Find the complex (or real) roots of $e^{\frac{3 x}{2}}+2 \cos \left(\frac{\sqrt{3} x}{2}\right)$

Define for natural $n\geq 2$ $$G(x,n)= \sum _{k=0}^\infty \frac{x^{k n}}{(k n)!}= \frac{\sum _{k=0}^{n-1} e^{x e^{\frac{2 i \pi k}{n}}}}{n}= G(x e^{\frac{2 i \pi}{n}},n)= \prod_{m=1}^\infty \left(1+\left(\frac{x}{r(n,m)}\right)^n\right)$$ where $r(n,m)$ is the $m$th biggest absolute value of a root of $G(x,n)$.

The complex roots for $G(x,2)$ and $G(x,4)$ are trivial to find. $r(2,m)=\left(m-\frac{1}{2}\right)\pi$ and $r(4,m)=\left(m-\frac{1}{2}\right)\pi\sqrt{2}$.

Because of rotational symmetry $G(x,3)=G(x e^{\frac{2 i \pi}{3}},3)$, we only need to find the roots on the real line. Note that $$G(x,3)=\frac{1}{3} \left(e^x+2 e^{-\frac{x}{2}} \cos \left(\frac{\sqrt{3} x}{2}\right)\right)$$ From this we can see that as $x\rightarrow -\infty$, the roots of $G(x,3)$ become arbitrarily close to the roots of $\cos \left(\frac{\sqrt{3} x}{2}\right)$

Because of this, we can use Newton's Method to find the roots of $G(x,3)$ by starting with the roots of $\cos \left(\frac{\sqrt{3} x}{2}\right)$.

The Mathematica code for $r(3,m)$ is

-x /. FindRoot[E^(3 x/2) + 2 Cos[(Sqrt[3] x)/2], {x, -2 Pi (-1/2 + m)/Sqrt[3]}]

This is not good enough though. I desire a formula for $r(3,m)$ which does not not involve a FindRoot. Also, because you could define Newton's method exactly with a limit, limits are not allowed in the final formula.

Because of input from the comments, we now have a new formula for $r(3,m)$: $$\frac{\pi(2m-1)}{\sqrt{3}}-\sum _{n=1}^{\infty } \frac{\left(-e^{\frac{1}{2} \sqrt{3} (\pi -2 \pi m)}\right)^n \left(\lim_{x\to \frac{\pi -2 \pi m}{\sqrt{3}}} \, \frac{\partial ^{n-1}}{\partial x^{n-1}}\left(\frac{x-\frac{\pi -2 \pi m}{\sqrt{3}}}{-e^{\frac{1}{2} \sqrt{3} (\pi -2 \pi m)}+e^{\frac{3 x}{2}}+2 \cos \left(\frac{\sqrt{3} x}{2}\right)}\right)^n\right)}{n!}$$ Though Mathematica cannot calculate this, the corresponding Mathematica code is:

-(([Pi] - 2 m [Pi])/Sqrt[3]) - Sum[(-E^(1/2 Sqrt[3] ([Pi] - 2 m [Pi])))^n/n!* Assuming[{Element[m, Integers]}, Limit[D[((x - ([Pi] - 2 m [Pi])/Sqrt[3])/(E^(3 x/2) + 2 Cos[(Sqrt[3] x)/2] - E^( 1/2 Sqrt[3] ([Pi] - 2 m [Pi]))))^n, {x, n - 1}], x -> ([Pi] - 2 m [Pi])/Sqrt[3]]], {n, 1, Infinity}]

The remaining work is to find the general $(n-1)$th derivative in the sum and to get rid of the limit in the sum.

Here is a contour plot showing the roots of $G(x,3)$. The orange curve is $\Im(G(x,3))=0$. The blue curve is $\Re(G(x,3))=0$. The plot shows $-10<\Re(x)<10,-10<\Im(x)<10$.

Also, by comparing the sum and product forms of $G(x,n)$, we can see $$\frac{1}{n!}=\sum_{m=1}^\infty\frac{1}{r(n,m)^n}$$

• You don't want a FindRoot. What sort of non-algebraic operations will you permit for this transcendental root? Nov 23 '15 at 2:31
• @Eric Pretty much anything goes. Could be infinite sums/products, any documented math functions, integrals. What I don't like about FindRoot is that it is an algorithm, rather than a function. Nov 23 '15 at 2:35
• This boils down to solving $\cos\bigg(\frac{x\sqrt{3}}{2}\bigg) = -\frac{1}{2} e^{\frac{3x}{2}}$. Nov 25 '15 at 9:33
• @Nazgand WA is telling me that $cos(x)=-\frac{1}{2}e^{x\sqrt 3}$ has different solutions than you are looking for... how did you get to that conclusion? The equation I give in my comment above still yields the correct solutions. Nov 25 '15 at 10:21
• Infinite sum representations for the roots, expressed in terms of the zeros of the cosine term, aren't too hard to obtain using the Lagrange inversion theorem but they aren't pretty. Nov 25 '15 at 10:53