$F(x)=1+\dfrac{x^2}{2!\cdot1!}+\dfrac{x^4}{4!\cdot2!}+\dfrac{x^6}{6!\cdot3!}+\dfrac{x^8}{8!\cdot4!}\cdots$
I wanted to know the function which could be represented as the Taylor series mentioned.
Every $2n$-th derivative gives $\dfrac{1}{n!}$ at $x=0$
It seems analogous to ${\cos(ix)}=1+\dfrac{x^2}{2!}+\dfrac{x^4}{4!}+\dfrac{x^6}{6!}+\dfrac{x^8}{8!}\cdots$
But I don't see how I can transform the expression into a familiar function.