Find zero of function $\sum_{n=0}^{\infty}\frac{cos(x(n+1))}{n!}$

Question

I'm interested in finding the smallest positive zero of function $$\sum_{n=0}^{\infty}\frac{\cos(x(n+1))}{n!},\qquad x\in\mathbb R$$ It is approximately equal to $0.832$. I've calculated Taylor series expansion of this sum, which is: $$\sum_{k=0}^{\infty}e \cdot\frac{(-1)^k \ B_{2k}\ x^{2k}}{(2k)!}$$ Where $B_n$is n-th Bell number.

But I don't know, whether it helps.
Thanks for all the help in solving this problem.

Answer 1

Your sum is equal to $$\Re\bigg(\sum_{n=0}^\infty \frac{e^{ix(n+1)}}{n!}\bigg)$$ or $$\Re\exp\big(ix+e^{ix}\big)=\Re\exp\big(\cos x+i\sin x+ix\big)$$ Using Euler’s formula, this is equal to $$e^{\cos x}\cos(x+\sin x)$$ To find when this is equal to zero, you must compute the zeroes of the function $$\cos(x+\sin x)$$ or the values of $x$ for which $$x+\sin x = \pi(n+1/2)$$ for some $n\in\mathbb Z$. I do not suspect an elementary solution will exist, but you can use this explicit form to calculate some pretty good approximate solutions.

The smallest of the zeroes will occur when $$x+\sin x = \pi/2$$ which is at approximately $x\approx 0.832$. See @Raymond’s answer for a more accurate approximation and a representation in terms of the Dottie number.

Answer 2

The more exact value is $\;0.83171119357973597757600960396587803808517294078679544552179166\cdots $ which is equal to $\;\dfrac{\pi}2-D\;$ with $D$ the "Dottie Number".

The Dottie number is obtained by iteration of the $\cos$ function on a calculator (in 'radian' mode) as detailed in Mathworld.