<0$?